Question 5.19: Objective: Calculate the dc voltages at each node and the dc...
Objective: Calculate the dc voltages at each node and the dc currents through the elements in a multistage circuit.
For the circuit in Figure 5.61, assume the B–E turn-on voltage is 0.7 V and β = 100 for each transistor
The Thevenin equivalent circuit of the base circuit of Q_{1} is shown in Figure 5.62. The various currents and nodal voltages are defined as shown. The Thevenin resistance and voltage are
R_{T H} = R_{1} || R_{2} = 100 || 50 = 33.3 k \Omega
and
V_{T H} = \left( \frac{R_{2}}{R_{1} + R_{2}} \right) (10) − 5 = \left( \frac{50}{150} \right) (10) − 5 = −1.67 V
Kirchhoff’s voltage law equation around the B–E loop of Q_{1} is
V_{T H} = I_{B1} R_{T H} + V_{B E} (on) + I_{E1} R_{E1} − 5
Noting that I_{E1} = (1 + β)I_{B1} , we have
I_{B1} = \frac{−1.67 + 5 − 0.7}{33.3 + (101)(2)} ⇒ 11.2 μA
Therefore,
I_{C1} = 1.12 mA
and
I_{E1} = 1.13 mA
Summing the currents at the collector of Q_{1} , we obtain
I_{R1} + I_{B2} = I_{C1}
which can be written as
\frac{5 − V_{C1}}{R_{C1}} + I_{B2} = I_{C1} (5.47)
The base current I_{B2} can be written in terms of the emitter current I_{E2}, as
follows:
I_{B2} = \frac{I_{E2}}{1 + β} = \frac{5 − V_{E2}}{(1 + β)R_{E2}} = \frac{5 − (V_{C1} + 0.7)}{(1 + β)R_{E2}} (5.48)
Substituting Equation (5.48) into (5.47), we obtain
\frac{5 − V_{C1}}{R_{C1}} + \frac{5 − (V_{C1} + 0.7)}{(1 + β)R_{E2}} = I_{C1} = 1.12 mA
which can be solved for V_{C1} to yield
V_{C1} = − 0.482 V
Then,
I_{R1} = \frac{5 − (−0.482)}{5} = 1.10 mA
To find V_{E2}, we have
V_{E2} = V_{C1} + V_{E B} (on) = −0.482 + 0.7 = 0.218 V
The emitter current I_{E2} is
I_{E2} = \frac{5 − 0.218}{2} = 2.39 mA
Then,
I_{C2} = \left( \frac{β}{1 + β} \right) I_{E2} = \left( \frac{100}{101} \right) (2.39) = 2.37 mA
and
I_{B2} = \frac{I_{E2}}{1 + β} = \frac{2.39}{101} ⇒ 23.7 μA
The remaining nodal voltages are
V_{E1} = I_{E1} R_{E1} − 5 = (1.13)(2) − 5 ⇒ V_{E1} = −2.74 V
and
V_{C2} = I_{C2} R_{C2} − 5 = (2.37)(1.5) − 5 = −1.45 V
We then find that
V_{C E1} = V_{C1} − V_{E1} = −0.482 − (−2.74) = 2.26 V
and that
V_{EC2} = V_{E2} − V_{C2} = 0.218 − (−1.45) = 1.67 V
Comment: These results show that both Q_{1} and Q_{1} are biased in the forward-active mode, as originally assumed. However, when we consider the ac operation of this circuit as an amplifier in the next chapter, we will see that a better design would increase the value of V_{EC2}.
