Question 5 of 8: Op-Amp Frequency Response — Bode Sketch and Regional Behaviour

Question 5: Op-Amp Frequency Response — Bode Sketch and Regional Behaviour (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. An ideal inverting op-amp stage. The input arm is $R_1$ in series with $C_1$; the feedback arm is $R_2$ in parallel with $C_2$; the non-inverting terminal is grounded.

Find. [a] the straight-line (asymptotic) magnitude plot from 0.001 Hz to 10 MHz with every gain level, corner frequency and unity-gain crossing identified and calculated; [b] the function performed at 0.01 Hz, 1000 Hz and 1 MHz, with the corresponding time-domain input-output relation in each case.

[Figure not reproduced: Figure 5 redrawn. The input arm is $R_1$ in series with $C_1$; the feedback arm is $R_2$ in parallel with $C_2$; the non-inverting input is grounded. See the official exam paper.]

Approach. Write the two arm impedances in the Laplace variable, form the inverting-amplifier gain $-Z_f/Z_{in}$, factor it into a single zero at the origin and two real poles, then read the corners and asymptotic slopes straight off the factored form.

1. Write the two arm impedances. For the series input arm and the parallel feedback arm, $$Z_{in}(s)=R_1+\frac{1}{sC_1}=\frac{1+sR_1C_1}{sC_1},\qquad Z_f(s)=\frac{R_2\cdot\dfrac{1}{sC_2}}{R_2+\dfrac{1}{sC_2}}=\frac{R_2}{1+sR_2C_2}$$
2. Form the transfer function. An ideal inverting stage has $V_o/V_i=-Z_f/Z_{in}$, so $$\frac{V_o}{V_i}(s)=-\frac{R_2}{1+sR_2C_2}\cdot\frac{sC_1}{1+sR_1C_1}$$ $$\boxed{\;\frac{V_o}{V_i}(s)=\frac{-\,s\,R_2C_1}{\left(1+sR_1C_1\right)\left(1+sR_2C_2\right)}\;}$$ This is a band-pass: one zero at the origin (hence rising response at low frequency) and two real poles (hence falling response at high frequency).
3. Compute the two corner frequencies from the time constants. $$\tau_1=R_1C_1=\left(160\times10^{3}\right)\left(1.0\times10^{-6}\right)=0.160\ \text{s}\;\Rightarrow\; f_1=\frac{1}{2\pi\tau_1}=\frac{1}{2\pi(0.160)}=0.995\ \text{Hz}$$ $$\tau_2=R_2C_2=\left(16\times10^{6}\right)\left(1\times10^{-12}\right)=16.0\ \mu\text{s}\;\Rightarrow\; f_2=\frac{1}{2\pi\tau_2}=\frac{1}{2\pi\left(16\times10^{-6}\right)}=9947\ \text{Hz}$$ $$\boxed{\;f_1=0.995\ \text{Hz}\ (\approx 1\ \text{Hz}),\qquad f_2=9.95\ \text{kHz}\ (\approx 10\ \text{kHz})\;}$$
4. Compute the midband gain. Between the two corners the zero has already flattened and neither pole has yet acted, so the capacitors drop out and the circuit is a plain inverting amplifier: $$\left|\frac{V_o}{V_i}\right|_{\text{mid}}=\frac{R_2}{R_1}=\frac{16\times10^{6}}{160\times10^{3}}=100$$ $$\boxed{\;\text{midband gain}=100=20\log_{10}(100)=40\ \text{dB}\;}$$
5. Compute the two unity-gain crossings. Below $f_1$ the response is set by the zero alone, $|V_o/V_i|\to\omega R_2C_1$, which reaches unity when $$\omega=\frac{1}{R_2C_1}\;\Rightarrow\; f_{u,\text{lo}}=\frac{1}{2\pi R_2C_1}=\frac{1}{2\pi\left(16\times10^{6}\right)\left(1\times10^{-6}\right)}=9.95\times10^{-3}\ \text{Hz}$$ Above $f_2$ both poles are active and the zero survives, leaving $|V_o/V_i|\to 1/(\omega R_1C_2)$, which reaches unity when $$\omega=\frac{1}{R_1C_2}\;\Rightarrow\; f_{u,\text{hi}}=\frac{1}{2\pi R_1C_2}=\frac{1}{2\pi\left(160\times10^{3}\right)\left(1\times10^{-12}\right)}=9.95\times10^{5}\ \text{Hz}$$ $$\boxed{\;f_{u,\text{lo}}=0.00995\ \text{Hz}\ (\approx 0.01\ \text{Hz}),\qquad f_{u,\text{hi}}=995\ \text{kHz}\ (\approx 1\ \text{MHz})\;}$$ Each unity-gain point sits two decades outside its corner, exactly as a 40 dB plateau falling at 20 dB per decade requires — a useful self-check on the arithmetic.
6. Assemble the straight-line plot. Four line segments describe the whole response: a $+20$ dB/decade rise up to $f_1$, a flat 40 dB plateau from $f_1$ to $f_2$, and a $-20$ dB/decade fall beyond $f_2$. At the plot limits the asymptotes read $-20$ dB at 0.001 Hz and $-20$ dB at 10 MHz.

[b] Function and time-domain relation at each test frequency. Each of the three specified frequencies falls in a different region of the plot, which is plainly the examiner's intent.

1. [1] At 0.01 Hz — well below $f_1$ — the circuit is a differentiator. Here $\omega R_1C_1\ll 1$ and $\omega R_2C_2\ll 1$, so both poles are inactive and the transfer function collapses to its zero: $$\frac{V_o}{V_i}(s)\to -sR_2C_1 \quad\Longrightarrow\quad \boxed{\;v_o(t)=-R_2C_1\,\frac{dv_i}{dt}=-16.0\,\frac{dv_i}{dt}\;}$$ with the coefficient $R_2C_1 = (16\times 10^{6})(1\times 10^{-6}) = 16.0$ s. Note that 0.01 Hz is essentially the lower unity-gain frequency, so the magnitude here is very close to 1 (computed exactly: 1.005).
2. [2] At 1000 Hz — between the corners — the circuit is an inverting amplifier of gain 100. $C_1$ is effectively a short and $C_2$ effectively an open, so $$\frac{V_o}{V_i}\to -\frac{R_2}{R_1}=-100 \quad\Longrightarrow\quad \boxed{\;v_o(t)=-\frac{R_2}{R_1}\,v_i(t)=-100\,v_i(t)\;}$$ The output tracks the input with no differentiation or integration, inverted and scaled by 100 (40 dB), which the exact transfer function confirms to within 0.01 dB at this frequency.
3. [3] At 1 MHz — well above $f_2$ — the circuit is an integrator. Now both $\omega R_1C_1\gg 1$ and $\omega R_2C_2\gg 1$, so $$\frac{V_o}{V_i}(s)\to\frac{-sR_2C_1}{\left(sR_1C_1\right)\left(sR_2C_2\right)}=-\frac{1}{sR_1C_2} \quad\Longrightarrow\quad \boxed{\;v_o(t)=-\frac{1}{R_1C_2}\int v_i(t)\,dt=-6.25\times10^{6}\!\int v_i(t)\,dt\;}$$ with $1/(R_1C_2) = 1/\!\left[(160\times10^{3})(1\times10^{-12})\right] = 6.25\times10^{6}\ \text{s}^{-1}$. As with the low end, 1 MHz is essentially the upper unity-gain frequency, so the magnitude here is again close to 1 (computed exactly: 0.995).