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Question 30

A Zener diode is connected to a battery and a load as shown below:


The currents, I, I$$_Z$$ and I$$_L$$ are respectively.

Solution

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$$I_L = \frac{V_Z}{R_L} = \frac{10\text{ V}}{2\text{ k}\Omega} = 5\text{ mA}$$

$$V_{\text{drop}} = V_{\text{source}} - V_Z = 60\text{ V} - 10\text{ V} = 50\text{ V}$$

$$I = \frac{V_{\text{drop}}}{R_s} = \frac{50\text{ V}}{4\text{ k}\Omega} = 12.5\text{ mA}$$

$$I = I_Z + I_L$$

$$I_Z = I - I_L = 12.5\text{ mA} - 5\text{ mA} = 7.5\text{ mA}$$

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