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Electronics 2 EE 317 Electrical Engineering Majmaah University 1st Semester 1433/1434 H
Chapter 5 Transistor Bias Circuits
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
1
Chapter Outline 5–1 The DC Operating Point 5–2 Voltage-Divider Bias 5–3 Other Bias Methods
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
2
Chapter Objectives • Discuss and determine the dc operating point of a linear amplifier. • Analyze a voltage-divider biased circuit. • Analyze an emitter bias circuit, a base bias circuit, an emitter- bias circuit, and a collector- bias circuit.
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
3
Introduction • As you learned in Chapter 4, a transistor must be properly biased in order to operate as an amplifier. • DC biasing is used to establish fixed dc values for the transistor currents and voltages called the dc operating point or quiescent point (Q-point). • In this chapter, several types of bias circuits are discussed. • This material lays the groundwork for the study of amplifiers, and other circuits that require proper biasing.
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
4
The DC Operating Point
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
5
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
6
1.2 Electronic Devices, 9th edition Thomas L. Floyd
3.4
5.6 © 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
7
The DC Operating Point
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
8
Linear Operation
Bias establishes the operating point (Q-point) of a transistor amplifier; the ac signal I (mA) moves above and below Load line this point. C
For this example, the dc base current is 300 µA. When the input causes the base current to vary between 200 µA and 400 µA, the collector current varies between 20 mA and 40 mA. Electronic Devices, 9th edition Thomas L. Floyd
Ic 40 ICQ
Ib
A
400 µ A Q
30 20
0
300 µ A = IBQ B
1.2
3.4 VCEQ
200 µ A
VCE (V)
5.6 Vce
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
9
Waveform Distortion A signal that swings outside the active region will be clipped. For example, the bias has established a low Qpoint. As a result, the signal will be clipped because it is too close to cutoff.
IB
Q
IC
ICQ
Input signal
Q Cutoff
0
VCC
VCE
Cutoff
Vce VCEQ
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
10
Waveform Distortion
High Q- point. The signal will be clipped because it is too close to saturation. Electronic Devices, 9th edition Thomas L. Floyd
Input signal too large. The signal will be clipped from both sides.
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
11
Quiz
Q. A signal that swings outside the active area will be a. clamped b. clipped c. unstable d. all of the above
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
12
Voltage-Divider Bias A practical way to establish a Q-point is to form a voltagedivider from VCC. R1 and R2 are selected to establish VB. If the divider is stiff, IB is small compared to I2. Then, R2 VB ≈ VCC R1 + R2
+VCC
R1
RC
IB I2 R2
Electronic Devices, 9th edition Thomas L. Floyd
RE
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
13
Quiz
Q. A stiff voltage divider is one in which a. there is no load current b. divider current is small compared to load current c. the load is connected directly to the source voltage d. loading effects can be ignored
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
14
Voltage-Divider Bias
+VCC +15 V
Determine the base voltage for the circuit. R2 VB = VCC R1 + R2 12 kΩ = ( +15 V ) = 4.62 V Ω + Ω 27 k 12 k
Electronic Devices, 9th edition Thomas L. Floyd
R1 27 kΩ
RC 1.2 kΩ βDC = 200
R2 12 kΩ
RE 680 Ω
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
15
Voltage-Divider Bias
+VCC +15 V
What is the emitter voltage, VE, and current, IE?
VE is one diode drop less than VB: VE = 4.62 V – 0.7 V = 3.92 V Applying Ohm’s law: IE = Electronic Devices, 9th edition Thomas L. Floyd
R1 27 kΩ
4.62 V
RC 1.2 kΩ βDC = 200
3.92 V R2 12 kΩ
RE 680 Ω
VE 3.92 V = = 5.76 mA RE 680 Ω © 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
16
Quiz
Q. Assuming a stiff voltage-divider for the circuit shown, the emitter voltage is +VCC +15 V
a. 4.3 V b. 5.7 V
R1 20 kΩ
c. 6.8 V d. 9.3 V
RC 1.8 kΩ βDC = 200
R2 10 kΩ
RE 1.2 kΩ
VB = 10 / (10 + 20) * 15 = 5 V VE = 5 – 0.7 = 4.3 V Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
17
Quiz
Q. For the circuit shown, the dc load line will intersect the y-axis at (neglecting IB) +VCC +15 V
a. 5.0 mA b. 10.0 mA
R1 20 kΩ
c. 15.0 mA d. none of the above
RC 1.8 kΩ βDC = 200
R2 10 kΩ
RE 1.2 kΩ
@ VCE=0: IC = (15 – 0) / (1.8k + 1.2k) = 15 / 3k = 5 mA Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
18
Voltage-Divider Bias The unloaded voltage divider approximation for VB gives reasonable results. A more exact solution is to Thevenize the input circuit. Looking from the base to the left: +VCC
+ VCC +15 V
+15 V
VTH = VB(no load) RC 1.2 kΩ
= 4.62 V RTH = R1||R2 = = 8.31 kΩ The Thevenin input circuit can be drawn Electronic Devices, 9th edition Thomas L. Floyd
+V TH 4.62 V
R TH
+
–
+
+ –
RE 680 Ω
RC 1.2 kΩ βDC = 200
βDC = 200
VBE – IB 8.31 kΩ
IE
R1 27 kΩ
R2 12 kΩ
RE 680 Ω
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
19
Thevenin Resistance, RTH To find RTH, replace voltage sources by a short circuit. Which means VCC will be grounded. RTH = R1||R2 1
2
3
+VCC +15 V
R1 27 kΩ
RC 1.2 kΩ βDC = 200
R2 12 kΩ
Electronic Devices, 9th edition Thomas L. Floyd
RE 680 Ω
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
20
Quiz
Q. If you Thevenize the input voltage divider, the Thevenin resistance is +VCC +15 V
a. 5.0 kΩ b. 6.67 kΩ Ω
R1 20 kΩ
c. 10 kΩ d. 30 kΩ
RC 1.8 kΩ βDC = 200
R2 10 kΩ
RE 1.2 kΩ
RTH = R1 R2 / (R1 + R2) = 20 * 10 / (20 + 10) = 200 / 30 = 6.67 kΩ Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
21
Voltage-Divider Bias
IE = IB + IC = IB + βDC IB = (βDC + 1) IB ≈ βDC IB VTH – VBE = IE RE + IB RTH ≈ IE RE + IE RTH / βDC = IE ( RE + RTH / βDC )
Now write KVL around the base emitter circuit and solve for IE. VTH = I B RTH + VBE + I E RE VTH − VBE R RE + TH β DC Substituting and solving, 4.62 V − 0.7 V IE = = 5.43 mA 8.31 k Ω 680 Ω + 200 and VE = IERE = (5.43 mA)(0.68 kΩ)
+ VCC +15 V
IE =
RC 1.2 kΩ +V TH 4.62 V
R TH
+
–
+
βDC = 200
VBE – IB 8.31 kΩ
IE
+ –
RE 680 Ω
= 3.69 V Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
22
Voltage-Divider Bias A pnp transistor can be biased from either a positive or negative supply. Notice that (b) and (c) are the same circuit; both with a positive supply. − VEE
+ VEE
R1
RC
R1
RC
R2
RE
R2
RE
R2
RE
R1
RC
+ VEE
(a) Electronic Devices, 9th edition Thomas L. Floyd
(b)
(c) © 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
23
Voltage-Divider Bias Determine IE for the pnp circuit. Assume a stiff voltage divider (no loading effect). R1 VB = VEE R + R 2 1 27 kΩ = ( +15.0 V ) = 10.4 V 27 kΩ + 12 kΩ
VE = VB + VBE = 10.4 V + 0.7 V = 11.1 V IE = Electronic Devices, 9th edition Thomas L. Floyd
VEE − VE 15.0 V − 11.1 V = = 5.74 mA RE 680 Ω
+VEE +15 V R2 12 kΩ
RE 680 Ω
10.4 V
11.1 V
R1 27 kΩ
RC 1.2 kΩ
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
24
Quiz
Q. For the circuit shown, the emitter voltage is a. less than the base voltage
+VEE +15 V
b. less than the collector voltage c. both of the above
R2 12 kΩ
RE 680 Ω
R1 27 kΩ
RC 1.2 kΩ
d. none of the above
(higher than the base voltage by 0.7 V) pnp circuit Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
25
Emitter Bias Emitter bias has excellent stability but requires both a positive and a negative source. For troubleshooting analysis, assume that VE for an npn transistor is about −1 V. VCC +15 V
Assuming that VE is −1 V, what is IE?
IE =
RC 3.9 kΩ
−VEE − 1 V −15 V − (−1 V) = = −1.87 mA RE 7.5 kΩ The minus sign means the current is directed downwards (opposite to initial assumption).
Electronic Devices, 9th edition Thomas L. Floyd
RB 68 kΩ
−1 V
RE 7.5 kΩ VEE −15 V
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
26
Emitter Bias The approximation that VE ≈ −1 V and the neglect of βDC may not be accurate enough for design work or detailed analysis. In this case, KVL can be applied to develop a more detailed formula for IE. KVL applied around the base-emitter circuit in Figure 5–17(a), which has been redrawn in part (b) for analysis, gives the following equation:
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
27
Emitter Bias VRB + VBE + VRE − VEE = 0 IB RB + VBE + IE RE − VEE = 0
IE ( RB / βDC) + IE RE = VEE − VBE IE (RE + RB / βDC) = VEE − VBE − Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
28
Quiz
Q. Emitter bias a. is not good for linear circuits b. uses a voltage-divider on the input c. requires dual power supplies d. all of the above
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
29
Quiz
Q. With the emitter bias shown, a reasonable assumption for troubleshooting work is that the VCC +15 V
a. base voltage = +1 V b. emitter voltage = +5 V
RC 3.9 kΩ
c. emitter voltage = −1 V d. collector voltage = VCC
RB 68 kΩ
RE 7.5 kΩ VEE −15 V
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
30
Base Bias 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 +V +15CCV
IB
What is IB?
RC 1.8 kΩ
RB +
IB = Electronic Devices, 9th edition Thomas L. Floyd
VCC − 0.7 V 15 V − 0.7 V = = 25.5 µA RB 560 kΩ
560 kΩ − VBE
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
31
Base Bias Compare VCE for the case where β = 100 and β = 300. For β = 100: I C = βI B = (100 )( 25.5 µA ) = 2.55 mA
VCE = VCC − I C RC
+VCC +15 V
= 15 V − ( 2.55 mA )(1.8 kΩ ) = 10.4 V For β = 300: I C = βI B = ( 300 )( 25.5 µA ) = 7.65 mA
VCE = VCC − I C RC = 15 V − ( 7.65 mA )(1.8 kΩ ) = 1.23 V
IC RB 560 kΩ
RC 1.8 kΩ + VCE −
The Q-point is β dependent. Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
32
Quiz
Q. The circuit shown is an example of a. base bias +VCC
b. collector- bias c. emitter bias d. voltage-divider bias
Electronic Devices, 9th edition Thomas L. Floyd
RC RB
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
33
Emitter- Bias An emitter resistor (RE) changes base bias into emitter bias, which is more predictable. The emitter resistor (RE) is a form of negative . If IC tries to increase, VE increases, causing an increase in VB because VB = VE + VBE.
+VCC
This increase in VB reduces the voltage across RB (VRB = VCC − VB), thus reducing IB and keeping IC from increasing. A similar action occurs if IC tries to decrease.
Electronic Devices, 9th edition Thomas L. Floyd
RC RB
RE
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
34
Emitter- Bias The equation for emitter current (IE) is found by writing KVL around the base circuit.
IE RE + IB RB = VCC − VBE
+VCC
IE RE + (IE / βDC) RB = VCC − VBE IE (RE + RB / βDC) = VCC − VBE
RC +
RB
−
IB
+
−
IE
Electronic Devices, 9th edition Thomas L. Floyd
+
RE
−
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
35
Collector- Bias Collector bias uses another form of negative to increase stability. (more on next slide) Instead of returning the base resistor (RB) to VCC, it is returned to the collector. +VCC
The equation for collector current (IC) is found by writing KVL around the base circuit.
RB
RC
(details on page 247 in the book) The result is
Electronic Devices, 9th edition Thomas L. Floyd
VCC − VBE IC = R RC + B β DC © 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
36
Collector- Bias Collector bias uses another form of negative to increase stability. Instead of returning the base resistor to VCC, it is returned to the collector. The negative creates an “offsetting” effect that tends to keep the Q-point stable. If IC tries to increase, it drops more voltage across RC, thereby causing VC to decrease. When VC decreases, there is a decrease in voltage across RB, which decreases IB.
+VCC RB
RC
The decrease in IB produces less IC which, in turn, drops less voltage across RC and thus offsets the decrease in VC. Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
37
Collector- Bias Compare IC for the case when β = 100 with the case when β = 300. +VCC + 15 V
When β = 100,
VCC − VBE 15 V − 0.7 V = = 2.80 mA R 1.8 kΩ + 330 kΩ RC + B 100 β DC When β = 300, IC =
RB
RC 1.8 kΩ
330 kΩ
VCC − VBE 15 V − 0.7 V IC = = = 4.93 mA RB 330 k Ω 1.8 kΩ + RC + 300 β DC Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
38
Quiz
Q. The circuit shown is an example of a. base bias
+VCC
b. collector- bias c. emitter bias
RB
RC
d. voltage-divider bias
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
39
Key Q-point The dc operating (bias) point of an amplifier specified by voltage and current values. DC load line A straight line plot of IC and VCE for a transistor circuit. Linear region The region of operation along the load line between saturation and cutoff. Stiff voltage A voltage divider for which loading effects divider can be ignored. The process of returning a portion of a circuit’s output back to the input in such a way as to oppose or aid a change in the output. Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
40
Chapter 5 Transistor Bias Circuits
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
1
Chapter Outline 5–1 The DC Operating Point 5–2 Voltage-Divider Bias 5–3 Other Bias Methods
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
2
Chapter Objectives • Discuss and determine the dc operating point of a linear amplifier. • Analyze a voltage-divider biased circuit. • Analyze an emitter bias circuit, a base bias circuit, an emitter- bias circuit, and a collector- bias circuit.
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
3
Introduction • As you learned in Chapter 4, a transistor must be properly biased in order to operate as an amplifier. • DC biasing is used to establish fixed dc values for the transistor currents and voltages called the dc operating point or quiescent point (Q-point). • In this chapter, several types of bias circuits are discussed. • This material lays the groundwork for the study of amplifiers, and other circuits that require proper biasing.
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
4
The DC Operating Point
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
5
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
6
1.2 Electronic Devices, 9th edition Thomas L. Floyd
3.4
5.6 © 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
7
The DC Operating Point
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
8
Linear Operation
Bias establishes the operating point (Q-point) of a transistor amplifier; the ac signal I (mA) moves above and below Load line this point. C
For this example, the dc base current is 300 µA. When the input causes the base current to vary between 200 µA and 400 µA, the collector current varies between 20 mA and 40 mA. Electronic Devices, 9th edition Thomas L. Floyd
Ic 40 ICQ
Ib
A
400 µ A Q
30 20
0
300 µ A = IBQ B
1.2
3.4 VCEQ
200 µ A
VCE (V)
5.6 Vce
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
9
Waveform Distortion A signal that swings outside the active region will be clipped. For example, the bias has established a low Qpoint. As a result, the signal will be clipped because it is too close to cutoff.
IB
Q
IC
ICQ
Input signal
Q Cutoff
0
VCC
VCE
Cutoff
Vce VCEQ
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
10
Waveform Distortion
High Q- point. The signal will be clipped because it is too close to saturation. Electronic Devices, 9th edition Thomas L. Floyd
Input signal too large. The signal will be clipped from both sides.
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
11
Quiz
Q. A signal that swings outside the active area will be a. clamped b. clipped c. unstable d. all of the above
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
12
Voltage-Divider Bias A practical way to establish a Q-point is to form a voltagedivider from VCC. R1 and R2 are selected to establish VB. If the divider is stiff, IB is small compared to I2. Then, R2 VB ≈ VCC R1 + R2
+VCC
R1
RC
IB I2 R2
Electronic Devices, 9th edition Thomas L. Floyd
RE
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
13
Quiz
Q. A stiff voltage divider is one in which a. there is no load current b. divider current is small compared to load current c. the load is connected directly to the source voltage d. loading effects can be ignored
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
14
Voltage-Divider Bias
+VCC +15 V
Determine the base voltage for the circuit. R2 VB = VCC R1 + R2 12 kΩ = ( +15 V ) = 4.62 V Ω + Ω 27 k 12 k
Electronic Devices, 9th edition Thomas L. Floyd
R1 27 kΩ
RC 1.2 kΩ βDC = 200
R2 12 kΩ
RE 680 Ω
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
15
Voltage-Divider Bias
+VCC +15 V
What is the emitter voltage, VE, and current, IE?
VE is one diode drop less than VB: VE = 4.62 V – 0.7 V = 3.92 V Applying Ohm’s law: IE = Electronic Devices, 9th edition Thomas L. Floyd
R1 27 kΩ
4.62 V
RC 1.2 kΩ βDC = 200
3.92 V R2 12 kΩ
RE 680 Ω
VE 3.92 V = = 5.76 mA RE 680 Ω © 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
16
Quiz
Q. Assuming a stiff voltage-divider for the circuit shown, the emitter voltage is +VCC +15 V
a. 4.3 V b. 5.7 V
R1 20 kΩ
c. 6.8 V d. 9.3 V
RC 1.8 kΩ βDC = 200
R2 10 kΩ
RE 1.2 kΩ
VB = 10 / (10 + 20) * 15 = 5 V VE = 5 – 0.7 = 4.3 V Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
17
Quiz
Q. For the circuit shown, the dc load line will intersect the y-axis at (neglecting IB) +VCC +15 V
a. 5.0 mA b. 10.0 mA
R1 20 kΩ
c. 15.0 mA d. none of the above
RC 1.8 kΩ βDC = 200
R2 10 kΩ
RE 1.2 kΩ
@ VCE=0: IC = (15 – 0) / (1.8k + 1.2k) = 15 / 3k = 5 mA Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
18
Voltage-Divider Bias The unloaded voltage divider approximation for VB gives reasonable results. A more exact solution is to Thevenize the input circuit. Looking from the base to the left: +VCC
+ VCC +15 V
+15 V
VTH = VB(no load) RC 1.2 kΩ
= 4.62 V RTH = R1||R2 = = 8.31 kΩ The Thevenin input circuit can be drawn Electronic Devices, 9th edition Thomas L. Floyd
+V TH 4.62 V
R TH
+
–
+
+ –
RE 680 Ω
RC 1.2 kΩ βDC = 200
βDC = 200
VBE – IB 8.31 kΩ
IE
R1 27 kΩ
R2 12 kΩ
RE 680 Ω
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
19
Thevenin Resistance, RTH To find RTH, replace voltage sources by a short circuit. Which means VCC will be grounded. RTH = R1||R2 1
2
3
+VCC +15 V
R1 27 kΩ
RC 1.2 kΩ βDC = 200
R2 12 kΩ
Electronic Devices, 9th edition Thomas L. Floyd
RE 680 Ω
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
20
Quiz
Q. If you Thevenize the input voltage divider, the Thevenin resistance is +VCC +15 V
a. 5.0 kΩ b. 6.67 kΩ Ω
R1 20 kΩ
c. 10 kΩ d. 30 kΩ
RC 1.8 kΩ βDC = 200
R2 10 kΩ
RE 1.2 kΩ
RTH = R1 R2 / (R1 + R2) = 20 * 10 / (20 + 10) = 200 / 30 = 6.67 kΩ Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
21
Voltage-Divider Bias
IE = IB + IC = IB + βDC IB = (βDC + 1) IB ≈ βDC IB VTH – VBE = IE RE + IB RTH ≈ IE RE + IE RTH / βDC = IE ( RE + RTH / βDC )
Now write KVL around the base emitter circuit and solve for IE. VTH = I B RTH + VBE + I E RE VTH − VBE R RE + TH β DC Substituting and solving, 4.62 V − 0.7 V IE = = 5.43 mA 8.31 k Ω 680 Ω + 200 and VE = IERE = (5.43 mA)(0.68 kΩ)
+ VCC +15 V
IE =
RC 1.2 kΩ +V TH 4.62 V
R TH
+
–
+
βDC = 200
VBE – IB 8.31 kΩ
IE
+ –
RE 680 Ω
= 3.69 V Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
22
Voltage-Divider Bias A pnp transistor can be biased from either a positive or negative supply. Notice that (b) and (c) are the same circuit; both with a positive supply. − VEE
+ VEE
R1
RC
R1
RC
R2
RE
R2
RE
R2
RE
R1
RC
+ VEE
(a) Electronic Devices, 9th edition Thomas L. Floyd
(b)
(c) © 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
23
Voltage-Divider Bias Determine IE for the pnp circuit. Assume a stiff voltage divider (no loading effect). R1 VB = VEE R + R 2 1 27 kΩ = ( +15.0 V ) = 10.4 V 27 kΩ + 12 kΩ
VE = VB + VBE = 10.4 V + 0.7 V = 11.1 V IE = Electronic Devices, 9th edition Thomas L. Floyd
VEE − VE 15.0 V − 11.1 V = = 5.74 mA RE 680 Ω
+VEE +15 V R2 12 kΩ
RE 680 Ω
10.4 V
11.1 V
R1 27 kΩ
RC 1.2 kΩ
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
24
Quiz
Q. For the circuit shown, the emitter voltage is a. less than the base voltage
+VEE +15 V
b. less than the collector voltage c. both of the above
R2 12 kΩ
RE 680 Ω
R1 27 kΩ
RC 1.2 kΩ
d. none of the above
(higher than the base voltage by 0.7 V) pnp circuit Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
25
Emitter Bias Emitter bias has excellent stability but requires both a positive and a negative source. For troubleshooting analysis, assume that VE for an npn transistor is about −1 V. VCC +15 V
Assuming that VE is −1 V, what is IE?
IE =
RC 3.9 kΩ
−VEE − 1 V −15 V − (−1 V) = = −1.87 mA RE 7.5 kΩ The minus sign means the current is directed downwards (opposite to initial assumption).
Electronic Devices, 9th edition Thomas L. Floyd
RB 68 kΩ
−1 V
RE 7.5 kΩ VEE −15 V
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
26
Emitter Bias The approximation that VE ≈ −1 V and the neglect of βDC may not be accurate enough for design work or detailed analysis. In this case, KVL can be applied to develop a more detailed formula for IE. KVL applied around the base-emitter circuit in Figure 5–17(a), which has been redrawn in part (b) for analysis, gives the following equation:
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
27
Emitter Bias VRB + VBE + VRE − VEE = 0 IB RB + VBE + IE RE − VEE = 0
IE ( RB / βDC) + IE RE = VEE − VBE IE (RE + RB / βDC) = VEE − VBE − Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
28
Quiz
Q. Emitter bias a. is not good for linear circuits b. uses a voltage-divider on the input c. requires dual power supplies d. all of the above
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
29
Quiz
Q. With the emitter bias shown, a reasonable assumption for troubleshooting work is that the VCC +15 V
a. base voltage = +1 V b. emitter voltage = +5 V
RC 3.9 kΩ
c. emitter voltage = −1 V d. collector voltage = VCC
RB 68 kΩ
RE 7.5 kΩ VEE −15 V
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
30
Base Bias 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 +V +15CCV
IB
What is IB?
RC 1.8 kΩ
RB +
IB = Electronic Devices, 9th edition Thomas L. Floyd
VCC − 0.7 V 15 V − 0.7 V = = 25.5 µA RB 560 kΩ
560 kΩ − VBE
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
31
Base Bias Compare VCE for the case where β = 100 and β = 300. For β = 100: I C = βI B = (100 )( 25.5 µA ) = 2.55 mA
VCE = VCC − I C RC
+VCC +15 V
= 15 V − ( 2.55 mA )(1.8 kΩ ) = 10.4 V For β = 300: I C = βI B = ( 300 )( 25.5 µA ) = 7.65 mA
VCE = VCC − I C RC = 15 V − ( 7.65 mA )(1.8 kΩ ) = 1.23 V
IC RB 560 kΩ
RC 1.8 kΩ + VCE −
The Q-point is β dependent. Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
32
Quiz
Q. The circuit shown is an example of a. base bias +VCC
b. collector- bias c. emitter bias d. voltage-divider bias
Electronic Devices, 9th edition Thomas L. Floyd
RC RB
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
33
Emitter- Bias An emitter resistor (RE) changes base bias into emitter bias, which is more predictable. The emitter resistor (RE) is a form of negative . If IC tries to increase, VE increases, causing an increase in VB because VB = VE + VBE.
+VCC
This increase in VB reduces the voltage across RB (VRB = VCC − VB), thus reducing IB and keeping IC from increasing. A similar action occurs if IC tries to decrease.
Electronic Devices, 9th edition Thomas L. Floyd
RC RB
RE
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
34
Emitter- Bias The equation for emitter current (IE) is found by writing KVL around the base circuit.
IE RE + IB RB = VCC − VBE
+VCC
IE RE + (IE / βDC) RB = VCC − VBE IE (RE + RB / βDC) = VCC − VBE
RC +
RB
−
IB
+
−
IE
Electronic Devices, 9th edition Thomas L. Floyd
+
RE
−
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
35
Collector- Bias Collector bias uses another form of negative to increase stability. (more on next slide) Instead of returning the base resistor (RB) to VCC, it is returned to the collector. +VCC
The equation for collector current (IC) is found by writing KVL around the base circuit.
RB
RC
(details on page 247 in the book) The result is
Electronic Devices, 9th edition Thomas L. Floyd
VCC − VBE IC = R RC + B β DC © 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
36
Collector- Bias Collector bias uses another form of negative to increase stability. Instead of returning the base resistor to VCC, it is returned to the collector. The negative creates an “offsetting” effect that tends to keep the Q-point stable. If IC tries to increase, it drops more voltage across RC, thereby causing VC to decrease. When VC decreases, there is a decrease in voltage across RB, which decreases IB.
+VCC RB
RC
The decrease in IB produces less IC which, in turn, drops less voltage across RC and thus offsets the decrease in VC. Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
37
Collector- Bias Compare IC for the case when β = 100 with the case when β = 300. +VCC + 15 V
When β = 100,
VCC − VBE 15 V − 0.7 V = = 2.80 mA R 1.8 kΩ + 330 kΩ RC + B 100 β DC When β = 300, IC =
RB
RC 1.8 kΩ
330 kΩ
VCC − VBE 15 V − 0.7 V IC = = = 4.93 mA RB 330 k Ω 1.8 kΩ + RC + 300 β DC Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
38
Quiz
Q. The circuit shown is an example of a. base bias
+VCC
b. collector- bias c. emitter bias
RB
RC
d. voltage-divider bias
Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
39
Key Q-point The dc operating (bias) point of an amplifier specified by voltage and current values. DC load line A straight line plot of IC and VCE for a transistor circuit. Linear region The region of operation along the load line between saturation and cutoff. Stiff voltage A voltage divider for which loading effects divider can be ignored. The process of returning a portion of a circuit’s output back to the input in such a way as to oppose or aid a change in the output. Electronic Devices, 9th edition Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458. All rights reserved.
40
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