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Zener Diode

Draw the curr...

Question

Draw the current-voltage characteristics for the device show in figure between the terminals A and B.

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Solution

(a) If a battery is connected between terminals A and B, with positive terminal connected to point A and negative terminal connected to point B, then the diode will get forward biassed by the applied voltage. So, the current voltage graph for this circuit will be the same as that of the characteristic curves of a forward-biassed diode.

(b) If a battery is connected between terminals A and B, with positive terminal connected to point A and negative terminal connected to point B, then the upper diode will get forward biassed and the lower diode will get reverse biassed by the applied voltage. So, this lower branch can be replaced by an open circuit; hence, the current flow through this branch will be zero. The current flows only through the upper diode, so the circuit on simplification will become identical to the circuit in part (a). Hence, the current voltage graph for this circuit will be the same as that of the characteristic curves of a forward-biassed diode.

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V = V_{0} s i n ω T .

I = I_{0} s i n (ω t + \frac{π}{2})

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V = V_{0} sin ω t

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I = I_{0} sin (ω t + \frac{π}{2})

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(r_{i})

β

(R_{i}

= {[\frac{Change in emitter base voltage (△ V_{B E})}{Change in base current (△ I_{B})}]}_{at constant V_{C E}}

R_{i} = {(\frac{△ V_{B E}}{△ I_{B}})}_{V_{C E}}

(β) = {(\frac{△ I_{C}}{△ I_{B}})}_{V_{C E}}

\frac{△ I_{C}}{△ I_{E} - △ I_{C}}

\frac{\frac{△ I_{C}}{△ I_{E}}}{\frac{△ I_{C}}{△ I_{E}} - \frac{△ I_{C}}{△ I_{E}}}

\frac{α}{1 - α}

\Rightarrow β = \frac{α}{1 - α}

A, B

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