Question 1 of 7: Current-Source-Driven Full-Wave Bridge Rectifier

Question 1: Current-Source-Driven Full-Wave Bridge Rectifier (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 current source $I_s(t)=10\sin(2\pi ft)\text{ mA}$, $f=1\text{kHz}$, feeds node $V_a$. Two resistors, each $R=10\text{k}\Omega$, sit in parallel between $V_a$ and ground. Node $V_a$ also drives the AC terminals of a full-wave diode bridge (4 diodes, $V_F=0.7\text{V}$ each) whose DC output is taken across $R_L=5\text{k}\Omega$, labelled $V_o$.

Find. $V_a(t)$ and $V_o(t)$ with peaks; the diode reverse-voltage rating; the maximum average power rating for the load resistor; $I_s(t)$ with peaks.

Approach. Collapse the two parallel $10\text{k}\Omega$ resistors to $R_{par}=5\text{k}\Omega$. While $|V_a|$ is below the bridge's two-diode turn-on threshold the bridge is an open circuit and $V_a=I_s\cdot R_{par}$; once $|V_a|$ exceeds that threshold, the conducting diode pair adds a $1.4\text{V}$-offset $R_L$ path in parallel with $R_{par}$, and node KCL gives a linear relation between $V_a$ and $I_s$ that holds for the rest of each half-cycle.

1. Turn-on threshold and node equation. The bridge conducts once $|V_a|>2V_F=1.4\text{V}$, i.e. once $|I_s|>1.4\text{V}/R_{par}=0.28\text{mA}$ — at a 10mA peak this is a sliver near each zero-crossing, negligible for the sketch. Above threshold, KCL at node $V_a$ (taking the positive half; negative is symmetric) gives $$I_s=\frac{V_a}{R_{par}}+\frac{V_a-2V_F}{R_L}$$ Since $R_{par}=R_L=5\text{k}\Omega$, this reduces to $2V_a=2V_F+I_s R_{par}$, i.e. $$\boxed{V_a=V_F+\tfrac12 R_{par}\,I_s=0.7+2.5\,I_s\ \ (I_s\text{ in mA, }V_a\text{ in V})}$$
2. Peak values. At $I_s=10\text{mA}$ (peak): $V_{a,\text{peak}}=0.7+2.5(10)=\boxed{25.7\text{ V}}$, and $V_{o,\text{peak}}=V_{a,\text{peak}}-2V_F=25.7-1.4=\boxed{24.3\text{ V}}$. By symmetry the negative half of $I_s$ produces $V_a=-25.7\text{V}$ but the SAME $V_o=+24.3\text{V}$ peak (full-wave rectification), with $V_o$ dipping to (near) zero at each zero-crossing dead-zone. See Fig. Q1-a.
3. Part (b) — reverse voltage rating. In a full bridge, the two OFF diodes each see a reverse voltage equal to the DC output plus one forward drop: $V_{\text{rev}}=V_o+V_F$. At the peak, $V_{\text{rev}}=24.3+0.7=\boxed{25.0\text{ V}}$. Choosing a standard part with a healthy margin (e.g. a 1N4004-class diode rated 50V PIV, roughly $2\times$ this worst case) is a safe, standard choice.
4. Part (c) — resistor power rating. $V_o(t)$ is a piecewise (dead-zone + clamp-shifted) rectified sine, so its average power is found by numerically integrating $V_o(t)^2/R_L$ over one period: $P_{avg,R_L}\approx\boxed{58.1\text{ mW}}$. A standard $\tfrac14\text{W}$ (250mW) resistor gives over $4\times$ margin. If "this resistor" instead means one of the two $10\text{k}\Omega$ shunt resistors, each dissipates $P_{avg,R}=\tfrac12\cdot\overline{V_a^2/R_{par}}\approx\boxed{33.5\text{ mW}}$ (they split the shared $V_a^2/R_{par}$ power equally by symmetry) — a $\tfrac14\text{W}$ part clears this case too.
5. Part (d) — $I_s(t)$. $I_s(t)$ is given directly as $10\sin(2\pi ft)\text{ mA}$: a $1\text{kHz}$ ($T=1\text{ms}$) sine of peak $\pm10\text{mA}$, in phase with $V_a$'s driving waveform (see Fig. Q1-d).