Biasing Techniques for CE Amplifiers
Fixed Bias or Base Bias:
- In order for a transistor to amplify, it has to be properly biased.
- This means forward biasing the base emitter junction and reverse biasing collector base junction.
- For linear amplification, the transistor should operate in active region ( If IE increases, IC increases, VCE decreases proportionally).
- The source VBB, through a current limit resistor RB forward biases the emitter diode and VCC through resistor RC (load resistance) reverse biases the collector junction as shown in fig.
- The dc base current through RB is given by
IB = (VBB – VBE) / RB
or VBE = VBB – IB RB
- Normally VBE is taken 0.7V or 0.3V.
- If exact voltage is required, then the input characteristic ( IB vs VBE) of the transistor should be used to solve the above equation.
- The load line for the input circuit is drawn on input characteristic.
- The two points of the load line can be obtained as given below:
For IB = 0, VBE = VBB.
and For VBE = 0, IB = VBB/ RB.
- The intersection of this line with input characteristic gives the operating point Q as shown in fig.
- If an ac signal is connected to the base of the transistor, then variation in VBE is about Q – point. This gives variation in IB and hence IC.
- In the output circuit, the load equation can be written as
VCE = VCC– IC RC
- This equation involves two unknown VCE and IC and therefore can not be solved. To solve this equation output characteristic ( IC vs VCE) is used.
- The load equation is the equation of a straight line and given by two points:
IC= 0, VCE = VCC
& VCE = 0, IC= VCC / RC
- The intersection of this line which is also called dc load line and the characteristic gives the operating point Q as shown in fig.
- The point at which the load line intersects with IB = 0 characteristic is known as cut off point.
- At this point base current is zero and collector current is almost negligibly small.
- At cut off the emitter diode comes out of forward bias and normal transistor action is lost. To a close approximation.
VCE ( cut off)= VCC (approximately).
- The intersection of the load line and IB = IB(max) characteristic is known as saturation point .
- At this point IB= IB(max), IC= IC(sat).
- At this point collector diodes comes out of reverse bias and again transistor action is lost.
- To a close approximation,
IC(sat) = VCC / RC(approximately ).
- The IB(sat) is the minimum current required to operate the transistor in saturation region.
- If the IB is less than IB (sat), the transistor will operate in active region. If IB > IB (sat) it always operates in saturation region.
If the transistor operates at saturation or cut off points and no where else then it is operating as a switch is shown in fig.
VBB = IB RB+ VBE
IB = (VBB – VBE ) / RB
- If IB> IB(sat), then it operates at saturation, If IB = 0, then it operates at cut off.
- If a transistor is operating as an amplifier then Q point must be selected carefully.
- Although we can select the operating point any where in the active region by choosing different values of RB & RC but the various transistor ratings such as maximum collector dissipation PC(max) maximum collector voltage VC(max) and IC(max) & VBE(max) limit the operating range.
- Once the Q point is established an ac input is connected.
- Due to this the ac source the base current varies.
- As a result of this collector current and collector voltage also varies and the amplified output is obtained.
- If the Q-point is not selected properly then the output waveform will not be exactly the input waveform. i.e. It may be clipped from one side or both sides or it may be distorted one.
Example-1
Find the transistor current in the circuit shown in fig., if ICO= 20nA, Beta =100.
Stability of Operating Point
Let us consider three operating points of transistor operating in common emitter amplifier.
Near cut off
Near saturation
In the middle of active region
- If the operating point is selected near the cutoff region, the output is clipped in negative half cycle as shown in fig.
- If the operating point is selected near saturation region, then the output is clipped in positive cycle as shown in fig.
- If the operating point is selected in the middle of active region, then there is no clipping and the output follows input faithfully as shown in fig.
- In biasing circuit shown in fig.(a), two different power supplies are required.
- To avoid the use of two supplies the base resistance RB is connected to VCC as shown in fig.(b).
- Now VCC is still forward biasing emitter diode.
- In this circuit Q point is very unstable.
- The base resistance RB is selected by noting the required base current IB for operating point Q.
IB = (VCC – VBE ) / RB
- Voltage across base emitter junction is approximately 0.7 V. Since VCC is usually very high
i.e. IB = VCC/ RB
- Since IB is constant therefore it is called fixed bias circuit.
Stability of quiescent operating point:
- Let us assume that the transistor is replaced by an other transistor of same type.
- The Betadc of the two transistors of same type may not be same.
- Therefore, if βdc increases then for same IB, output characteristic shifts upward.
- If βdc decreases, the output characteristic shifts downward.
- Since IB is maintained constant, therefore the operating point shifts from Q to Q1 as shown in fig.
- The new operating point may be completely unsatisfactory.
- Therefore, to maintain operating point stable, IB should be allowed to change so as to maintain VCE and IC constant as βdc changes.
- A second cause for bias instability is a variation in temperature.
- The reverse saturation current changes with temperature.
- Specifically, ICO doubles for every 100C rise in temperature.
- The collector current IC causes the collector junction temperature to rise, which in turn increases ICO.
- As a result of this growth ICO, IC will increase ( βdc IB + (1+ βdc ) ICO) and so on.
- It may be possible that this process goes on and the ratings of the transistors are exceeded.
- This increase in IC changes the characteristic and hence the operating point.
Emitter Feedback Bias
- Fig. , shows the emitter feedback bias circuit.
- In this circuit, the voltage across resistor RE is used to offset the changes in βdc.
- If βdc increases, the collector current increases.
- This increases the emitter voltage which decrease the voltage across base resistor and reduces base current.
- The reduced base current result in less collector current, which partially offsets the original increase in βdc.
- The feedback term is used because output current (IC) produces a change in input current (IB).
- RE is common in input and output circuits.
Collector Feedback Bias:
- In this case, the base resistor is returned back to collector as shown in fig.
- If temperature increases. βdc increases.
- This produces more collectors current.
- As IC increases, collector emitter voltage decreases.
- It means less voltage across RB and causes a decrease in base current this decreasing IC, and compensating the effect of βdc.
Voltage Divider Bias:
- If the load resistance RC is very small, e.g. in a transformer coupled circuit, then there is no improvement in stabilization in the collector to base bias circuit over fixed bias circuit.
- A circuit which can be used even if there is no dc resistance in series with the collector, is the voltage divider bias or self bias.
- The current in the resistance RE in the emitter lead causes a voltage drop which is in the direction to reverse bias the emitter junction.
- Since this junction must be forward biased, the base voltage is obtained from the supply through R1, R2 network.
- If Rb = R1 parallel to R2 , equivalent resistance is very very small, then VBE voltage is independent of ICO and stability factor becomes 0.
- For best stability R1 and R2 must be kept small.
- If IC tends to increase, because of ICO, then the current in RC increases, hence base current is decreased because of more reverse biasing and it reduces IC .
- The smaller the value of Rb, the better is the stabilization but S cannot be reduced be unity.
- Hence IC always increases more than ICO.
- If Rb is reduced, then current drawn from the supply increases.
- Also if RE is increased then to operate at same Q-point, the magnitude of VCC must be increased.
- In both the cases the power loss increased and reduced efficiency.
- In order to avoid the loss of ac signal because of the feedback caused by RE, this resistance is often by passed by a large capacitance (> 10 m F) so that its reactance at the frequency under consideration is very small.
Emitter Bias:
- Fig. shows the emitter bias circuit.
- The circuit gets this name because the negative supply VEE is used to forward bias the emitter junction through resistor RE.
- VCC still reverse biases collector junction.
- This also gives the same stability as voltage divider circuit but it is used only if split supply is available.
Example-1
Determine the Q-point for the CE amplifier given in fig. 1, if R1 = 1.5Kohm and Rs = 7Kohm . A 2N3904 transistor is used with Beta = 180, RE = 100ohm and RC = Rload = 1Kohm . Also determine the Pout(ac) and the dc power delivered to the circuit by the source.