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For the circuit shown below ​

For the circuit shown below ​-example-1
User Heartcroft
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1 Answer

6 votes

Answer:

48.00 microamps

Step-by-step explanation:

The base voltage is limited by the zener to 5.5 V. If we assume the B-E voltage drop is 0.7 V, then the voltage across RE is 5.5-0.7 = 4.8 volts. That means the emitter current is 4.8/2.0k = 2.4 mA.

The base current is that amount divided by (1+β), so is 2.4 mA/(1+49) = 48 μA.

User Asteroid
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