"Chapter 1 - The Op Amp's Place in the World" - HTL Wien 10

Review of Circuit Theory
Transistor Amplifier
2 V (from 8 V to 10 V) there is still enough voltage dropped across R C to keep the transistor
on. Set the collector-emitter voltage at 4 V; when the collector voltage swings negative
2 V (from 8 V to 6 V) the transistor still has 2 V across it, so it stays linear. This sets the
emitter voltage (V E) at 4 V.
R C V 12 V C
I C
R E R E1 R E2 V E
I E
12 V 8V
10 mA
400
VE
IB IC VE
IC 4V
400
10 mA
Use Thevenin’s equivalent circuit to calculate R 1 and R 2 as shown in Figure 2–16.
R2
12
R1 + R2
R1 || R2
Figure 2–16. Thevenin Equivalent of the Base Circuit
I B I C
10 mA
100
V TH 12R 2
R 1 R 2
R TH R 1 R 2
R 1 R 2
IB
0.1 mA
VB = 4.6 V
(2–30)
(2–31)
(2–32)
(2–33)
(2–34)
We want the base voltage to be 4.6 V because the emitter voltage is then 4 V. Assume
a voltage drop of 0.4 V across R TH, so Equation 2–35 can be written. The drop across R TH
may not be exactly 0.4 V because of beta variations, but a few hundred mV does not matter
is this design. Now, calculate the ratio of R 1 and R 2 using the voltage divider rule (the
load current has been accounted for).
RTH 0.4
k 4k
0.1
VTH IBRTh VB 0.4 4.6 5 12
R1 R2 R 2 7
5 R 1
R 2
(2–35)
(2–36)
(2–37)
R 1 is almost equal to R 2, thus selecting R 1 as twice the Thevenin resistance yields approximately
4 K as shown in Equation 2–35. Hence, R 1 = 11.2 k and R 2 = 8 k. The ac gain is
2-11