1. Transistor Biasing What is meant by biasing the transistor?
Applying external dc voltages to ensure that transistor operates in the desired region
Which is the desired region?
For amplifier application, transistor should operate in active region
For switch application, it should operate in cut-off and sat.
What is meant by quiescent point (Q-point)?
The point we get by plotting the dc values of IC , IB and VCE (when ac input is zero) on the transistor characteristics
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2. Transistor Biasing Transistor characteristics showing Q-point
Q-point is in the middle of active region.
This is called “Class-A” operation. (Other classes are discussed later)
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3. Transistor Biasing Types of biasing: Fixed bias and Self bias
The value of IB is “fixed” by choosing proper value for RB
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4. Transistor Biasing Pros and Cons of Fixed bias:
Pros: 1) Simple circuit
2) Uses very few resistors
Cons: Q-point is unstable
1) If temperature increases, then β increases, and hence ICQ and VCEQ vary (effectively Q-point shifts)
2) If the transistor is replaced with another transistor having different β value, then also Q-point shifts
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5. Transistor Biasing Load Line We have:
This is an equation of straight line with points VCC/RC and VCC lying on y-axis and x-axis respectively
This line is called “Load line” because it depends on resistor RC considered as “Load” and VCC
Intersection of load line and transistor characteristic curve is the Q-point or operating point
This point is the common solution for characteristics and load line equation
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6. Transistor Biasing
Variation in load line with circuit parameters VCC, RC and RB
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7. Transistor Biasing Voltage divider bias or Self bias
Resistor RE connected between emitter and ground
Voltage-divider resistors R1 & R2 replace RB
Circuit can be analyzed in two methods:
Exact method (using Thevenin’s theorem)
Approximation method (neglecting base current)
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8. Transistor Biasing Exact method:
Input side of self-bias (Fig. a) transformed into Thevenin’s equivalent circuit (Fig. b) where, RTH is the resistance looking into the terminals A & B (Fig. c) and VTH is given by:
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10. Transistor Biasing Since β >> 1 and (β+1)RE >> RTH
Since IC is almost independent of β, Q-point is stable
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11. Any change of transistor parameter that causes IC to increase will cause IE to increase by almost the same amount. An increase in IE causes an increase in VE
VE = IE . RE
But, VB is essentially constant, so VBE reduces with an increase in VE.
VBE= VB – VE
The reduction in VBE reduces IB, which then reduces IC, thus compensating for the parameter change that tried to increase IC.
Changes in the bias value of IC automatically change the input voltage in a way that has opposite effect on IC, thus tending to restore IC to its original value. So, the bias circuit is known as self bias and emitter resistor (stabilizes bias point) is known as emitter stabilization.
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12. Advantages of voltage divider bias
Since equation for IC does not contain β, we say that IC is independent of temperature variation and transistor replacement.
Advantages of voltage divider bias
. Q-point is stable against variation in temperature and replacement of transistor.
Disadvantages of voltage divider bias
Analysis and design are complex
More circuit components required
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13. Tutorials
For a fixed bias circuit using Si transistor, RB = 500 kΩ, RC = 2 kΩ, VCC = 15 V, ICBO = 20 µA and β = 70. Find the collector current ICQ and VCEQ at Q-point. Take VBE as 0.7 V. (Ans: 3.422mA, 8.156V)
A Si transistor is biased for a constant base current. If β = 80, VCEQ = 8 V, RC = 3 kΩ and VCC = 15 V, find ICQ and the value of RB required (Ans: 2.33 mA, 493 K)
3. Repeat problem 2 if the transistor is a germanium device. (VBE=0.3V) (Ans: 2.33 mA, 507 K)
For a self bias circuit using silicon transistor, RE = 300 Ω, RC = 500 Ω, VCC = 15 V, β = 100 and . If 10R2= βRE, then determine the value of R1 to get VCEQ = VCC / (Ans: 9.85K)
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14. Tutorials
Draw the DC load line and mark the Q- point on fixed bias circuit Assume Beta DC=100 and neglect Base–Emitter voltage. (Vcc=30V,RB=1.5 Mohm, RC=5 Kohm).
Ans: VCE,max=30v,VCEQ=20v,ICQ=2mA
2. In a fixed bias cct. Find the base current required to establish VCE=6v, also find RB & IE, (VBE=0.7v, Beta DC=120, VCC=12v,RC=2.2.Kohm).
Ans: IB=22.75 uA, RB=497 Kohm
3. Determine the region in which the transistor operates. (VBE=0.2v,RB=120kohm,RC=1kohm,VCC=15v,Beta DC=120).
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15. Tutorials (IC =0.85 mA, VCE = 12.22 V)
Determine the DC bias voltage VCE and the current IC for the voltage divider configuration.
(IC =0.85 mA, VCE = V)
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16. Tutorials
5. For a self bias circuit using silicon transistor, RE = 300 Ω, RC = 500 Ω, VCC = 15 V, β = 100 and . If 10R2= βRE, then determine the value of R1 to get VCEQ = VCC / (Ans: 9.85K)
For a self bias circuit, the transistor is a Si device, RE = 200 Ω, R1 = 10R2 = 10 kΩ, RC = 2 kΩ, β = 100 and VCC = 15 V. Determine the values of ICQ and VCEQ.
7. Suppose if the transistor used in problem 6 failed, and was replaced with a new transistor with β = 75. Is the new transistor still biased for active region operation?
A self bias circuit uses Si transistor, RC = 330 Ω, RE = 100 Ω and VCC = 12 V. Estimate the values of R1 and R2 required to provide a base current of 0.3 mA, so as to locate the operating point at ICQ = 18 mA and VCEQ = 4.25 V.
For a fixed bias circuit, VCC = 12 V and RC = 4 kΩ. The Ge transistor used is characterized by β = 50, ICEO = 0 and VCE sat = 0.2 V. Find the value of RB that just results in saturation (Ans: 198.3K)
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17. Transistor Amplifier Amplifier
Device which gives larger swing in output voltage proportional to the input voltage swing
BJT basically amplifies current: Collector current equals beta times Base current
With proper circuit designs, we can get voltage amplification and power (both voltage and current) amplification
For faithful amplification (no distortion), BJT should operate in Active region throughout the input cycle (Class A)
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18. Transistor Amplifier
Without any bias, transistor is in cut-off (IC=0, VCE=0)
Biasing circuit fixes the operating point in the middle of active region required for faithful amplification
Figure shows common-emitter amplifier circuit employing fixed bias
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19. Transistor Amplifier
With reference to the fig in previous slide, as input voltage vin varies, iin varies, thus base current iB varies
This variation in base current is amplified beta times to get variation in collector current iC
Output voltage vout is VCC – iC RC
Note that if vin increases, there is proportional decrease in vout, but of greater magnitude
Similarly if vin decreases, vout increases proportionally
Thus output voltage of CE amplifier is 180o out of phase with input voltage
(Note that small letters are used to represent ac quantities)
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20. Transistor Amplifier
Voltage gain or voltage amplification factor is vout /vin which is dependent on β, RC and other physical parameters of the transistor
Figure shows input and output waveforms for the amplifier circuit shown previously
Note the dc shift in the output voltage waveform. i.e., when vin = 0, vout = VCEQ
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21. Transistor Amplifier
This animation shows the working of Common Emitter transistor circuit
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22. Transistor Amplifier
Gain of the amplifier is usually expressed in decibels
(AV)dB = 20 log10 | AV |
Usually a gain of 100 (i.e., 40 dB) can be obtained using single transistor. For higher gain requirement, two or more amplifier stages are to be cascaded
Overall gain is product of individual gains; But when expressed in dB, overall gain is sum of individual gains (in dB)
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23. Transistor Amplifier RC coupling
Fig shows CE amplifier employing self bias
Additional components are CC and CE
CC is called coupling capacitor – used to prevent dc component from entering or leaving amplifier stage
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24. Transistor Amplifier Frequency response of amplifier
CE is called emitter bypass capacitor – used to bypass the ac emitter current – preventing it from flowing through RE
If ac emitter current is allowed to pass through RE, then vBE reduces and hence output voltage reduces
Frequency response of amplifier
It’s important to know the behavior of amplifier at different frequencies
Gain is NOT constant at all frequencies – depends on various factors
Frequency response is a plot of gain versus frequency
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25. Transistor Amplifier Figure shows frequency response plot
At lower and higher frequencies, gain is less
Gain attains const value at mid frequencies
Bandwidth of amplifier is range of frequencies over which gain is not less than 3 dB of maximum gain
20 log | AVO | = 20 log | AVO | – 3
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26. Transistor Amplifier Analysis of frequency response curve
At very low frequencies, reactance of coupling capacitors is high, hence there is loss of signal voltage across capacitors, resulting in reduced gain
Also at low frequencies, emitter bypass capacitor does not fully bypass the ac emitter current, hence ac voltage drop develops across RE, resulting in reduced gain
At very high frequencies, shunt capacitances due to wiring and inter-layer junction capacitances will be prominent, hence resulting in signal loss
At mid frequencies, gain is maximum and constant
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27. Transistor Amplifier Classification of amplifiers:
Based on mode of operation:
Class A: collector current flows throughout the complete input cycle (360o); Q-point is in the centre of active region (no distortion)
Class B: collector current flows during (positive or negative) half cycle of input; Q-point is at “just cut-off” or “just saturation”
Class AB: collector current flows for more than half cycle, but less than full cycle of input waveform; Q-point is “near cut-off” or “near saturation”
Class C: collector current flows for less than half cycle of input waveform; Q-point is in “deep cutoff” or “deep saturation”
(For classes B, AB and C, output is distorted or clipped)
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28. Tutorials
A three-stage amplifier circuit has first stage gain of 45 dB, second stage gain of 50 dB and third stage gain of –5 dB. What is the overall gain? If input to the first stage is 0.1mV, what is the output of final stage?
An amplifier has maximum gain of 200 and bandwidth of 500 kHz. If lower cutoff freq is 50 Hz, what is the upper cutoff freq and gain at this frequency?
Design a self bias circuit, given the following parameters:
IC=1mA, VCC=12V, VCE=VCC/2, VB=VCC/10, beta=100, VBE=0.6V, R2=2K
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29. End of Module 3
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30. Transistor Biasing Approximate analysis:
Carried out only if βRE ≥ 10R2
IB is negligible compared to I1 and I2
So,
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