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17-Phys-A5 · December 2018

Question 2 of 6

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

17-Phys-A5-B — Analog and Digital Electronic Circuits — National Exams, December 2018
3 hours duration. Open book exam. Answer all SIX (6) questions.

Reference texts: A. S. Sedra & K. C. Smith, Microelectronic Circuits, 8th ed. (BJT/MOSFET biasing and small-signal amplifiers Ch. 6–7, current-mirror loads and cascodes Ch. 7–8, feedback amplifiers Ch. 10, high-frequency response and open-circuit time constants Ch. 9, D/A converters Ch. 17); M. M. Mano & M. D. Ciletti, Digital Design, 6th ed. (Boolean algebra Ch. 2, static CMOS logic gates Ch. 10).

Question 2 (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.

SymbolValue
$R_1=R_2=R_3$$1\,\text{k}\Omega$
$R_4$$10\,\text{k}\Omega$
$C_1$$1\,\mu\text{F}$

Two cascaded inverting op-amp stages: stage 1 has input impedance $R_1\parallel C_1$ and feedback $R_2$; stage 2 has input resistor $R_3$ and feedback $R_4$. Op-amps are ideal.

Find. (a) $H(s)=v_o/v_i$. (b) The asymptotic Bode magnitude plot of $|H(j\omega)|$ in dB vs. $\omega$ (rad/s).

[Figure not reproduced: Fig. 2 from the exam paper. See the official exam paper or the cited reference text.]

Fig. 2 (reproduced from the exam paper) — two-stage inverting op-amp cascade; stage 1's input branch is $R_1\parallel C_1$, stage 2 is a fixed inverter.

Approach. Write each inverting stage's transfer function ($H=-Z_f/Z_{in}$), then multiply.

  1. Stage 1 transfer function. $Z_{in,1}(s)=R_1\parallel\dfrac{1}{sC_1}=\dfrac{R_1}{1+sR_1C_1}$, and $Z_{f,1}=R_2$, so $$H_1(s)=-\frac{Z_{f,1}}{Z_{in,1}}=-\frac{R_2}{R_1}\left(1+sR_1C_1\right)$$
  2. Stage 2 transfer function. Stage 2 has no reactive element: $$H_2(s)=-\frac{R_4}{R_3}$$
  3. Combine. $$H(s)=H_1(s)H_2(s)=\frac{R_2R_4}{R_1R_3}\left(1+sR_1C_1\right)$$ Substituting the values ($R_2R_4/R_1R_3=10$, $R_1C_1=1\,\text{k}\Omega\times1\,\mu\text{F}=1\,\text{ms}$): $$\boxed{H(s)=\frac{v_o}{v_i}=10\left(1+\frac{s}{1000}\right)}$$ — a single real zero at $\omega_z=1/R_1C_1=1000\,\text{rad/s}$.
  4. Bode sketch (part b). At low frequency $|H|\to 10=20\,\text{dB}$, flat. Past the zero at $\omega_z=1000\,\text{rad/s}$ the magnitude rises at $+20\,\text{dB/decade}$ (a zero always adds rising slope, never flattens it) — sketched below.
101102103104105106ω (rad/s, log scale)-200204060|A| (dB)ωz = 1000 rad/s20 dB (=10 V/V)+20 dB/decade
Bode magnitude sketch of $|H(j\omega)|$: flat at 20 dB up to $\omega_z=1000$ rad/s, then rising at +20 dB/decade.
QuantityValue
DC/low-freq gain10 V/V (20 dB)
Zero frequency $\omega_z$1000 rad/s
High-frequency slope+20 dB/decade