A full wave rectifier circuit with diodes (D₁) and (D₂) is shown in the figure. If input supply voltage V in = 220sin (100π t) volt, then at t = 15 msec (a) D₁ and D₂ both are forward biased (b) D₁ and D₂ both are reverse biased (c) D₁ is forward biased, D₂ is reverse biased (d) D₁ is reverse biased, D₂ is forward biased.

Answer: (d). ω = 100π ⇒ T = (2π)/(ω) = 20 ms. At t = 15 ms: Vin = 220 sin (1.5π) = -220 V, the negative half cycle. Now the top end of the secondary (to D₁) is negative and the bottom end (to D₂) is positive relative to the centre tap. So D₁ is reverse biased and D₂ is forward biased. Answer (d).

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Rectifiers & Filters

A full wave rectifier circuit with diodes $(D_1)$ and $(D_2)$ is shown in the figure. If input supply voltage $\text{V}$ in $= 220\sin(100\pi t)$ volt, then at $t = 15\text{ msec}$

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Answer: (d)

$\omega = 100\pi \Rightarrow T = \frac{2\pi}{\omega} = 20$ ms.
At $t = 15$ ms: $V_{in} = 220 \sin(1.5\pi) = -220$ V, the negative half cycle.
Now the top end of the secondary (to $D_1$ ) is negative and the bottom end (to $D_2$ ) is positive relative to the centre tap.
So $D_1$ is reverse biased and $D_2$ is forward biased. Answer (d).

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