BJT Small-Signal Amplifiers — Solved Problems

About this problem set

Small-signal analysis of BJT stages using the re model and the hybrid h-parameter model: common emitter with and without emitter bypass, emitter follower, common base, and a two-stage RC-coupled cascade whose overall gain is expressed in decibels.

Each problem below is solved in full — the given data is listed first, the governing relation is stated, and the arithmetic is carried through to a boxed answer. Work through the statement yourself before opening the solution. The underlying theory is covered in the Electronic Devices lecture series.

PROBLEM 01

Common-Emitter Stage with Bypassed Emitter

Problem Statement

A voltage-divider biased CE amplifier has \(V_{CC} = 20~\text{V}\), \(R_1 = 56~\text{k}\Omega\), \(R_2 = 8.2~\text{k}\Omega\), \(R_C = 6.8~\text{k}\Omega\) and \(R_E = 1.5~\text{k}\Omega\). The emitter resistor is fully bypassed by a large capacitor; input and output are capacitively coupled and the output is unloaded. \(\beta = 90\), \(r_o = \infty\).

Find the DC emitter current and hence \(r_e\).
Draw on the \(r_e\) model to find \(Z_i\), \(Z_o\) and the no-load voltage gain \(A_v = v_o/v_i\).
Find the current gain \(A_i = i_o/i_i\).

Solution

\(V_{CC} = 20~\text{V}\), \(R_1 = 56~\text{k}\Omega\), \(R_2 = 8.2~\text{k}\Omega\), \(R_C = 6.8~\text{k}\Omega\), \(R_E = 1.5~\text{k}\Omega\) (bypassed), \(\beta = 90\)
Simplified \(r_e\) model, \(r_o = \infty\), all capacitors are short circuits at signal frequencies, \(V_T = 26~\text{mV}\).

Step 1 — DC analysis for \(r_e\)

\[\begin{aligned} R_{Th} &= R_1 \parallel R_2 = 7.1526~\text{k}\Omega, \qquad E_{Th} = \frac{R_2 V_{CC}}{R_1+R_2} = 2.555~\text{V} \\ I_B &= \frac{E_{Th} - V_{BE}}{R_{Th} + (\beta+1)R_E} = \frac{1.855~\text{V}}{7.153~\text{k} + 91(1.5~\text{k})} = 12.91~\mu\text{A} \\ I_E &= (\beta+1)I_B = 1.175~\text{mA} \\ r_e &= \frac{V_T}{I_E} = \frac{26~\text{mV}}{1.175~\text{mA}} = 22.13~\Omega \end{aligned}\]

\(I_E = 1.17~\text{mA}\), \(r_e = 22.1~\Omega\)

Step 2 — impedances and gain

With \(R_E\) bypassed the emitter is at signal ground, so the base sees \(\beta r_e\) in parallel with the divider:

\[\begin{aligned} \beta r_e &= 90 \times 22.13~\Omega = 1992~\Omega \\ Z_i &= R_1 \parallel R_2 \parallel \beta r_e = 7.153~\text{k} \parallel 1992~\Omega = 1558~\Omega \\ Z_o &= R_C \parallel r_o = R_C = 6.8~\text{k}\Omega \\ A_v &= -\frac{R_C}{r_e} = -\frac{6800~\Omega}{22.13~\Omega} = -307.3 \end{aligned}\]

\(Z_i = 1558~\Omega\), \(Z_o = 6.8~\text{k}\Omega\), \(A_v = -307\) (180\(^\circ\) phase inversion).

Step 3 — current gain

\[A_i = -A_v\,\frac{Z_i}{R_C} = 307.3 \times \frac{1558~\Omega}{6800~\Omega} = 70.40\]

\(A_i = 70.4\)

Note that \(A_i\) came out close to, but below, \(\beta = 90\): the divider steals part of the input current before it reaches the base. The gain \(-R_C/r_e\) depends on \(I_E\) only through \(r_e\), so it is really set by \(A_v = -I_C R_C / V_T\) — the DC drop across \(R_C\) divided by 26 mV.

PROBLEM 02

Gain Collapse with an Unbypassed Emitter, and Swamping Design

Problem Statement

The bypass capacitor is now removed from the amplifier of the previous problem (\(R_C = 6.8~\text{k}\Omega\), \(R_E = 1.5~\text{k}\Omega\), \(R_1 = 56~\text{k}\Omega\), \(R_2 = 8.2~\text{k}\Omega\), \(\beta = 90\), \(r_e = 22.1~\Omega\)), so the whole of \(R_E\) is in the signal path.

Find the new \(A_v\), \(Z_i\) and \(Z_o\), and state the factor by which the gain has collapsed.
Design: split \(R_E\) into an unbypassed part \(R_{E1}\) in series with a bypassed part \(R_{E2}\) (with \(R_{E1} + R_{E2} = 1.5~\text{k}\Omega\), so the DC bias is untouched) to obtain \(A_v = -40\). Use the nearest E24 value for \(R_{E1}\) and quote the gain and \(Z_i\) actually achieved.
Why is the swamped design preferred even though it throws gain away?

Solution

\(r_e = 22.13~\Omega\) (unchanged — removing the bypass capacitor does not alter the DC Q-point)
\(R_C = 6800~\Omega\), \(R_E = 1500~\Omega\), \(R_1 \parallel R_2 = 7.153~\text{k}\Omega\), \(\beta = 90\), \(r_o = \infty\)

Part (a) — fully unbypassed emitter

The emitter is no longer at signal ground, so the emitter current develops \(i_e R_E\) which subtracts from the input:

\[\begin{aligned} A_v &= -\frac{R_C}{r_e + R_E} = -\frac{6800}{22.13 + 1500} = -\frac{6800}{1522.1} = -4.467 \\ Z_b &= \beta(r_e + R_E) = 90(1522.1~\Omega) = 137.0~\text{k}\Omega \\ Z_i &= R_1 \parallel R_2 \parallel Z_b = 7.153~\text{k} \parallel 137.0~\text{k} = 6.798~\text{k}\Omega \\ Z_o &= R_C = 6.8~\text{k}\Omega \end{aligned}\]

\(A_v = -4.47\), \(Z_i = 6.80~\text{k}\Omega\), \(Z_o = 6.8~\text{k}\Omega\); the gain has fallen by a factor of 68.8.

\(Z_i\) has risen from 1558 \(\Omega\) to 6.80 k\(\Omega\) and is now pinned by the bias divider, because \(Z_b = 137~\text{k}\Omega\) swamps it.

Part (b) — design for \(A_v = -40\)

Only \(R_{E1}\) appears in the signal path, so set \(-R_C/(r_e + R_{E1}) = -40\):

\[\begin{aligned} R_{E1} &= \frac{R_C}{|A_v|} - r_e = \frac{6800~\Omega}{40} - 22.13~\Omega = 170.0 - 22.13 = 147.9~\Omega \\ &\Rightarrow R_{E1} = 150~\Omega\ \text{(E24)}, \qquad R_{E2} = 1500 - 150 = 1350~\Omega \end{aligned}\]

\[\begin{aligned} A_v &= -\frac{6800}{22.13 + 150} = -39.50 \\ Z_b &= 90(172.13) = 15.49~\text{k}\Omega, \qquad Z_i = 7.153~\text{k} \parallel 15.49~\text{k} = 4.893~\text{k}\Omega \end{aligned}\]

\(R_{E1} = 150~\Omega\) unbypassed, \(R_{E2} = 1350~\Omega\) bypassed; achieved \(A_v = -39.5\) and \(Z_i = 4.89~\text{k}\Omega\).

Part (c) — why swamp

With \(R_{E1} \gg r_e\) the gain tends to \(-R_C/R_{E1} = -45.3\), a ratio of two resistors. \(r_e\) is set by \(I_E\) and therefore drifts with temperature and varies from device to device; a gain that depends only weakly on it is repeatable. The price is the 68.8-fold gain loss seen in part (a) if you swamp completely, which is why only part of \(R_E\) is left unbypassed.

PROBLEM 03

Emitter Follower: Impedance Buffering

Problem Statement

An emitter follower (common collector) is fixed-biased: \(V_{CC} = 12~\text{V}\), a 220 k\(\Omega\) resistor from the rail to the base, \(R_E = 3.3~\text{k}\Omega\) from emitter to ground, output taken from the emitter through a coupling capacitor. \(\beta = 100\), \(V_{BE} = 0.7~\text{V}\), \(r_o = \infty\).

Find \(r_e\).
Find \(Z_i\), \(Z_o\) and \(A_v\) with no load.
Repeat \(Z_i\) and \(A_v\) with a \(1~\text{k}\Omega\) load capacitively coupled to the emitter.

Solution

\(V_{CC} = 12~\text{V}\), \(R_B = 220~\text{k}\Omega\), \(R_E = 3.3~\text{k}\Omega\), \(\beta = 100\)
Simplified \(r_e\) model, \(r_o = \infty\), \(V_T = 26~\text{mV}\).

Part (a) — \(r_e\)

\[\begin{aligned} I_B &= \frac{V_{CC} - V_{BE}}{R_B + (\beta+1)R_E} = \frac{11.3~\text{V}}{220~\text{k} + 101(3.3~\text{k})} = 20.42~\mu\text{A} \\ I_E &= (\beta+1)I_B = 2.063~\text{mA}, \qquad r_e = \frac{26~\text{mV}}{2.063~\text{mA}} = 12.60~\Omega \end{aligned}\]

\(r_e = 12.6~\Omega\)

Part (b) — unloaded small-signal parameters

Looking into the base, the emitter resistor is multiplied by \(\beta\); looking back into the emitter, \(r_e\) appears divided by \(\beta\) (i.e. in parallel with \(R_E\)):

\[\begin{aligned} Z_b &= \beta(r_e + R_E) = 100(12.60 + 3300) = 331.26~\text{k}\Omega \\ Z_i &= R_B \parallel Z_b = 220~\text{k} \parallel 331.3~\text{k} = 132.2~\text{k}\Omega \\ Z_o &= R_E \parallel r_e = 3300 \parallel 12.60 = 12.56~\Omega \\ A_v &= \frac{R_E}{R_E + r_e} = \frac{3300}{3312.6} = 0.99619 \end{aligned}\]

\(Z_i = 132~\text{k}\Omega\), \(Z_o = 12.6~\Omega\), \(A_v = 0.9962 \approx 1\), in phase.

Part (c) — with a 1 k\(\Omega\) load

\[\begin{aligned} R_E \parallel R_L &= 3300 \parallel 1000 = 767.4~\Omega \\ A_v &= \frac{767.4}{767.4 + 12.60} = 0.98384 \\ Z_i &= 220~\text{k} \parallel 100(780.05) = 220~\text{k} \parallel 78.00~\text{k} = 57.59~\text{k}\Omega \end{aligned}\]

Loaded: \(A_v = 0.9838\), \(Z_i = 57.6~\text{k}\Omega\).

The stage has voltage gain of essentially unity but transforms a 132 k\(\Omega\) source-facing impedance into a 12.6 \(\Omega\) source impedance — a power gain of the order of \(\beta\). That is the whole point of a follower: it buys current drive, not voltage.

PROBLEM 04

Common-Base Amplifier: Low Input Impedance

Problem Statement

A common-base stage uses split supplies. The emitter is returned to \(-V_{EE} = -4~\text{V}\) through \(R_E = 1~\text{k}\Omega\), the base is grounded, and the collector goes to \(+V_{CC} = 12~\text{V}\) through \(R_C = 2.2~\text{k}\Omega\). Signal is injected at the emitter and taken from the collector. \(\alpha = 0.98\), \(V_{BE} = 0.7~\text{V}\).

Find \(I_E\), \(r_e\) and the DC collector-to-base voltage.
Find \(Z_i\), \(Z_o\), \(A_v\) and \(A_i\).
Contrast the result with the CE stage of Problem 1.

Solution

\(V_{EE} = 4~\text{V}\), \(R_E = 1~\text{k}\Omega\), \(V_{CC} = 12~\text{V}\), \(R_C = 2.2~\text{k}\Omega\), \(\alpha = 0.98\)
Base grounded, so the whole of \(V_{EE} - V_{BE}\) appears across \(R_E\); \(V_T = 26~\text{mV}\).

Part (a) — DC operating point

\[\begin{aligned} I_E &= \frac{V_{EE} - V_{BE}}{R_E} = \frac{4 - 0.7}{1~\text{k}\Omega} = 3.30~\text{mA} \\ r_e &= \frac{26~\text{mV}}{3.30~\text{mA}} = 7.879~\Omega \\ I_C &= \alpha I_E = 3.234~\text{mA}, \qquad V_{CB} = V_{CC} - I_C R_C = 12 - 7.115 = 4.885~\text{V} \end{aligned}\]

\(I_E = 3.30~\text{mA}\), \(r_e = 7.88~\Omega\), \(V_{CB} = 4.89~\text{V}\)

Part (b) — small-signal parameters

The source drives the emitter directly, so it sees \(r_e\) in parallel with \(R_E\):

\[\begin{aligned} Z_i &= R_E \parallel r_e = 1000 \parallel 7.879 = 7.817~\Omega \\ Z_o &= R_C = 2.2~\text{k}\Omega \\ A_v &= \frac{\alpha R_C}{r_e} = \frac{0.98 \times 2200~\Omega}{7.879~\Omega} = +273.6 \\ A_i &= \frac{i_o}{i_i} = -\alpha = -0.98 \end{aligned}\]

\(Z_i = 7.82~\Omega\), \(Z_o = 2.2~\text{k}\Omega\), \(A_v = +274\) (no phase inversion), \(A_i \approx -1\).

Part (c) — comparison

Parameter	CE (bypassed)	CB
\(Z_i\)	\(1558~\Omega\)	\(7.82~\Omega\)
\(A_v\)	-307	+274
\(A_i\)	70.4	\(\approx -1\)
Phase	180\(^\circ\)	0\(^\circ\)

The CB stage gives comparable voltage gain with no current gain at all, and an input impedance of a few ohms. That is useless for a voltage source but ideal for accepting current from a low-impedance source such as a coaxial line, and it has no Miller capacitance, which is why CB is the high-frequency configuration.

PROBLEM 05

Hybrid-Parameter Analysis of a CE Stage

Problem Statement

A CE amplifier has an AC base-bias resistance \(R_B = R_1 \parallel R_2 = 4.7~\text{k}\Omega\), a collector resistor \(R_C = 3.3~\text{k}\Omega\), a capacitively coupled load \(R_L = 4.7~\text{k}\Omega\), and a fully bypassed emitter. The transistor data sheet quotes \(h_{ie} = 1.1~\text{k}\Omega\), \(h_{fe} = 120\), \(h_{oe} = 20~\mu\text{S}\); \(h_{re}\) is negligible.

Find \(Z_i\), \(Z_o\), \(A_v\) and \(A_i\) from the approximate hybrid model.
Extract \(r_e\) and \(I_E\) from the \(h\)-parameters.
Recompute \(A_v\) from the \(r_e\) model with \(r_o = \infty\) and explain the discrepancy.

Solution

\(h_{ie} = 1100~\Omega\), \(h_{fe} = 120\), \(h_{oe} = 20~\mu\text{S}\), \(h_{re} \approx 0\)
\(R_B = 4.7~\text{k}\Omega\), \(R_C = 3.3~\text{k}\Omega\), \(R_L = 4.7~\text{k}\Omega\)
Approximate hybrid model: the \(h_{re}v_{ce}\) generator in the input loop is dropped.

Part (a) — hybrid model

\(h_{oe}\) is a conductance across the output, so it contributes \(1/h_{oe} = 50.0~\text{k}\Omega\) in parallel with the collector load:

\[\begin{aligned} \frac{1}{h_{oe}} &= \frac{1}{20~\mu\text{S}} = 50.0~\text{k}\Omega \\ R_L' &= R_C \parallel R_L \parallel \frac{1}{h_{oe}} = 3.3~\text{k} \parallel 4.7~\text{k} \parallel 50.0~\text{k} = 1.8664~\text{k}\Omega \\ A_v &= -\frac{h_{fe}R_L'}{h_{ie}} = -\frac{120 \times 1866.4~\Omega}{1100~\Omega} = -203.6 \\ Z_i &= R_B \parallel h_{ie} = 4700 \parallel 1100 = 891.4~\Omega \\ Z_o &= \frac{1}{h_{oe}} \parallel R_C = 50.0~\text{k} \parallel 3.3~\text{k} = 3.096~\text{k}\Omega \\ A_i &= \frac{h_{fe}}{1 + h_{oe}(R_C \parallel R_L)} = \frac{120}{1 + (20~\mu\text{S})(1939~\Omega)} = 115.5 \end{aligned}\]

\(Z_i = 891~\Omega\), \(Z_o = 3.10~\text{k}\Omega\), \(A_v = -203.6\), \(A_i = 115.5\)

Part (b) — extracting \(r_e\)

The two models describe the same device, so \(h_{ie} = \beta r_e\) and \(h_{fe} = \beta\):

\[\begin{aligned} r_e &= \frac{h_{ie}}{h_{fe}} = \frac{1100~\Omega}{120} = 9.167~\Omega \\ I_E &= \frac{26~\text{mV}}{r_e} = \frac{26~\text{mV}}{9.167~\Omega} = 2.836~\text{mA} \end{aligned}\]

\(r_e = 9.17~\Omega\), corresponding to \(I_E = 2.84~\text{mA}\)

Part (c) — \(r_e\) model with \(r_o = \infty\)

\[\begin{aligned} R_L'' &= R_C \parallel R_L = 1.9387~\text{k}\Omega \\ A_v &= -\frac{R_L''}{r_e} = -\frac{1938.8}{9.167} = -211.5 \end{aligned}\]

\(r_e\) model gives \(A_v = -211.5\) against \(-203.6\) from the hybrid model — the magnitude is 3.88\% higher.

The two models are the same model. The only difference is that the \(h\)-parameter version keeps the output conductance \(h_{oe}\); dropping it (\(r_o = \infty\)) over-estimates the load and hence the gain. The rule of thumb \(r_o \ge 10 R_C\) makes the error negligible — here \(1/h_{oe} = 50.0~\text{k}\Omega\) is 15.2 times \(R_C\), so a 3.88\% error is expected.

PROBLEM 06

Two-Stage Cascaded Amplifier and Overall Gain in dB

Problem Statement

Two identical RC-coupled CE stages are cascaded. Each stage has \(\beta = 100\), a quiescent \(I_E = 2~\text{mA}\), \(R_C = 4.7~\text{k}\Omega\), a bias network presenting \(R_1 \parallel R_2 = 6.8~\text{k}\Omega\), and a fully bypassed emitter. A \(10~\text{k}\Omega\) load is coupled to the second collector and the source has an internal resistance \(R_s = 600~\Omega\).

Find \(r_e\) and the input impedance of one stage.
Find the loaded gain of each stage, remembering that stage 1 is loaded by \(Z_{i2}\).
Find the overall gain \(A_v = v_o/v_i\) and express it in dB.
Find \(A_{vs} = v_o/v_s\) in dB and state the loss caused by the source resistance.

Solution

Per stage: \(\beta = 100\), \(I_E = 2~\text{mA}\), \(R_C = 4.7~\text{k}\Omega\), \(R_1 \parallel R_2 = 6.8~\text{k}\Omega\), \(R_E\) bypassed
\(R_L = 10~\text{k}\Omega\), \(R_s = 600~\Omega\), \(r_o = \infty\), \(V_T = 26~\text{mV}\)

Step 1 — \(r_e\) and stage input impedance

\[\begin{aligned} r_e &= \frac{26~\text{mV}}{2~\text{mA}} = 13.0~\Omega, \qquad \beta r_e = 1300~\Omega \\ Z_i &= 6.8~\text{k} \parallel 1.300~\text{k} = 1091~\Omega \end{aligned}\]

\(r_e = 13.0~\Omega\), \(Z_{i} = 1091~\Omega\) for each stage.

Step 2 — per-stage loaded gains

Stage 1 does not see \(R_C\) alone: the input impedance of stage 2 hangs on its collector through the coupling capacitor.

\[\begin{aligned} R_{L1} &= R_C \parallel Z_{i2} = 4.7~\text{k} \parallel 1091~\Omega = 885.7~\Omega \\ A_{v1} &= -\frac{R_{L1}}{r_e} = -\frac{885.7}{13.0} = -68.13 \\ R_{L2} &= R_C \parallel R_L = 4.7~\text{k} \parallel 10~\text{k} = 3197.3~\Omega \\ A_{v2} &= -\frac{3197.3}{13.0} = -245.9 \end{aligned}\]

\(A_{v1} = -68.1\), \(A_{v2} = -245.9\). The first stage is heavily loaded by the second.

Step 3 — overall gain

\[\begin{aligned} A_v &= A_{v1} \times A_{v2} = (-68.13)(-245.9) = +16756 \\ A_v\big|_{\text{dB}} &= 20\log_{10}|A_v| = 20\log_{10}(16756) = 84.48~\text{dB} \end{aligned}\]

Stage	\(\|A_v\|\)	Gain (dB)
1	68.13	36.67
2	245.9	47.82
Overall	16756	84.48

\(A_v = +16756\), i.e. 84.48 dB. The output is in phase with the input: two inversions cancel.

Check: 36.67 + 47.82 = 84.48 dB. Cascading multiplies gains, so in dB they simply add.

Step 4 — source loading

\[\begin{aligned} A_{vs} &= A_v \frac{Z_i}{Z_i + R_s} = 16756 \times \frac{1091}{1091 + 600} = 16756 \times 0.6453 = 10812 \\ A_{vs}\big|_{\text{dB}} &= 20\log_{10}(10812) = 80.68~\text{dB} \end{aligned}\]

\(A_{vs} = 80.68~\text{dB}\); the source resistance costs 3.81 dB.

The 3.81 dB is lost purely in the divider formed by \(R_s\) and the 1091 \(\Omega\) input impedance. An emitter follower placed in front would raise \(Z_i\) into the tens of k\(\Omega\) and recover almost all of it — which is exactly why real cascades start with a buffer.