1. Bipolar Junction Transistor Amplifier
Chapter 3
Bipolar Junction Transistor Amplifier

2. 3.0 BIPOLAR JUNCTION TRANSISTOR AMPLIFIER
3.1 Amplifier Operation
3.2 Transistor ac Model
3.3 Common-Emitter Amplifier
3.4 Common-Collector Amplifier
3.5 Common-Base Amplifier
3.6 Multistage Amplifier

3. 3.1 Amplifier Operation
Biasing is to establish Q-point for small ac signal from antenna, microscopes, sensors etc to amplifies, often referred as small-signal amplifier.
Recall that small changes in Ib causes large changes in Ic.
The small ac voltage causes Ib to increase and decrease accordingly and with this small change in Ic mimic the input only with greater amplitude.
Fig 4-20a & b (stacked)

4. AC Quantities
As Ic inc, Vc dec. Ic varies above and below Q-point in phase with Ib. Vce varies above and below Q-point value 1800 out of phase with Vb. Transistor always produces a phase inversion between Vb and Vc.
Vce can represents rms, average, peak, or peak to peak, but rms will be assumed unless stated otherwise. vce can be any instantaneous value on the curve.

5. Boundary between cutoff and saturation is called linear region
Boundary between cutoff and saturation is called linear region. Transistor which operates in the linear region is called a linear amplifier.
A voltage-divider biased transistor with a sinusoidal ac source capacitively coupled to base through C1 and load capacitively coupled to the collector through C2.
C1 and C2 used to block dc and thus prevent the internal source resistance, Rs and load resistance, RL from changing the dc bias voltages at the base and collector. Only ac component reaches load because of the capacitive coupling.

6. A Graphical Picture
Vb produce Ib that varies above and below Q-point on ac load line, also Ic and Vce.
ac load line differ from dc load line because the effective ac collector resistance is RL in parallel with RC and is less than dc collector resistance RC.

7. Example :
Let Q as the selected operating point (quiescent point).
If Ib varies about 10 µA, find Vce and Ic variations

8. 3.2 Transistor ac model
To visualize the operation of a transistor in an amplifier circuit, it is often useful to represent the device by a model circuit.
A transistor model circuit uses various internal transistor parameter to represents its operation.
Transistor circuits can view by use of resistance or r parameters for better understanding.

9. r Paramaters r parameter description αac ac alpha (Ic/Ie) βac
ac beta (Ic/Ib)
r’e
ac emitter resistance
r’b
ac base resistance
r’c
ac collector resistance

10. Since the base resistance, r’b is small it is normally is not considered (can be replaced by a short) and since the collector resistance, r’c is fairly high and consider it as an open.
The emitter and base resistance, r’e is the main parameter that is viewed. Ic = αacIe = βacIb

11. Determining r’e by a Formula
r’e is most important, and it’s value is
r’e = 25 mV/IE
Example :
Calculate r’e, if IE = 5mA
r’e = 25mV/5mA = 5 Ω
Fig 6-6 r-parameter comparison
Relationship of transistor symbol to r-parameter model

12. Comparison of AC Beta (ac) to DC Beta (DC)
A graph of IC versus IB is nonlinear.
The two graphs best illustrate the difference between DC and ac. The two only differ slightly.
Fig 6-7 IC vs IB curves

13. h parameter h parameter Description Condition hi input impedance
output short hr
voltage feedback ratio
input open hf
forward current gain ho
output admittance
configuration
h parameters
Common-emitter
hie, hre, hfe, hoe
Common-base
hib, hrb, hfb, hob
Common-collector
hic, hrc, hfc, hoc

14. Relationships of h-Parameters and
r-Parameters
The ac current ratios, αac and βac, convert directly from h parameter as follows:
αac = hfb
βac = hfe
r’e = hre/hoe
r’c = (hre + 1)/hoe
r’b = hie - (1+ hfe) hre/hoe

15. 3.3 Common-Emitter Amplifier
Common-emitter amplifier exhibits high voltage and current gain. The Vout is 180º out of phase with Vin. Now lets use our dc and ac analysis methods to view this type of transistor circuit. It used voltage divider bias and coupling capacitor, C1 and C3.
Fig. 6-8 ce amp
Figure 3.0 : A common-emitter amplifier

16. DC Analysis
DC component of the circuit “sees” only the part of the circuit that is within the boundaries of C1, C2, and C3 as the dc will not pass through these components and consider open.
RINbase = βDCRE = 150 x 560Ω = 84KΩ
Since RINbase is 10 x R2, so neglect.
VB = (R2/(R1+R2)) x VCC
= (6.8KΩ/28.8KΩ) x 12V = 2.83V
VE = VB – VBE = 2.83V – 0.7V = 2.13V
IE = VE/RE = 2.13V/560Ω = 3.80mA = IC
VC = VCC – ICRC = 12V – 3.8mAx1KΩ
= 8.2V
VCE = VC – VE = 8.2V – 2.13V = 6.07V
Fig 6-9 dc eq. ce amp

17. AC equivalent circuit for the amplifier in Figure 3.0
AC Analysis
Basically replaces the capacitors C1, C2 and C3 with shorts because their value are selected so Xc is negligible at signal frequency and can be considered to be 0 Ω.
dc source being replace with ground, assume, Vs has internal resistance, rint= 0Ω, so V ac across ac source = 0. So Vcc are also effectively shorts to ground for ac analysis.
Fig 6-10&6-11 ac eq. ce circuit
AC equivalent circuit for the amplifier in Figure 3.0

18. RC and R1 have one end connected to ac ground, because in actual circuit they are connected to VCC which is in fact ac ground.
It a common-emitter amplifier because, C2 keeps emitter at ac ground which is the common point in circuit.
If internal resistance (re), of ac source is 0Ω, then all source voltage appears at base terminal.
If ac source has a nonzero internal resistance, 3 factor determine base voltage, source resistance, Rs, bias resistance R1||R2, and input resistance, Rinbase.
Total input resistance,
Rin(tot) = R1||R2||Rin(base)
Vb = (Rin(tot)/(Rs + Rin(tot))) x Vs
If Rs << Rin(tot), then Vb = Vs
where Vb is input voltage, Vin to amplifier.

19. Input Resistance and Output Resistance
Input resistance looking at base, Rin(base) = Vin/Iin = Vb/Ib
Base voltage, Vb = Ier’e and since Ie = Ic, then Ib = Ie/βac
So, Rin(base) = Ier’e/Ie/β ac = β ac r’e
Total input resistance, Rin(tot) = R1||R2||Rin(base)
Output resistance, looking at collector,
Rout = RC || r’c, but since r’c >> RC
So, Rout = RC
Fig 6-10&6-11 ac eq. ce circuit

20. Example : If ac input with Vs = 10mV rms, source resistance of 300Ω and IE of 3.8mA, find the signal voltage at base.
r’e = 25mV/IE = 25mV/3.8mA = 6.58Ω
Rin(base) = βr’e = 160 x 6.58Ω = 1.05KΩ
Rin(tot) = R1||R2||Rin(base) = 873 Ω
Vb = (Rin(tot)/(Rs + Rin(tot))) x Vs = 7.44mV

21. Voltage Gain
Ac voltage gain for common-emitter amplifier is developed using the model circuit below
Voltage gain is;
Av = Vout/Vin = Vc/Vb
Since, VC = α IeRC = IeRC
and Vb = =Ier’e
So Av = IeRC/Ier’e
Av = RC/r’e
Fig 6-14

22. Attenuation = Vs/Vb = (Rs + Rin(total)) / Rin(total)
Attenuation is the reduction in signal voltage as it passes through a circuit and correspond to a gain of less than 1.
Attenuation from the ac supply internal resistance and input resistance must be determined since it affects the overall gain.
Attenuation = Vs/Vb = (Rs + Rin(total)) / Rin(total)
The overall voltage gain of the amplifier, A’v is the voltage gain from base to collector, Av times the reciprocal attenuation.
A’v = (Vb/Vs)Av = Vc/Vs
Fig 6-19

23. Example: Suppose we have a source, Vs of 10mV
Example: Suppose we have a source, Vs of 10mV. Due to resistances, the voltage felt at base (amplifier input),Vb is 5mV. Determine the overall gain. Attenuation factor = 10mV/5mV = 2 The overall gain, A’v = Av x reciprocal of attenuation = 20 x (2)- = 10

24. Effect of the Emitter Bypass Capasitor on Voltage Gain
Emitter bypass capacitor, C2 provides an effective short to ac signal around RE, thus keep emitter at ac ground.
With bypass capacitor, the gain of a given amplifier is maximum and equal to RC/r’e.
The value of bypass capacitor must be large enough so that the reactance over the frequency range of the amplifier is very small (ideally 0Ω) compared to RE. The XC(bypass) should be 10 Xc << RE at minimum frequency for which the amplifier must operate.
Fig 6-19

25. Example : Select min C2 value if amp freq is 2kHZ to 10kHz
Since RE = 560Ω = 10XC, so XC = 56Ω
So C2 at 2kHz = 1/(2πfXC) = 1/(2(2 kHz)( 56Ω)) = 1.42µF
This is a minimum value for the bypass capacitor for this circuit. A large value can be use, although cost and physical size usually impose limitations.
Fig 6-8 ce amp

26. Voltage Gain without the bypass Capacitor
Without C2, emitter is no longer at ac ground, instead RE is now seen by the signal between emitter and ground and adds to internal re’ .
Av = RC/(re’+RE).
Example : Calculate base-to-collector voltage gain of the amplifier both without and with an emitter bypass capacitor if there is no load resistor.
Without C2 gain is, Av = RC/(r’e+RE) = 1.0 kΩ/6.58Ω =1.76
With C2 gain is, Av = RC/r’e = 1.0kΩ/6.58Ω = 152
The effect of RE is to decrease the ac voltage gain
Fig 6-18

27. Effect of a Load on the Voltage Gain
When load RL is added at the output capacitor end C3, effective resistance seen by the signal is RC in parallel with RL.
Rc = RC//RL
Replace RC with Rc, Av = Rc/re’
When Rc < RC, Av reduced, if RL >> RC, then Rc = RC and RL have little effect on Av.
Fig 6-18

28. Therefore, Av = Rc/r’e = 833 Ω/6.58 Ω = 127
Example : Calculate the base-to-collector voltage gain of the amplifier when load resistance of 5 kΩ in connected to output and RC is 1 k Ω. emitter is effectively bypassed and r’e = 6.58Ω.
ac collector resistance, Rc = RC||RL =RCRL/(RC+RL) = ((1.0 kΩ)(5 kΩ))/6 kΩ = 833 Ω
Therefore, Av = Rc/r’e = 833 Ω/6.58 Ω = 127
unloaded gain was found to be 152
Fig 6-18

29. Stability of the Voltage Gain
Stability is a measure of how well an amplifier maintains its design values over changes in temperature or for a transistor with a different β.
Bypassing RE with capacitor causes the gain to be more dependent on re’ since (Av = RC/r’e) ; r’e depends on IE and on temperature; reduced stability.
Without bypass capacitor, the gain is Av=RC/(re’+RE) and if RE is big enough. Av = RC/RE…. Which is less dependent on re’ or IE.

30. Swamping r’e to stabilize the voltage gain
Swamping is a method used to minimize the effect of r’e without reducing the voltage gain to it minimum value.
In a swamp amplifier, RE is partially bypassed so that a reasonable gain can be achieved and the effect of r’e on the gain is greaty reduced or eliminated.
The total external emitter resistance, RE is formed with two separate emitter resistors, RE1 and RE2 (RE2 is bypassed and the other is not).
Av= RC/(r’e + RE1)
= RC/RE1 (if RE1 is at least ten times larger than r’e then
the effect of r’e is minimized)

31. Both resistors (RE1 and RE2) affects the dc bias while only RE1 affects the ac voltage gain.
Fig 6-18

32. Examples :
RE2 bypassed by C2. Assume r’e = 20 Ω and since RE1 is more than 10 times r’e, the approximate voltage gain is
So, Av = RC/RE1 = 3.3kΩ/330Ω = 10

33. The Effect of Swamping on the Amplifier’s Input Resistance
ac input resistance, with RE completely bypassed, is
Rin = βacr’e
When emitter resistance, partially bypassed,
Rin = βac(re’+ RE1)
Vout at collector is 1800 out of phase with Vin at base, shown by – Av.

34. Total current signal produce by source is, Is = Vs/(Rin(tot) + Rs)
Current Gain
Overall current gain of the common-emitter, Ai = Ic/Is
Is is total signal current produced by the source, part of which is base current (Ib) and part of which (Ibias) goes through the bias circuit (R1||R2).
Total current signal produce by source is, Is = Vs/(Rin(tot) + Rs)
Power gain is, Ap = A’vAi, where A’v = Vc/Vs

35. 3.4 Common-Collector Amplifier
Common-collector amplifier is usually referred to as the emitter follower because there is no phase inversion or voltage gain. The output is taken from the emitter. The common-collector amplifier’s main advantages are it’s high current gain and high input resistance.
Fig 6-25 Emitter-follower

36. No phase inversion and Vout approx same as Vin.
Voltage Gain
Input signal is capacitive coupled to base, output signal is capacitive coupled to emitter, and collector is at ac ground.
No phase inversion and Vout approx same as Vin.
Av = Vout/Vin, Xc are negligible at freq
operation
Re = RE ||RL
Vout = IeRe and Vin = Ie(r’e + Re)
Av = Re/(re’ +Re)
if no load, Re = RE
Gain is always less than 1.
If Re >> r ’e than Av = 1
And since no inversion
So called emitter-follower
Emitter-follower model for voltage gain derivation

37. Input Resistance
Because of the high input resistance it used as a buffer to reduce the loading effect when a circuit is driving a low impedance loads.
Emitter resistor is never bypassed because Vout is taken across Re.
Rin(base) = Vin/Iin = Vb/Ib = Ie(re’+Re)/Ib
Since Ie = Ic = βIb
Rin(base) = βIb(re’+Re)/Ib = β(re’+Re)
If Re >> re’ so Rin(base) = βRe
Rin(tot) = R1 || R2 || Rin(base)
Fig 6-26 emitter-follower equivalent

38. Output resistance Current Gain Power Gain
With the load removed, Rout looking at emitter of emitter follower:
Rout = (Rs/ β) || RE (Rs - resistance of input source)
Output resistance is very low and this makes it useful for driving low impedance loads.
Current Gain
Current gain for emitter follower , Ai = Ie/Iin
where Iin = Vin/Rin(tot)
Power Gain
Power gain, Ap = AvAi
Since Av = 1, so Ap = Ai

39. Darlington Pair
Darlington pair is used to boost the input resistance to reduce loading of high output impedance circuits.
Collectors are joined together and emitter of the input transistor is connected to the base of the output transistor.
Ie1 = βac1Ib1
Ie1 = Ib2, so Ie2 = βac2Ie1
= βac1βac2Ib1
βac = βac1βac2
Neglecting r’e by assuming it is much
smaller than RE (r’e<<RE)
So Rin = βac1βac2RE
Fig 6-28 darlington pair

40. Application
Emitter-follower or buffer often used as interface between a circuit with high Rout and low RL.
Example : common-emitter amp with RC = 1.0kΩ and RL = 8Ω
If re’ = 5Ω, so Av = RC/re’ = 200 (with no load)
With load;
Rc = RC||RL = 1.0k Ω || 8 Ω = 7.94Ω,
Av = Rc/re’
= 7.49/5 = 1.59Ω (so not acceptable because Av is lost)
So it can be used to interface the amplifier and speaker.

41. 3.5 Common-Base Amplifier
Common-base amplifier has high voltage gain with a current gain no higher than 1. It has a low input resistance making it ideal for low impedance input sources. ac signal is applied to emitter and output is taken from the collector.
Fig 6-30

42. Base is the common terminal and is at ac ground because of C2
Base is the common terminal and is at ac ground because of C2. Input signal is capacitive coupled to emitter and output is capacitive coupled from collector to RL.
Voltage gain:
Av = Vout/Vin = Vc/Ve = IcRc/(Ie(r’e||RE)) = IeRc/(Ie(r’e||RE))
If RE >> r’e, then Av = Rc/r’e where Rc = RC||RL
No phase inversion from emitter to collector
Input resistance (looking in emitter): Rin(emitter) = Vin/Iin = Ve/Ie = (Ie(r’e||RE))/Ie
If RE >> r’e then Rin(emitter) = r’e this is typical
Output resistance (looking in collector): Rout = r’c||RC
Typical value rc >> RC, so Rout = RC
Current gain, Ai = 1
Power gain, Ap = Av

43. 3.6 Multistage Amplifiers
Two or more amplifiers can be connected to increase the gain of an ac signal. Basic objective is to increase overall voltage gain, Av.
Overall gain can be calculated by simply multiplying each gain.
A’v = Av1Av2Av3 ……
Voltage gain often expressed in decibels dB, Av (dB) = 20logAv
So, A’v(dB) = A’v1(dB) + A’v2(dB) + A’v3(dB) +…
Fig 6-32

44. Capacitively-Coupled Multistage Amplifier
Use two stage capacitive coupled amplifier where capacitive coupling keeps dc bias voltages separate but allows the ac to pass through to the next stage because Xc = 0 at frequency operation.
Av1 have loading effect on stage 2. Rin(tot)2 presents an ac load to first stage because coupling capacitor C3 appear short at signal freq.
From collector Q1, R5 and R6 appear parallel with Rin(base2). In other word, signal at collector Q1 see R3, R5, R6 and Rin(base2) of stage 2 all parallel to ac ground. Thus, effective ac collector resistance, Rc(Q1) is total of all this resistor in parallel.
Fig stage ce amp

45. A two-stage common-emitter amplifier

46. Av stage1 decrease by loading of stage 2 because effective collector resistance of stage1 is less than actual value of collector R3. Remember, Av = Rc/re’.
The ac collector resistance of the first stage, Rc1 = R3||R5||R6||Rinbase2
The voltage gain of the first stage, Av1 = Rc1/re’
The voltage gain of the second stage, Av2 = R7/re’
Overall voltage gain, Av’ = Av1 x Av2

47. DC voltages in the capacitively coupled multistage amplifiers
Since Q1 and Q2 identical, the dc voltage is same.
βdcR4 >> R2 and βdcR8 >> R6, so ;
VB = R2/(R1 + R2) x VCC
VE = VB – 0.7
IE = VE/R4
IC = IE
VC = VCC – ICR3

48. Direct coupled Multistage Amplifiers
For direct-coupled multi stage amplifiers, there are no coupling and no bypass capacitor in the circuit. dc collector voltage of stage 1, provide base-bias voltage for stage 2.
It’s has better low-frequency response than capacitive coupled type in which reactance of coupling and by-pass capacitor are increase. The increased reactance of capacitors at lower frequencies produces gain reduction in capacitively coupled amplifiers.
So the amplifiers are used to amplify low frequency without gain loss because no capacitive reactance in circuit.
The disadvantage is that small changes in dc bias from temperature changes or supply variations are amplifier by succeeding stage, which result in drift in dc level in circuit.
Fig 6-35 Direct-coupled amp