Electronics 2 Laboratory
Lab 1: BJT Transistor Bias Circuits
Remember from previous labs and lectures that the transistor has three regions of operation,
1. Active or linear region.
2. Saturation region.
3. Cut-off region.
The active or linear operation region is used for amplifier applications. In this lab, we will study an
effective method to make the transistor operate in the linear region.
DC Bias
For make the transistor operate in the linear region the DC voltages and current in the amplifier must be
set properly. The operating point of the transistor is called the Q-point (quiescent point).
The signal to be amplified will disrupt the Q-point enough from the input side which will cause a
similar but amplified disruption at the output. This is depicted in figure 1.
Figure 1
In figure 1 the Q-point is located at the center of the characteristic curve of the transistor which gives
enough room for the output signal to swing (positive and negative swing), this leads to proper linear
amplification action.
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If the Q-point slides to the point A or B, the output signal will be distorted due to the transistor going to
the cutoff or saturation regions. This is showing in figure 2.
Figure 2
The previous discussion shows the importance of the setting the Q-point for proper amplifier operaion.
To set the Q-point you need what is knowing as biasing circuits.
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Direct Biasing Circuit
The first method that comes to mind is to apply the biasing voltage directly by means of a voltage to
the base to get the current IB and hence get a current IC = β×IB as a result which in turn will set the
voltage VCE therefore setting the Q-point.
There is one important step in this approach which is to find the value of β for your particular
transistor.
Step 1: Find the value of β for the given transistor. Refer to the datasheet. Record your observations.
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Step 2: Construct the circuit in figure 3, select the values of RB and RC so that IB will cause a current IC
which will set the value of the VCE to be in the middle of the VCC, to allow for maximum voltage swing
at the output (for both the positive and negative portions of the signal) as seen in figure 1.
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Figure 3
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Step 3: Change the transistor with another transistor of the same part number record the values of the
Q-point for each transistor. Do it three times and fill the table below.
Q-point Transistor 1 Transistor 2 Transistor 3
IC
VCE
Write down your observations on the direct method of biasing transistors.
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Voltage-Divider Biasing Circuit
Voltage-divider biasing is the most widely used method of biasing, in this method only one voltage
source is used to set both IB and IC, a voltage divider network is used to set the value of VB hence get
the required IB. Figure 4 shows the general structure of this circuit.
Figure 4
One obvious advantage of this method is the use of a single source instead of two source in the direct
method. The calculations for R1, R2, RC and RE will be more complicated due of the transistor drawing
current from the voltage divider.
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If however the voltage divider (that is the voltage at point A) is unaffected by the transistor (IB is very
small compared to the current I2) then the calculations will be much simpler. Why?
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Following the steps below to design a biasing circuit that will make the voltage VC in the middle of VCC
to allow for maximum swing as seen in figure 1, for now assume any value for IC.
V CC
1. Step 1: Use the equation VC = VCC – IC×RC To calculate RC, here we want VC = = 5 V,
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select IC = 2 mA (This value is assumed),
Calculate the value of RC = ……….
2. Now we can assume a value for VE to make the amplifier stable against temperature effects,
assume a value of VE = 1 V and remember that I E≃I C using this information calculate the
value of RE.
RE = …………...
3. Now select the values of R1 and R2 to make VB = VE + VBE, Remember that for the voltage
divider to be stiff (i.e to be stable regardless of the current IB) the current I2 must be
R1 R 2 1
I 2 ≥10×I B for this we use the equation ⩽ R , where Ri(base), DC = β R E
R 1+ R 2 10 i (base), DC
you can make precise calculation, however this might be not be necessary since you can assume
that R2 low enough to make sure that I 2 ≥10×I B . Then use voltage divider formula to
calculate R1.
R2 = ………… R1 = ………….
4. Now that we calculated the RC, RE, R1 and R2 construct the circuit of figure 4, and measure the
values of VC and IC.
IC = ………. and VC = ………….
Those values represent the Q-point.
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5. Change the transistor with another transistor of the same part number record the values of the
Q-point for each transistor. Do it three times and fill the table below.
Q-point Transistor 1 Transistor 2 Transistor 3
IC
VCE
Write down your observations.
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