Question 4.8: Figure 4–21 shows a current-source biasing circuit. What is ...

Analog Fundamentals: A Systems Approach

Figure 4–21 shows a current-source biasing circuit. What is $I_{D}$ ?

4.8

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You should recognize emitter bias for the bipolar junction transistor but with no base resistor. Since the base is connected directly to ground, the emitter voltage is -0.7 V due to the forward-biased base-emitter pn junction. This means that the voltage across $R_{E}$ is 14.3 V and the current in $R_{E}$ is constant. From Ohm’s law, find the emitter current as

$I_{E} = \frac{V_{R_{E} } }{R_{E} } = \frac{14.3 V}{15 k\Omega } = 0.95 mA$

This current is provided to the JFET. Therefore, $I_{D}$ = 0.95 mA.

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