Question 7 of 7

Question 7 (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.

Approach. Negative feedback through the conducting diodes forces $V_R=V_A(t)$ (ideal op-amp, virtual short). For $V_A>0$ the current path is $V_c\to$ D1 $\to V_1\to M\to V_2\to$ D3 $\to V_R\to R\to$ ground; for $V_A<0$ it is ground $\to R\to V_R\to$ D4 $\to V_1\to M\to V_2\to$ D2 $\to V_c$. Either way the meter current flows from $V_1$ to $V_2$ and equals $|V_A(t)|/R$, independent of the diode drops and of $r$. (a) apply formula (44) to that full-wave current. (b) $V_c$ must sit two diode drops plus the meter’s $Ir$ drop beyond $V_R$. (c) read the OFF diodes’ reverse bias off the same node voltages.

1. Part (a) — sizing R. The meter current is $i(t)=|V_A(t)|/R$, a full-wave-rectified sine of peak $I_p=V_{A,peak}/R$: both conducting diodes and the coil resistance are inside the feedback loop, so they drop out. Using the exam’s full-wave average formula (44), $I_{avg}=(2/\pi)I_p$; setting this to the full-scale rating: $$I_{avg}=\frac{2}{\pi}\cdot\frac{V_{A,peak}}{R}=2\,\text{mA}\ \Rightarrow\ R=\frac{2V_{A,peak}}{\pi(2\,\text{mA})}=\frac{2(5)}{\pi(0.002)}=1591.5\,\Omega$$ $$R=\boxed{1.59\ \text{k}\Omega}\qquad I_p=\frac{5}{1591.5}=3.14\,\text{mA}$$ The coil resistance $r$ does not enter. Subtracting it, as if $R$ and $r$ formed a series divider across $V_A$, would give $1.49\,\text{k}\Omega$ and a meter that reads 6.7% high.
2. Part (b) — headroom of $V_c$ to saturation. At the positive peak ($V_A=5\,\text{V}$) D1 and D3 conduct $I_p=3.14\,\text{mA}$, so $V_c$ must sit two diode drops plus the coil drop above $V_R=V_A$: $$V_{c,max}=V_R+V_{D3}+I_pr+V_{D1}=5+0.5+(3.14\,\text{mA})(100\,\Omega)+0.5=\boxed{6.31\ \text{V}}$$ Taking the op-amp’s saturation level as its $\pm9\,\text{V}$ supply rails (ideal op-amp), the headroom to $+9\,\text{V}$ is $$9-6.31=\boxed{2.69\ \text{V}}$$ — the op-amp stays in its linear range at full scale. By symmetry $V_{c,min}=-6.31\,\text{V}$ at the negative peak, also 2.69 V inside the $-9\,\text{V}$ rail. A real op-amp that saturates about 1 V inside its rails would still leave about 1.7 V.
3. Part (c) — is 1 V PRV enough? At the positive peak (D1, D3 ON; D2, D4 OFF) the node voltages are $$V_R=5\,\text{V},\quad V_2=V_R+V_D=5.5\,\text{V},\quad V_1=V_2+I_pr=5.814\,\text{V},\quad V_c=V_1+V_D=6.314\,\text{V}$$ The OFF diode D2 (anode $V_2$, cathode $V_c$) is reverse-biased by $V_c-V_2=0.814\,\text{V}$, and D4 (anode $V_R$, cathode $V_1$) by $V_1-V_R=0.814\,\text{V}$. Both equal $V_D+I_pr=0.5+0.314$. The negative half-cycle is the mirror image, with D1 and D3 then blocking the same 0.814 V. $$V_{rev,max}=V_D+I_pr=\boxed{0.81\ \text{V}}\ <\ \text{PRV rating}=1\,\text{V}$$ The 1 V PRV rating IS sufficient, with about 0.19 V (19%) of margin. Because the op-amp drives the bridge from inside a feedback loop, each OFF diode only ever sees one diode drop plus the small coil drop, far below the 5 V amplitude of $V_A$.