Biasing Circuits and Transistor Amplifier Operation
Base Bias Circuit
As was previously seen the circuit below turns a transistor on. The circuit now can be referred to as
Base Bias.
Operation
Initially with no base current flowing the output or collector voltage VOUT = VC = VCC the supply voltage.
The resistor RB is connected from VCC to the base of the transistor. Base current will flow in the
transistor causing the transistor to turn on. The base-emitter junction in the transistor is just like a
diode.
Transistor On
Base-Emitter voltage VBE = 0.7 V
The base current causes a collector current to flow IC = β x IB
Current flow through the collector resistor RC will cause a voltage drop on this resistor of VRC and the
VOUT = VC will start to decrease from the value of VCC the supply voltage. This follows from KVL.
VCC - VRC - VC = 0
so VC = VCC - VRC
Q Point
This term is used to describe the values of IC and VCE for a biased-on transistor and they are referred to
as ICQ and VCEQ.
Q Point Values
ICQ and VCEQ
Example
Calculate the values of IB, VBE, IC and VC.
Β = 150
IB = (12 V - 0.7 V)/ 270 kΩ = 42 µA
IC = β x IB = 150 x 42 µA = 6.28 mA
VRC = IC x RC = 6.28 mA x 1 kΩ = 6.28 V
VC = 12 V – 6.28 V = 5.72 V
The Q Point values for this circuit are
ICQ = 6.28 mA and VCEQ = 5.72 V
The analysis and example above show clearly that the Q Point values are dependent on the value of β.
Recall from previous notes that the value of β is highly variable.
The result is that if we were to construct a large number of the above Base Bias circuits (in
manufacturing) the Q Point values ICQ and VCEQ for the circuit would vary widely.
Transistor Amplifier Basics
What does an amplifier do?
An amplifier takes a very small AC input or signal voltage and makes it much larger at the output.
Consider the case where you plug your iPOD into an amplifier so you can play its music through external
speakers at a party. The audio output signal from the iPOD may only be a few hundred mV, not nearly
enough to drive a pair of speakers. The amplifiers job is raise the few hundred mV to 10 or 20 V or more
without changing the shape of the waveform in any way. Most audio amplifiers are integrated circuit
types these days, but we will look at a basic transistor amplifier (which is what you have inside an
Integrated Circuit).
Amplifier Distortion
The other important job that an amplifier is concerned with (other than making the output signal larger
than the input signal) is to be sure that the input and output signals look the same – that there is no
distortion in the output signal. Distortion in the output signal – if it is music - can cause the music to
sound badly and not as the artist intended it to.
No Bias Circuit Operation
Let’s examine the case where the transistor is not turned on by a Bias circuit and then apply an AC
waveform at the input. Notice that there is no bias resistor connected to a DC voltage source so the DC
base current IB will be 0, the DC collector current IC will be 0, the DC output voltage VCE will be VCC.
What happens if an AC signal voltage is applied at VIN as shown? The positive half cycle of the input
voltage will cause base current to flow in RB and the Base-Emitter junction. This will cause collector
current flow and the collector-emitter voltage will decrease from a value of VCC. On the negative half
cycle of the input voltage the transistor already being in the Off state will not turn on and the collectoremitter voltage will stay at a value of VCC.
Input and Output Amplifier Waveforms
The most significant observation from these waveforms is that the Input and Output waveforms do not
look the same – the output is distorted because the transistor has not been turned on by a Bias circuit.
Amplifier Operation with a Bias Circuit
Consider the circuit below that uses base bias and has an AC signal of 200 mV P-P connected at VIN.
β = 100
Here is a very basic transistor amplifier using a Base Bias circuit that we discussed earlier. It has some
serious limitations as a practical amplifier but will serve to illustrate how the transistor amplifies a small
input signal.
Recall that the base current IB was determined as:
IB = (VCC - VBE)/RB = (10 V - 0.7 V)/186 kΩ = 50 µA
And
IC = β x IB = 100 x 50 µa = 5.0 mA
And
Vout = VC = VCE = VCC – IC x RC = 10 V – (5 mA)(1 kΩ) = 5V
And
VBE = 0.7 V
The Q Point values for this circuit are
ICQ = 5.0 mA and VCEQ = 5.0 V
These values are the DC Bias conditions for the circuit. Biasing is required so that the transistor can
react to small input voltages which are not large enough to turn on the transistor (recall that the
transistor starts to conduct at approximately 0.5 V - the voltage at the base emitter - and the normal
base emitter operating voltage is about 0.7 V).
An approximation of the IB vs VBE response curve looks something like this for the transistor input
IB
50µa
0.7V
VBE
If a 200 mV P-P sine wave is applied to the base emitter junction and when the sine wave is at its
positive peak of 100 mV
VBE will now be 0.8 V (0.7 V + 0.1 V)
The input curve shows that base current will rise. Let us assume it rises to 80 µA. The collector current IC
will rise to β x IB = 100 x 80 µA = 8 mA.
The output voltage VCE will decrease to VCE = VCC – (IC x RC) = 10V – (8 mA)(1kΩ) = 2V
By a similar procedure, when VIN goes to -100 mV, VBE drops to 0.6 V so that IB falls to about 20 µA.
Therefore IC will decrease to 2 mA and VCE will increase to 8 V.
To put this together we have for the AC voltages and currents:
VIN,P-P = 200 mV
IB,P-P = 60 µA
IC,P-P = 6 mA
VCE,P-P = 6 V
Amplifier Waveforms
Current, Voltage and Power Gain
Gain represents how much bigger (or smaller) and output quantity is than the corresponding input
quantity. Gains can be greater than 1 or less than 1 (a loss).
Gain is represented by the letter A and is subscripted according to type – current, voltage or power
The AC Current gain βAC (also called AI) is found as
AI = βAC = IC/IB = 6 mA/60 µA = 100
The Voltage Gain AV is found as
AV = VCE,P-P/VIN,P-P = 6 V/200 mV = 30
The Power Gain AP = Current Gain x Voltage Gain
AP = 100 x 30 = 3000
There are a few things you should notice about this basic amplifier:
•
•
The output voltage is 180° out of phase with the input voltage (not a problem)
The input and output waveforms have the same shape – no distortion. This is a good thing
•
•
•
•
The transistor is a current controlled device, we apply an input voltage, but it is the resulting
change in base current that causes the collector current to change. The output voltage is
changed as a result of the changed collector current flowing through collector resistor RC
The output voltage is a mixture of AC and DC since the average VCE is the bias voltage of 5V.
VOUT is a 6 V P-P sine wave added to 5 V DC. The output sine wave changes from 2 V to 8 V, not
from -3 V to +3 V (this IS a problem)
What the transistor is actually doing is controlling the flow of current from the DC supply as the
input signal varies.
The maximum peak to peak output voltage possible is 10 V P-P (or in other words it is set by
VCC). We set the bias point at 5 V in the middle of the output range so that the transistor output
can swing an equal amount in both directions (from 5V up to 10V, and from 5V down to 0V)
Importance of the Location of Q Point on DC Load Line
Recall from before the drawing of the DC load line. This graph of VCE vs IC shows the Cut-Off and
Saturation points.
VCE
Cut-Off (VCE=VCC, IC=0)
Q Point for smaller β
Q Point (centered)
Q Point for larger β
IC
Saturation (VCE ≈ 0, Ic,Max)
Also shown on the graph is the location of the Q Point for a bias circuit which represents a particular set
of VCEQ and ICq values. This location of the Q Point is not unique – if the value of β changes the Q Point
will move up or down the load line.
The ideal location for the Q Point is centered on the DC load line. When an AC signal is applied to the
bias circuit - to use the circuit as an amplifier - the instantaneous value of VOUT (VCE) moves up and down
the DC load line from the location of the Q Point. The output voltage is a symmetrical signal (to avoid
clipping distortion) and must move equally up or down the load line. A maximum output voltage will be
achieved when the position of the Q Point is centered on the load line.
Base Bias Dependence on β
The bias circuit we used is called a fixed Base Bias circuit because the base current is
IB = (VCC - VBE)/RB
All of these are constant so if a replacement transistor with a different value of β is plugged in the base
current is still 50 µA but the collector current IC and VCE Q point bias values will be different because of
the variability in β. If, for example a different transistor where β = 180 is used the new IC will be 180 x 50
µA = 9 mA and the new VCE will be 10 V – (9 mA)(1 kΩ) = 1V. This means that VOUT can only fall 1 V
and because we must have symmetry in output for most waveforms, we are limited to 1V P-P as our
maximum output. We need a bias circuit that is not affected by variations in β.
Temperature Sensitivity
The other big problem is that transistors are quite sensitive to variations in temperature and if
temperature rises, for example, the collector current will increase as more electrons in the transistor
structure will pick up enough energy to break away from atoms and add to the current flow. The effect
is the same as an increase in β and the bias values will drift upward. A bias circuit is needed which is
immune to temperature change as well.
Note: This discussion about Q Point, the DC load line and the maximum output voltage is not taking into
account the concept of an AC load line.