10. (10 pts) During this class we developed a Zener regulated power supply. The

Question

10. (10 pts) During this class we developed a Zener regulated power supply. The circuit to the right improves the operation by adding a transistor. Assume the Zener diode has a breakdown voltage of 8.3V. D1 D2 Vin RL a) What is the voltage across the load resistor? b) Assume the voltage at the capacitor is 22V. If R is 120, RL is 100, what is the current through the load resistor and the collector current? c) Give a short explanation as to why this circuit is an improvement over the one you built in the lab.

Explanation / Answer

Solution :-

a) Voltage (V)   = RL Vi/ R+ RL

= 100 x 22/ 220

= 10 volt .

As voltage V is greater than  Vz i.e 8.3 volt , diode is in on - state ,therefore

Voltage across load resistor (VL) = VZ = 8.3 volts .

B) Current across Load Resisitor (IL) = VL/RL = 8.3 /100 = 83 mA.

Voltage across resistor (VR)= Vi - VL

= 22-8.3 = 13.7 volt

IR = VR/R = 13.7/ 0.12Kohm = 114.16 mA

Therefore collector current i.e current across zener (IZ)= IR-IL = 114.16 - 83 = 31.16 mA.

C) This circuit is similar to the Lab circuit, except that the zener diode is connected to the base of a npn transistor.

Note that in a transistor the current required by the base is only 1/hFE times the emitter to collector current. So a low power rated zener diode can regulate the base voltage of the BJT that can pass a huge current through it.

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