1. Lecture :
TRANSISTOR BIAS
CIRCUITS
Code: Semester II
Apr 5, 2012 1
EEE2213 2010/2011

2. THE DC OPERATING POINT
A transistor must be properly biased with a dc
voltage in order to operate as a linear amplifier.
A dc operating point must be set so that signal
variations at the input terminal are implified &
accurately reproduced at the output terminal.
Note that when we bias a transistor, the dc voltage
and current values will be established.
Thus, at the dc operating point, IC & VCE have
specified values.
Q-point: the dc operating point (quiescent point)
Apr 5, 2012 2

3. DC Bias
Bias establishes the dc operating point (Q-point)
for proper linear operation of an amplifier.
If an amplifier is not biased with correct dc
voltages on the input and output, it can go into
saturation or cutoff when an input signal is applied.
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4. FIGURE 5–1 Examples of linear and nonlinear operation of an inverting
amplifier (the triangle symbol).
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5. Figure 5-1 shows the effects of proper & improper
dc biasing of an inverting amplifier.
Part (a): Linear operation
The output signal is an amplified replica of the
input signal except that it is inverted, which means
that it is 180° out of phase with the input.
The output signal swings equally above & below
the dc bias level of the output, V DC(out).
Apr 5, 2012 5

6. Part (b & c): Nonlinear operation
Improper biasing can cause distortion in the output
signal.
Part (b): shows the limiting of the positive portion
of the output voltage as a result of a Q-point (dc
operating point) being too close to cutoff.
Part (c): shows the limiting of the negative portion
of the output voltage as a result of a dc operating
point being too close to saturation.
Apr 5, 2012 6

7. Graphical Analysis
The transistor in Figure is biased with VCC & VBB to obtain certain
values of IB, IC, IE, and VCE.
The collector characteristic curves for this particular transistor are
shown in Figure.
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8. Figure 5-3 Illustration of Q-point adjustment.
Apr 5, 2012 8

9. Refer to Figure 5-3;
Assign 3 values to IB & observe what happens to IC & VCE.
First, VBB is adjusted to produce an IB of 200µA.
Since IC=βDCIB, the collector current is 20 mA, & VCE;
VCE=VCC – ICRC = 10 V – (20 mA)(220Ω) = 10 V – 4.4 V = 5.6 V
(this Q-point is shown in (a) as Q1.)
Next, in (b);
VBB is increased to produce an IB of 300µA & an IC of 30 mA.
VCE=VCC – ICRC = 10 V – (30 mA)(220Ω) = 10 V – 6.6 V = 3.4 V
(this Q-point is shown in (a) as Q2.)
Next, in (c);
VBB is increased to produce an IB of 400µA & an IC of 40 mA.
VCE=VCC – ICRC = 10 V – (40 mA)(220Ω) = 10 V – 8.8 V = 1.2 V
Apr 5, 2012 9

10. DC Load Line
The dc operation of a transistor circuit can be described graphically
using a dc load line.
DC Load Line: a straight line drawn on the characteristic curves from
the saturation value where IC=IC(sat) on the y-axis to the cutoff value
where VCE = VCC on the x-axis.
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11. Load line is determined by the external circuit (VCC & RC), not
determined by the transistor itself, which is described by the
characteristic curve.
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12. Refer to Figure 5-3,
the equation for IC is;
This is the equation of a straight line with a slope of -1/RC, an x intercept of
VCE=VCC, & a y intercept of VCC/RC, which is IC(sat).
The point at which the load line intersects a characteristic curve represents the
Q-point for that particular value of IB.
Figure 5-4(b): shows the Q-point on the load line for each value of IB in Figure
5-3. 2012
Apr 5, 12

13. Linear Operation
The region along the load line including all points between saturation &
cutoff is generally known as the linear region of the transistor’s
operation.
As long as the transistor is operated in this region, the output voltage is
ideally a linear reproduction of the input.
Apr 5, 2012 13

14. Figure shows an example of the linear operation of a transistor.
AC quantities are indicated by lowercase italic subscripts.
Assume a sinusoidal voltage, Vin, is superimposed on VBB, causing the base
current to vary sinusoidally 100 µA above and below its Q-point value of 300
µA.
This will causes the collector current to vary 10 mA above and below its Q-point
value of 30 mA.
From 2012 variation in collector current, the collector-to-emitter voltage varies
Apr 5,
the 14

15. Point A on the load line: corresponds to the positive peak of the sinusoidal input
voltage.
Point B on the load line: corresponds to the negative peak.
Point Q: corresponds to the zero value of the sine wave.
VCEQ, ICQ, & IBQ are dc Q-point values with no input sinusoidal voltage applied.
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16. Waveform Distortion
Under certain input signal conditions, the location of the Q-point on the
load line can cause one peak of the VCE waveform to be limited or clipped
(Figure (a) & (b).
Apr 5, 2012 16

17. In each case, the input signal is too large for the Q-point location & is
driving the transistor into cutoff or saturation during a portion of the input
cycle.
When both peaks are limited as in (c), the transistor is being driven into
both saturation and cutoff by an excessively large input signal.
When only the positive peak is
limited, the transistor is being
driven into cutoff but not
saturation.
When only the negative peak is
limited, the transistor is being
driven into saturation but not
cutoff.
Apr 5, 2012 17

18. Example 5-1
Apr 5, 2012 18

19. Lecture 9-2:
TRANSISTOR BIAS
CIRCUITS
Code: Semester II
Apr 5, 2012 19
EEE2213 2010/2011

20. VOLTAGE-DIVIDER BIAS
A method of biasing a transistor for linear
operation using a single-source resistive voltage
divider.
Earlier, a separate dc source, VBB, was used to bias
the base-emitter junction because it could be
varied independently of VCC, & it helped to
illustrate transistor operation.
More practical bias method: use VCC as the single
bias source (refer next Figure).
Apr 5, 2012 20

21. Apr 5, 2012 21

22. A dc bias voltage at the base of the transistor can be developed by a
resistive voltage divider that consists of R1 & R2.
VCC: the dc collector supply voltage
2 current paths are between point A & ground;
(2)through R2, and
(3)through the base-emitter junction of the transistor & RE
Generally, voltage-divider bias circuits are designed so that the
base current is much smaller than the current (I2) through R2.
For this case, the voltage-divider circuit is very straightforward to
analyze because the loading effect of the base current can be
ignored.
Stiff voltage divider: A voltage divider in which the base current
is small compared to the current in R2 (the base voltage is relatively
Apr 5, 2012 22
independent of different transistors & temperature effects).

23. To analyze a voltage-divider circuit in which IB is small
compared to R2, first calculate the voltage on the base;
R1 and R2 are selected to establish VB. If the divider is stiff,
IB is small compared to I2. Then,
+VCC
 R2  +15 V
VB ≈   VCC
 R1 + R2 
R1 +VCCRC
27 kΩ 1.2 kΩ
Determine the base voltage for the βDC = 200
R1 RC
circuit.
IB
R2 RE
12 kΩ 680 Ω
I2
 R2 
VB =   VCC
 R1 + R2  R2 RE
 12 kΩ 
=  ( +15 V ) = 4.62 V
 27 kΩ + 12 kΩ 

24. • Once we have the base voltage, we can find the
voltages & currents in the circuit;
• and
• Then,
• From the VC & VE, we can determine VCE;
Apr 5, 2012 24

25. What is the emitter voltage, VE, and current, IE?
+VCC
+15 V
VE is one diode drop less than VB:
R1 RC
27 kΩ 1.2 kΩ
VE = 4.62 V – 0.7 V = 3.92 V 4.62 V βDC = 200
Applying Ohm’s law: 3.92 V
R2 RE
12 kΩ 680 Ω
VE 3.92 V
IE = = = 5.76 mA
RE 680 Ω

26. Example 5-2
Apr 5, 2012 26

27. Ideally, a voltage-divider circuit is stiff; where transistor does
not appear as a significant load.
To determine if the divider is stiff or not, then need to examine
the dc input resistance looking in at the base.
Apr 5, 2012 27

28. Loading Effects of Voltage-Divider Bias
(1) DC Input Resistance at the
Transistor Base
(2) Stability of Voltage-
Divider Bias
(3) Voltage-Divider Biased
PNP Transistor
Apr 5, 2012 28

29. DC Input Resistance at the
Transistor Base
 The dc input resistance of the transistor is proportional to
βDC, thus it will change for different transistors.
 When a transistor is operating in its linear region, the
emitter current is βDC IB.
 When the emitter resistor is viewed from the base circuit,
the resistor appears to be larger than its actual value by a
factor of is βDC because of the current gain in the transistor.
Thus, β DCVB
RIN (BASE ) =
IE
 This is the effective load on the voltage divider illustrated
in earlier Figure.
Apr 5, 2012 29

30.  We can estimate the loading effect by comparing RIN(BASE) to
the resistor R2 in the voltage divider.
 As long as RIN(BASE) is at least 10x larger than R2, the
loading effect will be 10% or less & the voltage divider is
stiff.
 If RIN(BASE) is less than 10x R2, it should be combined in
parallel with R2.
Apr 5, 2012 30

31. Example 5-3
Apr 5, 2012 31

32. Stability of Voltage-Divider Bias
 Thevenin’s theorem is applied when
analyze a voltage-divider biased
transistor circuit for base current
loading effects.
 1st, get an equivalent base-emitter
circuit for this circuit using Thevenin’s
theorem.
Apr 5, 2012 32

33.  Looking out from the base terminal, the bias circuit can be
redrawn as shown in this figure.
 Apply Thevenin’s theorem to the
circuit left of point A, with VCC replaced
by a short to ground & the transistor
disconnected from the circuit.
 The voltage at point A with respect to
ground is,
 and the resistance is,
Apr 5, 2012 33

34.  The Thevenin equivalent of the bias circuit, connected to the
transistor base, is as shown in the grey box in the figure.
 Applying Kirchhoff’s voltage law
around the equivalent base-emitter loop
gives,
 Substituting, using Ohm’s law, &
solving for VTH,
 Substituting IE / βDC for IB,
Apr 5, 2012 34

35.  Solving for IE,
 If RTH / βDC is small compared to RE, the
result is the same as for unloaded voltage
divider.
 Reason of why a voltage-divider bias is
widely used;
 Reasonably good bias stability is
achieved with a single supply voltage.
Apr 5, 2012 35

36. The unloaded voltage divider approximation for VB
gives reasonable results. A more exact solution is to
Thevenize the input circuit.
+ VCC +VCC
+15 V +15 V
VTH = VB(no load)
RC R1 RC
= 4.62 V 1.2 kΩ 27 kΩ 1.2 kΩ
RTH = R1||R2 = +V TH
4.62 V
+
R TH
– +
βDC = 200 βDC = 200
IB VBE –
= 8.31 kΩ 8.31 kΩ
+ R2 RE
RE
The Thevenin IE
–
680 Ω 12 kΩ 680 Ω
input circuit can
be drawn

37. Now write KVL around the base emitter circuit and
solve for IE.
VTH = I B RTH + VBE + I E RE + VCC
+15 V
VTH − VBE
IE =
R
RE + TH RC
β DC 1.2 kΩ
Substituting and solving, +V TH R TH
+ – +
4.62 V − 0.7 V 4.62 V βDC = 200
IE = = 5.43 mA IB VBE –
680 Ω + 8.31 kΩ 8.31 kΩ
200 + RE
IE
and –
680 Ω
VE = IERE = (5.43 mA)(0.68 kΩ)
= 3.69 V

38. Voltage-Divider Biased PNP
Transistor
 pnp transistor requires bias polarities opposite to the npn.
 This can be accomplished by using a negative collector supply
voltage: (a), or with a positive emitter supply voltage (b). 3rd
figure is same as (b), only that it is drawn upside down.
Apr 5, 2012 38

39.  Analysis procedure is the same as for an npn transistor circuit.
 For a stiff voltage divider (ignoring loading
effects),
 By Ohm’s law,
 Thus,
Apr 5, 2012 39

40. Determine IE for the pnp circuit. Assume a
stiff voltage divider (no loading effect).
+VEE
+15 V
 R1 
VB =   VEE R2 RE
 R1 + R2  12 kΩ 680 Ω
 27 kΩ  10.4 V 11.1 V
=  ( +15.0 V ) = 10.4 V
 27 kΩ + 12 kΩ 
VE = VB + VBE = 10.4 V + 0.7 V = 11.1 V R1 RC
27 kΩ 1.2 kΩ
VEE − VE 15.0 V − 11.1 V
IE = = = 5.74
RE 680 Ω
mA

41. Example 5-4 & 5-5
Apr 5, 2012 41

42. EMITTER BIAS
Provides excellent bias stability in spite of changes in β
or temperature.
Use both a positive & a negative supply voltage.
In an npn circuit, the small
base current causes the base
voltage to be slightly below
ground.
The emitter voltage is one
diode drop less than this.
Apr 5, 2012 42

43. The combination of this small drop across RB & VBE forces
the emitter to be at approximately -1 V.
Thus, emitter current is,
Apr 5, 2012 43

44. Applying the approximation that
to calculate the collector voltage, thus
Apr 5, 2012 44

45. Apr 5, 2012 45

46. Example 5-6
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47. • Approximation that & the neglect of βDC may not be
accurate enough for design work or detailed analysis.
• Kirchhoff’s voltage law (KVL) can be applied to develop a
more detailed formula for IE.
• In (a): KVL applied around the
base-emitter circuit
• In (b): (a) is redrawn for
analysis
• Thus,
• Substitute &
transposing VEE,
• Using Ohm’s law,
Apr 5, 2012 47

48. • Thus,
• Voltages with respect to ground are indicated by a single
subscript.
• Thus, the emitter voltage with respect to the ground is,
• The base voltage with respect to ground is,
• The collector with respect to the ground is,
Apr 5, 2012 48

49. Example 5-7
Apr 5, 2012 49

50. BASE BIAS
 This method of biasing is common in switching circuits.
 Analysis of this circuit for the linear region shows that it is
directly dependent on βDC.
 Starts with KVL around the base circuit,
 Substitutes IBRB for VRB,
Thus,
Apr 5, 2012 50

51.  When KVL is applied around the collector circuit, we get,
 Thus,
 Substitute IC = βDCIB, we get,
Apr 5, 2012 51

52. Base bias is used in switching circuits because of
its simplicity, but not widely used in linear
applications because the Q-point is β dependent.
Base current is derived from the collector
supply through a large base resistor.
+VCC
+VCC
+15 V
What is IB? RC
1.8 kΩ
RB
560 kΩ
VCC − 0.7 V 15 V − 0.7 V
IB = = = 25.5 µA
RB 560 kΩ

53. Q-Point Stability of Base Bias
 Equation
shows that IC is dependent on βDC. Thus, a variation in βDC causes
IC and, as a result, VCE to change, thus changing the Q-point of the
transistor. Therefore, the base bias circuit is extremely beta-
dependent & unpredictable.
 βDC varies with temperature & collector current.
 There is a large spread of βDC values from one transistor to
another of the same type due to manufacturing variations.
Apr 5, 2012 53
 Therefore, base bias is rarely used in linear circuits.

54. Example 5-8
Apr 5, 2012 54

55. EMITTER-FEEDBACK BIAS
 If an emitter resistor is added to the base-circuit, we get an
emitter-feedback bias.
 Help make base bias more predictable with
negative feedback, which negates any
attempted change in collector current with an
opposing change in base voltage.
 If the collector current tries to increase, the
emitter voltage increases, causing an increase in
base voltage due to VB = VE + VBE
 This increase in base voltage reduces the
voltage across RB, thus reducing the base
current & keeping the collector current from
increasing.
Apr 5, 2012 55

56. The same scenario happens if the collector
current tries to decrease.
 This circuit is better for linear circuits than
base bias, but still dependent on βDC & is not
predictable as voltage-divider bias.
 To calculate IE, write KVL around the base
circuit.
Substitute IE/βDC for IB, can see that IE is still
dependent on βDC.
Apr 5, 2012 56

57. Example 5-9
Apr 5, 2012 57

58. COLLECTOR-FEEDBACK BIAS
 The base resistor RB is connected to the collector rather than to
VCC.
 The collector voltage provides the bias for the base-
emitter junction.
 The negative feedback creates an “offsetting” effect
that tends to keep the Q-point stable.
 If IC tries to increase, it drops more voltage across RC,
and thus causing VC to decrease.
 When VC decreases, there is a decrease in voltage
across RB, which deecreases IB.
 The decrease in IB produces less IC which, in turn,
drops less voltage across RC & thus offsets the decrease
in VC.
Apr 5, 2012 58

59.  By Ohm’s law,
 Assume that IC >> IB, thus the collector
voltage,
 Also,
 Thus,
Apr 5, 2012 59

60.  Thus,
 Since the emitter is ground, VCE = VC, then
Apr 5, 2012 60

61. Q-Point Stability Over
Temperature
 Equation
shows that IC is dependent to some extent on βDC and VBE .
 The dependency can be minimized by making RC >> RB/βDC &
VCE >>VBE.
 The collector-feedback bias is essentially eliminates the βDC &
VBE dependency even if the stated conditions are met.
 βDC varies directly with temperature, & VBE varies inversely with
temperature. As the temperature goes up in a collector-feedback 61
Apr 5, 2012 β
DC

62.  The increase in βDC acts to increase IC. The decrease in
VBE acts to increase IB which, in turn also acts to increase
IC. As IC tries to increase, the voltage drop across RC also
tries to increase. This tends to reduce the collector voltage
& therefore the voltage across RB, thus reducing IB &
offsetting the attempted increase in IC & the attempted
decrease in VC.
 The result is that the collector-feedback circuit
maintains a relatively stable Q-point. The reverse action
occurs when the temperature decreases.
Apr 5, 2012 βDC 62

63. Compare IC for the case when β = 100 with the case
when β = 300.
+VCC
When β = 100, + 15 V
VCC − VBE 15 V − 0.7 V
IC = = = 2.80 mA RC
RB 1.8 kΩ + 330 kΩ RB
RC + 100 1.8 kΩ
β DC
330 kΩ
When β = 300,
VCC − VBE 15 V − 0.7 V
IC = = = 4.93 mA
RB 1.8 kΩ + 330 kΩ
RC + 300
β DC

64. Example 5-10
Apr 5, 2012 64