Collector Feedback Bias - AC | DC Load Line & Q Point Calculations
Learning Objectives:
- What is collector feedback bias?
- Evaluate Q point, AC load line, DC load line
- How to select Q point according to the application?
- Centred Q point
- Q point near saturation
- Q point near the cutoff
- Maximum possible peak to peak voltage
This topic helps you to design collector feedback bias circuits.
This is another biasing technique of BJT based circuits. In this technique, a resistor is connected in between base and collector terminals. The collector terminal provides biasing voltage for the base emitter terminal.
There are a similar number of components required as it is in base bias. But it is far more stable than the base bias circuit. The circuit has better stability that is thermal or bias stability.
The important aspect of this bias is that it is designed to work only in the active region.
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| Fig 1: A Collector feedback bias |
DC Load Line:
KVL at the output circuit.
-VCC + ICRC + IBRC + VCE = 0 ..eq 1
IB = IC/β
Substitute IB, equation 1 becomes
-VCC + ICRC + (IC/β)*RC + VCE = 0
-VCC + ICRC (1+ 1/β) + VCE = 0..eq2
Substitute VCE = 0 in ..eq2
VCC = ICRC(1+1/β)
IC = VCC/RC(1+1/β)
With the help of approximation 1+1/β = 1
IC = VCC/RC ..eq3
Substitute IC = 0 in ..eq2
VCE = VCC ..eq4
Q Point (VCEQ , ICQ):
Apply KVL to the input circuit.
-VCC + ICRC + IBRB + VBE = 0 ..eq5
IB = IC/β
Substitute IB and solve for IC
-VCC + ICRC + ICRB/β+ VBE = 0
-VCC + IC(RC + RB/β)+ VBE = 0
ICQ = (VCC - VBE) / (RC + RB/β)..eq6
KVL at output circuit.
-VCC + ICRC + IBRC + VCE = 0 ..eq7
IB = IC/β
Substitute IB and solve for VCE
VCE = VCC - ICRC - ICRC/β
VCE = VCC - ICRC(1+ 1/β)
1+1/β ~ 1
VCEQ = VCC - ICRC..eq8
AC Load Line:
I am not going into the details of AC load line.
iC(sat) = ICQ + VCEQ/rC..eq9
vce(cut) = VCEQ + ICQ*rC..eq10
Example 1: RB = 40k
Step 1: DC Load Line
From eq3 and eq4
VCE = 10V
IC = 10mA
Step 2: Q Point (VCEQ , ICQ)
From eq6
ICQ = (VCC - VBE) / (RC + RB/β)
ICQ = (10 - 0.7) / (1k + 40k/100)
ICQ = 6.6mA
From eq8
VCEQ = VCC - ICRC
VCEQ = 10 - 6.6m*1k
VCEQ = 3.4V
Step 3: AC Load Line
From eq9
iC(sat) = ICQ + VCEQ/rC
iC(sat) = 6.6m + 3.4/500
iC(sat) = 6.6m + 6.8m
iC(sat) = 13.4mA
From eq10
vce(cut) = VCEQ + ICQ*rC
vce(cut) = 3.4 + 6.6m*500
vce(cut) = 6.7V
Step 4: Maximum Peak To Peak Voltage:
We will check for both compliances. Amplifier compliance will be the smaller value.
PP =2* VCEQ = 6.8V
PP = 2*ICQ*rC = 6.6V
Since the Q point is almost in the middle of the AC load line. Hence both compliances are almost the same. It is also visible in the figure below
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| Fig 2: Example 1 |
Example 2: RB = 100k
Step 1: DC Load Line
From eq3 and eq4
VCE = 10V
IC = 10mA
Step 2: Q Point (VCEQ , ICQ)
From eq6
ICQ = (VCC - VBE) / (RC + RB/β)
ICQ = (10 - 0.7) / (1k + 100k/100)
ICQ = 4.65mA
From eq8
VCEQ = VCC - ICRC
VCEQ = 10 - 4.65m*1k
VCEQ = 5.35V
Step 3: AC Load Line
From eq9
iC(sat) = ICQ + VCEQ/rC
iC(sat) = 4.65m + 5.35/500
iC(sat) = 4.65m + 10.7m
iC(sat) = 15.35mA
From eq10
vce(cut) = VCEQ + ICQ*rC
vce(cut) = 5.35 + 4.65m*500
vce(cut) = 7.6V
Step 4: Maximum Peak To Peak Voltage:
We will check for both compliances. Amplifier compliance will be the smaller value.
PP =2* VCEQ = 10.7V
PP = 2*ICQ*rC = 4.65V
Output compliance is the smaller value and hence it is 4.65V. Or you can say maximum peak to peak voltage swing should not be greater than 4.65V.
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| Fig 3: Example 2 |
Example 3: RB = 150k
Step 1: DC Load Line
From eq3 and eq4
VCE = 10V
IC = 10mA
Step 2: Q Point (VCEQ , ICQ)
From eq6
ICQ = (VCC - VBE) / (RC + RB/β)
ICQ = (10 - 0.7) / (1k + 150k/100)
ICQ = 3.72mA
From eq8
VCEQ = VCC - ICRC
VCEQ = 10 - 3.72m*1k
VCEQ = 6.28V
Step 3: AC Load Line
From eq9
iC(sat) = ICQ + VCEQ/rC
iC(sat) = 3.72m + 6.28/500
iC(sat) = 3.72m + 12.56m
iC(sat) = 16.28mA
From eq10
vce(cut) = VCEQ + ICQ*rC
vce(cut) = 6.28 + 3.72m*500
vce(cut) = 8.14V
Step 4: Maximum Peak To Peak Voltage:
We will check for both compliances. Amplifier compliance will be the smaller value.
PP =2* VCEQ = 12.56V
PP = 2*ICQ*rC = 3.72V
Output compliance is the smaller value and hence it is 3.72V. Or you can say maximum peak to peak voltage swing should not be greater than 3.72V.
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| Fig 4: Example 3 |
Conclusion:
You have observed the different values of RB and Q point position on the load line. Q point remains in the active region for a wide range of RB. In collector feedback bias, as you decrease RB, the Q point shifts towards saturation but doesn't reach the saturation region.
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