Question 19.3: Determine the quiescent voltages and currents in the followi......
Determine the quiescent voltages and currents in the following feedback circuit.
Quiescent base voltage
If we assume that the base current is negligible, as no constant current can flow through the input capacitor, the quiescent base voltage is determined simply by the supply voltage V_{CC} and potential divider formed by R_{1} and R_{2} . Hence
V_{B} \approx V_{CC} \frac{R_{2}}{R_{1} + R_{2}}Therefore in our example
V_{B} \approx 10 \frac{10 kΩ}{27 kΩ + 10 kΩ} \approx 2.7 VQuiescent emitter voltage
As the base-to-emitter voltage V_{BE} is assumed to be constant, it is simple to determine the emitter voltage from the base voltage. Thus, the quiescent emitter voltage is simply
V_{E} = V_{B} – V_{BE}and in our circuit
V_{E} = 2.7 – 0.7 = 2.0 VQuiescent emitter current
Knowing the voltage across the emitter resistor and its value gives us the emitter current
I_{E} = \frac{V_{E}}{R_{E}}and therefore
I_{E} = \frac{2.0 V}{1 kΩ} = 2 mAQuiescent collector current
If the base current is negligible, it follows that the collector current is equal to the emitter current
I_{C} \approx I_{E}Therefore in our circuit
I_{C} \approx I_{E} = 2 mAQuiescent collector (output) voltage
In this circuit the output voltage is simply the collector voltage. This is determined by the supply voltage V_{CC} and voltage across the collector resistor R_{C} . The voltage across R_{C} is simply the product of its resistance and the collector current, so
V_{o (quiescent)} = V_{C} = V_{CC} – I_{C}R_{C}In this case
V_{o (quiescent)} = 10 V – 2 mA × 2.2 Ω = 5.6 V