1. electronics fundamentals
circuits, devices, and applications
THOMAS L. FLOYD
DAVID M. BUCHLA
chapter 17

2. Bipolar junction transistors (BJTs)
The BJT is a transistor with three regions and two pn junctions. The regions are named the emitter, the base, and the collector and each is connected to a lead.
There are two types of BJTs – npn and pnp.
Separating the regions are two junctions.
C (Collector) C
Base-Collector junction n p
B (Base) p B n n
Base-Emitter junction p
E (Emitter) E

3. BJT biasing
For normal operation, the base-emitter junction is forward-biased and the base collector junction is reverse-biased.
For the pnp transistor, this condition requires that the base is more negative than the emitter and the collector is more negative than the base.
For the npn transistor, this condition requires that the base is more positive than the emitter and the collector is more positive than the base.
BC reverse- biased + +
pnp
npn +
BE forward- biased +

4. BJT currents
A small base current (IB) is able to control a larger collector current (IC). Some important current relationships for a BJT are: IC I IB I IE I

5. Because the base current is small, the approximation
Voltage-divider bias
Because the base current is small, the approximation
is useful for calculating the base voltage.
After calculating VB, you can find VE by subtracting 0.7 V for VBE. R1 RC VC
Next, calculate IE by applying Ohm’s law to RE: VB VE R2 RE
Then apply the approximation
Finally, you can find the collector voltage from

6. Example: Solution: Voltage-divider bias
Calculate VB, VE, and VC for the circuit.
Solution:
3.02 V
+15 V
VE = VB V =
2.32 V R1 RC
27 kW
2.2 kW
2N3904 R2 RE
6.8 kW
1.0 kW
9.90 V

7. Collector characteristic curves
The collector characteristic curves are a family of curves that show how collector current varies with the collector-emitter IC
voltage for a given IB.
IB6
The curves are divided into three regions:
IB5
The breakdown region is after the active region and is is characterized by rapid increase in collector current. Operation in this region may destroy the transistor.
The active region is after the saturation region. This is the region for operation of class-A operation.
IB4
The saturation region occurs when the base-emitter and the base-collector junctions are both forward biased.
IB3
IB2
IB1
IB = 0
VCE

8. Load lines
A load line is an IV curve that represents the response of a circuit that is external to a specified load.
For example, the load line for the Thevenin circuit can be found by calculating the two end points: the current with a shorted load, and the output voltage with no load.
I (mA) 6
Load line
2.0 kW 4
+12 V 2
INL = 0 mA
VNL = 12 V
ISL = 6.0 mA
VSL = 0 V
V (V) 4 8 12

9. Load lines
The IV response for any load will intersect the load line and enables you to read the load current and load voltage directly from the graph.
I (mA)
Example:
Read the load current and load voltage from the graph if a 3.0 kW resistor is the load.
IV curve for
3.0 kW resistor 6
2.0 kW 4
Q-point
RL =
+12 V 2
3.0 kW
V (V) 4 8 12
VL = 7.2 V
IL = 2.4 mA

10. curve and the base current establish the Q-point.
Load lines
The load line concept can be extended to a transistor circuit. For example, if the transistor is connected as a load, the transistor characteristic
I (mA)
curve and the base current establish the Q-point. 6
2.0 kW 4
+12 V 2
V (V) 4 8 12

11. Assume the IV curves are as shown:
Load lines
Load lines can illustrate the operating conditions for a transistor circuit.
Assume the IV curves are as shown:
If you add a transistor load to the last circuit, the base current will establish the Q-point. Assume the base current is represented by the blue line.
I (mA)
For this base current, the Q-point is: 6
2.0 kW 4
+12 V 2
V (V) 4 8 12
The load voltage (VCE) and current (IC) can be read from the graph.

12. Load lines
Example:
For the transistor, assume the base current is established at 10 mA by the bias circuit. Show the Q-point and read the value of VCE and IC.
The Q-point is the intersection of the load line with the 10 mA base current.
IC (mA)
2.0 kW
IB = 25 mA
+12 V 6
IB = 20 mA
IB = 15 mA 4
IB = 10 mA 2
IB = 5.0 mA
Bias circuit
VCE (V) 4 8 12
VCE = 7.0 V; IC = 2.4 mA

13. Example: Solution: Signal operation
When a signal is applied to a transistor circuit, the output can have a larger amplitude because the small base current controls a larger collector current.
IC (mA)
Example:
For the load line and characteristic curves from the last example (Q-point shown) assume IB varies between 5.0 mA and 15 mA due to the input signal. What is the change in the collector current?
IB = 25 mA 6
IB = 20 mA
IB = 15 mA 4
IB = 10 mA 2
Solution:
IB = 5.0 mA
The operation along the load line is shown in red.
VCE (V) 4 8 12
Reading the collector current, IC varies from 1.2 mA to 3.8 mA.

14. CE amplifier
In a common-emitter amplifier, the input signal is applied to the base and the output is taken from the collector. The signal is larger but inverted at the output.
VCC
Output coupling capacitor RC R1 C2
Input coupling capacitor C1 R2
Bypass capacitor C3 RE

15. Example: Solution: Voltage gain of a CE amplifier
Calculate the voltage gain of the CE amplifier. The dc conditions were calculated earlier; IE was found to be 2.32 mA.
Solution:
VCC = +15 V RC R1 C2
2.2 kW C1
27 kW
2N3904
1.0 mF
2.2 mF R2 RE C3
204
6.8 kW
1.0 kW
100 mF
Sometimes the gain will be shown with a negative sign to indicate phase inversion.

16. Input resistance of a CE amplifier
The input resistance of a CE amplifier is an ac resistance that includes the bias resistors and the resistance of the emitter circuit as seen by the base.
VCC = +15 V
Because IB << IE, the emitter resistance appears to be much larger when viewed from the base circuit. The factor is (bac+1), which is approximately equal to bac. RC R1 C2
2.2 kW C1
27 kW
2N3904
1.0 mF
2.2 mF R2 RE C3
Using this approximation, Rin(tot) = R1||R2||bacre.
6.8 kW
1.0 kW
100 mF

17. Example: Solution: Input resistance of a CE amplifier
Calculate the input resistance of the CE amplifier. The transistor is a 2N3904 with an average bac of 200. The value of re was found previously to be 10.8 W. Thus, bacre = 2.16 kW.
VCC = +15 V
Solution: RC R1 C2
Rin(tot) = R1||R2||bacre
2.2 kW C1
27 kW
= 27 kW||6.8 kW||2.16 kW
2N3904
1.0 mF
= 1.55 kW
2.2 mF R2 RE C3
Notice that the input resistance of this configuration is dependent on the value of bac, which can vary.
6.8 kW
1.0 kW
100 mF

18. CC amplifier
In a common-collector amplifier, the input signal is applied to the base and the output is taken from the emitter. There is no voltage gain, but there is power gain.
The output voltage is nearly the same as the input; there is no phase reversal as in the CE amplifier.
VCC R1 C1
The input resistance is larger than in the equivalent CE amplifier because the emitter resistor is not bypassed. R2 RE

19. Example: Solution: CC amplifier
Calculate re and Rin(tot) for the CC amplifier. Use b = 200.
Solution:
VCC = +15 V R1 C1
22 kW
3.2 W
2N3904
Rin(tot) = R1||R2||bac(re+ RE)
10 mF R2 RE
27 kW
= 22 kW||27 kW|| 200 (1.0 kW) =
9.15 kW
1.0 kW
Because re is small compared to RE, it has almost no affect on Rin(tot).

20. Class B amplifiers
The class B amplifier is more efficient than the class A amplifier. It is widely used in power amplifiers.
VCC
The amplifier shown uses complementary transistors – one is an npn and the other is a pnp. R1 C1 Q1
The bias method shown avoids cross-over distortion by bringing the transistors just above cutoff using diodes. D1 C3 C2 D2 Q2 RL R2

21. The BJT as a switch
BJTs are used in switching applications when it is necessary to provide current drive to a load.
VCC
VCC
In switching applications, the transistor is either in cutoff or in saturation. RC RC
In cutoff, the input voltage is too small to forward-bias the transistor. The output (collector) voltage will be equal to VCC.
VOUT
IIN = 0
= VCC
IIN > IC(sat)/bDC
= 0 V
When IIN is sufficient to saturate the transistor, the transistor acts like a closed switch. The output is near 0 V.

22. Feedback oscillators
An oscillator is a circuit that generates a repetitive waveform on its output. A feedback oscillator uses positive feedback from the output to sustain oscillations.
Conditions for oscillations are
The phase shift around the loop must be 0o. Av
Acl = AvB = 1
2. The closed loop gain must be 1 (unity gain).
Feedback circuit B

23. Feedback oscillators
The Colpitts and Hartley oscillators are examples of a feedback oscillator. The amplifier sections are nearly identical, but the LC feedback network is different.
To obtain the required signal reinforcement (positive feedback), the amplifier inverts the signal (180o) and the feedback network shifts the phase another 180o.
An additional capacitor is in the amplifier to block dc.
Hartley oscillator feedback network
Colpitts oscillator feedback network.

24. Selected Key Terms Bipolar junction transistor (BJT) Class A amplifier
Saturation
A transistor with three doped semiconductor regions separated by two pn junctions.
An amplifier that conducts for the entire input cycle and produces an output signal that is a replica of the input signal in terms of its waveshape.
The state of a transistor in which the output current is maximum and further increases of the input variable have no effect on the output.

25. Selected Key Terms Cutoff Q-point Amplification Common-emitter (CE)
Class B amplifier
The non-conducting state of a transistor.
The dc operating (bias) point of an amplifier.
The process of producing a larger voltage, current or power using a smaller input signal as a pattern.
A BJT amplifier configuration in which the emitter is the common terminal.
An amplifier that conducts for half the input cycle.

26. Selected Key Terms
Oscillator
Feedback
A circuit that produces a repetitive waveform on its output with only a dc supply voltage as an input.
The process of returning a portion of a circuit’s output signal to the input in such a way as to create certain specified operating conditions.

27. Quiz
1. The Thevenin circuit shown has a load line that crosses the y-axis at
a. +10 V.
b. +5 V.
c. 2 mA.
d. the origin.
5.0 kW
+10 V

28. Quiz
2. In a common-emitter amplifier, the output ac signal will normally
have greater voltage than the input.
have greater power than the input.
be inverted.
all of the above.

29. Quiz
3. In a common-collector amplifier, the output ac signal will normally
have greater voltage than the input.
have greater power than the input.
be inverted.
have all of the above.

30. Quiz 4. The type of amplifier shown is a a. common-collector.
b. common-emitter.
c. common-drain.
d. none of the above.
VCC R1 C1 R2 RE

31. Quiz
5. Crossover distortion in a Class B amplifier happens during the time interval between positive and negative alternations of the input when
a. one transistor is conducting while the other is off.
b. both transistors are conducting.
c. neither transistor is conducting.
d. all of the above.

32. Quiz 6. The bypass capacitor in a Common-Emitter amplifier will
a. short AC voltages across RE.
b. decreases the signal voltage gain.
c. all of the above.
d. none of the above.

33. Quiz 7. The conditions for oscillation are:
a. feedback loop phase shift must be 0 degrees.
b. loop gain must be 1.
c. both a and b.
d. none of the above.

34. Quiz 8. The quartz in a crystal oscillator
a. is used to fine tune the resonant frequency.
b. is used as the resonant tank circuit.
c. provides mediocre stability
d. all of the above.

35. Quiz 9. A Colpitts or Hartley oscillator both have
a. positive feedback.
b. amplification.
c. a closed loop gain of 1.
d. all of the above.

36. Quiz 10. The following oscillators use feedback:
a. RC phase-shift oscillator.
b. Hartley oscillator.
c. Colpitts oscillator.
d. all of the above.

37. Quiz
Answers:
1. c
2. d
3. b
4. a
5. c
6. a
7. c
8. b
9. d
10. d