Chapter 7.
DC Biasing Circuit
7.1 DC load line
7.2 Base Bias
7.3 Voltage-Divider Bias
DC biasing circuit
Typical amplifier operation
As IB is varied from an initial value,
IC and VCE are amplified.
DC Load Line
Typical amplifier operation
As IB is varied from an initial value,
IC and VCE are amplified.
VCE VCC VRC VCC I C RC
VCC VCE
IC
RC
I C 0 when VCE VCC ~ " Cut - off "
VCC
Maximum I C when VCE 0 ~ " Saturation "
RC
A graph of all possible combination of VCE and IC for
an amplifier.
VCE ,max VCC
Note that RC
I C ,max VCC
RC
Example 7.2
Q. Plot the dc load line for the circuit shown in the figure. Then,
find the load line VCE values for IC = 1 mA, 2 mA, and 5 mA
The Q Point
“Quiescent” ~ quiet and inactive.
A point on the dc load line that intersect with the IC vs. VCE characteristic curve ~ valid
for the dc input case
Optimum amplifier operation is expected
at the Q-point for a midpoint-biased circuit
(중간점 바이어스)
Q. What if not the midpoint bias?
Circuit Analysis of the Base Bias Circuit
Starting from the calculation of I B
VRB VCC VBE VCC 0.7(V)
IB
RB RB RB
I C hFE I B (where hFE DC )
VCE VCC I C RC
We can find the Q-point values of IC and VCE
Example 7.3
Determine the Q-point values…
VCC VBE 8 0.7
IB 20.3 A
RB 360k
I C h FE I B (100)(20.3 A) 2.03mA
VCE VCC I C RC 8V 2.03mA 2k 3.94V
Q-point Shift
Base-bias circuits are extremely susceptible to a problem called Q-point shift
Q-point shift: A condition where a change in operating temperature
results in a change in IC and VCE.
Q-point Shift
Temperature increases
→ hFE (or ) increases
→ IC increases
→ VCE decreases ~ too much sensitive to
Voltage-Divider Bias Circuit
• The most commonly used biasing scheme
• A biasing circuit that contains a voltage
divider in its base circuit
• -independent circuit: ICQ & VCEQ are highly
stable against the change in dc (or hFE)
R1 and R2 : voltage divider
R2
VB VCC
R1 R2
VE VB VBE VB 0.7(V)
VE
IE I CQ (assume)
RE
VCEQ VCC I CQ ( RC RE )
IE I CQ
IB
hFE 1 hFE
R2
CC
V VBE
VE VB VBE
2
R R
I CQ 1
RE RE RE
Example 7.7
Determine the value of ICQ & VCEQ.
* Note the voltage instead the current!
With assuming
IE I CQ
IB
hFE 1 hFE
Base Input Resistance or Transistor Loading
In the equivalent circuit of the base-emitter junction, there occurs an effective increase of RE being Rbase ,
where Rbase hFE RE , corresponding to the amplification of IB being IE ~ IC = hFEIB
When Rbase hFE RE 10 R2 , I 2 10 I B ~ Transistor loading effect is neglected and I1 I 2
When Rbase hFE RE 10 R2 , I B is not so small. We apply Thevenin's equivalent circuit model
for the precise circuit analysis
Thevenin’s Equivalent Circuit for the Base Input
R2
VTH VB VCC
R1 R2
Apply the voltage divider rule with assuming
an infinite resistance R replacing the transistor
1
1 1
RTH R1 || R2
R1 R2
Make a short circuit replacing the battery of VCC
and find a total resistance from the base to the ground
[Proof of Eq. 7-15]
1
VTH I B RTH VBE I E RE & VCC I C RC VCE I E RE & I E I B I C (1 hFE ) I B 1 IC
hFE
h h 1 hFE 1 hFE 1 I
I C FE I E FE TH B TH BE
V I R V TH BE
V V
C
RTH
hFE 1 FE
h 1 E
R FE
h 1 E
R h
FE 1 R
E FEh
hFE 1
VTH VBE
hFE 1 VTH VBE V VBE
VBE I C FE E
RTH h 1 R
1
C
I VTH TH
hFE 1 RE FE
h 1 E
R RTH hFE 1 RE RTH R RTH
1
E
hFE 1 RE
hFE hFE
hFE
Example 7.9
Determine the value of ICQ & VCEQ.
R2 10k
VTH VCC =(20V) =2.56V
R1 R2 78k
RTH R1 || R2 (10k ) || (68k ) 8.72k
VTH VBE 2.56V 0.7V 1.86V
I CQ 1.46mA
RTH 8.72k 1.27k
RE 1.1k
hFE 50
VCEQ VCC I CQ ( RC RE ) 20V (1.46mA)(7.3k ) 9.34V
When we neglect the transistor loading effect,
R2 10k
VB VCC (20V ) 2.56V
R1 R2 78k
VE VB VBE 2.56V 0.7V 1.86V
VE 1.86V
IE 1.69mA I CQ
RE 1.1k
VCEQ VCC I CQ ( RC RE ) 20V (1.69mA)(7.3k ) 7.66V
Load Line for Voltage-Divider Bias Circuit
When TR is saturated, VCE 0
VCC
I C ( sat )
RC RE
When TR is in cut-off,
VCE ( off ) VCC
