Question 3 of 7: Active Bandpass Amplifier with Bias-Current Compensation

Question 3: Active Bandpass Amplifier with Bias-Current Compensation (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. Inverting op-amp with input branch $R_1=10\text{k}\Omega$ in series with $C_1=0.1\mu F$, feedback branch $R_2=10\text{k}\Omega$ in parallel with $C_2=1\text{nF}$, and a compensating resistor $R_2=10\text{k}\Omega$ from the (+) input to ground. Input bias current $I_B=10\mu A$ at each input terminal; saturation at $\pm10\text{V}$.

Find. $A_v(j\omega)$, its sketch, $f_L$, $f_H$, bandwidth, passband gain, and the DC output offset caused by the bias currents.

Approach. This is an inverting amplifier whose input and feedback impedances are each an $R$-$C$ combination, so $A_v(j\omega)=-Z_2/Z_1$ is a bandpass shape: $C_1$ blocks DC/low frequencies (high-pass corner from $R_1C_1$), $C_2$ shorts out $R_2$ at high frequencies (low-pass corner from $R_2C_2$), and the flat passband in between is set by $R_2/R_1$. For the offset, note $C_1$ blocks DC entirely, so only $R_2$ (feedback) carries the inverting input's bias current at DC — compare that to the compensating $R_2$ at the (+) input.

1. Part (a) — AC gain. With $Z_1=R_1+\dfrac{1}{j\omega C_1}$ and $Z_2=R_2\|\dfrac{1}{j\omega C_2}=\dfrac{R_2}{1+j\omega R_2C_2}$: $$A_v(j\omega)=-\frac{Z_2}{Z_1}=-\frac{R_2}{(1+j\omega R_2C_2)\left(R_1+\dfrac{1}{j\omega C_1}\right)}$$ $$=-\frac{R_2}{R_1}\cdot\frac{j\omega R_1C_1}{(1+j\omega R_1C_1)(1+j\omega R_2C_2)}$$ which is a single-pole-single-zero bandpass response with a zero at DC, a low-frequency pole from $R_1C_1$ and a high-frequency pole from $R_2C_2$.
2. Part (c) — corner frequencies, bandwidth, passband gain. $$f_L=\frac{1}{2\pi R_1C_1}=\frac{1}{2\pi(10\text{k}\Omega)(0.1\mu F)}=\boxed{159.2\text{ Hz}}$$ $$f_H=\frac{1}{2\pi R_2C_2}=\frac{1}{2\pi(10\text{k}\Omega)(1\text{nF})}=\boxed{15.92\text{ kHz}}$$ $$\text{BW}=f_H-f_L=15.92\text{k}-0.159\text{k}=\boxed{15.76\text{ kHz}}$$ $$\left|A_{v,\text{passband}}\right|=\frac{R_2}{R_1}=\boxed{1.00\ (0\text{ dB, inverting})}$$
3. Part (b) — frequency response shape. Below $f_L$, $|A_v|$ rises at $+20\text{dB/decade}$ (the input zero at DC combined with the $R_1C_1$ pole); it is flat at $0\text{dB}$ between $f_L$ and $f_H$; above $f_H$ it falls at $-20\text{dB/decade}$ as $C_2$ shorts the feedback resistor — a standard single bandpass hump, inverting (180°) throughout the passband.
4. Part (d) — DC offset from input bias current. $C_1$ is open at DC, so no DC current can flow through $R_1$ — the only DC path to the inverting input is the feedback resistor $R_2$. With $I_{B-}$ flowing into the (−) input through $R_2$, and the standard inverting-amplifier offset formula (DC gain $R_1\to\infty$, so the noise gain $\to1$): $$V_{o,\text{offset}}=I_{B-}R_2-(1)\,I_{B+}R_{\text{comp}}$$ Here $R_{\text{comp}}=10\text{k}\Omega$ (the resistor placed from (+) to ground) exactly equals the feedback $R_2=10\text{k}\Omega$ that the (−) input sees at DC. Since $I_{B+}=I_{B-}=10\mu A$ (matched, as given): $$V_{o,\text{offset}}=(10\mu A)(10\text{k}\Omega)-(10\mu A)(10\text{k}\Omega)=\boxed{0\text{ V}}$$ The compensation resistor is sized on purpose to match the DC resistance the inverting input sees ($R_2$, since $C_1$ blocks $R_1$) — with matched bias currents the offsets cancel exactly.