What is the collector-emitter voltage in Fig. 33-19?
Chapter 33
Q. 33.6
Step-by-Step
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 VSubtract 0.7 V from this to get
V_E = 1.8 V – 0.7 V = 1.1 VThe emitter current is
I_E=\frac{1.1 V}{1 K\Omega } =1.1 mASince 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 VThe collector-emitter voltage is
V_{CE} = 6.04 – 1.1 V = 4.94 VHere 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.