Question 1 of 6: PNP common-emitter amplifier — gain and bandwidth

Question 1: PNP common-emitter amplifier — gain and bandwidth (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. A single PNP transistor is biased by a 1 mA current source feeding its emitter from the positive rail; the emitter is tied to signal ground by an infinite bypass capacitor, so the stage is a common-emitter amplifier driven at the base through a coupling capacitor and loaded by $R_{L}$ to the negative rail.

Find. The mid-band voltage gain referred to the source $v_{OUT}/v_{S}$, the lower and upper 3 dB corner frequencies, and the second (non-dominant) high-frequency pole.

Approach. Fix the bias current, convert it to $g_{m}$ and $r_{\pi}$, evaluate the mid-band gain as an input divider followed by $-g_{m}R_{L}$, then treat the two frequency ends separately: the single coupling capacitor gives $f_{L}$, and a Miller split of $C_{\mu}$ gives the input and output high-frequency poles.

1. Fix the d.c. operating point and confirm the transistor is active. The current source sets the emitter current directly, so $$I_{E}=I_{bias}=1\ \text{mA},\qquad I_{C}=\alpha I_{E}=\frac{\beta}{\beta+1}I_{E}=\frac{100}{101}(1\ \text{mA})=0.990\ \text{mA},$$ and the base current is $I_{B}=I_{E}/(\beta+1)=9.90\ \mu\text{A}$. For a PNP device that current leaves the base and flows through $R_{1}$ to ground, so $V_{B}=-I_{B}R_{1}=-9.9\ \text{mV}$, $V_{E}=V_{B}+V_{EB}=0.690\ \text{V}$ and $V_{C}=-V_{EE}+I_{C}R_{L}=-10+4.95=-5.05\ \text{V}$. Hence $$\boxed{V_{EC}=V_{E}-V_{C}=5.74\ \text{V}\ \gg\ V_{EC(\text{sat})}=0.3\ \text{V}}$$ so the transistor is comfortably in the forward-active region and the bias source still has $V_{CC}-V_{E}=9.31\ \text{V}$ of compliance.
2. Convert the bias current into small-signal parameters. Because the question asks for an estimate, take $I_{C}\approx I_{E}=1\ \text{mA}$ (the $\alpha$ correction is checked at the end of this answer and moves the gain by less than 1 %). Then $$g_{m}=\frac{I_{C}}{V_{T}}=\frac{1\ \text{mA}}{25\ \text{mV}}=40\ \text{mA/V},\qquad r_{\pi}=\frac{\beta}{g_{m}}=\frac{100}{40\ \text{mA/V}}=2.5\ \text{k}\Omega .$$ The paper instructs us to neglect $r_{x}$ and $r_{o}$; for the record $r_{o}=V_{A}/I_{C}=100\ \text{k}\Omega$, twenty times $R_{L}$, which is why dropping it is legitimate here.
3. Part (a) — split the mid-band gain into an input divider and a transconductance stage. In the mid-band $C_{1}$ is a short and $C_{2}$ holds the emitter at signal ground, so the stage is a textbook common-emitter amplifier. Looking into the base the source sees $$R_{in}=R_{1}\parallel r_{\pi}=\frac{(1\ \text{k}\Omega)(2.5\ \text{k}\Omega)} {1\ \text{k}\Omega+2.5\ \text{k}\Omega}=714.3\ \Omega ,$$ so only part of the source voltage reaches the base: $$\frac{v_{b}}{v_{S}}=\frac{R_{in}}{R_{S}+R_{in}}=\frac{714.3}{600+714.3}=0.5435 .$$ The collector then delivers $v_{OUT}/v_{b}=-g_{m}R_{L}=-(40\ \text{mA/V})(5\ \text{k}\Omega)=-200$, and multiplying the two factors gives $$\boxed{\frac{v_{OUT}}{v_{S}}=-200\times0.5435=-108.7\ \text{V/V}}$$ an inverting gain of magnitude 108.7, or 40.7 dB. Note how much gain the source resistance costs: driving the base from an ideal source would have given the full 200.
4. Part (b) — find the single low-frequency pole. Only $C_{1}$ can produce a low-frequency pole, because $C_{2}=\infty$ is a perfect short at every frequency of interest and the bias current source is an open circuit to signal. Killing the source and looking out from the terminals of $C_{1}$, the resistance is the series combination of what lies on each side: $$R_{C1}=R_{S}+\left(R_{1}\parallel r_{\pi}\right)=600+714.3=1314.3\ \Omega .$$ A single pole set by one capacitor and one resistance puts the 3 dB point at $$\boxed{f_{L}=\frac{1}{2\pi R_{C1}C_{1}} =\frac{1}{2\pi(1314.3\ \Omega)(10\ \mu\text{F})}=12.1\ \text{Hz}}$$ comfortably below the audio band, which is exactly what a 10 $\mu$F coupling capacitor is chosen to achieve.
5. Part (c) — Miller-split the collector-base capacitance and find the dominant high-frequency pole. With $r_{x}$ neglected and no $C_{\pi}$ specified, $C_{\mu}$ bridges base and collector across a voltage gain of $-g_{m}R_{L}=-200$. The Miller theorem replaces it by a capacitance to ground at each node, $$C_{M,in}=C_{\mu}\left(1+g_{m}R_{L}\right)=2\ \text{pF}\times201=402\ \text{pF},\qquad C_{M,out}=C_{\mu}\left(1+\frac{1}{g_{m}R_{L}}\right)=2.01\ \text{pF}.$$ At the base the driving resistance is now everything in parallel, because $C_{1}$ is a short at high frequency: $$R_{B}=R_{S}\parallel R_{1}\parallel r_{\pi}=600\parallel1000\parallel2500=326.1\ \Omega,$$ which places the input pole at $$\boxed{f_{H}=\frac{1}{2\pi R_{B}C_{M,in}} =\frac{1}{2\pi(326.1\ \Omega)(402\ \text{pF})}=1.21\ \text{MHz}}$$ This is the upper 3 dB frequency, since the second pole computed next is more than ten times higher.
6. Part (d) — evaluate the pole at the output node. The output node sees $C_{M,out}$ against the collector load alone, $R_{L}=5\ \text{k}\Omega$ (recall $r_{o}$ is neglected), so $$\boxed{f_{p2}=\frac{1}{2\pi R_{L}C_{M,out}} =\frac{1}{2\pi(5\ \text{k}\Omega)(2.01\ \text{pF})}=15.8\ \text{MHz}}$$ Because $f_{p2}/f_{H}=13$, the dominant-pole approximation used in part (c) is self consistent: the second pole contributes only about 0.03 dB of extra roll-off at $f_{H}$, so quoting $f_{H}$ as the upper 3 dB frequency is sound.

The stage therefore has a mid-band gain of about 109 and a usable band from roughly 12 Hz to 1.2 MHz, a gain–bandwidth product of about 132 MHz.