By the end of this chapter you should be able to:
- State the small-signal assumption, derive the 10 mV limit on \(v_{be}\), and justify separating the d.c. and a.c. analyses.
- Derive \(r_e = V_T/I_E\) and construct the \(r_e\) small-signal model of a transistor.
- Construct the hybrid-\(\pi\) model, define \(g_m\), \(r_\pi\) and \(r_o\), and show the exact relationships to the \(r_e\) model.
- Define the four h-parameters, relate them to the other two models, and judge when \(h_{re}\) may be neglected.
- Derive and compute \(A_v\), \(Z_i\) and \(Z_o\) for a loaded common-emitter stage.
- Explain the role of the emitter bypass capacitor and design a swamping resistor to trade gain for stability.
- Analyse the common-base and emitter-follower configurations and choose between all three for a given job.
Chapter 18 left a transistor sitting quietly at a chosen operating point, drawing 2 mA of collector current and dropping 6 V. Nothing about that arrangement amplifies anything: it is a d.c. circuit, and a d.c. circuit that turns 24 mW of supply power into 12 mW of heat in a transistor and 12 mW in two resistors is not obviously an achievement. The point of establishing the Q-point was to make the device ready to respond to a signal, and this chapter is about that response.
The obstacle is that the transistor is not a linear device. Its collector current depends exponentially on \(V_{BE}\), and an exponential has no gain — it has a different gain at every point. The way round it is the oldest trick in engineering: if the signal is small enough, replace the curve by its tangent at the Q-point and treat the device as linear. The reward is that the whole apparatus of linear circuit analysis becomes available — superposition, Thévenin equivalents, impedances, transfer functions — and a transistor stage can be reduced to a handful of resistors and one dependent source. This chapter builds two versions of that reduction, the \(r_e\) model and the hybrid-\(\pi\) model, shows they are the same model in different clothes, connects both to the h-parameters a data sheet quotes, and then uses them to compute the gain and the two impedances of all three configurations of Chapter 17.
1 What "Small Signal" Buys, and What It Costs
The collector current of a transistor in the active region obeys \(I_C = I_S e^{V_{BE}/V_T}\). Suppose the base-emitter voltage is the sum of a steady bias \(V_{BEQ}\) and a signal \(v_{be}\). Then
The exponential expanded as a series about the operating point. The first term is the bias current, the second is the wanted linear response, and every term after that is distortion.
Keeping only the linear term gives \(i_c = I_{CQ}v_{be}/V_T = g_m v_{be}\), and the transistor has become a linear voltage-controlled current source. The cost is stated by the first term discarded: the ratio of the second-harmonic term to the fundamental is \(v_{be}/2V_T\). With \(V_T = 26\) mV, a signal amplitude of 5.2 mV gives 10 per cent distortion, 2.6 mV gives 5 per cent, and 0.52 mV gives 1 per cent.
Ten millivolts of base-emitter signal gives roughly 19 per cent second-harmonic distortion at the collector, which sounds a great deal and is the usual textbook boundary because it is where the error in the gain is still only a few per cent. For genuinely linear work the limit is nearer 1 or 2 mV, and any stage that must handle more than that needs either negative feedback around it or the swamping resistor of Section 5, which raises the permitted input to 88 mV in the design worked here.
The assumption also permits a second, purely practical simplification. Because the circuit is now linear, superposition applies, and the d.c. and a.c. analyses can be done separately and added. The rules for drawing the a.c. equivalent circuit follow directly:
- Replace every d.c. supply by a short circuit to ground. A supply rail holds a constant voltage, so it has no signal on it, so it is at signal ground. The collector resistor, which is returned to \(V_{CC}\), therefore appears to the signal as a resistor to ground.
- Replace every capacitor that is large enough by a short circuit. Coupling and bypass capacitors are chosen so that their reactance at the lowest signal frequency of interest is small compared with the resistance they work into. In the mid-band they are short circuits, and Chapter 25 examines what happens at the low-frequency end where they are not.
- Replace the transistor by its small-signal model. That is what the next two sections construct.
Notice what the bypass capacitor \(C_E\) achieves. The emitter resistor is essential to the d.c. stability of Chapter 18 and is fatal to the a.c. gain, since it applies exactly the same negative feedback to the signal that it applies to the drift. Putting a large capacitor across it leaves it in place for d.c. and removes it for a.c. — one component that lets a single circuit have two different behaviours in two different frequency ranges.
2 The \(r_e\) Model
Looking into the emitter of a transistor whose base is held at a fixed potential, the signal sees the forward-biased base-emitter junction. Its small-signal resistance is obtained by differentiating the diode equation, exactly as in Chapter 6:
The single most useful number in transistor amplifier design. With \(V_T = 26\) mV it is conveniently written \(r_e = 26\ \text{mV}/I_E(\text{mA})\), so 26 Ω at 1 mA, 13 Ω at 2 mA, 2.6 Ω at 10 mA. For the amplifier of Figure 19.1, \(I_E = 2.029\) mA gives \(r_e = 26/2.029 = 12.81\ \Omega\).
Three things about \(r_e\) are worth fixing in the mind. It is a small-signal resistance, the slope of the characteristic and not the ratio \(V_{BE}/I_E\), which at this operating point would be \(0.7/2.029\ \text{mA} = 345\ \Omega\), twenty-seven times larger and of no use for anything. It is not a physical resistor, so it dissipates nothing and cannot be measured with an ohmmeter. And it is set entirely by the bias current, which means a designer controls it directly through the design of Chapter 18.
Looking into the base, however, the signal sees something different, because the base carries only \(1/(1+\beta)\) of the emitter current. The same voltage \(v_{be}\) produces a base current \((1+\beta)\) times smaller, so the resistance looking into the base is \((1+\beta)\) times larger:
For the worked amplifier, \((1+\beta)r_e = 101 \times 12.81 = 1294\ \Omega\). This is the same reflection rule that put \((1+\beta)R_E\) into the bias equations of Chapter 18, and it works for any impedance in the emitter, real or incremental.
The \(r_e\) model is then simply these two facts drawn as a circuit: a resistance \((1+\beta)r_e\) between base and emitter, and a current source \(\beta i_b\) from collector to emitter, with the Early-effect resistance \(r_o\) of Chapter 17 in parallel with the source. That is all there is to it, and its virtue is that every element has an obvious physical meaning.
3 The Hybrid-\(\pi\) Model and the h-Parameters
The hybrid-\(\pi\) model describes the same transistor but takes \(v_{be}\) rather than \(i_b\) as the controlling variable, which is closer to the physics: the collector current is set by the base-emitter voltage through the exponential law, and the base current is an incidental consequence. Its central parameter is the transconductance
For \(I_C = 2.009\) mA and \(V_T = 26\) mV, \(g_m = 77.27\) mS. Like \(r_e\), it depends only on the bias current, and it is the one transistor parameter that is genuinely predictable to within a per cent or two.
The input resistance of the model is \(r_\pi = v_{be}/i_b\). Since \(i_c = g_mv_{be}\) and \(i_b = i_c/\beta\),
which is the same 1294 Ω the \(r_e\) model produced. The two models are identical, and the translations are exact:
Note the small subtlety in the translation. \(g_m = I_C/V_T\) while \(1/r_e = I_E/V_T\), so \(g_m r_e = I_C/I_E = \alpha = 0.990\), not exactly one. The difference is a per cent and is normally ignored, but it is the reason \(r_\pi\) equals \((1+\beta)r_e\) exactly and \(\beta r_e\) only approximately.
The h-parameters are a third description, and they exist because they can be measured without needing either terminal to be open-circuit at high frequency. Treating the transistor as a two-port with input current and output voltage as the independent variables:
| Parameter | Definition | Equivalent | Typical value |
|---|---|---|---|
| \(h_{ie}\) | \(v_{be}/i_b\) with \(v_{ce} = 0\) | \(r_\pi = (1+\beta)r_e\) | 1.29 kΩ at 2 mA |
| \(h_{fe}\) | \(i_c/i_b\) with \(v_{ce} = 0\) | \(\beta_{ac}\) | 100 |
| \(h_{oe}\) | \(i_c/v_{ce}\) with \(i_b = 0\) | \(1/r_o\) | 20.1 µS, i.e. \(r_o = 49.8\) kΩ |
| \(h_{re}\) | \(v_{be}/v_{ce}\) with \(i_b = 0\) | reverse transfer; no counterpart in the simple models | \(2\times10^{-4}\) |
The exact h-parameter expressions for a loaded stage are
Applying them to the amplifier of Figure 19.1, where \(R_L' = R_C\|R_L = 1.667\ \text{k}\Omega\): \(\Delta_h = 1294(20.09\times10^{-6}) - (2\times10^{-4})(100) = 0.026 - 0.020 = 0.006\), so \(A_v = -100(1667)/(1294 + 0.006 \times 1667) = -166\,700/1304 = -127.8\) and \(Z_i = 1294 - (0.02 \times 1667)/(1.0335) = 1294 - 32.3 = 1262\ \Omega\). The reverse-transfer term has cost 2.5 per cent of the input impedance and nothing much else, which is why \(h_{re}\) is set to zero in every hand calculation and why the \(r_e\) model, which has no \(h_{re}\) at all, gives answers within a few per cent of the exact ones.
4 The Common-Emitter Stage: Gain and Impedances
Draw the a.c. equivalent of Figure 19.1 by the three rules of Section 1: \(V_{CC}\) becomes ground, so \(R_1\) and \(R_C\) go to ground; \(C_1\), \(C_2\) and \(C_E\) become shorts, so the emitter goes to ground and the source and load connect directly. Replace the transistor by the \(r_e\) model and the whole circuit has four nodes.
Take the stage of Figure 19.1 with \(\beta = 100\), \(V_A = 100\) V, \(V_T = 26\) mV, and the Q-point from Chapter 18: \(I_C = 2.009\) mA, \(I_E = 2.029\) mA.
Device parameters. \(r_e = 26/2.029 = 12.81\ \Omega\); \(g_m = 2.009/26 = 77.27\) mS; \(r_\pi = (1+\beta)r_e = 1294\ \Omega\); \(r_o = V_A/I_C = 100/2.009\ \text{mA} = 49.78\ \text{k}\Omega\).
Input impedance. Looking in at the coupling capacitor, the bias divider and the base appear in parallel:
The transistor dominates. Doubling the divider resistances would raise \(Z_i\) only to 1160 Ω, because \((1+\beta)r_e\) is the smaller term; the way to a high input impedance is the emitter follower of Section 6, not a lighter divider.
Output impedance. With the source killed, \(i_b = 0\), so the controlled source is an open circuit and
essentially \(R_C\). This is a high output impedance, and it is why a common-emitter stage cannot drive a low-resistance load without a follower after it.
Voltage gain. The total collector load is \(R_C\|R_L\|r_o = 2000\|10\,000\|49\,780 = 1613\ \Omega\), and \(v_o = -\beta i_b(1613)\) while \(v_i = i_b(1+\beta)r_e\), so
The alternative hybrid-\(\pi\) route gives \(A_v = -g_mR_L' = -(0.07727)(1613) = -124.6\), the one per cent difference being the factor \(\alpha\) noted earlier. The exact h-parameter formula gave \(-127.8\). All three agree to within two per cent, which is far better than the components are specified to.
Gain from the source. The 600 Ω source resistance forms a divider with \(Z_i\):
A third of the gain has been lost at the input terminals. This is what a low input impedance costs, and it is the argument for the emitter follower as a buffer.
Current and power gain. \(A_i = A_vZ_i/R_L = -125.9(1036)/10\,000 = -13.0\), and the power gain is \(|A_vA_i| = 1641\), or 32.2 dB. The voltage gain alone is 42.0 dB.
Maximum input. The 10 mV rule limits \(v_{be}\), which here is the whole input, so \(v_i \le 10\) mV and the output is limited to 1.26 V peak — well within the available swing of about 4 V. The stage is distortion-limited, not swing-limited, and Section 5 fixes that.
5 The Bypass Capacitor and Swamping
Two questions were deferred: how large \(C_E\) must be, and what happens if the emitter is not fully bypassed.
Sizing the bypass capacitor. The capacitor must be a short circuit compared with the resistance it sees, and that resistance is not \(R_E\). Looking up into the emitter from the capacitor's terminals, the signal sees the transistor's emitter, whose resistance is \(r_e\) plus whatever is in the base circuit divided by \((1+\beta)\):
For the worked stage, \(R_s\|R_1\|R_2 = 600\|5194 = 537.9\ \Omega\), so the bracket is \(12.81 + 537.9/101 = 12.81 + 5.33 = 18.14\ \Omega\), and in parallel with \(R_E = 1\) kΩ that gives 17.82 Ω.
A cut-off frequency of 20 Hz therefore needs
which is why Figure 19.1 shows 470 µF, giving \(f_c = 19.0\) Hz. A student who used \(R_E = 1\) kΩ in that formula would have specified 8 µF and built an amplifier with almost no gain below a kilohertz. The coupling capacitors, by contrast, work into much larger resistances — \(Z_i + R_s = 1.64\) kΩ at the input and \(R_C + R_L = 12\) kΩ at the output — and 10 µF and 4.7 µF are ample. Chapter 25 treats all three break frequencies together.
Swamping. The gain \(-R_L'/r_e\) has an unpleasant property: \(r_e\) depends on the bias current, which Chapter 18 could hold only to a few per cent, and it depends on temperature through \(V_T\). Leaving a small part of the emitter resistance unbypassed fixes this. Split \(R_E\) into \(R_{E1}\), unbypassed, and \(R_{E2}\), bypassed, keeping \(R_{E1}+R_{E2}\) at the value the d.c. design requires. The emitter is no longer at signal ground, and the reflection rule gives
Split the 1 kΩ emitter resistor into \(R_{E1} = 100\ \Omega\) unbypassed and \(R_{E2} = 900\ \Omega\) bypassed. The d.c. conditions are unchanged, so \(r_e\) is still 12.81 Ω. Take \(R_L' = R_C\|R_L = 1667\ \Omega\).
The gain has fallen by a factor of 8.8 and the input impedance has risen by a factor of 3.44. What has been bought is stability. Suppose the bias current is 20 per cent low or 25 per cent high, so that \(r_e\) ranges from 10.25 to 16.02 Ω:
| \(I_E\) | \(r_e\) | Unswamped \(A_v = -R_L'/r_e\) | Swamped \(A_v = -R_L'/(r_e+100)\) |
|---|---|---|---|
| 1.623 mA | 16.02 Ω | \(-104.1\) | \(-14.37\) |
| 2.029 mA | 12.81 Ω | \(-130.1\) | \(-14.77\) |
| 2.536 mA | 10.25 Ω | \(-162.6\) | \(-15.12\) |
The unswamped gain varies by 45 per cent over that range; the swamped gain varies by 5.1 per cent. And because the input can now be \(88\) mV before \(v_{be}\) reaches 10 mV — the ratio \((r_e+R_{E1})/r_e = 8.8\) again — the maximum undistorted output has actually risen, from 1.26 V to 1.30 V, while the distortion at any given output level has fallen by a factor of 8.8.
That is what negative feedback always does: it trades gain, which is cheap and can be recovered by adding another stage, for linearity, predictability and input impedance, which cannot.
6 Common Base and the Emitter Follower
The other two configurations follow from the same model with different terminals grounded, and each answers a need the common-emitter stage cannot.
Common base. Ground the base, drive the emitter, take the output at the collector. The input current is now \(i_e\) rather than \(i_b\), so the input impedance is \(r_e\) itself, in parallel with whatever bias resistance sits at the emitter:
The voltage gain has the same magnitude as the common-emitter stage but no phase inversion, because a rising input at the emitter reduces \(v_{be}\) and hence reduces the collector current, letting the collector rise. The current gain is \(\alpha = 0.990\), just below unity.
A thirteen-ohm input impedance is useless for a voltage amplifier and exactly right for two other jobs: terminating a 50 or 75 Ω transmission line, and accepting current from a source that is effectively a current source, such as a photodiode. Common base also has by far the best high-frequency response of the three, because the collector-base capacitance is not multiplied by the gain — there is no Miller effect, since the base is grounded — which is why the cascode of Chapter 25 puts a common-base stage on top of a common-emitter one.
Common collector, the emitter follower. Ground the collector for signals, drive the base, take the output at the emitter. Now the whole of \(R_E\|R_L\) is unbypassed, so it appears in the reflection rule:
Always positive and always just below one, because \(r_e\) is small compared with \(R_E'\). The stage has no voltage gain and no phase inversion — the emitter "follows" the base, one diode drop below it.
Take the same bias network, remove \(R_C\), and drive a 1 kΩ load from the emitter through a coupling capacitor. The Q-point is unchanged — \(R_C\) never affected \(I_E\) — so \(r_e = 12.81\ \Omega\) and \(V_{CE} = 12 - 2.029 = 9.97\) V.
Gain. \(R_E' = 1000\|1000 = 500\ \Omega\), so \(A_v = 500/512.81 = 0.975\).
Input impedance. Looking into the base, the emitter circuit is reflected up by \((1+\beta)\):
and the bias divider then loads it: \(Z_i = 5194\|51\,790 = 4.72\ \text{k}\Omega\). The divider is now the limiting element, so a follower designed for a high input impedance uses bootstrapping or a much lighter divider.
Output impedance. Looking back into the emitter with the source killed, the base circuit is reflected down by \((1+\beta)\):
What has been achieved. A source of 600 Ω that could not usefully drive a 1 kΩ load now sees 4.72 kΩ, and the load is driven from 17.8 Ω. No voltage gain, but a current gain of \(A_vZ_i/R_L = 0.975(4720)/1000 = 4.60\) and a power gain of 4.49. The follower is not an amplifier; it is an impedance transformer, and the correct place for it is between a common-emitter stage and anything demanding.
7 Choosing Between the Three
All the numbers in the table below come from the same transistor at the same Q-point — \(I_E = 2.029\) mA, \(r_e = 12.81\ \Omega\), \(\beta = 100\) — with the same 2 kΩ collector resistor and 1 kΩ emitter resistor where they apply. The differences are entirely a matter of which terminal is common.
| Common emitter | Common base | Common collector (follower) | |
|---|---|---|---|
| Voltage gain \(A_v\) | \(-R_L'/r_e = -125.9\) | \(+R_L'/r_e = +130.1\) | \(R_E'/(r_e+R_E') = 0.975\) |
| Current gain \(A_i\) | \(\beta\) into the base; \(-13.0\) into a 10 kΩ load | \(\alpha = 0.990\) | \(1+\beta = 101\); 4.60 into a 1 kΩ load |
| Phase | inverting, \(180^\circ\) | non-inverting | non-inverting |
| Input impedance \(Z_i\) | \((1+\beta)r_e\| \text{bias} = 1.04\ \text{k}\Omega\) | \(r_e\|R_E = 12.7\ \Omega\) | \((1+\beta)(r_e+R_E')\|\text{bias} = 4.72\ \text{k}\Omega\) |
| Output impedance \(Z_o\) | \(R_C\|r_o = 1.92\ \text{k}\Omega\) | \(\approx R_C = 2\ \text{k}\Omega\) | \(R_E\|(r_e+R_s'/(1+\beta)) = 17.8\ \Omega\) |
| Power gain | 1641 (32.2 dB) — the largest of the three | 129 (21.1 dB) | 4.49 (6.5 dB) |
| High-frequency behaviour | poor: Miller multiplies \(C_{bc}\) by \((1+|A_v|)\) | excellent: no Miller effect | good |
| Used for | General-purpose voltage and power gain | Line termination, current input, cascode upper device | Buffering, impedance transformation, output stages |
The pattern is easy to remember once the reflection rule is internalised. Whichever terminal is the input, the impedance seen there is \(r_e\) multiplied or divided by \((1+\beta)\) according to how far the terminal is from the emitter: the emitter itself gives \(r_e\), the base gives \((1+\beta)r_e\). Whichever terminal is the output, the impedance seen there is either the collector resistor, because the collector is a current source, or \(r_e\) plus the source resistance divided by \((1+\beta)\), because the emitter is a voltage follower.
Two of these appear again almost immediately. Chapter 20 uses the emitter follower as the output stage of every power amplifier it builds, because a loudspeaker is a few ohms and only a follower can drive it. Chapter 25 uses the common-base stage on top of a common-emitter stage to make the cascode, which has the gain of the one and the bandwidth of the other. The common-emitter stage on its own, meanwhile, is the stage in every textbook and in rather fewer real circuits than one might expect, because in practice it is nearly always wrapped in feedback — the swamping of Section 5 being the simplest form of it, and the operational amplifier of Chapter 27 the most complete.
8 Summary and Key Results
| Quantity | Expression | Value |
|---|---|---|
| Small-signal limit | second harmonic \(= v_{be}/2V_T\) | 10 mV gives 19 % distortion; 1 mV gives 1.9 % |
| Emitter resistance | \(r_e = V_T/I_E\) | 12.81 Ω |
| Transconductance | \(g_m = I_C/V_T\) | 77.27 mS; \(g_mr_e = \alpha = 0.990\) |
| Base input resistance | \(r_\pi = (1+\beta)r_e = \beta/g_m = h_{ie}\) | 1294 Ω |
| Output resistance | \(r_o = V_A/I_C = 1/h_{oe}\) | 49.78 kΩ |
| h-parameters | \(h_{ie}, h_{fe}, h_{oe}, h_{re}\) | 1294 Ω, 100, 20.1 µS, \(2\times10^{-4}\); \(\Delta_h = 0.006\) |
| CE voltage gain | \(A_v = -R_L'/r_e = -g_mR_L'\) | \(-125.9\) loaded; \(-150.1\) unloaded; \(-127.8\) from the exact h-formula |
| CE input impedance | \(Z_i = R_1\|R_2\|(1+\beta)r_e\) | 1036 Ω |
| CE output impedance | \(Z_o = R_C\|r_o\) | 1923 Ω |
| Gain from the source | \(A_{vs} = A_vZ_i/(Z_i+R_s)\) | \(-79.7\); a third of the gain lost at the input |
| Bypass capacitor | \(R_{\text{seen}} = R_E\|(r_e + R_s'/(1+\beta))\) | 17.82 Ω, so \(C_E = 470\ \mu\)F for \(f_c = 19\) Hz |
| Swamping | \(A_v = -R_L'/(r_e+R_{E1})\) | \(R_{E1} = 100\ \Omega\): \(A_v = -14.77\), gain spread 45 % → 5.1 % |
| Common base | \(A_v = +R_L'/r_e\), \(Z_i = R_E\|r_e\) | \(+130.1\), 12.7 Ω; no Miller effect |
| Emitter follower | \(A_v = R_E'/(r_e+R_E')\) | 0.975; \(Z_i = 4.72\) kΩ, \(Z_o = 17.8\ \Omega\) |
9 Common Mistakes
The capacitor across the emitter resistor does not work into \(R_E\); it works into \(R_E\) in parallel with the resistance looking up into the emitter, which is \(r_e + R_s'/(1+\beta)\) and is typically a few tens of ohms. For the amplifier here that is 17.82 Ω against \(R_E = 1\) kΩ — a factor of 56. Sizing \(C_E\) from the wrong resistance gives 8 µF where 470 µF is needed, and the resulting amplifier has its low-frequency corner at 1.1 kHz instead of 19 Hz. The symptom in the laboratory is a stage whose gain is correct at 10 kHz and hopeless at 100 Hz, and the usual first guess — that the coupling capacitors are too small — is wrong.
At \(I_E = 2.029\) mA the junction drops 0.7 V, so its d.c. resistance is 345 Ω; its small-signal resistance is \(V_T/I_E = 12.81\ \Omega\), twenty-seven times smaller. Using 345 Ω in the gain formula gives \(A_v = -4.8\) instead of \(-125.9\). The distinction is the same one that separated the static and dynamic resistances of a diode in Chapter 6: one is the ratio of the coordinates of the operating point, the other is the slope of the curve there, and only the slope tells you anything about a small change. The same trap catches students computing \(r_o\) from \(V_{CE}/I_C\) rather than from \(V_A/I_C\).
Writing \(A_v = -\beta R_L'/h_{ie}\) is correct, and it invites the conclusion that a higher-\(\beta\) transistor gives more gain. It does not, because \(h_{ie} = (1+\beta)r_e\) contains the same \(\beta\), and the two cancel to leave \(-R_L'/r_e\). Swapping a \(\beta = 100\) device for a \(\beta = 300\) one at the same bias current changes the voltage gain by under one per cent; it changes the input impedance by a factor of three, and the gain measured from a real source therefore does change, but through \(Z_i\) and not through \(A_v\). Understanding which quantities contain \(\beta\) and which do not is what separates a design that works with any transistor from one that works with the transistor it was built with.
10 Chapter Review
1. A common-emitter stage has \(V_{CC} = 15\) V, \(R_1 = 33\) kΩ, \(R_2 = 8.2\) kΩ, \(R_C = 3.3\) kΩ, \(R_E = 1.2\) kΩ, fully bypassed, \(\beta = 150\), \(R_L = 22\) kΩ. Find the Q-point, \(r_e\), \(A_v\), \(Z_i\) and \(Z_o\), neglecting \(r_o\).
First the bias. \(V_{th} = 15(8.2/41.2) = 2.985\) V and \(R_{th} = (33\times8.2)/41.2 = 6.568\) kΩ. Then \(I_B = (2.985-0.7)/(6568 + 151 \times 1200) = 2.285/187\,768 = 12.17\ \mu\text{A}\), giving \(I_C = 1.826\) mA and \(I_E = 1.838\) mA. The collector voltage is \(15 - 1.826(3.3) = 8.97\) V, the emitter is at \(1.838(1.2) = 2.21\) V, so \(V_{CE} = 6.77\) V — a healthy Q-point near the middle of the load line. Next \(r_e = 26/1.838 = 14.15\ \Omega\) and \((1+\beta)r_e = 151(14.15) = 2137\ \Omega\). The a.c. collector load is \(R_C\|R_L = (3.3 \times 22)/25.3 = 2.870\ \text{k}\Omega\), so \(A_v = -2870/14.15 = -202.8\). The input impedance is \(R_{th}\|(1+\beta)r_e = 6568\|2137 = 1612\ \Omega\), and the output impedance is \(R_C = 3.3\ \text{k}\Omega\). Including \(r_o = V_A/I_C = 100/1.826\ \text{mA} = 54.8\ \text{k}\Omega\) would reduce the a.c. load to 2.727 kΩ and the gain to \(-192.7\), a 5 per cent correction.
2. Show that the mid-band voltage gain of a common-emitter stage is independent of \(\beta\), and explain why the gain measured with a real source is not.
Write the gain in its h-parameter form, \(A_v = -h_{fe}R_L'/h_{ie}\), and substitute \(h_{fe} = \beta\) and \(h_{ie} = (1+\beta)r_e\). Then \(A_v = -\beta R_L'/[(1+\beta)r_e] = -\alpha R_L'/r_e\), and since \(\alpha\) lies between 0.98 and 0.999 for any usable transistor, \(A_v \approx -R_L'/r_e\) with \(\beta\) gone. The reason is that \(\beta\) enters twice and cancels: a higher-\(\beta\) device produces more collector current per unit of base current, but it also draws proportionally less base current for the same base-emitter voltage. Expressed in hybrid-\(\pi\) terms the cancellation is invisible from the start, since \(A_v = -g_mR_L'\) and \(g_m = I_C/V_T\) contains no \(\beta\) at all. The gain from a source is different because it includes the input divider: \(A_{vs} = A_vZ_i/(Z_i+R_s)\), and \(Z_i\) contains \((1+\beta)r_e\). For the worked amplifier with \(R_s = 600\ \Omega\), \(\beta = 100\) gives \(Z_i = 1036\ \Omega\) and \(A_{vs} = -79.7\), while \(\beta = 300\) gives \((1+\beta)r_e = 3856\ \Omega\), \(Z_i = 5194\|3856 = 2213\ \Omega\) and \(A_{vs} = -125.9(2213/2813) = -99.0\), 24 per cent higher. The device parameter reaches the answer through the impedance, not through the gain.
3. An amplifier must have a voltage gain of \(-20 \pm 5\) per cent from a stage whose bias current is known only to \(\pm 20\) per cent. Design the swamping, taking \(R_C\|R_L = 2\) kΩ and a nominal \(I_E = 2\) mA.
At \(I_E = 2\) mA, \(r_e = 13.0\ \Omega\), and the required gain fixes the total emitter impedance: \(|A_v| = R_L'/(r_e+R_{E1}) = 20\) needs \(r_e + R_{E1} = 2000/20 = 100\ \Omega\), so \(R_{E1} = 87\ \Omega\); take the preferred value 82 Ω, which gives a nominal gain of \(2000/95.0 = -21.1\). Now test the tolerance. With \(I_E\) 20 per cent low, \(I_E = 1.6\) mA and \(r_e = 16.25\ \Omega\), so \(|A_v| = 2000/98.25 = 20.4\); with \(I_E\) 20 per cent high, \(I_E = 2.4\) mA and \(r_e = 10.83\ \Omega\), so \(|A_v| = 2000/92.83 = 21.5\). The spread is 20.4 to 21.5, that is \(\pm 2.8\) per cent about 20.95 — comfortably inside the \(\pm 5\) per cent requirement. Without swamping the same current spread would give gains from \(2000/16.25 = 123\) to \(2000/10.83 = 185\), a spread of \(\pm 20\) per cent, since the gain would then be inversely proportional to \(r_e\) alone. The general rule falls out of the algebra: the fractional gain variation is the fractional \(r_e\) variation multiplied by \(r_e/(r_e+R_{E1})\), so the swamping resistor reduces the sensitivity by exactly the factor by which it reduces the gain.
4. Compare the load a 600 Ω source sees, and the signal that reaches the transistor, for (a) a common-emitter stage with \(Z_i = 1.04\) kΩ, (b) a common-base stage with \(Z_i = 12.7\ \Omega\), and (c) an emitter follower with \(Z_i = 4.72\) kΩ.
In each case the fraction of the source voltage that reaches the stage is \(Z_i/(Z_i+R_s)\). For the common-emitter stage, \(1036/1636 = 0.633\), so 63 per cent of the signal survives and the overall gain is \(-125.9(0.633) = -79.7\). For the common-base stage, \(12.7/612.7 = 0.0207\), so only 2.1 per cent survives and the overall gain collapses from \(+130\) to \(+2.7\) — the configuration is useless when driven from a voltage source of any appreciable resistance. For the follower, \(4720/5320 = 0.887\), so 89 per cent survives and the overall gain is \(0.975(0.887) = 0.865\). Three conclusions follow. A common-base stage should be driven from a current source or from a matched transmission line, never from a voltage source: its virtue is that it terminates a line correctly, not that it accepts a voltage. A follower placed in front of the common-emitter stage would deliver 89 per cent instead of 63 per cent of the signal to the following stage's input, recovering 40 per cent of the lost gain at the cost of one transistor. And impedance matching in the power sense is rarely wanted in a small-signal amplifier: matching \(Z_i\) to \(R_s\) would deliver the most power to the stage but only half the voltage, and voltage is what the next stage responds to.
5. An emitter follower has \(V_{CC} = 12\) V, \(R_1 = 47\) kΩ, \(R_2 = 15\) kΩ, \(R_E = 470\ \Omega\), \(\beta = 120\), driving a 100 Ω load through a coupling capacitor from a 2 kΩ source. Find \(A_v\), \(Z_i\), \(Z_o\) and comment on whether the design is suitable for driving a headphone.
Bias first: \(V_{th} = 12(15/62) = 2.903\) V, \(R_{th} = (47\times15)/62 = 11.37\) kΩ, so \(I_B = (2.903-0.7)/(11\,370 + 121 \times 470) = 2.203/68\,240 = 32.28\ \mu\text{A}\), giving \(I_E = 121(32.28\ \mu\text{A}) = 3.906\) mA and \(r_e = 26/3.906 = 6.66\ \Omega\). The a.c. emitter load is \(R_E\|R_L = 470\|100 = 82.5\ \Omega\), so \(A_v = 82.5/(6.66+82.5) = 0.925\) — noticeably below unity, because the 100 Ω load has pulled \(R_E'\) down to 82.5 Ω, only twelve times \(r_e\). The input impedance is \(Z_{i(\text{base})} = 121(89.16) = 10.79\ \text{k}\Omega\) in parallel with \(R_{th} = 11.37\ \text{k}\Omega\), giving \(Z_i = 5.54\ \text{k}\Omega\); the source divider then passes \(5540/7540 = 0.735\) of the signal, for an overall gain of 0.680. The output impedance is \(R_E\|(r_e + R_s\|R_{th}/121) = 470\|(6.66 + 1701/121) = 470\|20.72 = 19.8\ \Omega\). For a headphone the design is workable but not good: 19.8 Ω driving 100 Ω is an acceptable ratio, but a third of the signal is lost at the input because the bias divider has pulled \(Z_i\) down to half of what the transistor alone would give. Raising \(R_1\) and \(R_2\) tenfold, or bootstrapping the divider from the emitter, would recover most of that. The larger issue is the quiescent current: 3.906 mA flows continuously, so the supply provides \(12 \times 3.906\ \text{mA} = 46.9\) mW and the transistor dissipates \((12-1.836)(3.906\ \text{mA}) = 39.7\) mW of it, while the largest signal the stage can pass without cutting off — a peak emitter swing equal to \(I_{CQ}\), or \(3.906\ \text{mA}\times 82.5\ \Omega = 0.322\) V — delivers only 0.52 mW to the 100 Ω load, an efficiency of 1.1 per cent. That is the class A problem in miniature, and Chapter 20 exists to solve it.