BIPOLAR JUNCTION TRANSISTORS
*** INCOMPLETE ***Introduction
The invention of the bipolar transistor in 1948 ushered in a revolution in electronics. Technical feats previously requiring relatively large, mechanically fragile, power-hungry vacuum tubes were suddenly achievable with tiny, mechanically rugged, power-thrifty specks of crystalline silicon. This revolution made possible the design and manufacture of lightweight, inexpensive electronic devices that we now take for granted. Understanding how transistors function is of paramount importance to anyone interested in understanding modern electronics.My intent here is to focus as exclusively as possible on the practical function and application of bipolar transistors, rather than to explore the quantum world of semiconductor theory. Discussions of holes and electrons are better left to another chapter in my opinion. Here I want to explore how to use these components, not analyze their intimate internal details. I don't mean to downplay the importance of understanding semiconductor physics, but sometimes an intense focus on solid-state physics detracts from understanding these devices' functions on a component level. In taking this approach, however, I assume that the reader possesses a certain minimum knowledge of semiconductors: the difference between "P" and "N" doped semiconductors, the functional characteristics of a PN (diode) junction, and the meanings of the terms "reverse biased" and "forward biased." If these concepts are unclear to you, it is best to refer to earlier chapters in this book before proceeding with this one.
A bipolar transistor consists of a three-layer "sandwich" of doped (extrinsic) semiconductor materials, either P-N-P or N-P-N. Each layer forming the transistor has a specific name, and each layer is provided with a wire contact for connection to a circuit. Shown here are schematic symbols and physical diagrams of these two transistor types:
The only functional difference between a PNP transistor and an NPN transistor is the proper biasing (polarity) of the junctions when operating. For any given state of operation, the current directions and voltage polarities for each type of transistor are exactly opposite each other.
Bipolar transistors work as current-controlled current regulators. In other words, they restrict the amount of current that can go through them according to a smaller, controlling current. The main current that is controlled goes from collector to emitter, or from emitter to collector, depending on the type of transistor it is (PNP or NPN, respectively). The small current that controls the main current goes from base to emitter, or from emitter to base, once again depending on the type of transistor it is (PNP or NPN, respectively). According to the confusing standards of semiconductor symbology, the arrow always points against the direction of electron flow:
Bipolar transistors are called bipolar because the main flow of electrons through them takes place in two types of semiconductor material: P and N, as the main current goes from emitter to collector (or visa-versa). In other words, two types of charge carriers -- electrons and holes -- comprise this main current through the transistor.
As you can see, the controlling current and the controlled current always mesh together through the emitter wire, and their electrons always flow against the direction of the transistor's arrow. This is the first and foremost rule in the use of transistors: all currents must be going in the proper directions for the device to work as a current regulator. The small, controlling current is usually referred to simply as the base current because it is the only current that goes through the base wire of the transistor. Conversely, the large, controlled current is referred to as the collector current because it is the only current that goes through the collector wire. The emitter current is the sum of the base and collector currents, in compliance with Kirchhoff's Current Law.
If there is no current through the base of the transistor, it shuts off like an open switch and prevents current through the collector. If there is a base current, then the transistor turns on like a closed switch and allows a proportional amount of current through the collector. Collector current is primarily limited by the base current, regardless of the amount of voltage available to push it. The next section will explore in more detail the use of bipolar transistors as switching elements.
- REVIEW:
- Bipolar transistors are so named because the controlled current must go through two types of semiconductor material: P and N. The current consists of both electron and hole flow, in different parts of the transistor.
- Bipolar transistors consist of either a P-N-P or an N-P-N semiconductor "sandwich" structure.
- The three leads of a bipolar transistor are called the Emitter, Base, and Collector.
- Transistors function as current regulators by allowing a small current to control a larger current. The amount of current allowed between collector and emitter is primarily determined by the amount of current moving between base and emitter.
- In order for a transistor to properly function as a current regulator, the controlling (base) current and the controlled (collector) currents must be going in the proper directions: meshing additively at the emitter and going against the emitter arrow symbol.
The transistor as a switch
Because a transistor's collector current is proportionally limited by its base current, it can be used as a sort of current-controlled switch. A relatively small flow of electrons sent through the base of the transistor has the ability to exert control over a much larger flow of electrons through the collector.Suppose we had a lamp that we wanted to turn on and off by means of a switch. Such a circuit would be extremely simple:
For the sake of illustration, let's insert a transistor in place of the switch to show how it can control the flow of electrons through the lamp. Remember that the controlled current through a transistor must go between collector and emitter. Since it's the current through the lamp that we want to control, we must position the collector and emitter of our transistor where the two contacts of the switch are now. We must also make sure that the lamp's current will move against the direction of the emitter arrow symbol to ensure that the transistor's junction bias will be correct:
In this example I happened to choose an NPN transistor. A PNP transistor could also have been chosen for the job, and its application would look like this:
The choice between NPN and PNP is really arbitrary. All that matters is that the proper current directions are maintained for the sake of correct junction biasing (electron flow going against the transistor symbol's arrow).
Going back to the NPN transistor in our example circuit, we are faced with the need to add something more so that we can have base current. Without a connection to the base wire of the transistor, base current will be zero, and the transistor cannot turn on, resulting in a lamp that is always off. Remember that for an NPN transistor, base current must consist of electrons flowing from emitter to base (against the emitter arrow symbol, just like the lamp current). Perhaps the simplest thing to do would be to connect a switch between the base and collector wires of the transistor like this:
If the switch is open, the base wire of the transistor will be left "floating" (not connected to anything) and there will be no current through it. In this state, the transistor is said to be cutoff. If the switch is closed, however, electrons will be able to flow from the emitter through to the base of the transistor, through the switch and up to the left side of the lamp, back to the positive side of the battery. This base current will enable a much larger flow of electrons from the emitter through to the collector, thus lighting up the lamp. In this state of maximum circuit current, the transistor is said to be saturated.
Of course, it may seem pointless to use a transistor in this capacity to control the lamp. After all, we're still using a switch in the circuit, aren't we? If we're still using a switch to control the lamp -- if only indirectly -- then what's the point of having a transistor to control the current? Why not just go back to our original circuit and use the switch directly to control the lamp current?
There are a couple of points to be made here, actually. First is the fact that when used in this manner, the switch contacts need only handle what little base current is necessary to turn the transistor on, while the transistor itself handles the majority of the lamp's current. This may be an important advantage if the switch has a low current rating: a small switch may be used to control a relatively high-current load. Perhaps more importantly, though, is the fact that the current-controlling behavior of the transistor enables us to use something completely different to turn the lamp on or off. Consider this example, where a solar cell is used to control the transistor, which in turn controls the lamp:
Or, we could use a thermocouple to provide the necessary base current to turn the transistor on:
Even a microphone of sufficient voltage and current output could be used to turn the transistor on, provided its output is rectified from AC to DC so that the emitter-base PN junction within the transistor will always be forward-biased:
The point should be quite apparent by now: any sufficient source of DC current may be used to turn the transistor on, and that source of current need only be a fraction of the amount of current needed to energize the lamp. Here we see the transistor functioning not only as a switch, but as a true amplifier: using a relatively low-power signal to control a relatively large amount of power. Please note that the actual power for lighting up the lamp comes from the battery to the right of the schematic. It is not as though the small signal current from the solar cell, thermocouple, or microphone is being magically transformed into a greater amount of power. Rather, those small power sources are simply controlling the battery's power to light up the lamp.
- REVIEW:
- Transistors may be used as switching elements to control DC power to a load. The switched (controlled) current goes between emitter and collector, while the controlling current goes between emitter and base.
- When a transistor has zero current through it, it is said to be in a state of cutoff (fully nonconducting).
- When a transistor has maximum current through it, it is said to be in a state of saturation (fully conducting).
Meter check of a transistor
Bipolar transistors are constructed of a three-layer semiconductor "sandwich," either PNP or NPN. As such, they register as two diodes connected back-to-back when tested with a multimeter's "resistance" or "diode check" functions:
Here I'm assuming the use of a multimeter with only a single continuity range (resistance) function to check the PN junctions. Some multimeters are equipped with two separate continuity check functions: resistance and "diode check," each with its own purpose. If your meter has a designated "diode check" function, use that rather than the "resistance" range, and the meter will display the actual forward voltage of the PN junction and not just whether or not it conducts current.
Meter readings will be exactly opposite, of course, for an NPN transistor, with both PN junctions facing the other way. If a multimeter with a "diode check" function is used in this test, it will be found that the emitter-base junction possesses a slightly greater forward voltage drop than the collector-base junction. This forward voltage difference is due to the disparity in doping concentration between the emitter and collector regions of the transistor: the emitter is a much more heavily doped piece of semiconductor material than the collector, causing its junction with the base to produce a higher forward voltage drop.
Knowing this, it becomes possible to determine which wire is which on an unmarked transistor. This is important because transistor packaging, unfortunately, is not standardized. All bipolar transistors have three wires, of course, but the positions of the three wires on the actual physical package are not arranged in any universal, standardized order.
Suppose a technician finds a bipolar transistor and proceeds to measure continuity with a multimeter set in the "diode check" mode. Measuring between pairs of wires and recording the values displayed by the meter, the technician obtains the following data:
- Meter touching wire 1 (+) and 2 (-): "OL"
- Meter touching wire 1 (-) and 2 (+): "OL"
- Meter touching wire 1 (+) and 3 (-): 0.655 volts
- Meter touching wire 1 (-) and 3 (+): "OL"
- Meter touching wire 2 (+) and 3 (-): 0.621 volts
- Meter touching wire 2 (-) and 3 (+): "OL"
Now we look for the one wire common to both sets of conductive readings. It must be the base connection of the transistor, because the base is the only layer of the three-layer device common to both sets of PN junctions (emitter-base and collector-base). In this example, that wire is number 3, being common to both the 1-3 and the 2-3 test point combinations. In both those sets of meter readings, the black (-) meter test lead was touching wire 3, which tells us that the base of this transistor is made of N-type semiconductor material (black = negative). Thus, the transistor is an PNP type with base on wire 3, emitter on wire 1 and collector on wire 2:
Please note that the base wire in this example is not the middle lead of the transistor, as one might expect from the three-layer "sandwich" model of a bipolar transistor. This is quite often the case, and tends to confuse new students of electronics. The only way to be sure which lead is which is by a meter check, or by referencing the manufacturer's "data sheet" documentation on that particular part number of transistor.
Knowing that a bipolar transistor behaves as two back-to-back diodes when tested with a conductivity meter is helpful for identifying an unknown transistor purely by meter readings. It is also helpful for a quick functional check of the transistor. If the technician were to measure continuity in any more than two or any less than two of the six test lead combinations, he or she would immediately know that the transistor was defective (or else that it wasn't a bipolar transistor but rather something else -- a distinct possibility if no part numbers can be referenced for sure identification!). However, the "two diode" model of the transistor fails to explain how or why it acts as an amplifying device.
To better illustrate this paradox, let's examine one of the transistor switch circuits using the physical diagram rather than the schematic symbol to represent the transistor. This way the two PN junctions will be easier to see:
A grey-colored diagonal arrow shows the direction of electron flow through the emitter-base junction. This part makes sense, since the electrons are flowing from the N-type emitter to the P-type base: the junction is obviously forward-biased. However, the base-collector junction is another matter entirely. Notice how the grey-colored thick arrow is pointing in the direction of electron flow (upwards) from base to collector. With the base made of P-type material and the collector of N-type material, this direction of electron flow is clearly backwards to the direction normally associated with a PN junction! A normal PN junction wouldn't permit this "backward" direction of flow, at least not without offering significant opposition. However, when the transistor is saturated, there is very little opposition to electrons all the way from emitter to collector, as evidenced by the lamp's illumination!
Clearly then, something is going on here that defies the simple "two-diode" explanatory model of the bipolar transistor. When I was first learning about transistor operation, I tried to construct my own transistor from two back-to-back diodes, like this:
My circuit didn't work, and I was mystified. However useful the "two diode" description of a transistor might be for testing purposes, it doesn't explain how a transistor can behave as a controlled switch.
What happens in a transistor is this: the reverse bias of the base-collector junction prevents collector current when the transistor is in cutoff mode (that is, when there is no base current). However, when the base-emitter junction is forward biased by the controlling signal, the normally-blocking action of the base-collector junction is overridden and current is permitted through the collector, despite the fact that electrons are going the "wrong way" through that PN junction. This action is dependent on the quantum physics of semiconductor junctions, and can only take place when the two junctions are properly spaced and the doping concentrations of the three layers are properly proportioned. Two diodes wired in series fail to meet these criteria, and so the top diode can never "turn on" when it is reversed biased, no matter how much current goes through the bottom diode in the base wire loop.
That doping concentrations play a crucial part in the special abilities of the transistor is further evidenced by the fact that collector and emitter are not interchangeable. If the transistor is merely viewed as two back-to-back PN junctions, or merely as a plain N-P-N or P-N-P sandwich of materials, it may seem as though either end of the transistor could serve as collector or emitter. This, however, is not true. If connected "backwards" in a circuit, a base-collector current will fail to control current between collector and emitter. Despite the fact that both the emitter and collector layers of a bipolar transistor are of the same doping type (either N or P), they are definitely not identical!
So, current through the emitter-base junction allows current through the reverse-biased base-collector junction. The action of base current can be thought of as "opening a gate" for current through the collector. More specifically, any given amount of emitter-to-base current permits a limited amount of base-to-collector current. For every electron that passes through the emitter-base junction and on through the base wire, there is allowed a certain, restricted number of electrons to pass through the base-collector junction and no more.
In the next section, this current-limiting behavior of the transistor will be investigated in more detail.
- REVIEW:
- Tested with a multimeter in the "resistance" or "diode check" modes, a transistor behaves like two back-to-back PN (diode) junctions.
- The emitter-base PN junction has a slightly greater forward voltage drop than the collector-base PN junction, due to more concentrated doping of the emitter semiconductor layer.
- The reverse-biased base-collector junction normally blocks any current from going through the transistor between emitter and collector. However, that junction begins to conduct if current is drawn through the base wire. Base current can be thought of as "opening a gate" for a certain, limited amount of current through the collector.
Active mode operation
When a transistor is in the fully-off state (like an open switch), it is said to be cutoff. Conversely, when it is fully conductive between emitter and collector (passing as much current through the collector as the collector power supply and load will allow), it is said to be saturated. These are the two modes of operation explored thus far in using the transistor as a switch.However, bipolar transistors don't have to be restricted to these two extreme modes of operation. As we learned in the previous section, base current "opens a gate" for a limited amount of current through the collector. If this limit for the controlled current is greater than zero but less than the maximum allowed by the power supply and load circuit, the transistor will "throttle" the collector current in a mode somewhere between cutoff and saturation. This mode of operation is called the active mode.
An automotive analogy for transistor operation is as follows: cutoff is the condition where there is no motive force generated by the mechanical parts of the car to make it move. In cutoff mode, the brake is engaged (zero base current), preventing motion (collector current). Active mode is when the automobile is cruising at a constant, controlled speed (constant, controlled collector current) as dictated by the driver. Saturation is when the automobile is driving up a steep hill that prevents it from going as fast as the driver would wish. In other words, a "saturated" automobile is one where the accelerator pedal is pushed all the way down (base current calling for more collector current than can be provided by the power supply/load circuit).
I'll set up a circuit for SPICE simulation to demonstrate what happens when a transistor is in its active mode of operation:
"Q" is the standard letter designation for a transistor in a schematic diagram, just as "R" is for resistor and "C" is for capacitor. In this circuit, we have an NPN transistor powered by a battery (V1) and controlled by current through a current source (I1). A current source is a device that outputs a specific amount of current, generating as much or as little voltage as necessary across its terminals to ensure that exact amount of current through it. Current sources are notoriously difficult to find in nature (unlike voltage sources, which by contrast attempt to maintain a constant voltage, outputting as much or as little current in the fulfillment of that task), but can be simulated with a small collection of electronic components. As we are about to see, transistors themselves tend to mimic the constant-current behavior of a current source in their ability to regulate current at a fixed value.
In the SPICE simulation, I'll set the current source at a constant value of 20 µA, then vary the voltage source (V1) over a range of 0 to 2 volts and monitor how much current goes through it. The "dummy" battery (Vammeter) with its output of 0 volts serves merely to provide SPICE with a circuit element for current measurement.
bipolar transistor simulation i1 0 1 dc 20u q1 2 1 0 mod1 vammeter 3 2 dc 0 v1 3 0 dc .model mod1 npn .dc v1 0 2 0.05 .plot dc i(vammeter) .end
type npn is 1.00E-16 bf 100.000 nf 1.000 br 1.000 nr 1.000
v1 i(ammeter) -1.000E-03 0.000E+00 1.000E-03 2.000E-03 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 0.000E+00 -1.980E-05 . * . . 5.000E-02 9.188E-05 . .* . . 1.000E-01 6.195E-04 . . * . . 1.500E-01 1.526E-03 . . . * . 2.000E-01 1.914E-03 . . . *. 2.500E-01 1.987E-03 . . . * 3.000E-01 1.998E-03 . . . * 3.500E-01 2.000E-03 . . . * 4.000E-01 2.000E-03 . . . * 4.500E-01 2.000E-03 . . . * 5.000E-01 2.000E-03 . . . * 5.500E-01 2.000E-03 . . . * 6.000E-01 2.000E-03 . . . * 6.500E-01 2.000E-03 . . . * 7.000E-01 2.000E-03 . . . * 7.500E-01 2.000E-03 . . . * 8.000E-01 2.000E-03 . . . * 8.500E-01 2.000E-03 . . . * 9.000E-01 2.000E-03 . . . * 9.500E-01 2.000E-03 . . . * 1.000E+00 2.000E-03 . . . * 1.050E+00 2.000E-03 . . . * 1.100E+00 2.000E-03 . . . * 1.150E+00 2.000E-03 . . . * 1.200E+00 2.000E-03 . . . * 1.250E+00 2.000E-03 . . . * 1.300E+00 2.000E-03 . . . * 1.350E+00 2.000E-03 . . . * 1.400E+00 2.000E-03 . . . * 1.450E+00 2.000E-03 . . . * 1.500E+00 2.000E-03 . . . * 1.550E+00 2.000E-03 . . . * 1.600E+00 2.000E-03 . . . * 1.650E+00 2.000E-03 . . . * 1.700E+00 2.000E-03 . . . * 1.750E+00 2.000E-03 . . . * 1.800E+00 2.000E-03 . . . * 1.850E+00 2.000E-03 . . . * 1.900E+00 2.000E-03 . . . * 1.950E+00 2.000E-03 . . . * 2.000E+00 2.000E-03 . . . * - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
The constant base current of 20 µA sets a collector current limit of 2 mA, exactly 100 times as much. Notice how flat the curve is for collector current over the range of battery voltage from 0 to 2 volts. The only exception to this featureless plot is at the very beginning, where the battery increases from 0 volts to 0.25 volts. There, the collector current increases rapidly from 0 amps to its limit of 2 mA.
Let's see what happens if we vary the battery voltage over a wider range, this time from 0 to 50 volts. We'll keep the base current steady at 20 µA:
bipolar transistor simulation i1 0 1 dc 20u q1 2 1 0 mod1 vammeter 3 2 dc 0 v1 3 0 dc .model mod1 npn .dc v1 0 50 2 .plot dc i(vammeter) .end
type npn is 1.00E-16 bf 100.000 nf 1.000 br 1.000 nr 1.000
v1 i(ammeter) -1.000E-03 0.000E+00 1.000E-03 2.000E-03 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 0.000E+00 -1.980E-05 . * . . 2.000E+00 2.000E-03 . . . * 4.000E+00 2.000E-03 . . . * 6.000E+00 2.000E-03 . . . * 8.000E+00 2.000E-03 . . . * 1.000E+01 2.000E-03 . . . * 1.200E+01 2.000E-03 . . . * 1.400E+01 2.000E-03 . . . * 1.600E+01 2.000E-03 . . . * 1.800E+01 2.000E-03 . . . * 2.000E+01 2.000E-03 . . . * 2.200E+01 2.000E-03 . . . * 2.400E+01 2.000E-03 . . . * 2.600E+01 2.000E-03 . . . * 2.800E+01 2.000E-03 . . . * 3.000E+01 2.000E-03 . . . * 3.200E+01 2.000E-03 . . . * 3.400E+01 2.000E-03 . . . * 3.600E+01 2.000E-03 . . . * 3.800E+01 2.000E-03 . . . * 4.000E+01 2.000E-03 . . . * 4.200E+01 2.000E-03 . . . * 4.400E+01 2.000E-03 . . . * 4.600E+01 2.000E-03 . . . * 4.800E+01 2.000E-03 . . . * 5.000E+01 2.000E-03 . . . * - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Same result! The collector current holds absolutely steady at 2 mA despite the fact that the battery (v1) voltage varies all the way from 0 to 50 volts. It would appear from our simulation that collector-to-emitter voltage has little effect over collector current, except at very low levels (just above 0 volts). The transistor is acting as a current regulator, allowing exactly 2 mA through the collector and no more.
Now let's see what happens if we increase the controlling (I1) current from 20 µA to 75 µA, once again sweeping the battery (V1) voltage from 0 to 50 volts and graphing the collector current:
bipolar transistor simulation i1 0 1 dc 75u q1 2 1 0 mod1 vammeter 3 2 dc 0 v1 3 0 dc .model mod1 npn .dc v1 0 50 2 .plot dc i(vammeter) .end
type npn is 1.00E-16 bf 100.000 nf 1.000 br 1.000 nr 1.000
v1 i(ammeter) -5.000E-03 0.000E+00 5.000E-03 1.000E-02 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 0.000E+00 -7.426E-05 . * . . 2.000E+00 7.500E-03 . . . * . 4.000E+00 7.500E-03 . . . * . 6.000E+00 7.500E-03 . . . * . 8.000E+00 7.500E-03 . . . * . 1.000E+01 7.500E-03 . . . * . 1.200E+01 7.500E-03 . . . * . 1.400E+01 7.500E-03 . . . * . 1.600E+01 7.500E-03 . . . * . 1.800E+01 7.500E-03 . . . * . 2.000E+01 7.500E-03 . . . * . 2.200E+01 7.500E-03 . . . * . 2.400E+01 7.500E-03 . . . * . 2.600E+01 7.500E-03 . . . * . 2.800E+01 7.500E-03 . . . * . 3.000E+01 7.500E-03 . . . * . 3.200E+01 7.500E-03 . . . * . 3.400E+01 7.500E-03 . . . * . 3.600E+01 7.500E-03 . . . * . 3.800E+01 7.500E-03 . . . * . 4.000E+01 7.500E-03 . . . * . 4.200E+01 7.500E-03 . . . * . 4.400E+01 7.500E-03 . . . * . 4.600E+01 7.500E-03 . . . * . 4.800E+01 7.500E-03 . . . * . 5.000E+01 7.500E-03 . . . * . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Not surprisingly, SPICE gives us a similar plot: a flat line, holding steady this time at 7.5 mA -- exactly 100 times the base current -- over the range of battery voltages from just above 0 volts to 50 volts. It appears that the base current is the deciding factor for collector current, the V1 battery voltage being irrelevant so long as it's above a certain minimum level.
This voltage/current relationship is entirely different from what we're used to seeing across a resistor. With a resistor, current increases linearly as the voltage across it increases. Here, with a transistor, current from emitter to collector stays limited at a fixed, maximum value no matter how high the voltage across emitter and collector increases.
Often it is useful to superimpose several collector current/voltage graphs for different base currents on the same graph. A collection of curves like this -- one curve plotted for each distinct level of base current -- for a particular transistor is called the transistor's characteristic curves:
Each curve on the graph reflects the collector current of the transistor, plotted over a range of collector-to-emitter voltages, for a given amount of base current. Since a transistor tends to act as a current regulator, limiting collector current to a proportion set by the base current, it is useful to express this proportion as a standard transistor performance measure. Specifically, the ratio of collector current to base current is known as the Beta ratio (symbolized by the Greek letter β):
Sometimes the β ratio is designated as "hfe," a label used in a branch of mathematical semiconductor analysis known as "hybrid parameters" which strives to achieve very precise predictions of transistor performance with detailed equations. Hybrid parameter variables are many, but they are all labeled with the general letter "h" and a specific subscript. The variable "hfe" is just another (standardized) way of expressing the ratio of collector current to base current, and is interchangeable with "β." Like all ratios, β is unitless.
β for any transistor is determined by its design: it cannot be altered after manufacture. However, there are so many physical variables impacting β that it is rare to have two transistors of the same design exactly match. If a circuit design relies on equal β ratios between multiple transistors, "matched sets" of transistors may be purchased at extra cost. However, it is generally considered bad design practice to engineer circuits with such dependencies.
It would be nice if the β of a transistor remained stable for all operating conditions, but this is not true in real life. For an actual transistor, the β ratio may vary by a factor of over 3 within its operating current limits. For example, a transistor with advertised β of 50 may actually test with Ic/Ib ratios as low as 30 and as high as 100, depending on the amount of collector current, the transistor's temperature, and frequency of amplified signal, among other factors. For tutorial purposes it is adequate to assume a constant β for any given transistor (which is what SPICE tends to do in a simulation), but just realize that real life is not that simple!
Sometimes it is helpful for comprehension to "model" complex electronic components with a collection of simpler, better-understood components. The following is a popular model shown in many introductory electronics texts:
This model casts the transistor as a combination of diode and rheostat (variable resistor). Current through the base-emitter diode controls the resistance of the collector-emitter rheostat (as implied by the dashed line connecting the two components), thus controlling collector current. An NPN transistor is modeled in the figure shown, but a PNP transistor would be only slightly different (only the base-emitter diode would be reversed). This model succeeds in illustrating the basic concept of transistor amplification: how the base current signal can exert control over the collector current. However, I personally don't like this model because it tends to miscommunicate the notion of a set amount of collector-emitter resistance for a given amount of base current. If this were true, the transistor wouldn't regulate collector current at all like the characteristic curves show. Instead of the collector current curves flattening out after their brief rise as the collector-emitter voltage increases, the collector current would be directly proportional to collector-emitter voltage, rising steadily in a straight line on the graph.
A better transistor model, often seen in more advanced textbooks, is this:
It casts the transistor as a combination of diode and current source, the output of the current source being set at a multiple (β ratio) of the base current. This model is far more accurate in depicting the true input/output characteristics of a transistor: base current establishes a certain amount of collector current, rather than a certain amount of collector-emitter resistance as the first model implies. Also, this model is favored when performing network analysis on transistor circuits, the current source being a well-understood theoretical component. Unfortunately, using a current source to model the transistor's current-controlling behavior can be misleading: in no way will the transistor ever act as a source of electrical energy, which the current source symbol implies is a possibility.
My own personal suggestion for a transistor model substitutes a constant-current diode for the current source:
Since no diode ever acts as a source of electrical energy, this analogy escapes the false implication of the current source model as a source of power, while depicting the transistor's constant-current behavior better than the rheostat model. Another way to describe the constant-current diode's action would be to refer to it as a current regulator, so this transistor illustration of mine might also be described as a diode-current regulator model. The greatest disadvantage I see to this model is the relative obscurity of constant-current diodes. Many people may be unfamiliar with their symbology or even of their existence, unlike either rheostats or current sources, which are commonly known.
- REVIEW:
- A transistor is said to be in its active mode if it is operating somewhere between fully on (saturated) and fully off (cutoff).
- Base current tends to regulate collector current. By regulate, we mean that no more collector current may exist than what is allowed by the base current.
- The ratio between collector current and base current is called "Beta" (β) or "hfe".
- β ratios are different for every transistor, and they tend to change for different operating conditions.
The common-emitter amplifier
At the beginning of this chapter we saw how transistors could be used as switches, operating in either their "saturation" or "cutoff" modes. In the last section we saw how transistors behave within their "active" modes, between the far limits of saturation and cutoff. Because transistors are able to control current in an analog (infinitely divisible) fashion, they find use as amplifiers for analog signals.One of the simpler transistor amplifier circuits to study is the one used previously for illustrating the transistor's switching ability:
It is called the common-emitter configuration because (ignoring the power supply battery) both the signal source and the load share the emitter lead as a common connection point. This is not the only way in which a transistor may be used as an amplifier, as we will see in later sections of this chapter:
Before, this circuit was shown to illustrate how a relatively small current from a solar cell could be used to saturate a transistor, resulting in the illumination of a lamp. Knowing now that transistors are able to "throttle" their collector currents according to the amount of base current supplied by an input signal source, we should be able to see that the brightness of the lamp in this circuit is controllable by the solar cell's light exposure. When there is just a little light shone on the solar cell, the lamp will glow dimly. The lamp's brightness will steadily increase as more light falls on the solar cell.
Suppose that we were interested in using the solar cell as a light intensity instrument. We want to be able to measure the intensity of incident light with the solar cell by using its output current to drive a meter movement. It is possible to directly connect a meter movement to a solar cell for this purpose. In fact, the simplest light-exposure meters for photography work are designed like this:
While this approach might work for moderate light intensity measurements, it would not work as well for low light intensity measurements. Because the solar cell has to supply the meter movement's power needs, the system is necessarily limited in its sensitivity. Supposing that our need here is to measure very low-level light intensities, we are pressed to find another solution.
Perhaps the most direct solution to this measurement problem is to use a transistor to amplify the solar cell's current so that more meter movement needle deflection may be obtained for less incident light. Consider this approach:
Current through the meter movement in this circuit will be β times the solar cell current. With a transistor β of 100, this represents a substantial increase in measurement sensitivity. It is prudent to point out that the additional power to move the meter needle comes from the battery on the far right of the circuit, not the solar cell itself. All the solar cell's current does is control battery current to the meter to provide a greater meter reading than the solar cell could provide unaided.
Because the transistor is a current-regulating device, and because meter movement indications are based on the amount of current through their movement coils, meter indication in this circuit should depend only on the amount of current from the solar cell, not on the amount of voltage provided by the battery. This means the accuracy of the circuit will be independent of battery condition, a significant feature! All that is required of the battery is a certain minimum voltage and current output ability to be able to drive the meter full-scale if needed.
Another way in which the common-emitter configuration may be used is to produce an output voltage derived from the input signal, rather than a specific output current. Let's replace the meter movement with a plain resistor and measure voltage between collector and emitter:
With the solar cell darkened (no current), the transistor will be in cutoff mode and behave as an open switch between collector and emitter. This will produce maximum voltage drop between collector and emitter for maximum Voutput, equal to the full voltage of the battery.
At full power (maximum light exposure), the solar cell will drive the transistor into saturation mode, making it behave like a closed switch between collector and emitter. The result will be minimum voltage drop between collector and emitter, or almost zero output voltage. In actuality, a saturated transistor can never achieve zero voltage drop between collector and emitter due to the two PN junctions through which collector current must travel. However, this "collector-emitter saturation voltage" will be fairly low, around several tenths of a volt, depending on the specific transistor used.
For light exposure levels somewhere between zero and maximum solar cell output, the transistor will be in its active mode, and the output voltage will be somewhere between zero and full battery voltage. An important quality to note here about the common-emitter configuration is that the output voltage is inversely proportional to the input signal strength. That is, the output voltage decreases as the input signal increases. For this reason, the common-emitter amplifier configuration is referred to as an inverting amplifier.
A quick SPICE simulation will verify our qualitative conclusions about this amplifier circuit:
common-emitter amplifier i1 0 1 dc q1 2 1 0 mod1 r 3 2 5000 v1 3 0 dc 15 .model mod1 npn .dc i1 0 50u 2u .plot dc v(2,0) .end
type npn is 1.00E-16 bf 100.000 nf 1.000 br 1.000 nr 1.000
i1 v(2) 0.000E+00 5.000E+00 1.000E+01 1.500E+01 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 0.000E+00 1.500E+01 . . . * 2.000E-06 1.400E+01 . . . * . 4.000E-06 1.300E+01 . . . * . 6.000E-06 1.200E+01 . . . * . 8.000E-06 1.100E+01 . . . * . 1.000E-05 1.000E+01 . . * . 1.200E-05 9.000E+00 . . * . . 1.400E-05 8.000E+00 . . * . . 1.600E-05 7.000E+00 . . * . . 1.800E-05 6.000E+00 . . * . . 2.000E-05 5.000E+00 . * . . 2.200E-05 4.000E+00 . * . . . 2.400E-05 3.000E+00 . * . . . 2.600E-05 2.000E+00 . * . . . 2.800E-05 1.000E+00 . * . . . 3.000E-05 2.261E-01 .* . . . 3.200E-05 1.850E-01 .* . . . 3.400E-05 1.694E-01 * . . . 3.600E-05 1.597E-01 * . . . 3.800E-05 1.527E-01 * . . . 4.000E-05 1.472E-01 * . . . 4.200E-05 1.427E-01 * . . . 4.400E-05 1.388E-01 * . . . 4.600E-05 1.355E-01 * . . . 4.800E-05 1.325E-01 * . . . 5.000E-05 1.299E-01 * . . . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
At the beginning of the simulation where the current source (solar cell) is outputting zero current, the transistor is in cutoff mode and the full 15 volts from the battery is shown at the amplifier output (between nodes 2 and 0). As the solar cell's current begins to increase, the output voltage proportionally decreases, until the transistor reaches saturation at 30 µA of base current (3 mA of collector current). Notice how the output voltage trace on the graph is perfectly linear (1 volt steps from 15 volts to 1 volt) until the point of saturation, where it never quite reaches zero. This is the effect mentioned earlier, where a saturated transistor can never achieve exactly zero voltage drop between collector and emitter due to internal junction effects. What we do see is a sharp output voltage decrease from 1 volt to 0.2261 volts as the input current increases from 28 µA to 30 µA, and then a continuing decrease in output voltage from then on (albeit in progressively smaller steps). The lowest the output voltage ever gets in this simulation is 0.1299 volts, asymptotically approaching zero.
So far, we've seen the transistor used as an amplifier for DC signals. In the solar cell light meter example, we were interested in amplifying the DC output of the solar cell to drive a DC meter movement, or to produce a DC output voltage. However, this is not the only way in which a transistor may be employed as an amplifier. In many cases, what is desired is an AC amplifier for amplifying alternating current and voltage signals. One common application of this is in audio electronics (radios, televisions, and public-address systems). Earlier, we saw an example where the audio output of a tuning fork could be used to activate a transistor as a switch. Let's see if we can modify that circuit to send power to a speaker rather than to a lamp:
In the original circuit, a full-wave bridge rectifier was used to convert the microphone's AC output signal into a DC voltage to drive the input of the transistor. All we cared about here was turning the lamp on with a sound signal from the microphone, and this arrangement sufficed for that purpose. But now we want to actually reproduce the AC signal and drive a speaker. This means we cannot rectify the microphone's output anymore, because we need undistorted AC signal to drive the transistor! Let's remove the bridge rectifier and replace the lamp with a speaker:
Since the microphone may produce voltages exceeding the forward voltage drop of the base-emitter PN (diode) junction, I've placed a resistor in series with the microphone. Let's simulate this circuit now in SPICE and see what happens:
common-emitter amplifier vinput 1 0 sin (0 1.5 2000 0 0) r1 1 2 1k q1 3 2 0 mod1 rspkr 3 4 8 v1 4 0 dc 15 .model mod1 npn .tran 0.02m 0.74m .plot tran v(1,0) i(v1) .end
legend: *: v(1) +: i(v1) v(1) (*)--- -2.000E+00 -1.000E+00 0.000E+00 1.000E+00 2.000E+00 (+)--- -8.000E-02 -6.000E-02 -4.000E-02 -2.000E-02 0.000E+00 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 0.000E+00 . . * . + 3.725E-01 . . . * . + 7.195E-01 . . . * . +. 1.024E+00 . . . *+ . 1.264E+00 . . + . * . 1.420E+00 . . + . . * . 1.493E+00 . +. . . * . 1.470E+00 . .+ . . * . 1.351E+00 . . + . . * . 1.154E+00 . . . + . * . 8.791E-01 . . . * . + . 5.498E-01 . . . * . + 1.877E-01 . . . * . + -1.872E-01 . . * . . + -5.501E-01 . . * . . + -8.815E-01 . . * . . + -1.151E+00 . * . . . + -1.352E+00 . * . . . + -1.472E+00 . * . . . + -1.491E+00 . * . . . + -1.422E+00 . * . . . + -1.265E+00 . * . . . + -1.022E+00 . * . . + -7.205E-01 . . * . . + -3.723E-01 . . * . . + 3.040E-06 . . * . + 3.724E-01 . . . * . + 7.205E-01 . . . * . +. 1.022E+00 . . . * + . 1.265E+00 . . + . * . 1.422E+00 . . + . . * . 1.491E+00 . +. . . * . 1.473E+00 . .+ . . * . 1.352E+00 . . + . . * . 1.151E+00 . . . + . * . 8.814E-01 . . . * . + . 5.501E-01 . . . * . + 1.880E-01 . . . * . + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
The simulation plots both the input voltage (an AC signal of 1.5 volt peak amplitude and 2000 Hz frequency) and the current through the 15 volt battery, which is the same as the current through the speaker. What we see here is a full AC sine wave alternating in both positive and negative directions, and a half-wave output current waveform that only pulses in one direction. If we were actually driving a speaker with this waveform, the sound produced would be horribly distorted.
What's wrong with the circuit? Why won't it faithfully reproduce the entire AC waveform from the microphone? The answer to this question is found by close inspection of the transistor diode-regulating diode model:
Collector current is controlled, or regulated, through the constant-current mechanism according to the pace set by the current through the base-emitter diode. Note that both current paths through the transistor are monodirectional: one way only! Despite our intent to use the transistor to amplify an AC signal, it is essentially a DC device, capable of handling currents in a single direction only. We may apply an AC voltage input signal between the base and emitter, but electrons cannot flow in that circuit during the part of the cycle that reverse-biases the base-emitter diode junction. Therefore, the transistor will remain in cutoff mode throughout that portion of the cycle. It will "turn on" in its active mode only when the input voltage is of the correct polarity to forward-bias the base-emitter diode, and only when that voltage is sufficiently high to overcome the diode's forward voltage drop. Remember that bipolar transistors are current-controlled devices: they regulate collector current based on the existence of base-to-emitter current, not base-to-emitter voltage.
The only way we can get the transistor to reproduce the entire waveform as current through the speaker is to keep the transistor in its active mode the entire time. This means we must maintain current through the base during the entire input waveform cycle. Consequently, the base-emitter diode junction must be kept forward-biased at all times. Fortunately, this can be accomplished with the aid of a DC bias voltage added to the input signal. By connecting a sufficient DC voltage in series with the AC signal source, forward-bias can be maintained at all points throughout the wave cycle:
common-emitter amplifier vinput 1 5 sin (0 1.5 2000 0 0) vbias 5 0 dc 2.3 r1 1 2 1k q1 3 2 0 mod1 rspkr 3 4 8 v1 4 0 dc 15 .model mod1 npn .tran 0.02m 0.78m .plot tran v(1,0) i(v1) .end
legend: *: v(1) +: i(v1) v(1) (*)--- 0.000E+00 1.000E+00 2.000E+00 3.000E+00 4.000E+00 (+)--- -3.000E-01 -2.000E-01 -1.000E-01 0.000E+00 1.000E-01 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2.300E+00 . . + . * . . 2.673E+00 . . + . * . . 3.020E+00 . +. . * . 3.322E+00 . + . . . * . 3.563E+00 . + . . . * . 3.723E+00 . + . . . * . 3.790E+00 . + . . . *. 3.767E+00 . + . . . *. 3.657E+00 . + . . . * . 3.452E+00 . + . . . * . 3.177E+00 . + . . . * . 2.850E+00 . .+ . * . . 2.488E+00 . . + . * . . 2.113E+00 . . + . * . . 1.750E+00 . . * . + . . 1.419E+00 . . * . + . . 1.148E+00 . . * . + . . 9.493E-01 . *. . + . . 8.311E-01 . * . . + . 8.050E-01 . * . . + . 8.797E-01 . * . . +. . 1.039E+00 . .* . + . . 1.275E+00 . . * . + . . 1.579E+00 . . * . + . . 1.929E+00 . . *+ . . 2.300E+00 . . + . * . . 2.673E+00 . . + . * . . 3.019E+00 . +. . * . 3.322E+00 . + . . . * . 3.564E+00 . + . . . * . 3.722E+00 . + . . . * . 3.790E+00 . + . . . *. 3.768E+00 . + . . . *. 3.657E+00 . + . . . * . 3.451E+00 . + . . . * . 3.178E+00 . + . . . * . 2.851E+00 . .+ . * . . 2.488E+00 . . + . * . . 2.113E+00 . . + . * . . 1.748E+00 . . * . + . . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
With the bias voltage source of 2.3 volts in place, the transistor remains in its active mode throughout the entire cycle of the wave, faithfully reproducing the waveform at the speaker. Notice that the input voltage (measured between nodes 1 and 0) fluctuates between about 0.8 volts and 3.8 volts, a peak-to-peak voltage of 3 volts just as expected (source voltage = 1.5 volts peak). The output (speaker) current varies between zero and almost 300 mA, 180o out of phase with the input (microphone) signal.
The following illustration is another view of the same circuit, this time with a few oscilloscopes ("scopemeters") connected at crucial points to display all the pertinent signals:
The need for biasing a transistor amplifier circuit to obtain full waveform reproduction is an important consideration. A separate section of this chapter will be devoted entirely to the subject biasing and biasing techniques. For now, it is enough to understand that biasing may be necessary for proper voltage and current output from the amplifier.
Now that we have a functioning amplifier circuit, we can investigate its voltage, current, and power gains. The generic transistor used in these SPICE analyses has a β of 100, as indicated by the short transistor statistics printout included in the text output (these statistics were cut from the last two analyses for brevity's sake):
type npn is 1.00E-16 bf 100.000 nf 1.000 br 1.000 nr 1.000
β is listed under the abbreviation "bf," which actually stands for "beta, forward". If we wanted to insert our own β ratio for an analysis, we could have done so on the .model line of the SPICE netlist.
Since β is the ratio of collector current to base current, and we have our load connected in series with the collector terminal of the transistor and our source connected in series with the base, the ratio of output current to input current is equal to beta. Thus, our current gain for this example amplifier is 100, or 40 dB.
Voltage gain is a little more complicated to figure than current gain for this circuit. As always, voltage gain is defined as the ratio of output voltage divided by input voltage. In order to experimentally determine this, we need to modify our last SPICE analysis to plot output voltage rather than output current so we have two voltage plots to compare:
common-emitter amplifier vinput 1 5 sin (0 1.5 2000 0 0) vbias 5 0 dc 2.3 r1 1 2 1k q1 3 2 0 mod1 rspkr 3 4 8 v1 4 0 dc 15 .model mod1 npn .tran 0.02m 0.78m .plot tran v(1,0) v(4,3) .end
legend: *: v(1) +: v(4,3) v(1) (*+)- 0.000E+00 1.000E+00 2.000E+00 3.000E+00 4.000E+00 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2.300E+00 . . + . * . . 2.673E+00 . . + . * . . 3.020E+00 . . + . * . 3.322E+00 . . +. . * . 3.563E+00 . . . + . * . 3.723E+00 . . . + . * . 3.790E+00 . . . + . * . 3.767E+00 . . . + . * . 3.657E+00 . . . + . * . 3.452E+00 . . + . * . 3.177E+00 . . + . . * . 2.850E+00 . . + . * . . 2.488E+00 . . + . * . . 2.113E+00 . + . * . . 1.750E+00 . + . * . . . 1.419E+00 . + . * . . . 1.148E+00 . + . * . . . 9.493E-01 .+ *. . . . 8.311E-01 + * . . . . 8.050E-01 + * . . . . 8.797E-01 .+ * . . . . 1.039E+00 . + .* . . . 1.275E+00 . + . * . . . 1.579E+00 . + . * . . . 1.929E+00 . + . *. . . 2.300E+00 . . + . * . . 2.673E+00 . . + . * . . 3.019E+00 . . + . * . 3.322E+00 . . +. . * . 3.564E+00 . . . + . * . 3.722E+00 . . . + . * . 3.790E+00 . . . + . * . 3.768E+00 . . . + . * . 3.657E+00 . . . + . * . 3.451E+00 . . + . * . 3.178E+00 . . + . . * . 2.851E+00 . . + . * . . 2.488E+00 . . + . * . . 2.113E+00 . + . * . . 1.748E+00 . + . * . . . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -Plotted on the same scale (from 0 to 4 volts), we see that the output waveform ("+") has a smaller peak-to-peak amplitude than the input waveform ("*"), in addition to being at a lower bias voltage, not elevated up from 0 volts like the input. Since voltage gain for an AC amplifier is defined by the ratio of AC amplitudes, we can ignore any DC bias separating the two waveforms. Even so, the input waveform is still larger than the output, which tells us that the voltage gain is less than 1 (a negative dB figure).
To be honest, this low voltage gain is not characteristic to all common-emitter amplifiers. In this case it is a consequence of the great disparity between the input and load resistances. Our input resistance (R1) here is 1000 Ω, while the load (speaker) is only 8 Ω. Because the current gain of this amplifier is determined solely by the β of the transistor, and because that β figure is fixed, the current gain for this amplifier won't change with variations in either of these resistances. However, voltage gain is dependent on these resistances. If we alter the load resistance, making it a larger value, it will drop a proportionately greater voltage for its range of load currents, resulting in a larger output waveform. Let's try another simulation, only this time with a 30 Ω load instead of an 8 Ω load:
common-emitter amplifier vinput 1 5 sin (0 1.5 2000 0 0) vbias 5 0 dc 2.3 r1 1 2 1k q1 3 2 0 mod1 rspkr 3 4 30 v1 4 0 dc 15 .model mod1 npn .tran 0.02m 0.78m .plot tran v(1,0) v(4,3) .end
legend: *: v(1) +: v(4,3) v(1) (*)-- 0.000E+00 1.000E+00 2.000E+00 3.000E+00 4.000E+00 (+)-- -5.000E+00 0.000E+00 5.000E+00 1.000E+01 1.500E+01 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2.300E+00 . . + . * . . 2.673E+00 . . .+ * . . 3.020E+00 . . . + * . 3.322E+00 . . . + . * . 3.563E+00 . . . + . * . 3.723E+00 . . . + . * . 3.790E+00 . . . + . * . 3.767E+00 . . . + . * . 3.657E+00 . . . + . * . 3.452E+00 . . . + . * . 3.177E+00 . . . + . * . 2.850E+00 . . . + * . . 2.488E+00 . . +. * . . 2.113E+00 . . + . * . . 1.750E+00 . . + * . . . 1.419E+00 . . +* . . . 1.148E+00 . . x . . . 9.493E-01 . *.+ . . . 8.311E-01 . * + . . . 8.050E-01 . * + . . . 8.797E-01 . * .+ . . . 1.039E+00 . .*+ . . . 1.275E+00 . . +* . . . 1.579E+00 . . + * . . . 1.929E+00 . . + *. . . 2.300E+00 . . + . * . . 2.673E+00 . . .+ * . . 3.019E+00 . . . + * . 3.322E+00 . . . + . * . 3.564E+00 . . . + . * . 3.722E+00 . . . + . * . 3.790E+00 . . . + . * . 3.768E+00 . . . + . * . 3.657E+00 . . . + . * . 3.451E+00 . . . + . * . 3.178E+00 . . . + . * . 2.851E+00 . . . + * . . 2.488E+00 . . +. * . . 2.113E+00 . . + . * . . 1.748E+00 . . + * . . . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
This time the output voltage waveform is significantly greater in amplitude than the input waveform. This may not be obvious at first, since the two waveforms are plotted on different scales: the input on a scale of 0 to 4 volts and the output on a scale of -5 to 15 volts. Looking closely, we can see that the output waveform ("+") crests between 0 and about 9 volts: approximately 3 times the amplitude of the input voltage.
We can perform another computer analysis of this circuit, only this time instructing SPICE to analyze it from an AC point of view, giving us peak voltage figures for input and output instead of a time-based plot of the waveforms:
common-emitter amplifier vinput 1 5 ac 1.5 vbias 5 0 dc 2.3 r1 1 2 1k q1 3 2 0 mod1 rspkr 3 4 30 v1 4 0 dc 15 .model mod1 npn .ac lin 1 2000 2000 .print ac v(1,0) v(4,3) .end
freq v(1) v(4,3) 2.000E+03 1.500E+00 4.418E+00
Peak voltage measurements of input and output show an input of 1.5 volts and an output of 4.418 volts. This gives us a voltage gain ratio of 2.9453 (4.418 V / 1.5 V), or 9.3827 dB.
Because the current gain of the common-emitter amplifier is fixed by β, and since the input and output voltages will be equal to the input and output currents multiplied by their respective resistors, we can derive an equation for approximate voltage gain:
As you can see, the predicted results for voltage gain are quite close to the simulated results. With perfectly linear transistor behavior, the two sets of figures would exactly match. SPICE does a reasonable job of accounting for the many "quirks" of bipolar transistor function in its analysis, hence the slight mismatch in voltage gain based on SPICE's output.
These voltage gains remain the same regardless of where we measure output voltage in the circuit: across collector and emitter, or across the series load resistor as we did in the last analysis. The amount of output voltage change for any given amount of input voltage will remain the same. Consider the two following SPICE analyses as proof of this. The first simulation is time-based, to provide a plot of input and output voltages. You will notice that the two signals are 180o out of phase with each other. The second simulation is an AC analysis, to provide simple, peak voltage readings for input and output:
common-emitter amplifier vinput 1 5 sin (0 1.5 2000 0 0) vbias 5 0 dc 2.3 r1 1 2 1k q1 3 2 0 mod1 rspkr 3 4 30 v1 4 0 dc 15 .model mod1 npn .tran 0.02m 0.74m .plot tran v(1,0) v(3,0) .end
legend: *: v(1) +: v(3) v(1) (*)-- 0.000E+00 1.000E+00 2.000E+00 3.000E+00 4.000E+00 (+)-- 0.000E+00 5.000E+00 1.000E+01 1.500E+01 2.000E+01 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2.300E+00 . . . + * . . 2.673E+00 . . +. * . . 3.020E+00 . . + . * . 3.324E+00 . . + . . * . 3.564E+00 . . + . . * . 3.720E+00 . . + . . * . 3.793E+00 . . + . . * . 3.770E+00 . . + . . * . 3.651E+00 . . + . . * . 3.454E+00 . . + . . * . 3.179E+00 . . + . . * . 2.850E+00 . . + . * . . 2.488E+00 . . .+ * . . 2.113E+00 . . . * + . . 1.750E+00 . . * . + . . 1.418E+00 . . * . + . . 1.149E+00 . . * . + . . 9.477E-01 . *. . +. . 8.277E-01 . * . . + . 8.091E-01 . * . . + . 8.781E-01 . * . . +. . 1.035E+00 . * . + . . 1.278E+00 . . * . + . . 1.579E+00 . . * . + . . 1.928E+00 . . *. + . . 2.300E+00 . . . + * . . 2.672E+00 . . +. * . . 3.020E+00 . . + . * . 3.322E+00 . . + . . * . 3.565E+00 . . + . . * . 3.722E+00 . . + . . * . 3.791E+00 . . + . . * . 3.773E+00 . . + . . * . 3.652E+00 . . + . . * . 3.451E+00 . . + . . * . 3.181E+00 . . + . . * . 2.850E+00 . . + . * . . 2.488E+00 . . .+ * . . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
common-emitter amplifier vinput 1 5 ac 1.5 vbias 5 0 dc 2.3 r1 1 2 1k q1 3 2 0 mod1 rspkr 3 4 30 v1 4 0 dc 15 .model mod1 npn .ac lin 1 2000 2000 .print ac v(1,0) v(3,0) .end
freq v(1) v(3) 2.000E+03 1.500E+00 4.418E+00
We still have a peak output voltage of 4.418 volts with a peak input voltage of 1.5 volts. The only difference from the last set of simulations is the phase of the output voltage.
So far, the example circuits shown in this section have all used NPN transistors. PNP transistors are just as valid to use as NPN in any amplifier configuration, so long as the proper polarity and current directions are maintained, and the common-emitter amplifier is no exception. The inverting behavior and gain properties of a PNP transistor amplifier are the same as its NPN counterpart, just the polarities are different:
- REVIEW:
- Common-emitter transistor amplifiers are so-called because the input and output voltage points share the emitter lead of the transistor in common with each other, not considering any power supplies.
- Transistors are essentially DC devices: they cannot directly handle voltages or currents that reverse direction. In order to make them work for amplifying AC signals, the input signal must be offset with a DC voltage to keep the transistor in its active mode throughout the entire cycle of the wave. This is called biasing.
- If the output voltage is measured between emitter and collector on a common-emitter amplifier, it will be 180o out of phase with the input voltage waveform. For this reason, the common-emitter amplifier is called an inverting amplifier circuit.
- The current gain of a common-emitter transistor amplifier with the load connected in series with the collector is equal to β. The voltage gain of a common-emitter transistor amplifier is approximately given here:
-
- Where "Rout" is the resistor connected in series with the collector and "Rin" is the resistor connected in series with the base.
The common-collector amplifier
Our next transistor configuration to study is a bit simpler in terms of gain calculations. Called the common-collector configuration, its schematic diagram looks like this:
It is called the common-collector configuration because (ignoring the power supply battery) both the signal source and the load share the collector lead as a common connection point:
It should be apparent that the load resistor in the common-collector amplifier circuit receives both the base and collector currents, being placed in series with the emitter. Since the emitter lead of a transistor is the one handling the most current (the sum of base and collector currents, since base and collector currents always mesh together to form the emitter current), it would be reasonable to presume that this amplifier will have a very large current gain (maximum output current for minimum input current). This presumption is indeed correct: the current gain for a common-collector amplifier is quite large, larger than any other transistor amplifier configuration. However, this is not necessarily what sets it apart from other amplifier designs.
Let's proceed immediately to a SPICE analysis of this amplifier circuit, and you will be able to immediately see what is unique about this amplifier:
common-collector amplifier vin 1 0 q1 2 1 3 mod1 v1 2 0 dc 15 rload 3 0 5k .model mod1 npn .dc vin 0 5 0.2 .plot dc v(3,0) .end
type npn is 1.00E-16 bf 100.000 nf 1.000 br 1.000 nr 1.000
vin v(3) 0.000E+00 2.000E+00 4.000E+00 6.000E+00 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 0.000E+00 7.500E-08 * . . . 2.000E-01 7.501E-08 * . . . 4.000E-01 2.704E-06 * . . . 6.000E-01 4.954E-03 * . . . 8.000E-01 1.221E-01 .* . . . 1.000E+00 2.989E-01 . * . . . 1.200E+00 4.863E-01 . * . . . 1.400E+00 6.777E-01 . * . . . 1.600E+00 8.712E-01 . * . . . 1.800E+00 1.066E+00 . * . . . 2.000E+00 1.262E+00 . * . . . 2.200E+00 1.458E+00 . * . . . 2.400E+00 1.655E+00 . * . . . 2.600E+00 1.852E+00 . *. . . 2.800E+00 2.049E+00 . * . . 3.000E+00 2.247E+00 . . * . . 3.200E+00 2.445E+00 . . * . . 3.400E+00 2.643E+00 . . * . . 3.600E+00 2.841E+00 . . * . . 3.800E+00 3.039E+00 . . * . . 4.000E+00 3.237E+00 . . * . . 4.200E+00 3.436E+00 . . * . . 4.400E+00 3.634E+00 . . * . . 4.600E+00 3.833E+00 . . *. . 4.800E+00 4.032E+00 . . * . 5.000E+00 4.230E+00 . . . * . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Unlike the common-emitter amplifier from the previous section, the common-collector produces an output voltage in direct rather than inverse proportion to the rising input voltage. As the input voltage increases, so does the output voltage. More than that, a close examination reveals that the output voltage is nearly identical to the input voltage, lagging behind only about 0.77 volts.
This is the unique quality of the common-collector amplifier: an output voltage that is nearly equal to the input voltage. Examined from the perspective of output voltage change for a given amount of input voltage change, this amplifier has a voltage gain of almost exactly unity (1), or 0 dB. This holds true for transistors of any β value, and for load resistors of any resistance value.
It is simple to understand why the output voltage of a common-collector amplifier is always nearly equal to the input voltage. Referring back to the diode-regulating diode transistor model, we see that the base current must go through the base-emitter PN junction, which is equivalent to a normal rectifying diode. So long as this junction is forward-biased (the transistor conducting current in either its active or saturated modes), it will have a voltage drop of approximately 0.7 volts, assuming silicon construction. This 0.7 volt drop is largely irrespective of the actual magnitude of base current, so we can regard it as being constant:
Given the voltage polarities across the base-emitter PN junction and the load resistor, we see that they must add together to equal the input voltage, in accordance with Kirchhoff's Voltage Law. In other words, the load voltage will always be about 0.7 volts less than the input voltage for all conditions where the transistor is conducting. Cutoff occurs at input voltages below 0.7 volts, and saturation at input voltages in excess of battery (supply) voltage plus 0.7 volts.
Because of this behavior, the common-collector amplifier circuit is also known as the voltage-follower or emitter-follower amplifier, in reference to the fact that the input and load voltages follow each other so closely.
Applying the common-collector circuit to the amplification of AC signals requires the same input "biasing" used in the common-emitter circuit: a DC voltage must be added to the AC input signal to keep the transistor in its active mode during the entire cycle. When this is done, the result is a non-inverting amplifier:
common-collector amplifier vin 1 4 sin(0 1.5 2000 0 0) vbias 4 0 dc 2.3 q1 2 1 3 mod1 v1 2 0 dc 15 rload 3 0 5k .model mod1 npn .tran .02m .78m .plot tran v(1,0) v(3,0) .end
legend: *: v(1) +: v(3) v(1) 0.000E+00 1.000E+00 2.000E+00 3.000E+00 4.000E+00 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2.300E+00 . . + . * . . 2.673E+00 . . +. * . . 3.020E+00 . . . + * . 3.322E+00 . . . + . * . 3.563E+00 . . . + . * . 3.723E+00 . . . +. * . 3.790E+00 . . . + * . 3.767E+00 . . . + * . 3.657E+00 . . . +. * . 3.452E+00 . . . + . * . 3.177E+00 . . . + . * . 2.850E+00 . . .+ * . . 2.488E+00 . . + . * . . 2.113E+00 . . + . * . . 1.750E+00 . + * . . . 1.419E+00 . + . * . . . 1.148E+00 . + . * . . . 9.493E-01 . + *. . . . 8.311E-01 .+ * . . . . 8.050E-01 .+ * . . . . 8.797E-01 . + * . . . . 1.039E+00 . + .* . . . 1.275E+00 . + . * . . . 1.579E+00 . + . * . . . 1.929E+00 . . + *. . . 2.300E+00 . . + . * . . 2.673E+00 . . +. * . . 3.019E+00 . . . + * . 3.322E+00 . . . + . * . 3.564E+00 . . . + . * . 3.722E+00 . . . +. * . 3.790E+00 . . . + * . 3.768E+00 . . . + * . 3.657E+00 . . . +. * . 3.451E+00 . . . + . * . 3.178E+00 . . . + . * . 2.851E+00 . . .+ * . . 2.488E+00 . . + . * . . 2.113E+00 . . + . * . . 1.748E+00 . + * . . . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Here's another view of the circuit, this time with oscilloscopes connected to several points of interest:
Since this amplifier configuration doesn't provide any voltage gain (in fact, in practice it actually has a voltage gain of slightly less than 1), its only amplifying factor is current. The common-emitter amplifier configuration examined in the previous section had a current gain equal to the β of the transistor, being that the input current went through the base and the output (load) current went through the collector, and β by definition is the ratio between the collector and emitter currents. In the common-collector configuration, though, the load is situated in series with the emitter, and thus its current is equal to the emitter current. With the emitter carrying collector current and base current, the load in this type of amplifier has all the current of the collector running through it plus the input current of the base. This yields a current gain of β plus 1:
Once again, PNP transistors are just as valid to use in the common-collector configuration as NPN transistors. The gain calculations are all the same, as is the non-inverting behavior of the amplifier. The only difference is in voltage polarities and current directions:
A popular application of the common-collector amplifier is for regulated DC power supplies, where an unregulated (varying) source of DC voltage is clipped at a specified level to supply regulated (steady) voltage to a load. Of course, zener diodes already provide this function of voltage regulation:
However, when used in this direct fashion, the amount of current that may be supplied to the load is usually quite limited. In essence, this circuit regulates voltage across the load by keeping current through the series resistor at a high enough level to drop all the excess power source voltage across it, the zener diode drawing more or less current as necessary to keep the voltage across itself steady. For high-current loads, an plain zener diode voltage regulator would have to be capable of shunting a lot of current through the diode in order to be effective at regulating load voltage in the event of large load resistance or voltage source changes.
One popular way to increase the current-handling ability of a regulator circuit like this is to use a common-collector transistor to amplify current to the load, so that the zener diode circuit only has to handle the amount of current necessary to drive the base of the transistor:
There's really only one caveat to this approach: the load voltage will be approximately 0.7 volts less than the zener diode voltage, due to the transistor's 0.7 volt base-emitter drop. However, since this 0.7 volt difference is fairly constant over a wide range of load currents, a zener diode with a 0.7 volt higher rating can be chosen for the application.
Sometimes the high current gain of a single-transistor, common-collector configuration isn't enough for a particular application. If this is the case, multiple transistors may be staged together in a popular configuration known as a Darlington pair, just an extension of the common-collector concept:
Darlington pairs essentially place one transistor as the common-collector load for another transistor, thus multiplying their individual current gains. Base current through the upper-left transistor is amplified through that transistor's emitter, which is directly connected to the base of the lower-right transistor, where the current is again amplified. The overall current gain is as follows:
Voltage gain is still nearly equal to 1 if the entire assembly is connected to a load in common-collector fashion, although the load voltage will be a full 1.4 volts less than the input voltage:
Darlington pairs may be purchased as discrete units (two transistors in the same package), or may be built up from a pair of individual transistors. Of course, if even more current gain is desired than what may be obtained with a pair, Darlington triplet or quadruplet assemblies may be constructed.
- REVIEW:
- Common-collector transistor amplifiers are so-called because the input and output voltage points share the collector lead of the transistor in common with each other, not considering any power supplies.
- The output voltage on a common-collector amplifier will be in phase with the input voltage, making the common-collector a non-inverting amplifier circuit.
- The current gain of a common-collector amplifier is equal to β plus 1. The voltage gain is approximately equal to 1 (in practice, just a little bit less).
- A Darlington pair is a pair of transistors "piggybacked" on one another so that the emitter of one feeds current to the base of the other in common-collector form. The result is an overall current gain equal to the product (multiplication) of their individual common-collector current gains (β plus 1).
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