BJT Common-Emitter Amplifier with an Ideal Current-Source Emitter: 98-Comp-A1, December 2017, Q4

Question 4: BJT Common-Emitter Amplifier with an Ideal Current-Source Emitter (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $I=1\text{mA}$ (sets $I_C\approx I_E=1\text{mA}$); $\beta=100$; $V_A=100\text{V}$; $V_T=25\text{mV}$; $R_S=100\,\Omega$; $R_B=100\text{k}\Omega$; $R_C=5\text{k}\Omega$; $R_L=5\text{k}\Omega$; supplies $+10\text{V}/-10\text{V}$; the emitter has no external resistor or bypass capacitor — only the ideal current source.

Given data
Quantity	Value
$I$	$1$ mA
$\beta$	$100$
$V_A$	$100$ V
$V_T$	$25$ mV
$R_S$	$100\,\Omega$
$R_B$	$100\,\text{k}\Omega$
$R_C=R_L$	$5\,\text{k}\Omega$

Find. $V_C$, $V_B$, $V_E$; small-signal model; $R_i$, $R_o$; open-circuit and loaded voltage gain.

Check: this circuit is deliberately different from the "standard" current-source-biased CE stage in one crucial respect: there, a bypass capacitor sits across the current source, AC-grounding the emitter and giving the usual $A_v=-g_m(R_C\parallel r_o)$ result. Here the figure shows only the bare current source at the emitter — no resistor, no bypass cap. An ideal current source presents infinite impedance to a small AC signal (open circuit), so the emitter has no external AC return path at all. This changes the answer qualitatively, as Part (d) shows.

Approach. DC bias: $I$ fixes $I_E\approx1\text{mA}$; $R_B$ carries the (small) base current to ground, and $R_C$ sets $V_C$. For the small-signal analysis, the ideal current source is replaced by an open circuit (its AC output resistance is infinite) — so the emitter node's only AC connections are the transistor's own $r_\pi$ (to the base) and $r_o$/dependent source (to the collector). Solve the resulting 3-node ($B$, $E$, $C$) hybrid-$\pi$ circuit exactly rather than assuming a textbook emitter-bypassed result.

Part (a) — DC operating point. With $I_E\approx1\text{mA}$, $I_B=I/(\beta+1)=1\text{mA}/101=9.90\,\mu\text{A}$. This current returns to ground through $R_B$, so $$V_B=-I_BR_B=-(9.90\,\mu\text{A})(100\text{k}\Omega)=\boxed{-0.990\text{ V}}$$ $$V_E=V_B-V_{BE}=-0.990-0.7=\boxed{-1.690\text{ V}}$$ $$V_C=V_{CC}-I_CR_C\approx10-(1\text{mA})(5\text{k}\Omega)=\boxed{5.0\text{ V}}$$
Part (b) — small-signal model. $$g_m=\frac{I_C}{V_T}=\frac{1\text{mA}}{25\text{mV}}=\boxed{40\text{ mA/V}},\qquad r_\pi=\frac{\beta}{g_m}=\frac{100}{40\text{mA/V}}=\boxed{2.5\text{ k}\Omega},\qquad r_o=\frac{V_A}{I_C}=\frac{100}{1\text{mA}}=\boxed{100\text{ k}\Omega}$$ Model: $v_i$ drives the base through $R_S$, in parallel with $R_B$ to ground; $r_\pi$ sits between base and emitter; the collector has $g_mv_\pi$ and $r_o$ (both referenced to the emitter, not ground) loaded by $R_C\parallel R_L$; crucially, the emitter node has no other connection — the current source is an open circuit for AC (Fig. Q4(b)).
Part (c) — $R_i$, $R_o$. Solving the 3-node hybrid-$\pi$ circuit exactly shows the floating emitter acts as an enormous local-feedback (degeneration) resistance: the pure-transistor input resistance at the base balloons to $r_\pi(1+g_mr_o)\approx10.1\text{M}\Omega$, so with $R_B=100\text{k}\Omega$ in parallel, $$R_i=R_B\parallel\big[r_\pi(1+g_mr_o)\big]=100\text{k}\parallel10.1\text{M}=\boxed{99.0\text{ k}\Omega}$$ (dominated entirely by $R_B$, since the transistor's own input resistance is 100$\times$ larger). Symmetrically, the output resistance looking into the collector is boosted to $\approx10.1\text{M}\Omega$ by the same open-emitter degeneration, so in parallel with $R_C$: $$R_o=R_C\parallel10.1\text{M}\approx\boxed{5.0\text{ k}\Omega}\quad\text{(essentially just }R_C\text{)}$$
Part (d) — open-circuit and loaded gain. With the emitter truly open for AC, the hybrid-$\pi$ node equation at $E$ forces $i_e\to0$: any $v_{be}$ that developed would need to push current through $r_\pi$ into $E$ with nowhere external to go, which self-consistently drives $v_{be}\to0$ (the emitter "floats" up to follow the base almost exactly). Since $i_c=g_mv_{be}$, the collector current — and hence the output voltage — collapses to essentially zero: $$A_{v,\text{oc}}=\boxed{\approx0}\quad(4.9\times10^{-4}\text{ V/V exactly, from the finite-}r_o\text{ nodal solve})$$ $$A_{v,\text{ld}}=\boxed{\approx0}\quad(2.5\times10^{-4}\text{ V/V with }R_L\text{ connected})$$ This is the extreme limit of emitter degeneration: for a resistively-degenerated CE stage, $A_v=-g_mR_C/(1+g_mR_E)\to0$ as $R_E\to\infty$, and an ideal current source is that limit ($R_E\to\infty$ for AC). The tiny non-zero residual above comes only from the finite $r_o$, which provides the sole (very weak) AC path from collector back to the emitter.

Fig. Q4(b) — hybrid-$\pi$ model with the emitter node isolated (current source open for AC, red): $r_\pi$, $g_mv_\pi$ and $r_o$ all reference the floating emitter node $E$, not ground, so no signal current can ultimately flow.

Final Results — Question 4
Quantity	Value
$V_B$	$-0.990$ V
$V_E$	$-1.690$ V
$V_C$	$5.0$ V
$g_m$	$40$ mA/V
$r_\pi$	$2.5$ k$\Omega$
$r_o$	$100$ k$\Omega$
$R_i$	$99.0$ k$\Omega$ (set by $R_B$)
$R_o$	$5.0$ k$\Omega$ (set by $R_C$)
$A_{v,\text{oc}}$	$\approx0$ ($4.9\times10^{-4}$)
$A_{v,\text{ld}}$	$\approx0$ ($2.5\times10^{-4}$)

Topic: Common-emitter amplifier with an un-bypassed ideal current-source emitter (infinite degeneration).

Key relations: Degenerated CE gain $A_v=-g_mR_C/(1+g_mR_E)$; as $R_E\to\infty$ (an ideal current source, open for AC), $A_v\to0$; input resistance at the base is boosted to $r_\pi(1+g_mr_o)$ when $R_E$ is this large.

Why this works: An ideal current source fixes DC bias current perfectly (no sensitivity to $\beta$ or supply drift) but, having infinite small-signal output resistance, provides no AC return path — it is the textbook degeneration resistor taken to its extreme. All of the local negative feedback that a finite $R_E$ would provide is maximized, driving the voltage gain to essentially zero; only the transistor's own finite $r_o$ (Early effect) prevents it from being exactly zero.

Common pitfall: Assuming any "current-source-biased" CE stage automatically gets the ideal $-g_m(R_C\parallel r_o)$ gain — that result requires an explicit bypass capacitor (or equivalent AC ground) across the current source. Without one, as here, the stage is completely degenerated and delivers essentially unity-gain emitter-follower-like tracking between base and emitter, with almost no signal reaching the collector.

Source: Sedra & Smith, Microelectronic Circuits, Ch.6 (BJT small-signal model, emitter degeneration and its effect on gain and input resistance).

Question 4 of 7: BJT Common-Emitter Amplifier with an Ideal Current-Source Emitter

Question 4: BJT Common-Emitter Amplifier with an Ideal Current-Source Emitter (20 marks)