Question

Consider the circuit shown. Assume that the diode \(D\) is ideal. The supply voltage \(v_s=325\sin(2\pi50t)\) V, \(L=500\ \mu H\), and \(R=10\ \Omega\).

The peak diode current (in amperes) is (round off to one decimal place).

Hint:

Since wL is tiny next to R here, the RL transient dies out almost instantly, so the peak current is close to Vm/Z, occurring near the middle of the positive half cycle.

Answer 1

Correct Answer: 32.5

Step 1: Set up the circuit equation.
The source, diode, inductor and resistor form one loop. Once the diode turns on, the current \(i(t)\) obeys
\[ L\frac{di}{dt}+Ri=V_m\sin(\omega t) \]
with \(V_m=325\) V and \(\omega=2\pi\times50=314.16\) rad/s.

Step 2: Find the impedance and the phase angle.
\[ \omega L=314.16\times500\times10^{-6}=0.1571\ \Omega \]
\[ Z=\sqrt{R^2+(\omega L)^2}=\sqrt{10^2+0.1571^2}=10.0012\ \Omega \]
\[ \varphi=\tan^{-1}\left(\frac{\omega L}{R}\right)=\tan^{-1}(0.01571)=0.9^{\circ} \]

Step 3: Write the general current expression.
For a half wave rectifier feeding an R-L load, the current while the diode conducts is
\[ i(\theta)=\frac{V_m}{Z}\left[\sin(\theta-\varphi)+\sin\varphi\ e^{-\theta\cot\varphi}\right],\qquad \theta=\omega t \]

Step 4: Check how fast the extra term dies out.
Here \(\cot\varphi=R/(\omega L)=63.66\), a very large number. So the term \(e^{-\theta\cot\varphi}\) collapses to almost zero within a fraction of a degree of the diode turning on. Practically, for the rest of the conduction period the current just tracks
\[ i(\theta)\approx\frac{V_m}{Z}\sin(\theta-\varphi) \]

Step 5: Locate the peak.
This sine term peaks when \(\theta-\varphi=90^{\circ}\), which falls well inside the conduction interval (the diode conducts for almost the whole positive half cycle here, since the tiny inductance barely delays the current zero crossing). At that point the decaying term has already vanished, so the peak value is simply
\[ i_{peak}=\frac{V_m}{Z}=\frac{325}{10.0012}=32.496\text{ A} \]

Step 6: Round the answer.
\[ \boxed{32.5} \]
The peak diode current works out to about 32.5 A.