Expanded Answer: Transistor Amplifier Problem in January/February 2008 Morseman Column Here s what I asked: This month s problem: Figure 4(a) shows a simple npn transistor amplifier. The transistor has = 220, and its base-emitter voltage is 0.7 V. What are the quiescent base and collector currents, and the collector voltage? Can you estimate the voltage gain, v 2 /v 1? Can you estimate the input impedance of the amplifier? What are the advantages and disadvantages of deriving the base biasing current from the collector rather than the positive line? Here is the Figure: Figure 1: Simple amplifier schematic. First, the dc conditions: We find the dc operating point by relating the voltage drop across the two resistors to the currents flowing through them: 6.7 0.7 = (I c + I b ) R L + I b R b but I b = I c ( so 6 = I c 1 + 1 ) R L + I c R b and 1 6 so I c R L + R b and I b = I c = 3 220 = 6 1 + 220 220 = 13.63 µa. 3 ma also V c = 6.7 I c R L = 6.7 3 1 = 3.7 V Small-signal Analysis Figure 2 (a) shows the circuit again, with input and output voltages renamed for convenience. The simplest useful small-signal equivalent circuit of a bi-polar transistor is shown in figure 2 (b). Small- 1
signal means that the circuit is linearized, and the two parameters involved ( and ) are evaluated at the transistor s dc operating point. More complex equivalent circuits are used in SPICE analysis, but this one is almost always good enough for simple estimates. Figure 2: (a) Circuit as drawn. (b) Equivalent small-signal transistor circuit. (c) Small-signal circuit of complete amplifier is now the small-signal beta, also known as h fe, 1 and is the slope of the versus I c characteristic of the transistor. Both the small and large-signal values are usually similar however, and is a very variable parameter anyway, so we will assume the same value. The resistance is the slope of the diode characteristic formed by the base-emitter junction, evaluated again at the operating point. If the diode is assumed to obey the ideal diode equation, its value is approximately 2 Figure 2 (c) shows the small-signal circuit of the complete amplifier. original circuit, and is derived with these assumptions: This looks nothing like the Both coupling capacitors have negligible impedance at the operating frequency, and can be replaced by short-circuits. Since we are considering only small-signal, ac excitation, and no ac voltage can exist on a perfect dc power supply, the power supply acts like a short-circuit at signal frequencies, and its positive and negative terminals can both be connected to ground (the bottom node). For convenience, we define two more currents, I in and I out, as shown. Then: v 2 = i 0 R L (1) i o = v 1 v 2 R b i b (2) i b = v 1 (3) 1 R eliminating i 0 and I b, A v = b 1 + 1 (4) R L R b 1 h fe is so named because it is one of the parameters of the four-element h-parameter equivalent circuit, much used in the early days. The two-element circuit used here is an approximate form of this. 2 This expression for is proved in my text, Transistor Electronics, and other places. See also the appendix to this document. 2
= ( 1 ) R p Rb (5) where R p = R L R b (6) so A v R p since R b >> (7) and R b R L (8) so simplifying further, A v R L (9) but (10) ( ) 40Ic so A v R L 40V L (11) where V L = I c R L = the dc voltage across the load resistor. (12) whence A v = 120 (13) The approximate expression for the voltage gain given by equation (11) is the same as that of a standard bi-polar amplifier. In fact, the approximation is quite good, overestimating the gain by a few percent. To find the input impedance, consider the input current, i in. But this is the admittance of two resistors in parallel, or i in = v 1 + v 1 v 2 (14) R b i in = y in = 1 ( + 1 v ) 2 1 (15) v 1 v 1 R b assume v 2 v 1 R L (16) then y in 1 + 1 R b + R L R b (17) y in 1 R b + 2 (18) Summarizing, we estimate: r in 1 R b y in 2 2 (19) r in 916 Ω 80I c (20) I c = 3 ma (21) I b = 13.6 µa (22) V c = 3.7 V (23) A v = 120 (24) r in = 916 Ω (25) We see that the voltage gain is very slightly less than a standard amplifier, the input admittance is about half that of a standard amplifier, 3
For typical component values, it is found that the fractional increase in collector current, I c, is about half that of the fractional increase in. Thus this circuit is best used to amplify very small signals, where operating point variations caused by variation will not cause non-linearity. An alternative viewpoint is to assume that the value of R b, the feedback resistor, is so large that we can ignore its effect on the gain, which becomes that of a standard common-emitter amplifier without feedback - as justified above. But its effect must be considered in the input impedance, particularly if the gain is high. We can include it using Miller s Theorem. This theorem says that if an amplifier has voltage gain A v, with an impedance z b connected between the input and output terminals, this is equivalent to removing z b, and connecting an impedance of z b /A v directly across the input terminals. If you work through this, you get the same approximate expressions as those for the voltage gain and input impedance as above. Advantages and Disadvantages You design a circuit like this by assuming some value of, and then choosing R b R L. This puts the quiescent dc collector voltage about half-way between the power supply voltage and ground, giving maximum signal swing. But as stated above, if is not quite right, it doesn t matter too much, because the negative dc feedback through R b partially compensates. This is an advantage. Another is that no biasing chain resistors nor associated by-pass capacitors are necessary. However, you can design an amplifier using one of the standard circuits with much better immunity to variation. This is a disadvantage. Another is that the input impedance is lowered by the negative feedback caused by R b. This can also be shown using Miller s theorem. An LTSpice Simulation Figure 3 shows (lower panel) an LTSpice model of the circuit and (upper panel) the waveform at the collector, for an input sinewave voltage of 1 mv peak-to-peak at 1 khz. I chose a BCW60B transistor (one supplied in the default LTSpice download) since it has nominal = 240, close enough to the design value. This is simulated using a transient analysis, and shows A v = 106 (estimated as -120 above) (26) V c = 3.5 V (estimated as 3.7 V above) (27) The input impedance can t be plotted using a transient analysis, since this requires dividing the input voltage, V 1, by the current through C 1, and both waveforms pass through zero, so the division blows up. However, plotting both and diving the peak value of V 1 by the peak current through C 1, which is 60 na, we get r in = 1.09 kω (estimated as 916 Ω above). (28) 4
A small-signal, AC analysis can plot the input impedance directly, since this uses a linearized model with transistor parameters chosen at the dc operating point. This shows the same voltage gain, and r in = 1.11 kω. Figure 3: Top: Waveform at the collector. Bottom: The simulated circuit. Design Equations Practical rules for designing an amplifier like this are Set the dc voltage at the collector, V c, halfway between the supply voltage and ground. Decide on a suitable collector current, I c. The value of the load resistor required is then R L = V c 2I c. Estimate (or find from some spec sheet) for this transistor. The value of the bias resistor required is then R b = R L. Choose the value of the input coupling capacitor (C 1 in the simulation diagram) such that C 100I c (29) f where f = the lowest operating frequency required. (30) 5
Appendix: Transistor Parameters I have used the relationship = : The least technically obscure reference leading to the derivation of this expression starts on page 80 of the excellent practical reference by Horowitz and Hill. 3 The collector current, I c, of a bi-polar transistor is almost exactly related to the voltage applied across the base-emitter junction, V be, by the ideal diode equation. This is I c = ( )] Vbe I s [exp 1 (31) where I s = diode leakage current, (32) V be = appplied base-emitter diode voltage, (33) = threshold voltage, = kt q (34) k = Boltmann s constant, (35) T = absolute (Kelvin) temperature, (36) q = charge on electron. (37) evaluating, 25.3mV at room temperature (38) For convenience, I round this off to = 25 mv. The second term ( 1) dominates in the reverse region, where V be is negative, making the exponential term very small. It represents the saturation current, which is just I s. For forward currents in excess of a few tens of microamp, it can be neglected. Then, to a very good approximation, differentiating, I c = I s [exp I c V be = I s [ exp ( )] Vbe V ( T )] Vbe (39) = I c (40) therefore inverting, V be I c = I c (41) But the input resistance seen between the base and emitter is taken with respect to the base current, I b. This is times less, so this input resistance will be times higher. thus = I b (42) substituting I b = I c (43) = (I c /) = = 25 10 3 I c I c (44) or more conveniently, = (45) 3 The Art of Electronics, Paul Horowitz and Winfield Hill, Cambridge, second edition. 6