Consider the circuit shown. Assume that the diode $D$ is ideal. The supply voltage $v_s = 325 \sin(2\pi 50t) \text{ V}$, $L = 500 \mu H$, and $R = 10 \text{ }\Omega$. The peak diode current (in amperes) is ______________ (Round off to one decimal place)
To find the peak diode current, we first determine the peak voltage and then the current through the resistor. The supply voltage is given by \(v_s = 325 \sin(2\pi 50t)\). Thus, the peak voltage \(V_m = 325\,\text{V}\).
Since the diode is ideal, it only conducts during the positive half-cycle of the input \(v_s\). The inductor and the resistor form an RL circuit. The impedance \(Z\) is given by:
\(Z = \sqrt{R^2 + (\omega L)^2}\)
\(\omega = 2\pi f = 2\pi \times 50 = 100\pi\,\text{rad/s}\)
\(L = 500\,\mu H = 0.0005\,\text{H}\)
Calculate \(Z\):
\(Z = \sqrt{10^2 + (100\pi \times 0.0005)^2} = \sqrt{100 + (0.157)^2} = \sqrt{100 + 0.024649} = 10.0012\,\Omega\)
Now calculate the peak current \(I_m = \frac{V_m}{Z}\):
\(I_m = \frac{325}{10.0012} \approx 32.496\,A\)
The peak diode current is approximately \(32.5\,A\), fitting within the range of 32.3 to 32.5.
Important Questions from Characteristics of Power Diodes
The barrier voltage for germanium is _________ at 25°C.
When a diode is switching from conducting state to blocking state, it has stored charge which needs to be discharged before, it blocks reverse current, this time required is known as _______.
