The common-base amplifier
The final transistor amplifier configuration we need to study is the common-base. This configuration is more complex than the other two, and is less common due to its strange operating characteristics.
It is called the common-base configuration because (DC power source aside), the signal source and the load share the base of the transistor as a common connection point:
Perhaps the most striking characteristic of this configuration is that the input signal source must carry the full emitter current of the transistor, as indicated by the heavy arrows in the first illustration. As we know, the emitter current is greater than any other current in the transistor, being the sum of base and collector currents. In the last two amplifier configurations, the signal source was connected to the base lead of the transistor, thus handling the least current possible.
Because the input current exceeds all other currents in the circuit, including the output current, the current gain of this amplifier is actually less than 1 (notice how Rload is connected to the collector, thus carrying slightly less current than the signal source). In other words, it attenuates current rather than amplifying it. With common-emitter and common-collector amplifier configurations, the transistor parameter most closely associated with gain was β. In the common-base circuit, we follow another basic transistor parameter: the ratio between collector current and emitter current, which is a fraction always less than 1. This fractional value for any transistor is called the alpha ratio, or α ratio.
Since it obviously can't boost signal current, it only seems reasonable to expect it to boost signal voltage. A SPICE simulation will vindicate that assumption:
common-base amplifier vin 0 1 r1 1 2 100 q1 4 0 2 mod1 v1 3 0 dc 15 rload 3 4 5k .model mod1 npn .dc vin 0.6 1.2 .02 .plot dc v(3,4) .end
v(3,4) 0.000E+00 5.000E+00 1.000E+01 1.500E+01 2.000E+01 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 5.913E-03 * . . . . 1.274E-02 * . . . . 2.730E-02 * . . . . 5.776E-02 * . . . . 1.193E-01 * . . . . 2.358E-01 .* . . . . 4.370E-01 .* . . . . 7.447E-01 . * . . . . 1.163E+00 . * . . . . 1.682E+00 . * . . . . 2.281E+00 . * . . . . 2.945E+00 . * . . . . 3.657E+00 . * . . . . 4.408E+00 . * . . . . 5.189E+00 . .* . . . 5.995E+00 . . * . . . 6.820E+00 . . * . . . 7.661E+00 . . * . . . 8.516E+00 . . * . . . 9.382E+00 . . * . . . 1.026E+01 . . .* . . 1.114E+01 . . . * . . 1.203E+01 . . . * . . 1.293E+01 . . . * . . 1.384E+01 . . . * . . 1.474E+01 . . . *. . 1.563E+01 . . . . * . 1.573E+01 . . . . * . 1.575E+01 . . . . * . 1.576E+01 . . . . * . 1.576E+01 . . . . * . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Notice how in this simulation the output voltage goes from practically nothing (cutoff) to 15.75 volts (saturation) with the input voltage being swept over a range of 0.6 volts to 1.2 volts. In fact, the output voltage plot doesn't show a rise until about 0.7 volts at the input, and cuts off (flattens) at about 1.12 volts input. This represents a rather large voltage gain with an output voltage span of 15.75 volts and an input voltage span of only 0.42 volts: a gain ratio of 37.5, or 31.48 dB. Notice also how the output voltage (measured across Rload) actually exceeds the power supply (15 volts) at saturation, due to the series-aiding effect of the the input voltage source.
A second set of SPICE analyses with an AC signal source (and DC bias voltage) tells the same story: a high voltage gain.
common-base amplifier vin 0 1 sin (0 0.12 2000 0 0) vbias 1 5 dc 0.95 r1 5 2 100 q1 4 0 2 mod1 v1 3 0 dc 15 rload 3 4 5k .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) (*)-- -2.000E-01 -1.000E-01 0.000E+00 1.000E-01 2.000E-01 (+)-- -1.500E+01 -1.000E+01 -5.000E+00 0.000E+00 5.000E+00 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 0.000E+00 . . + * . . -2.984E-02 . . + * . . . -5.757E-02 . + . * . . . -8.176E-02 . + . * . . . -1.011E-01 . + * . . . -1.139E-01 . + * . . . . -1.192E-01 . + * . . . . -1.174E-01 . + * . . . . -1.085E-01 . + *. . . . -9.213E-02 . + .* . . . -7.020E-02 . + . * . . . -4.404E-02 . + * . . . -1.502E-02 . . + * . . . 1.496E-02 . . + . * . . 4.400E-02 . . + . * . . 7.048E-02 . . + * . . 9.214E-02 . . . + *. . 1.081E-01 . . . + .* . 1.175E-01 . . . + . * . 1.196E-01 . . . + . * . 1.136E-01 . . . + . * . 1.009E-01 . . . + * . 8.203E-02 . . .+ * . . 5.764E-02 . . + . * . . 2.970E-02 . . + . * . . -1.440E-05 . . + * . . -2.981E-02 . . + * . . . -5.755E-02 . + . * . . . -8.178E-02 . + . * . . . -1.011E-01 . + * . . . -1.138E-01 . + * . . . . -1.192E-01 . + * . . . . -1.174E-01 . + * . . . . -1.085E-01 . + *. . . . -9.209E-02 . + .* . . . -7.020E-02 . + . * . . . -4.407E-02 . + * . . . -1.502E-02 . . + * . . . 1.496E-02 . . + . * . . 4.417E-02 . . + . * . . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
As you can see, the input and output waveforms are in phase with each other. This tells us that the common-base amplifier is non-inverting.
common-base amplifier vin 0 1 ac 0.12 vbias 1 5 dc 0.95 r1 5 2 100 q1 4 0 2 mod1 v1 3 0 dc 15 rload 3 4 5k .model mod1 npn .ac lin 1 2000 2000 .print ac v(1,0) v(3,4) .end
freq v(1) v(3,4) 2.000E+03 1.200E-01 5.129E+00
Voltage figures from the second analysis (AC mode) show a voltage gain of 42.742 (5.129 V / 0.12 V), or 32.617 dB:
Here's another view of the circuit, showing the phase relations and DC offsets of various signals in the circuit just simulated:
. . . and for a PNP transistor:
Predicting voltage gain for the common-base amplifier configuration is quite difficult, and involves approximations of transistor behavior that are difficult to measure directly. Unlike the other amplifier configurations, where voltage gain was either set by the ratio of two resistors (common-emitter), or fixed at an unchangeable value (common-collector), the voltage gain of the common-base amplifier depends largely on the amount of DC bias on the input signal. As it turns out, the internal transistor resistance between emitter and base plays a major role in determining voltage gain, and this resistance changes with different levels of current through the emitter.
While this phenomenon is difficult to explain, it is rather easy to demonstrate through the use of computer simulation. What I'm going to do here is run several SPICE simulations on a common-base amplifier circuit, changing the DC bias voltage slightly while keeping the AC signal amplitude and all other circuit parameters constant. As the voltage gain changes from one simulation to another, different output voltage amplitudes will be noticed as a result.
Although these analyses will all be conducted in the AC mode, they were first "proofed" in the transient analysis mode (voltage plotted over time) to ensure that the entire wave was being faithfully reproduced and not "clipped" due to improper biasing. No meaningful calculations of gain can be based on waveforms that are distorted:
common-base amplifier DC bias = 0.85 volts vin 0 1 ac 0.08 vbias 1 5 dc 0.85 r1 5 2 100 q1 4 0 2 mod1 v1 3 0 dc 15 rload 3 4 5k .model mod1 npn .ac lin 1 2000 2000 .print ac v(1,0) v(3,4) .end
freq v(1) v(3,4) 2.000E+03 8.000E-02 3.005E+00
common-base amplifier dc bias = 0.9 volts vin 0 1 ac 0.08 vbias 1 5 dc 0.90 r1 5 2 100 q1 4 0 2 mod1 v1 3 0 dc 15 rload 3 4 5k .model mod1 npn .ac lin 1 2000 2000 .print ac v(1,0) v(3,4) .end
freq v(1) v(3,4) 2.000E+03 8.000E-02 3.264E+00
common-base amplifier dc bias = 0.95 volts vin 0 1 ac 0.08 vbias 1 5 dc 0.95 r1 5 2 100 q1 4 0 2 mod1 v1 3 0 dc 15 rload 3 4 5k .model mod1 npn .ac lin 1 2000 2000 .print ac v(1,0) v(3,4) .end
freq v(1) v(3,4) 2.000E+03 8.000E-02 3.419E+00
A trend should be evident here: with increases in DC bias voltage, voltage gain increases as well. We can see that the voltage gain is increasing because each subsequent simulation produces greater output voltage for the exact same input signal voltage (0.08 volts). As you can see, the changes are quite large, and they are caused by miniscule variations in bias voltage!
The combination of very low current gain (always less than 1) and somewhat unpredictable voltage gain conspire against the common-base design, relegating it to few practical applications.
- REVIEW:
- Common-base transistor amplifiers are so-called because the input and output voltage points share the base lead of the transistor in common with each other, not considering any power supplies.
- The current gain of a common-base amplifier is always less than 1. The voltage gain is a function of input and output resistances, and also the internal resistance of the emitter-base junction, which is subject to change with variations in DC bias voltage. Suffice to say that the voltage gain of a common-base amplifier can be very high.
- The ratio of a transistor's collector current to emitter current is called α. The α value for any transistor is always less than unity, or in other words, less than 1.
Biasing techniques
In the common-emitter section of this chapter, we saw a SPICE analysis where the output waveform resembled a half-wave rectified shape: only half of the input waveform was reproduced, with the other half being completely cut off. Since our purpose at that time was to reproduce the entire waveshape, this constituted a problem. The solution to this problem was to add a small bias voltage to the amplifier input so that the transistor stayed in active mode throughout the entire wave cycle. This addition was called a bias voltage.There are applications, though, where a half-wave output is not problematic. In fact, some applications may necessitate this very type of amplification. Because it is possible to operate an amplifier in modes other than full-wave reproduction, and because there are specific applications requiring different ranges of reproduction, it is useful to describe the degree to which an amplifier reproduces the input waveform by designating it according to class. Amplifier class operation is categorized by means of alphabetical letters: A, B, C, and AB.
Class A operation is where the entire input waveform is faithfully reproduced. Although I didn't introduce this concept back in the common-emitter section, this is what we were hoping to attain in our simulations. Class A operation can only be obtained when the transistor spends its entire time in the active mode, never reaching either cutoff or saturation. To achieve this, sufficient DC bias voltage is usually set at the level necessary to drive the transistor exactly halfway between cutoff and saturation. This way, the AC input signal will be perfectly "centered" between the amplifier's high and low signal limit levels.
Class B operation is what we had the first time an AC signal was applied to the common-emitter amplifier with no DC bias voltage. The transistor spent half its time in active mode and the other half in cutoff with the input voltage too low (or even of the wrong polarity!) to forward-bias its base-emitter junction.
By itself, an amplifier operating in class B mode is not very useful. In most circumstances, the severe distortion introduced into the waveshape by eliminating half of it would be unacceptable. However, class B operation is a useful mode of biasing if two amplifiers are operated as a push-pull pair, each amplifier handling only half of the waveform at a time:
Transistor Q1 "pushes" (drives the output voltage in a positive direction with respect to ground), while transistor Q2 "pulls" the output voltage (in a negative direction, toward 0 volts with respect to ground). Individually, each of these transistors is operating in class B mode, active only for one-half of the input waveform cycle. Together, however, they function as a team to produce an output waveform identical in shape to the input waveform.
A decided advantage of the class B (push-pull) amplifier design over the class A design is greater output power capability. With a class A design, the transistor dissipates a lot of energy in the form of heat because it never stops conducting current. At all points in the wave cycle it is in the active (conducting) mode, conducting substantial current and dropping substantial voltage. This means there is substantial power dissipated by the transistor throughout the cycle. In a class B design, each transistor spends half the time in cutoff mode, where it dissipates zero power (zero current = zero power dissipation). This gives each transistor a time to "rest" and cool while the other transistor carries the burden of the load. Class A amplifiers are simpler in design, but tend to be limited to low-power signal applications for the simple reason of transistor heat dissipation.
There is another class of amplifier operation known as class AB, which is somewhere between class A and class B: the transistor spends more than 50% but less than 100% of the time conducting current.
If the input signal bias for an amplifier is slightly negative (opposite of the bias polarity for class A operation), the output waveform will be further "clipped" than it was with class B biasing, resulting in an operation where the transistor spends the majority of the time in cutoff mode:
At first, this scheme may seem utterly pointless. After all, how useful could an amplifier be if it clips the waveform as badly as this? If the output is used directly with no conditioning of any kind, it would indeed be of questionable utility. However, with the application of a tank circuit (parallel resonant inductor-capacitor combination) to the output, the occasional output surge produced by the amplifier can set in motion a higher-frequency oscillation maintained by the tank circuit. This may be likened to a machine where a heavy flywheel is given an occasional "kick" to keep it spinning:
Called class C operation, this scheme also enjoys high power efficiency due to the fact that the transistor(s) spend the vast majority of time in the cutoff mode, where they dissipate zero power. The rate of output waveform decay (decreasing oscillation amplitude between "kicks" from the amplifier) is exaggerated here for the benefit of illustration. Because of the tuned tank circuit on the output, this type of circuit is usable only for amplifying signals of definite, fixed frequency.
Another type of amplifier operation, significantly different from Class A, B, AB, or C, is called Class D. It is not obtained by applying a specific measure of bias voltage as are the other classes of operation, but requires a radical re-design of the amplifier circuit itself. It's a little too early in this chapter to investigate exactly how a class D amplifier is built, but not too early to discuss its basic principle of operation.
A class D amplifier reproduces the profile of the input voltage waveform by generating a rapidly-pulsing squarewave output. The duty cycle of this output waveform (time "on" versus total cycle time) varies with the instantaneous amplitude of the input signal. The following plots demonstrate this principle:
The greater the instantaneous voltage of the input signal, the greater the duty cycle of the output squarewave pulse. If there can be any goal stated of the class D design, it is to avoid active-mode transistor operation. Since the output transistor of a class D amplifier is never in the active mode, only cutoff or saturated, there will be little heat energy dissipated by it. This results in very high power efficiency for the amplifier. Of course, the disadvantage of this strategy is the overwhelming presence of harmonics on the output. Fortunately, since these harmonic frequencies are typically much greater than the frequency of the input signal, they can be filtered out by a low-pass filter with relative ease, resulting in an output more closely resembling the original input signal waveform. Class D technology is typically seen where extremely high power levels and relatively low frequencies are encountered, such as in industrial inverters (devices converting DC into AC power to run motors and other large devices) and high-performance audio amplifiers.
A term you will likely come across in your studies of electronics is something called quiescent, which is a modifier designating the normal, or zero input signal, condition of a circuit. Quiescent current, for example, is the amount of current in a circuit with zero input signal voltage applied. Bias voltage in a transistor circuit forces the transistor to operate at a different level of collector current with zero input signal voltage than it would without that bias voltage. Therefore, the amount of bias in an amplifier circuit determines its quiescent values.
In a class A amplifier, the quiescent current should be exactly half of its saturation value (halfway between saturation and cutoff, cutoff by definition being zero). Class B and class C amplifiers have quiescent current values of zero, since they are supposed to be cutoff with no signal applied. Class AB amplifiers have very low quiescent current values, just above cutoff. To illustrate this graphically, a "load line" is sometimes plotted over a transistor's characteristic curves to illustrate its range of operation while connected to a load resistance of specific value:
A load line is a plot of collector-to-emitter voltage over a range of base currents. At the lower-right corner of the load line, voltage is at maximum and current is at zero, representing a condition of cutoff. At the upper-left corner of the line, voltage is at zero while current is at a maximum, representing a condition of saturation. Dots marking where the load line intersects the various transistor curves represent realistic operating conditions for those base currents given.
Quiescent operating conditions may be shown on this type of graph in the form of a single dot along the load line. For a class A amplifier, the quiescent point will be in the middle of the load line, like this:
In this illustration, the quiescent point happens to fall on the curve representing a base current of 40 µA. If we were to change the load resistance in this circuit to a greater value, it would affect the slope of the load line, since a greater load resistance would limit the maximum collector current at saturation, but would not change the collector-emitter voltage at cutoff. Graphically, the result is a load line with a different upper-left point and the same lower-right point:
Note how the new load line doesn't intercept the 75 µA curve along its flat portion as before. This is very important to realize because the non-horizontal portion of a characteristic curve represents a condition of saturation. Having the load line intercept the 75 µA curve outside of the curve's horizontal range means that the amplifier will be saturated at that amount of base current. Increasing the load resistor value is what caused the load line to intercept the 75 µA curve at this new point, and it indicates that saturation will occur at a lesser value of base current than before.
With the old, lower-value load resistor in the circuit, a base current of 75 µA would yield a proportional collector current (base current multiplied by β). In the first load line graph, a base current of 75 µA gave a collector current almost twice what was obtained at 40 µA, as the β ratio would predict. Now, however, there is only a marginal increase in collector current between base current values of 75 µA and 40 µA, because the transistor begins to lose sufficient collector-emitter voltage to continue to regulate collector current.
In order to maintain linear (no-distortion) operation, transistor amplifiers shouldn't be operated at points where the transistor will saturate; that is, in any case where the load line will not potentially fall on the horizontal portion of a collector current curve. In this case, we'd have to add a few more curves to the graph before we could tell just how far we could "push" this transistor with increased base currents before it saturates.
It appears in this graph that the highest-current point on the load line falling on the straight portion of a curve is the point on the 50 µA curve. This new point should be considered the maximum allowable input signal level for class A operation. Also for class A operation, the bias should be set so that the quiescent point is halfway between this new maximum point and cutoff:
Now that we know a little more about the consequences of different DC bias voltage levels, it is time to investigate practical biasing techniques. So far, I've shown a small DC voltage source (battery) connected in series with the AC input signal to bias the amplifier for whatever desired class of operation. In real life, the connection of a precisely-calibrated battery to the input of an amplifier is simply not practical. Even if it were possible to customize a battery to produce just the right amount of voltage for any given bias requirement, that battery would not remain at its manufactured voltage indefinitely. Once it started to discharge and its output voltage drooped, the amplifier would begin to drift in the direction of class B operation.
Take this circuit, illustrated in the common-emitter section for a SPICE simulation, for instance:
That 2.3 volt "Vbias" battery would not be practical to include in a real amplifier circuit. A far more practical method of obtaining bias voltage for this amplifier would be to develop the necessary 2.3 volts using a voltage divider network connected across the 15 volt battery. After all, the 15 volt battery is already there by necessity, and voltage divider circuits are very easy to design and build. Let's see how this might look:
If we choose a pair of resistor values for R2 and R3 that will produce 2.3 volts across R3 from a total of 15 volts (such as 8466 Ω for R2 and 1533 Ω for R3), we should have our desired value of 2.3 volts between base and emitter for biasing with no signal input. The only problem is, this circuit configuration places the AC input signal source directly in parallel with R3 of our voltage divider. This is not acceptable, as the AC source will tend to overpower any DC voltage dropped across R3. Parallel components must have the same voltage, so if an AC voltage source is directly connected across one resistor of a DC voltage divider, the AC source will "win" and there will be no DC bias voltage added to the signal.
One way to make this scheme work, although it may not be obvious why it will work, is to place a coupling capacitor between the AC voltage source and the voltage divider like this:
The capacitor forms a high-pass filter between the AC source and the DC voltage divider, passing almost all of the AC signal voltage on to the transistor while blocking all DC voltage from being shorted through the AC signal source. This makes much more sense if you understand the superposition theorem and how it works. According to superposition, any linear, bilateral circuit can be analyzed in a piecemeal fashion by only considering one power source at a time, then algebraically adding the effects of all power sources to find the final result. If we were to separate the capacitor and R2--R3 voltage divider circuit from the rest of the amplifier, it might be easier to understand how this superposition of AC and DC would work.
With only the AC signal source in effect, and a capacitor with an arbitrarily low impedance at signal frequency, almost all the AC voltage appears across R3:
With only the DC source in effect, the capacitor appears to be an open circuit, and thus neither it nor the shorted AC signal source will have any effect on the operation of the R2--R3 voltage divider:
Combining these two separate analyses, we get a superposition of (almost) 1.5 volts AC and 2.3 volts DC, ready to be connected to the base of the transistor:
Enough talk -- it's about time for a SPICE simulation of the whole amplifier circuit. I'll use a capacitor value of 100 µF to obtain an arbitrarily low (0.796 Ω) impedance at 2000 Hz:
voltage divider biasing vinput 1 0 sin (0 1.5 2000 0 0) c1 1 5 100u r1 5 2 1k r2 4 5 8466 r3 5 0 1533 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) (*)-- -2.000E+00 -1.000E+00 0.000E+00 1.000E+00 2.000E+00 (+)-- -3.000E-01 -2.000E-01 -1.000E-01 0.000E+00 1.000E-01 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 0.000E+00 . . * + . . 3.730E-01 . . + * . . 7.197E-01 . . + . * . . 1.022E+00 . . + . * . 1.263E+00 . . + . . * . 1.423E+00 . + . . * . 1.490E+00 . +. . . * . 1.467E+00 . +. . . * . 1.357E+00 . .+ . . * . 1.152E+00 . . + . . * . 8.774E-01 . . + . * . . 5.505E-01 . . + . * . . 1.878E-01 . . . x . . -1.870E-01 . . * . + . . -5.500E-01 . . * . + . . -8.810E-01 . . * . + . -1.152E+00 . * . . + . -1.351E+00 . * . . + . -1.469E+00 . * . . + . -1.495E+00 . * . . + . -1.420E+00 . * . . + . -1.261E+00 . * . . + . -1.025E+00 . * . + . -7.205E-01 . . * . +. . -3.713E-01 . . * . + . . 1.800E-04 . . * + . . 3.726E-01 . . + * . . 7.194E-01 . . + . * . . 1.022E+00 . . + . * . 1.264E+00 . . + . . * . 1.422E+00 . + . . * . 1.490E+00 . +. . . * . 1.468E+00 . +. . . * . 1.357E+00 . .+ . . * . 1.151E+00 . . + . . * . 8.775E-01 . . + . * . . 5.509E-01 . . + . * . . 1.877E-01 . . . x . . -1.871E-01 . . * . + . . -5.522E-01 . . * . + . . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Notice that there is substantial distortion in the output waveform here: the sine wave is being clipped during most of the input signal's negative half-cycle. This tells us the transistor is entering into cutoff mode when it shouldn't (I'm assuming a goal of class A operation as before). Why is this? This new biasing technique should give us exactly the same amount of DC bias voltage as before, right?
With the capacitor and R2--R3 resistor network unloaded, it will provide exactly 2.3 volts worth of DC bias. However, once we connect this network to the transistor, it is no longer loaded. Current drawn through the base of the transistor will load the voltage divider, thus reducing the DC bias voltage available for the transistor. Using the diode-regulating diode transistor model to illustrate, the bias problem becomes evident:
A voltage divider's output depends not only on the size of its constituent resistors, but also on how much current is being divided away from it through a load. In this case, the base-emitter PN junction of the transistor is a load that decreases the DC voltage dropped across R3, due to the fact that the bias current joins with R3's current to go through R2, upsetting the divider ratio formerly set by the resistance values of R2 and R3. In order to obtain a DC bias voltage of 2.3 volts, the values of R2 and/or R3 must be adjusted to compensate for the effect of base current loading. In this case, we want to increase the DC voltage dropped across R3, so we can lower the value of R2, raise the value of R3, or both.
voltage divider biasing vinput 1 0 sin (0 1.5 2000 0 0) c1 1 5 100u r1 5 2 1k r2 4 5 6k <--- R2 decreased to 6 k ohms r3 5 0 4k <--- R3 increased to 4 k ohms 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) (*)-- -2.000E+00 -1.000E+00 0.000E+00 1.000E+00 2.000E+00 (+)-- -3.000E-01 -2.000E-01 -1.000E-01 0.000E+00 1.000E-01 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 0.000E+00 . . + * . . 3.730E-01 . . + . * . . 7.197E-01 . + . . * . . 1.022E+00 . + . . * . 1.263E+00 . + . . . * . 1.423E+00 .+ . . . * . 1.490E+00 + . . . * . 1.467E+00 + . . . * . 1.357E+00 . + . . . * . 1.152E+00 . + . . . * . 8.774E-01 . + . . * . . 5.505E-01 . +. . * . . 1.878E-01 . . + . * . . -1.870E-01 . . + * . . . -5.500E-01 . . * + . . -8.810E-01 . . * . + . . -1.152E+00 . * . . + . . -1.351E+00 . * . . + . . -1.469E+00 . * . . + . . -1.495E+00 . * . . +. . -1.420E+00 . * . . + . . -1.261E+00 . * . . + . . -1.025E+00 . * . + . . -7.205E-01 . . * . + . . -3.713E-01 . . * + . . . 1.800E-04 . . + * . . 3.726E-01 . . + . * . . 7.194E-01 . + . . * . . 1.022E+00 . + . . * . 1.264E+00 . + . . . * . 1.422E+00 .+ . . . * . 1.490E+00 + . . . * . 1.468E+00 + . . . * . 1.357E+00 . + . . . * . 1.151E+00 . + . . . * . 8.775E-01 . + . . * . . 5.509E-01 . +. . * . . 1.877E-01 . . + . * . . -1.871E-01 . . + * . . . -5.522E-01 . . * + . . - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
As you can see, the new resistor values of 6 kΩ and 4 kΩ (R2 and R3, respectively) results in class A waveform reproduction, just the way we wanted.
- REVIEW:
- Class A operation is where an amplifier is biased so as to be in the active mode throughout the entire waveform cycle, thus faithfully reproducing the whole waveform.
- Class B operation is where an amplifier is biased so that only half of the input waveform gets reproduced: either the positive half or the negative half. The transistor spends half its time in the active mode and half its time cutoff. Complementary pairs of transistors running in class B operation are often used to deliver high power amplification in audio signal systems, each transistor of the pair handling a separate half of the waveform cycle. Class B operation delivers better power efficiency than a class A amplifier of similar output power.
- Class AB operation is where an amplifier is biased at a point somewhere between class A and class B.
- Class C operation is where an amplifier's bias forces it to amplify only a small portion of the waveform. A majority of the transistor's time is spent in cutoff mode. In order for there to be a complete waveform at the output, a resonant tank circuit is often used as a "flywheel" to maintain oscillations for a few cycles after each "kick" from the amplifier. Because the transistor is not conducting most of the time, power efficiencies are very high for a class C amplifier.
- Class D operation requires an advanced circuit design, and functions on the principle of representing instantaneous input signal amplitude by the duty cycle of a high-frequency squarewave. The output transistor(s) never operate in active mode, only cutoff and saturation. Thus, there is very little heat energy dissipated and energy efficiency is high.
- DC bias voltage on the input signal, necessary for certain classes of operation (especially class A and class C), may be obtained through the use of a voltage divider and coupling capacitor rather than a battery connected in series with the AC signal source.
Input and output coupling
To overcome the challenge of creating necessary DC bias voltage for an amplifier's input signal without resorting to the insertion of a battery in series with the AC signal source, we used a voltage divider connected across the DC power source. To make this work in conjunction with an AC input signal, we "coupled" the signal source to the divider through a capacitor, which acted as a high-pass filter. With that filtering in place, the low impedance of the AC signal source couldn't "short out" the DC voltage dropped across the bottom resistor of the voltage divider. A simple solution, but not without any disadvantages.Most obvious is the fact that using a high-pass filter capacitor to couple the signal source to the amplifier means that the amplifier can only amplify AC signals. A steady, DC voltage applied to the input would be blocked by the coupling capacitor just as much as the voltage divider bias voltage is blocked from the input source. Furthermore, since capacitive reactance is frequency-dependent, lower-frequency AC signals will not be amplified as much as higher-frequency signals. Non-sinusoidal signals will tend to be distorted, as the capacitor responds differently to each of the signal's constituent harmonics. An extreme example of this would be a low-frequency square-wave signal:
Incidentally, this same problem occurs when oscilloscope inputs are set to the "AC coupling" mode. In this mode, a coupling capacitor is inserted in series with the measured voltage signal to eliminate any vertical offset of the displayed waveform due to DC voltage combined with the signal. This works fine when the AC component of the measured signal is of a fairly high frequency, and the capacitor offers little impedance to the signal. However, if the signal is of a low frequency, and/or contains considerable levels of harmonics over a wide frequency range, the oscilloscope's display of the waveform will not be accurate.
In applications where the limitations of capacitive coupling would be intolerable, another solution may be used: direct coupling. Direct coupling avoids the use of capacitors or any other frequency-dependent coupling component in favor of resistors. A direct-coupled amplifier circuit might look something like this:
With no capacitor to filter the input signal, this form of coupling exhibits no frequency dependence. DC and AC signals alike will be amplified by the transistor with the same gain (the transistor itself may tend to amplify some frequencies better than others, but that is another subject entirely!).
If direct coupling works for DC as well as for AC signals, then why use capacitive coupling for any application? One reason might be to avoid any unwanted DC bias voltage naturally present in the signal to be amplified. Some AC signals may be superimposed on an uncontrolled DC voltage right from the source, and an uncontrolled DC voltage would make reliable transistor biasing impossible. The high-pass filtering offered by a coupling capacitor would work well here to avoid biasing problems.
Another reason to use capacitive coupling rather than direct is its relative lack of signal attenuation. Direct coupling through a resistor has the disadvantage of diminishing, or attenuating, the input signal so that only a fraction of it reaches the base of the transistor. In many applications, some attenuation is necessary anyway to prevent normal signal levels from "overdriving" the transistor into cutoff and saturation, so any attenuation inherent to the coupling network is useful anyway. However, some applications require that there be no signal loss from the input connection to the transistor's base for maximum voltage gain, and a direct coupling scheme with a voltage divider for bias simply won't suffice.
So far, we've discussed a couple of methods for coupling an input signal to an amplifier, but haven't addressed the issue of coupling an amplifier's output to a load. The example circuit used to illustrate input coupling will serve well to illustrate the issues involved with output coupling.
In our example circuit, the load is a speaker. Most speakers are electromagnetic in design: that is, they use the force generated by an lightweight electromagnet coil suspended within a strong permanent-magnet field to move a thin paper or plastic cone, producing vibrations in the air which our ears interpret as sound. An applied voltage of one polarity moves the cone outward, while a voltage of the opposite polarity will move the cone inward. To exploit cone's full freedom of motion, the speaker must receive true (unbiased) AC voltage. DC bias applied to the speaker coil tends to offset the cone from its natural center position, and this tends to limit the amount of back-and-forth motion it can sustain from the applied AC voltage without overtraveling. However, our example circuit applies a varying voltage of only one polarity across the speaker, because the speaker is connected in series with the transistor which can only conduct current one way. This situation would be unacceptable in the case of any high-power audio amplifier.
Somehow we need to isolate the speaker from the DC bias of the collector current so that it only receives AC voltage. One way to achieve this goal is to couple the transistor collector circuit to the speaker through a transformer:
Voltage induced in the secondary (speaker-side) of the transformer will be strictly due to variations in collector current, because the mutual inductance of a transformer only works on changes in winding current. In other words, only the AC portion of the collector current signal will be coupled to the secondary side for powering the speaker. The speaker will "see" true alternating current at its terminals, without any DC bias.
Transformer output coupling works, and has the added benefit of being able to provide impedance matching between the transistor circuit and the speaker coil with custom winding ratios. However, transformers tend to be large and heavy, especially for high-power applications. Also, it is difficult to engineer a transformer to handle signals over a wide range of frequencies, which is almost always required for audio applications. To make matters worse, DC current through the primary winding adds to the magnetization of the core in one polarity only, which tends to make the transformer core saturate more easily in one AC polarity cycle than the other. This problem is reminiscent of having the speaker directly connected in series with the transistor: a DC bias current tends to limit how much output signal amplitude the system can handle without distortion. Generally, though, a transformer can be designed to handle a lot more DC bias current than a speaker without running into trouble, so transformer coupling is still a viable solution in most cases.
Another method to isolate the speaker from DC bias in the output signal is to alter the circuit a bit and use a coupling capacitor in a manner similar to coupling the input signal to the amplifier:
This circuit resembles the more conventional form of common-emitter amplifier, with the transistor collector connected to the battery through a resistor. The capacitor acts as a high-pass filter, passing most of the AC voltage to the speaker while blocking all DC voltage. Again, the value of this coupling capacitor is chosen so that its impedance at the expected signal frequency will be arbitrarily low.
The blocking of DC voltage from an amplifier's output, be it via a transformer or a capacitor, is useful not only in coupling an amplifier to a load, but also in coupling one amplifier to another amplifier. "Staged" amplifiers are often used to achieve higher power gains than what would be possible using a single transistor:
While it is possible to directly couple each stage to the next (via a resistor rather than a capacitor), this makes the whole amplifier very sensitive to variations in the DC bias voltage of the first stage, since that DC voltage will be amplified along with the AC signal until the last stage. In other words, the biasing of the first stage will affect the biasing of the second stage, and so on. However, if the stages are capacitively coupled as shown in the above illustration, the biasing of one stage has no effect on the biasing of the next, because DC voltage is blocked from passing on to the next stage.
Transformer coupling between amplifier stages is also a possibility, but less often seen due to some of the problems inherent to transformers mentioned previously. One notable exception to this rule is in the case of radio-frequency amplifiers where coupling transformers are typically small, have air cores (making them immune to saturation effects), and can be made part of a resonant circuit so as to block unwanted harmonic frequencies from passing on to subsequent stages. The use of resonant circuits assumes that the signal frequency remains constant, of course, but this is typically the case in radio circuitry. Also, the "flywheel" effect of LC tank circuits allows for class C operation for high efficiency:
Having said all this, it must be mentioned that it is possible to use direct coupling within a multi-stage transistor amplifier circuit. In cases where the amplifier is expected to handle DC signals, this is the only alternative.
- REVIEW:
- Capacitive coupling acts like a high-pass filter on the input of an amplifier. This tends to make the amplifier's voltage gain decrease at lower signal frequencies. Capacitive-coupled amplifiers are all but unresponsive to DC input signals.
- Direct coupling with a series resistor instead of a series capacitor avoids the problem of frequency-dependent gain, but has the disadvantage of reducing amplifier gain for all signal frequencies by attenuating the input signal.
- Transformers and capacitors may be used to couple the output of an amplifier to a load, to eliminate DC voltage from getting to the load.
- Multi-stage amplifiers often make use of capacitive coupling between stages to eliminate problems with the bias from one stage affecting the bias of another.
Feedback
If some percentage of an amplifier's output signal is connected to the input, so that the amplifier amplifies part of its own output signal, we have what is known as feedback. Feedback comes in two varieties: positive (also called regenerative), and negative (also called degenerative). Positive feedback reinforces the direction of an amplifier's output voltage change, while negative feedback does just the opposite.A familiar example of feedback happens in public-address ("PA") systems where someone holds the microphone too close to a speaker: a high-pitched "whine" or "howl" ensues, because the audio amplifier system is detecting and amplifying its own noise. Specifically, this is an example of positive or regenerative feedback, as any sound detected by the microphone is amplified and turned into a louder sound by the speaker, which is then detected by the microphone again, and so on . . . the result being a noise of steadily increasing volume until the system becomes "saturated" and cannot produce any more volume.
One might wonder what possible benefit feedback is to an amplifier circuit, given such an annoying example as PA system "howl." If we introduce positive, or regenerative, feedback into an amplifier circuit, it has the tendency of creating and sustaining oscillations, the frequency of which determined by the values of components handling the feedback signal from output to input. This is one way to make an oscillator circuit to produce AC from a DC power supply. Oscillators are very useful circuits, and so feedback has a definite, practical application for us.
Negative feedback, on the other hand, has a "dampening" effect on an amplifier: if the output signal happens to increase in magnitude, the feedback signal introduces a decreasing influence into the input of the amplifier, thus opposing the change in output signal. While positive feedback drives an amplifier circuit toward a point of instability (oscillations), negative feedback drives it the opposite direction: toward a point of stability.
An amplifier circuit equipped with some amount of negative feedback is not only more stable, but it tends to distort the input waveform to a lesser degree and is generally capable of amplifying a wider range of frequencies. The tradeoff for these advantages (there just has to be a disadvantage to negative feedback, right?) is decreased gain. If a portion of an amplifier's output signal is "fed back" to the input in such a way as to oppose any changes in the output, it will require a greater input signal amplitude to drive the amplifier's output to the same amplitude as before. This constitutes a decreased gain. However, the advantages of stability, lower distortion, and greater bandwidth are worth the tradeoff in reduced gain for many applications.
Let's examine a simple amplifier circuit and see how we might introduce negative feedback into it:
The amplifier configuration shown here is a common-emitter, with a resistor bias network formed by R1 and R2. The capacitor couples Vinput to the amplifier so that the signal source doesn't have a DC voltage imposed on it by the R1/R2 divider network. Resistor R3 serves the purpose of controlling voltage gain. We could omit if for maximum voltage gain, but since base resistors like this are common in common-emitter amplifier circuits, we'll keep it in this schematic.
Like all common-emitter amplifiers, this one inverts the input signal as it is amplified. In other words, a positive-going input voltage causes the output voltage to decrease, or go in the direction of negative, and visa-versa. If we were to examine the waveforms with oscilloscopes, it would look something like this:
Because the output is an inverted, or mirror-image, reproduction of the input signal, any connection between the output (collector) wire and the input (base) wire of the transistor will result in negative feedback:
The resistances of R1, R2, R3, and Rfeedback function together as a signal-mixing network so that the voltage seen at the base of the transistor (in reference to ground) is a weighted average of the input voltage and the feedback voltage, resulting in signal of reduced amplitude going into the transistor. As a result, the amplifier circuit will have reduced voltage gain, but improved linearity (reduced distortion) and increased bandwidth.
A resistor connecting collector to base is not the only way to introduce negative feedback into this amplifier circuit, though. Another method, although more difficult to understand at first, involves the placement of a resistor between the transistor's emitter terminal and circuit ground, like this:
This new feedback resistor drops voltage proportional to the emitter current through the transistor, and it does so in such a way as to oppose the input signal's influence on the base-emitter junction of the transistor. Let's take a closer look at the emitter-base junction and see what difference this new resistor makes:
With no feedback resistor connecting the emitter to ground, whatever level of input signal (Vinput) makes it through the coupling capacitor and R1/R2/R3 resistor network will be impressed directly across the base-emitter junction as the transistor's input voltage (VB-E). In other words, with no feedback resistor, VB-E equals Vinput. Therefore, if Vinput increases by 100 mV, then VB-E likewise increases by 100 mV: a change in one is the same as a change in the other, since the two voltages are equal to each other.
Now let's consider the effects of inserting a resistor (Rfeedback) between the transistor's emitter lead and ground:
Note how the voltage dropped across Rfeedback adds with VB-E to equal Vinput. With Rfeedback in the Vinput -- VB-E loop, VB-E will no longer be equal to Vinput. We know that Rfeedback will drop a voltage proportional to emitter current, which is in turn controlled by the base current, which is in turn controlled by the voltage dropped across the base-emitter junction of the transistor (VB-E). Thus, if Vinput were to increase in a positive direction, it would increase VB-E, causing more base current, causing more collector (load) current, causing more emitter current, and causing more feedback voltage to be dropped across Rfeedback. This increase of voltage drop across the feedback resistor, though, subtracts from Vinput to reduce the VB-E, so that the actual voltage increase for VB-E will be less than the voltage increase of Vinput. No longer will a 100 mV increase in Vinput result in a full 100 mV increase for VB-E, because the two voltages are not equal to each other.
Consequently, the input voltage has less control over the transistor than before, and the voltage gain for the amplifier is reduced: just what we expected from negative feedback.
In practical common-emitter circuits, negative feedback isn't just a luxury; it's a necessity for stable operation. In a perfect world, we could build and operate a common-emitter transistor amplifier with no negative feedback, and have the full amplitude of Vinput impressed across the transistor's base-emitter junction. This would give us a large voltage gain. Unfortunately, though, the relationship between base-emitter voltage and base-emitter current changes with temperature, as predicted by the "diode equation." As the transistor heats up, there will be less of a forward voltage drop across the base-emitter junction for any given current. This causes a problem for us, as the R1/R2 voltage divider network is designed to provide the correct quiescent current through the base of the transistor so that it will operate in whatever class of operation we desire (in this example, I've shown the amplifier working in class-A mode). If the transistor's voltage/current relationship changes with temperature, the amount of DC bias voltage necessary for the desired class of operation will change. In this case, a hot transistor will draw more bias current for the same amount of bias voltage, making it heat up even more, drawing even more bias current. The result, if unchecked, is called thermal runaway.
Common-collector amplifiers, however, do not suffer from thermal runaway. Why is this? The answer has everything to do with negative feedback:
Note that the common-collector amplifier has its load resistor placed in exactly the same spot as we had the Rfeedback resistor in the last circuit: between emitter and ground. This means that the only voltage impressed across the transistor's base-emitter junction is the difference between Vinput and Voutput, resulting in a very low voltage gain (usually close to 1 for a common-collector amplifier). Thermal runaway is impossible for this amplifier: if base current happens to increase due to transistor heating, emitter current will likewise increase, dropping more voltage across the load, which in turn subtracts from Vinput to reduce the amount of voltage dropped between base and emitter. In other words, the negative feedback afforded by placement of the load resistor makes the problem of thermal runaway self-correcting. In exchange for a greatly reduced voltage gain, we get superb stability and immunity from thermal runaway.
By adding a "feedback" resistor between emitter and ground in a common-emitter amplifier, we make the amplifier behave a little less like an "ideal" common-emitter and a little more like a common-collector. The feedback resistor value is typically quite a bit less than the load, minimizing the amount of negative feedback and keeping the voltage gain fairly high.
Another benefit of negative feedback, seen clearly in the common-collector circuit, is that it tends to make the voltage gain of the amplifier less dependent on the characteristics of the transistor. Note that in a common-collector amplifier, voltage gain is nearly equal to unity (1), regardless of the transistor's β. This means, among other things, that we could replace the transistor in a common-collector amplifier with one having a different β and not see any significant changes in voltage gain. In a common-emitter circuit, the voltage gain is highly dependent on β. If we were to replace the transistor in a common-emitter circuit with another of differing β, the voltage gain for the amplifier would change significantly. In a common-emitter amplifier equipped with negative feedback, the voltage gain will still be dependent upon transistor β to some degree, but not as much as before, making the circuit more predictable despite variations in transistor β.
The fact that we have to introduce negative feedback into a common-emitter amplifier to avoid thermal runaway is an unsatisfying solution. It would be nice, after all, to avoid thermal runaway without having to suppress the amplifier's inherently high voltage gain. A best-of-both-worlds solution to this dilemma is available to us if we closely examine the nature of the problem: the voltage gain that we have to minimize in order to avoid thermal runaway is the DC voltage gain, not the AC voltage gain. After all, it isn't the AC input signal that fuels thermal runaway: it's the DC bias voltage required for a certain class of operation: that quiescent DC signal that we use to "trick" the transistor (fundamentally a DC device) into amplifying an AC signal. We can suppress DC voltage gain in a common-emitter amplifier circuit without suppressing AC voltage gain if we figure out a way to make the negative feedback function with DC only. That is, if we only feed back an inverted DC signal from output to input, but not an inverted AC signal.
The Rfeedback emitter resistor provides negative feedback by dropping a voltage proportional to load current. In other words, negative feedback is accomplished by inserting an impedance into the emitter current path. If we want to feed back DC but not AC, we need an impedance that is high for DC but low for AC. What kind of circuit presents a high impedance to DC but a low impedance to AC? A high-pass filter, of course!
By connecting a capacitor in parallel with the feedback resistor, we create the very situation we need: a path from emitter to ground that is easier for AC than it is for DC:
The new capacitor "bypasses" AC from the transistor's emitter to ground, so that no appreciable AC voltage will be dropped from emitter to ground to "feed back" to the input and suppress voltage gain. Direct current, on the other hand, cannot go through the bypass capacitor, and so must travel through the feedback resistor, dropping a DC voltage between emitter and ground which lowers the DC voltage gain and stabilizes the amplifier's DC response, preventing thermal runaway. Because we want the reactance of this capacitor (XC) to be as low as possible, Cbypass should be sized relatively large. Because the polarity across this capacitor will never change, it is safe to use a polarized (electrolytic) capacitor for the task.
Another approach to the problem of negative feedback reducing voltage gain is to use multi-stage amplifiers rather than single-transistor amplifiers. If the attenuated gain of a single transistor is insufficient for the task at hand, we can use more than one transistor to make up for the reduction caused by feedback. Here is an example circuit showing negative feedback in a three-stage common-emitter amplifier:
Note how there is but one "path" for feedback, from the final output to the input through a single resistor, Rfeedback. Since each stage is a common-emitter amplifier -- and thus inverting in nature -- and there are an odd number of stages from input to output, the output signal will be inverted with respect to the input signal, and the feedback will be negative (degenerative). Relatively large amounts of feedback may be used without sacrificing voltage gain, because the three amplifier stages provide so much gain to begin with.
At first, this design philosophy may seem inelegant and perhaps even counter-productive. Isn't this a rather crude way to overcome the loss in gain incurred through the use of negative feedback, to simply recover gain by adding stage after stage? What is the point of creating a huge voltage gain using three transistor stages if we're just going to attenuate all that gain anyway with negative feedback? The point, though perhaps not apparent at first, is increased predictability and stability from the circuit as a whole. If the three transistor stages are designed to provide an arbitrarily high voltage gain (in the tens of thousands, or greater) with no feedback, it will be found that the addition of negative feedback causes the overall voltage gain to become less dependent of the individual stage gains, and approximately equal to the simple ratio Rfeedback/Rin. The more voltage gain the circuit has (without feedback), the more closely the voltage gain will approximate Rfeedback/Rin once feedback is established. In other words, voltage gain in this circuit is fixed by the values of two resistors, and nothing more.
This advantage has profound impact on mass-production of electronic circuitry: if amplifiers of predictable gain may be constructed using transistors of widely varied β values, it makes the selection and replacement of components very easy and inexpensive. It also means the amplifier's gain varies little with changes in temperature. This principle of stable gain control through a high-gain amplifier "tamed" by negative feedback is elevated almost to an art form in electronic circuits called operational amplifiers, or op-amps. You may read much more about these circuits in a later chapter of this book!
- REVIEW:
- Feedback is the coupling of an amplifier's output to its input.
- Positive, or regenerative feedback has the tendency of making an amplifier circuit unstable, so that it produces oscillations (AC). The frequency of these oscillations is largely determined by the components in the feedback network.
- Negative, or degenerative feedback has the tendency of making an amplifier circuit more stable, so that its output changes less for a given input signal than without feedback. This reduces the gain of the amplifier, but has the advantage of decreasing distortion and increasing bandwidth (the range of frequencies the amplifier can handle).
- Negative feedback may be introduced into a common-emitter circuit by coupling collector to base, or by inserting a resistor between emitter and ground.
- An emitter-to-ground "feedback" resistor is usually found in common-emitter circuits as a preventative measure against thermal runaway.
- Negative feedback also has the advantage of making amplifier voltage gain more dependent on resistor values and less dependent on the transistor's characteristics.
- Common-collector amplifiers have a lot of negative feedback, due to the placement of the load resistor between emitter and ground. This feedback accounts for the extremely stable voltage gain of the amplifier, as well as its immunity against thermal runaway.
- Voltage gain for a common-emitter circuit may be re-established without sacrificing immunity to thermal runaway, by connecting a bypass capacitor in parallel with the emitter "feedback resistor."
- If the voltage gain of an amplifier is arbitrarily high (tens of thousands, or greater), and negative feedback is used to reduce the gain to reasonable levels, it will be found that the gain will approximately equal Rfeedback/Rin. Changes in transistor β or other internal component values will have comparatively little effect on voltage gain with feedback in operation, resulting in an amplifier that is stable and easy to design.
Amplifier impedances
*** PENDING ***- REVIEW:
Current mirrors
An interesting and often-used circuit applying the bipolar junction transistor is the so-called current mirror, which serves as a simple current regulator, supplying nearly constant current to a load over a wide range of load resistances.We know that in a transistor operating in its active mode, collector current is equal to base current multiplied by the ratio β. We also know that the ratio between collector current and emitter current is called α. Because collector current is equal to base current multiplied by β, and emitter current is the sum of the base and collector currents, α should be mathematically derivable from β. If you do the algebra, you'll find that α = β/(β+1) for any transistor.
We've seen already how maintaining a constant base current through an active transistor results in the regulation of collector current, according to the β ratio. Well, the α ratio works similarly: if emitter current is held constant, collector current will remain at a stable, regulated value so long as the transistor has enough collector-to-emitter voltage drop to maintain it in its active mode. Therefore, if we have a way of holding emitter current constant through a transistor, the transistor will work to regulate collector current at a constant value.
Remember that the base-emitter junction of a BJT is nothing more than a PN junction, just like a diode, and that the "diode equation" specifies how much current will go through a PN junction given forward voltage drop and junction temperature:
If both junction voltage and temperature are held constant, then the PN junction current will likewise be constant. Following this rationale, if we were to hold the base-emitter voltage of a transistor constant, then its emitter current should likewise be constant, given a constant temperature:
This constant emitter current, multiplied by a constant α ratio, gives a constant collector current through Rload, provided that there is enough battery voltage to keep the transistor in its active mode for any change in Rload's resistance.
Maintaining a constant voltage across the transistor's base-emitter junction is easy: use a forward-biased diode to establish a constant voltage of approximately 0.7 volts, and connect it in parallel with the base-emitter junction:
Now, here's where it gets interesting. The voltage dropped across the diode probably won't be 0.7 volts exactly. The exact amount of forward voltage dropped across it depends on the current through the diode, and the diode's temperature, all in accordance with the diode equation. If diode current is increased (say, by reducing the resistance of Rbias), its voltage drop will increase slightly, increasing the voltage drop across the transistor's base-emitter junction, which will increase the emitter current by the same proportion, assuming the diode's PN junction and the transistor's base-emitter junction are well-matched to each other. In other words, transistor emitter current will closely equal diode current at any given time. If you change the diode current by changing the resistance value of Rbias, then the transistor's emitter current will follow suit, because the emitter current is described by the same equation as the diode's, and both PN junctions experience the same voltage drop.
Remember, the transistor's collector current is almost equal to its emitter current, as the α ratio of a typical transistor is almost unity (1). If we have control over the transistor's emitter current by setting diode current with a simple resistor adjustment, then we likewise have control over the transistor's collector current. In other words, collector current mimics, or mirrors, diode current.
Current through resistor Rload is therefore a function of current set by the bias resistor, the two being nearly equal. This is the function of the current mirror circuit: to regulate current through the load resistor by conveniently adjusting the value of Rbias. It is very easy to create a set amount of diode current, as current through the diode is described by a simple equation: power supply voltage minus diode voltage (almost a constant value), divided by the resistance of Rbias.
To better match the characteristics of the two PN junctions (the diode junction and the transistor base-emitter junction), a transistor may be used in place of a regular diode, like this:
Because temperature is a factor in the "diode equation," and we want the two PN junctions to behave identically under all operating conditions, we should maintain the two transistors at exactly the same temperature. This is easily done using discrete components by gluing the two transistor cases back-to-back. If the transistors are manufactured together on a single chip of silicon (as a so-called integrated circuit, or IC), the designers should locate the two transistors very close to one another to facilitate heat transfer between them.
The current mirror circuit shown with two NPN transistors is sometimes called a current-sinking type, because the regulating transistor conducts current to the load from ground ("sinking" current), rather than from the positive side of the battery ("sourcing" current). If we wish to have a grounded load, and a current sourcing mirror circuit, we could use PNP transistors like this:
- REVIEW:
- A current mirror is a transistor circuit that regulates current through a load resistance, the regulation point being set by a simple resistor adjustment.
- Transistors in a current mirror circuit must be maintained at the same temperature for precise operation. When using discrete transistors, you may glue their cases together to help accomplish this.
- Current mirror circuits may be found in two basic varieties: the current sinking configuration, where the regulating transistor connects the load to ground; and the current sourcing configuration, where the regulating transistor connects the load to the positive terminal of the DC power supply.
Transistor ratings and packages
*** INCOMPLETE ***Like all electrical and electronic components, transistors are limited in the amounts of voltage and current they can handle without sustaining damage. Since transistors are a bit more complex than some of the other components you're used to seeing at this point, they tend to have more kinds of ratings. What follows is an itemized description of some typical transistor ratings.
Power dissipation: When a transistor conducts current between collector and emitter, it also drops voltage between those two points. At any given time, the power dissipated by a transistor is equal to the product (multiplication) of collector current and collector-emitter voltage. Just like resistors, transistors are rated in terms of how many watts they can safely dissipate without sustaining damage. High temperature is the mortal enemy of all semiconductor devices, and bipolar transistors tend to be more susceptible to thermal damage than most. Power ratings are always given in reference to the temperature of ambient (surrounding) air. When transistors are to be used in hotter-than-normal environments, their power ratings must be derated to avoid a shortened service life.
Reverse voltages: As with diodes, bipolar transistors are rated for maximum allowable reverse-bias voltage across their PN junctions. This includes voltage ratings for the base-emitter junction, base-collector junction, and also from collector to emitter. The rating for maximum collector-emitter voltage can be thought of in terms of the maximum voltage it can withstand while in full-cutoff mode (no base current). This rating is of particular importance when using a bipolar transistor as a switch.
Collector current: A maximum value for collector current will be given by the manufacturer in amps. Understand that this maximum figure assumes a saturated state (minimum collector-emitter voltage drop). If the transistor is not saturated, and in fact is dropping substantial voltage between collector and emitter, the maximum power dissipation rating will probably be exceeded before the maximum collector current rating will. Just something to keep in mind when designing a transistor circuit!
Saturation voltages: Ideally, a saturated transistor acts as a closed switch contact between collector and emitter, dropping zero voltage at full collector current. In reality this is never true. Manufacturers will specify the maximum voltage drop of a transistor at saturation, both between the collector and emitter, and also between base and emitter (forward voltage drop of that PN junction). Collector-emitter voltage drop at saturation is generally expected to be 0.3 volts or less, but this figure is of course dependent on the specific type of transistor. Base-emitter forward voltage drop is very similar to that of an equivalent diode, which should come as no surprise. Beta: The ratio of collector current to base current, β is the fundamental parameter characterizing the amplifying ability of a bipolar transistor. β is usually assumed to be a constant figure in circuit calculations, but unfortunately this is far from true in practice. As such, manufacturers provide a set of β (or "hfe") figures for a given transistor over a wide range of operating conditions, usually in the form of maximum/minimum/typical ratings. It may surprise you to see just how widely β can be expected to vary within normal operating limits. One popular small-signal transistor, the 2N3903, is advertised as having a β ranging from 15 to 150 depending on the amount of collector current. Generally, β is highest for medium collector currents, decreasing for very low and very high collector currents.
Alpha: the ratio of collector current to emitter current, α may be derived from β, being equal to β/(β+1).
Bipolar transistors come in a wide variety of physical packages. Package type is primarily dependent upon the power dissipation of the transistor, much like resistors: the greater the maximum power dissipation, the larger the device has to be to stay cool. There are several standardized package types for three-terminal semiconductor devices, any of which may be used to house a bipolar transistor. This is an important fact to consider: there are many other semiconductor devices other than bipolar transistors which have three connection points. It is impossible to positively identify a three-terminal semiconductor device without referencing the part number printed on it, and/or subjecting it to a set of electrical tests.
- REVIEW:
BJT quirks
*** PENDING ***Nonlinearity Temperature drift Thermal runaway Junction capacitance Noise Mismatch (problem with paralleling transistors) β cutoff frequency Alpha cutoff frequency
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