1. 2.5 Transistor Applications

2. BJT as an Amplifier
Amplification is the process of linearity increasing the amplitude of an electrical signal.
dc current
- VBE, VCB, VCE are dc voltages from one transistor terminal to another.
- VB, VC, VE are dc voltages from transistor terminal to ground.
- IB, IC and IE are the dc transistor currents.
- RE is an external dc emitter resistance
ac current
- Vbe, Vcb, Vce are ac voltages from one transistor terminal to another.
- Vb, Vc, Ve are ac voltages from transistor terminal to ground.
- Ib, Ic and Ie are the ac transistor currents.
- Internal ac resistances, r’
- Internal ac emitter resistances, r’e.
- Re is an external ac emitter resistance

3. Amplification of a relatively small ac voltage can be had by placing the ac signal source in the base circuit.
Recall that small changes in the base current circuit causes large changes in collector current circuit.
The small ac voltage causes the base current to increase and decrease accordingly and with this small change in current the collector current will mimic the input only with greater amplitude.

4. Forward bias b-e junction present low r’e, so
Ie = Ic = Vb/r’e
Vc = IcRC Vout = AvVb
= IeRC
Av = Vc/Vb = IeRC/Ier’e
= RC/r’e
Exercise :
Calculate Av and Vout if
r’e = 70 Ω

5. BJT as Switch
A transistor when used as a switch is simply being biased so that it is in cutoff (switched off) or saturation (switched on). Remember that the VCE in cutoff is VCC and 0V in saturation.
Fig 4-22

6. In cutoff condition, BE junction is not forward-bias, all current is zero and VCE(cutoff) = VCC.
In saturation condition, BE junction is forward-bias, so enough IB to produce max IC, and transistor is saturated.
IC(sat) = (VCC – VCE(sat)) /RC, VCE(sat) << VCC, so neglected.
IB(min) for saturation = IC(sat) / ß
Exercise :
Calculate VCE when Vin = 0V.
Calculate IB(min) to saturated if ß= 100, neglect VCE(sat)
Calculate RB max when Vin = 5V

7. A simple application of a transistor switch
A square wave input voltage with a period of 2 s is applied to the input.
When 0 V, the transistor is cutoff, no Ic, then the LED does not emit light.
When input is high level, the transistor saturates, LED is forward biased, Ic flow though the LED and causes it to emit light.

8. 2.6 Transistor Packages and Terminal Identification

9. General Purpose tr is used for low or medium power amplifier or switching circuit. Packages are either plastic or metal.

10. Typical multiple-transistor packages

11. Power Transistors
Used to handle large current more 1A and/or large voltage. Package are metal tab or case and a heat sink for heat dissipation

12. .
RF Transistors
Design to operate at extremely high frequencies and used in communication system. The shape and lead configuration are designed to optimize certain high-freq parameters.

13. 2.7 Typically Faults and Troubleshooting

15. Troubleshooting
Testing a transistor can be viewed more simply if you view it as testing two diode junctions. Forward bias having low resistance and reverse bias having infinite resistance.
Fig 4-32a&b

16. The diode test function of a multimeter is more reliable than using an ohmmeter. Make sure to note whether it is an npn or pnp and polarize the test leads accordingly.
Fig DMM test

17. Testing a defective npn transistor
Testing a defective npn transistor. Leads are reversed for a pnp transistor.

18. 2.8 DC Operating Point

19. The goal of amplification in most cases is to increase the amplitude of an ac signal without altering it. If not correctly bias, output can goes into saturation or cutoff
In (a), output signal is an amplified replica of the input signal (it is inverted which mean 1800 out of phase with the input).
In (b), illustrates limiting of the positive portion of the output voltage as a result of a Q-point (dc operating point) being too close to cutoff.
In (c), illustrates limiting of the negative portion of the output voltage as a result of a Q-point (dc operating point) being too close to saturation.
Fig 5-1a, b, & c
Q-Point to close to cutoff
Q-Point = dc operating point
Q-Point to close to saturation

20. -Graphical Analysis-
How to draw a dc load line for a given biased transistor circuit?
Figure (a) shows a transistor that is biased with VCC and VBB to obtain IB, IC, IE and VCE.
Figure (b) shows the characteristic curves. These curve will be used to graphically illustrate the effects of dc bias.
Figure : A dc-biased transistor circuit with variable bias voltage (VBB)
for generating the collector characteristics curves shown in part (b)

21. For a transistor circuit to amplify it must be properly biased with dc voltages. The dc operating point between saturation and cutoff is called the Q-point. The goal is to set the Q-point such that it does not go into saturation or cutoff when an ac signal is applied.
VBB is adjusted: (To illustrated of Q-point adjustment) If IB = 200µA, so IC = 20mA and VCE = VCC – ICRC = 10 – (20mA)(220Ω) = 5.6V, Q1 If IB = 300µA, so IC = 30mA and VCE = VCC – ICRC = 10 – (30mA)(220Ω) = 3.4V, Q2 If IB = 400µA, so IC = 40mA and VCE = VCC – ICRC = 10 – (40mA)(220Ω) = 1.2V, Q3 Q1, Q2 and Q3 is shown by diagram below. VCC – ICRC – VCE = 0 IC = -(1/RC)VCE + VCC/RC y = mx + b straight line equation.

22. Illustration of Q-point adjustments.

23. By joining the 3 points, the straight line established and can be drawn (known as load line)
Recall that the collector characteristic curves graphically show the relationship of IC and VCE for different IB. With the DC load line superimposed across the collector curves for this particular transistor we see that 30 mA of collector current is best for maximum amplification, giving equal amount above and below the Q-point. Note that this is three different scenarios of IC being viewed simultaneously.
Fig 5-2a & Fig 5-4

24. Linear region : Region along load-line between saturation and cutoff
Linear region : Region along load-line between saturation and cutoff. With a good Q-point established, IC swings do not exceed the limits of operation (saturation and cutoff). However, applying too much ac voltage to the base would result in driving the collector current into saturation or cutoff resulting in a distorted or clipped waveform.
Fig 5-5 circuit and load line w/signals

25. Transistor being driven into saturation and/or cutoff.

26. Exercise : Assume ß = 200, find IB (max peak) for linear operation.
IBQ = (VBB – VBE)/RB = (10 – 0.7)/47KΩ = 198µA
ICQ = ß IB = 200 x 198 µA = 39.6 mA
VCEQ = VCC – ICRC = 20V – 13.07V = 6.93V
Ideal saturation, IC(sat) = VCC/RC = 20V/330Ω = 60.6 mA
ICpeak = IC(sat) – ICQ = 60.6mA – 39.6mA = ±21mA
IB(max-peak) = ICpeak/ß = 21mA/200 = 105µA

27. 2.9 Voltage-Divider Bias

28. Voltage-divider bias is the most widely used type of bias circuit
Voltage-divider bias is the most widely used type of bias circuit. Only one power supply is needed and voltage-divider bias is more stable (independent) than other bias types.
Two current path to ground, I2 and IE. R1 and R2 are used to provide the needed voltage to point A (base).
If IB is not small enough, resistance to ground from the base, RINbase is significant enough to consider in most cases.
Fig 5-9 Voltage-Div. Bias

29. In the case where base to ground resistance (input resistance) RINbase is low enough to consider, so RIN(base) = βDCRE
We can view the voltage at point A of the circuit in two ways, with or without the input resistance(point A to ground) considered.
Fig 5-10a & b

30. Input Resistance at base RINbase = VIN/IIN VIN = VBE + IERE
Assume, VBE << IERE, VIN = IERE = ßIBRE
IIN = IB
RINbase = VIN/IIN = ßIBRE/IB = ßRE
Exercise : Calculate, RINbase if ß = 125, RC = 560Ω and RE = 1.0KΩ
RINbase = ßRE
= 125 x 1.0KΩ
= 125KΩ

31. Analysis of Voltage-Divider Bias Circuit
RINbase = VIN/IIN = ßIBRE/IB = ßRE
Total resistance to ground, R2 || RINbase = R2 || ßRE
Voltage divider R1 and ßRE parallel with R2
If ßRE << R2 (Normally 10 times) - Not stiff
VB = (R2 || ßRE/(R1 + R2 || ßRE) VCC
If ßRE >> R2 (Normally 10 times) – Stiff
VB = (R2/(R1 + R2))VCC
Stiff voltage divider –
a voltage divider in which the base current is small compared to the current in R2
Fig 5-9 Voltage-Div. Bias

32. VE = VB – VBE
IE = VE/RE and assume, IC = IE
VC = VCC – ICRC
VCE = VC – VE
VCC – ICRC – IERE – VCE = 0
Since IC ≈ IE
VCE = VCC – IC(RC + RE)

33. Exercise : Calculate VCE and IC if ß = 100
RINbase = ßRE = 100 x 560Ω = 56KΩ (10 times value R2, neglect)
VB = (R2/(R1 + R2))VCC = (5.6KΩ/ 15.6KΩ)10V = 3.59V
VE = VB – VBE = 3.59V – 0.7V = 2.89V
IC = IE = VE/RE = 2.89V/ 5.6KΩ = 5.16mA
VCE = VCC – IC(RC + RE)
= 10V – 5.16mA x 1.56KΩ = 1.95V

34. 2.10 Others Bias Methods

35. Substituting the expression for IB into the formula IC = ß , yields
Base Bias
VCC – VRB – VBE = 0
VCC – IBRB – VBE = 0
IB = (VCC – VBE)/RB
VCC – ICRC – VCE = 0
VCE = VCC – ICRC
Substituting the expression for IB into the formula IC = ß , yields
IC = ß ((VCC – VBE)/RB)
This type of circuit is very unstable and ß dependent, since ß changes with temperature and IC also from one transistor to another due to manufacturing variations.
Fig 5-19 Base bias circuit

36. Emitter Bias
The circuit is independent of ß making it as stable as the voltage-divider type, but it requires two power supplies.
IB = IE/ ß
IC = IE = (-VEE-VBE) /(RE + RB/ß)
VE = VEE + IERE
VB = VE + VBE
VC = VCC – ICRC
IE depend to VBE and β, both change with temperature and IB
This can improve by, RE >> RB/ β, IE = -(VEE – VBE)/RE
VEE >> VBE, IE = VEE/RE
Fig 5-21a npn emitter bias

37. Collector-Feedback Bias
RB is connected to the collector rather than to VCC. The negative feedback creates an ‘offsetting’ effect that tends to keep Q-point stable. IF IC↑ VC↓ VRB↓ IB ↓. The decrease of IB produces less IC, drop VRC and thus offset the decrease in VC.
IB = (VC – VBE)/RB
VC = VCC – ICRC ( Assume Ic>>IB)
IC = (VCC – VBE)/(RC + RB/b)
Since emitter is ground, VCE = VC
VCE = VCC – ICRC
IC depend on β and VBE, and can minimize by,
RC >> RB and VCC >> VBE
Fig 5-23 collector feedback

38. A man came to the Prophet and said, ‘O Messenger of God
A man came to the Prophet and said, ‘O Messenger of God! Who among the people is the most worthy of my good companionship? The Prophet said: Your mother. The man said, ‘Then who?' The Prophet said: Then your mother. The man further asked, ‘Then who?' The Prophet said: Then your mother. The man asked again, ‘Then who?' The Prophet said: Then your father.’ (Bukhari, Muslim).