Learning Objectives:
- What is voltage divider bias?
- How to select Q point according to the application?
- Centred Q point
- Q point near saturation point near the cutoff
- Maximum possible peak to peak voltage
This topic helps you to design a voltage divider biased amplifier.
Analysis of a voltage divider circuit:
Of course, you are familiar with the voltage divider circuit and voltage divider rule.
The potential divider bias is the most commonly used biasing technique.
Look at figure 1, R1, R2 and VCC form a voltage divider.
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| Fig 1 Voltage divider biased transistor |
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| Fig 2 DC equivalent |
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| Fig 3 AC equivalent |
Step 1: Calculate the base current IB and base voltage VB
The first step is a little bit difficult to understand as a beginner. But once you have fully understood the concept, it becomes really easy. I will discuss three different methods of calculating IB and VB.
First Method: Approximate Method
Look at fig 4, the current through R1 is equal to the current through R2. This only happens when the base current IB is very small. We assume IB is very small as compared to I (see fig 4) IB is so small and we can neglect it. And hence current flowing through R1 is equal to the current through R2. It is 20 times greater than IB (design rule).
IB = I/20
In this case voltage across the base is approximately equal to the voltage across R2 (see fig 3). From Ohm's law,
VB = I*R2
I = VCC/(R1+R2)
IB = I/20
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| Fig 4 |
Second Method: Approximate Method
RB = VB/IB ... equation 1
See fig 5, voltage across base is equal to voltage across RE
VB = IERE
IC = IE = β*IB
Substitute VB in equation 1
RB = IERE/IB
RB = β*IBRE/IB
RB = β*RE
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| Fig 5 |
Third Method | Exact Method | Thevenin's Equivalent Method
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| Fig 6 |
The exact method involves Thevenin's equivalent circuit. Replace R1 and R2 with its Thevenin equivalent circuit.
Without going into the details of Thevenin's equivalent method, consider this equivalent circuit shown in fig 6, and equation 1 and equation 2.
Rth = R1*R2/(R1+R2) eq 1
Eth = VR2 = R2*VCC/(R1+R2) eq 2
You can find VB and IB with the help of the above-mentioned methods.
Step 2:
Calculate emitter voltage with the help of base voltage VB.
VBE = VB - VE
Step 3:
After calculating emitter voltage VE, you can find emitter current IE and hence collector current IC.
IE ~ IC
IE = VE/RE
Step 4:
Now, you can calculate VC with the help of KVL at the output circuit.
VC = VCC - ICRC
Step 5:
You can find out the collector to emitter voltage VCE.
VCE = VC - VE
Step 6:
Now think about the load line. Load line is drawn over the characteristics curves. Apply KVL at the output circuit.
-VCC + ICRC + VCE + IERE = 0
IE ~ IC
-VCC + ICRC + VCE + ICRE = 0
-VCC + VCE + IC(RC + RE) = 0 equation 3
equation 3 is the equation of the load line.
Substitute IC = 0, for x-intercept
VCC = VCE … equation 4
Substitute VCE = 0, for y-intercept
IC = VCC/(RC + RE) ... equation 5
Q Point Variations:
Now you want to get the Q point on different locations of the load line. You already did this in base bias analysis. Like Q point near saturation or Q point in an active region. By assigning different values to RE we can get the Q point on the desired location. I have solved the circuit in fig 1 for three different values of RE.
Example 1: RE = 500
Find Q point (VCEQ, ICQ)
Do all the steps from 1 to 5.
Step 1:
You can choose any method for solving step 1. Let's try the first approximate method.
In this case, we need to assume current through R1 is equal to the current through R2 and the base current is very small and can be neglected.
I = VCC/(R1+R2)
I = 10/(10k+5k)
I = 0.66mA
VB = I*R2
VB = 3.3V
Step 2:
VBE = VB - VE
VE = 3.3 - 0.7
VE = 2.6V
Step 3:
IE = VE/RE
IE = 2.6/500
IE = 5.2mA
IE ~ ICQ ~ 5.2mA
Step 4:
VC = VCC - ICRC
VC = 10 - 5.2m*1k
VC = 4.8V
Step 5:
VCEQ = VC - VE
VCEQ = 4.8 - 2.6
VCEQ = 2.2V
Find DC load line
Step 6:
KVL at the output circuit
-VCC + VCE + IC(RC + RE) = 0 ..1.1
Substitute VCE = 0 in 1.1
IC = VCC/(RC + RE)
IC = 10/(1000+500)
IC = 6.7mA
Substitute IC = 0 in 1.1
VCE = VCC = 10V
Find AC load line
Step 7:
vce(cut) = VCEQ + ICQ*rC
rC = RC || RL
vce(cut) = 2.2 + 5.2m*500
vce(cut) = 2.2 + 2.6
vce(cut) = 4.8V
ic(sat) = ICQ + VCEQ/rC
ic(sat) = 5.2m + 2.2/500
ic(sat) = 9.6mA
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| Fig 7 |
Example 2: RE = 750
Find Q point (VCEQ, ICQ)
Do all the steps from 1 to 5.
Step 1:
You can choose any method for solving step 1. Let's try the first approximate method.
In this case, we need to assume current through R1 is equal to the current through R2 and the base current is very small and can be neglected.
I = VCC/(R1+R2)
I = 10/(10k+5k)
I = 0.66mA
VB = I*R2
VB = 3.3V
Step 2:
VBE = VB - VE
VE = 3.3 - 0.7
VE = 2.6V
Step 3:
IE = VE/RE
IE = 2.6/750
IE = 3.4mA
IE ~ ICQ ~ 3.4mA
Step 4:
VC = VCC - ICRC
VC = 10 - 3.4m*1k
VC = 6.6V
Step 5:
VCE = VC - VE
VCE = 6.6 - 2.6
VCEQ = 4V
Find DC load line
Step 6:
KVL at the output circuit
-VCC + VCE + IC(RC + RE) = 0 ..2.1
Substitute VCE = 0 in 2.1
IC = VCC/(RC + RE)
IC = 10/(1000+750)
IC = 5.7mA
Substitute IC = 0 in 2.1
VCE = VCC = 10V
Find AC load line
Step 7:
vce(cut) = VCEQ + ICQ*rC
rC = RC || RL
vce(cut) = 4 + 3.4m*500
vce(cut) = 4 + 1.7
vce(cut) = 5.7V
ic(sat) = ICQ + VCEQ/rC
ic(sat) = 3.4m + 4/500
ic(sat) = 11.4mA
Results:
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| Fig 8 |
Example 3: RE = 536
Find Q point (VCEQ, ICQ)
Do all the steps from 1 to 5.
Step 1:
I = VCC/(R1+R2)
I = 10/(10k+5k)
I = 0.66mA
VB = I*R2
VB = 3.3V
Step 2:
VBE = VB - VE
VE = 3.3 - 0.7
VE = 2.6V
Step 3:
IE = VE/RE
IE = 2.6/536
IE = 4.8mA
IE ~ ICQ ~ 4.8mA
Step 4:
VC = VCC - ICRC
VC = 10 - 4.8m*1k
VC = 5.2V
Step 5:
VCE = VC - VE
VCE = 5.2 - 2.6
VCEQ = 2.6V
Find DC load line
Step 6:
KVL at the output circuit
-VCC + VCE + IC(RC + RE) = 0 3.1
Substitute VCE = 0 in 3.1
IC = VCC/(RC + RE)
IC = 10/(1000+536)
IC = 6.5mA
Substitute IC = 0 in 3.1
VCE = VCC = 10V
Find AC load line
Step 7:
vce(cut) = VCEQ + ICQ*rC
rC = RC || RL
vce(cut) = 2.6 + 4.8m*500
vce(cut) = 2.6 + 2.4
vce(cut) = 5V
ic(sat) = ICQ + VCEQ/rC
ic(sat) = 4.8m + 2.6/500
ic(sat) = 10mA
Results:
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| Fig 9 |
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