The Junction Rectifier — Question 6

Question 6

Finite bridge drops reduce the conduction interval

A bridge rectifier has two conducting diodes in each path, each modeled by fixed drop $V_f$. With resistive load and sinusoidal source, $v_o=\max(V_m|\sin\theta|-2V_f,0)$. Let $V_m=10.0\,\mathrm V$ and $V_f=.700\,\mathrm V$.

1. Find the conduction angle during each rectified half cycle.

2. Derive and evaluate the average output voltage.

Question 6 - Solution

Finite bridge drops reduce the conduction interval

(a) Method and calculation

$$\theta_0=\sin^{-1}(2V_f/V_m)=\sin^{-1}(.14)=.14046\,\mathrm{rad}.$$$$\boxed{\Delta\theta=\pi-2\theta_0=2.86067\,\mathrm{rad}=163.9^\circ}.$$

(b) Interpretation and further calculation

$$V_{\mathrm{dc}}=\frac1\pi\int_{\theta_0}^{\pi-\theta_0}(V_m\sin\theta-2V_f)d\theta.$$$$\boxed{V_{\mathrm{dc}}=\frac{2V_m\cos\theta_0-2V_f(\pi-2\theta_0)}{\pi}=5.029\,\mathrm V}.$$

Physical check

Subtracting 2Vf from the ideal mean would incorrectly include negative output near each zero crossing. The clipped intervals must be removed from the integral.