Design Objectives for Four-Resistor Bias Network
V EQ = R EQ I B + V BE + R E I E
V EQ – V BE – R EQ I B
I E = ---------------------------------------------------- or
RE
V EQ – V BE
I E ≈ ---------------------------- for R EQ I B « V EQ – V BE
RE
R EQ is designed so that its voltage drop is
negligible.
VCC = 12V
R1
36 kΩ
22 kΩ
RC
16 kΩ
RE
I1
I2
IB
R2
18 kΩ
In this case, I E is determined by V EQ , V BE
V EQ – V BE
4 – 0.7
and R E . I E ≈ ---------------------------- = ---------------- = 206µA .
RE
16k
Another constraint is power dissipation. Choose I 2 ≤ I C ⁄ 5 . Now, power dissipated in R 1 and R 2 is less than 17% of total quiescent power. Also, I 2 » I B for
β F ≥ 50 .
Lecture 28
28 - 1
Basic Current Mirror
For VBE = 0.7V, βF = 100, VA = 0V and IS =
1.4x10-16A (redundant). Find IREF and IC2.
VB, 12V
Assume that both transistors are matched.
Both transistors operate in forward-active
region.
V B – V C1
12 – 0.7
I REF = ----------------------- = ------------------- = 202µA
R
56k
I C1
I REF = I C1 + I B1 + I B2 but -------- = I B1 = I B2
βF
IREF
VCC, 12V
R
56 kΩ
IC2
IC1
IB1
+ I
B2
VBE
-
∴I REF = I B2 ( β F + 2 ) ⇒ I B2 = 1.98µA
Therefore, I C2 = β F I B2 = 198µA ≈ I REF
IC
– 16
198µA
I S = ---------------------- = ---------------------- = 1.37 ×10 A
V ⁄V
e 0.7 / 0.025
e BE T
Lecture 28
28 - 2
Two-Resistor Bias Circuit
10V
Find the Q-point for the circuit shown if βF = 75 and VBE =
0.7V.
10 = 1.2k ( I C + I B ) + 5kI B + V BE
Assume transistor operates in forward-active region,
1.2 kΩ
5 kΩ
IC
I C = β F I B and 10 = 1.2k ( β F I B + I B ) + 5kI B + V BE .
IB
10 – V BE
-------------------------------------------IB =
1.2k ( β F + 1 ) + 5k
10 – V BE
10 – 0.7
I C = β F I B = β F --------------------------------------------- = 75 -------------------------------------------- = 7.25mA
1.2k ( β F + 1 ) + 5k
1.2k ( 75 + 1 ) + 5k
IC 

V CE = 10 – 1.2k ( I C + I B ) = 10 – 1.2k  I C + ------ = 1.18V
β F

Since V CE > V BE > 0 , our assumption is correct.
Lecture 28
28 - 3
A PNP Transistor Example
Find the Q-point of the shown circuit
since the emitter is connected to a higher potential than the
10V
base we have V
EB = 0.7V
The emitter current is found through the loop
10 = 0.7 + 2I E
to get I E = 4.65mA
Assuming forward active region we have I C = αI E = 4.6mA
2 kΩ
IE
β = 100
IB
1 kΩ
-10V
Checking the region of operation by calculating V CB we get
V CB = – 10 + I C × 1K = – 5.4V
it follows that the Collector-base junction is reverse-biased and our assumption of forward active region is correct
Lecture 28
28 - 4