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## Q. 33.6

What is the collector-emitter voltage in Fig. 33-19?

## Verified Solution

The voltage divider produces an unloaded output voltage of

$V_{BB}=\frac{2.2 K\Omega }{10 K\Omega +2.2 K\Omega }10V= 1.8 V$

Subtract 0.7 V from this to get

$V_E = 1.8 V – 0.7 V = 1.1 V$

The emitter current is

$I_E=\frac{1.1 V}{1 K\Omega } =1.1 mA$

Since the collector current almost equals the emitter current, we can calculate the collector-to-ground voltage like this

$V_C = 10 V – (1.1 mA)(3.6 k\Omega) = 6.04 V$

The collector-emitter voltage is

$V_{CE} = 6.04 – 1.1 V = 4.94 V$

Here is an important point: The calculations in this preliminary analysis do not depend on changes in the transistor, the collector current, or the temperature. This is why the Q point of this circuit is stable, almost rock solid.