Recall Lecture 13 Biasing of BJT Three types of biasing Fixed Bias Biasing Circuit Biasing using...
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Transcript of Recall Lecture 13 Biasing of BJT Three types of biasing Fixed Bias Biasing Circuit Biasing using...
Recall Lecture 13
Biasing of BJT Three types of biasing
Fixed Bias Biasing Circuit Biasing using Collector to Base Feedback Resistor Voltage Divider Biasing Circuit
Applications of BJT As digital logic gates
NOT NOR
CHAPTER 5BASIC BJT AMPLIFIERS
(AC ANALYSIS)
The Bipolar Linear AmplifierThe Bipolar Linear Amplifier
Bipolar transistors have been traditionally used in linear amplifier circuits because of their relatively high gain.
To use the circuit as an amplifier, the transistor needs to be biased with a dc voltage at a quiescent point (Q-point) such that the transistor is biased in the forward-active region.
If a time-varying signal is superimposed on the dc input voltage, the output voltage will change along the transfer curve producing a time-varying output voltage.
If the time-varying output voltage is directly proportional to and larger than the time-varying input voltage, then the circuit is a linear amplifier.
The linear amplifier applies superposition principle Response – sum of responses of the circuit for
each input signals alone So, for linear amplifier,
DC analysis is performed with AC source turns off or set to zero
AC analysis is performed with DC source set to zero
EXAMPLE
iC , iB and iE,
vCE and vBE
Sum of both ac and dc components
Graphical Analysis and ac Equivalent CircuitGraphical Analysis and ac Equivalent Circuit
From the concept of small signal, all the time-varying signals are superimposed on dc values. Then:
and
PERFORMING DC and AC analysis
DC ANALYSIS AC ANALYSIS
Turn off AC SUPPLY = short circuit
Turn off DC SUPPLY = short circuit
DO YOU STILL REMEMBER?
DC equivalent AC equivalent
rd
idIDQ
VDQ = V
Let’s assume that Model 2 is used
AC equivalent circuit –
Small-Signal Hybrid-Small-Signal Hybrid-ππ Equivalent Equivalent
ib
OR
THE SMALL SIGNAL PARAMETERS
The resistance rπ is called diffusion resistance or B-E input resistance. It is connected between Base and Emitter terminals
The term gm is called a transconductance
ro = VA / ICQrO = small signal transistor output resistance
VA is normally equals to , hence, if that is the case, rO = open circuit
Hence from the equation of the AC parameters, we HAVE to perform DC analysis first in order to calculate them.
EXAMPLE
The transistor parameter are = 125 and VA=200V. A value of gm = 200 mA/V is desired. Determine the collector current, ICQ and then find r and ro
ANSWERS: ICQ = 5.2 mA, r= 0.625 k and ro = 38.5 k
CALCULATION OF GAIN
Voltage Gain, AV = vo / vs
Current Gain, Ai = iout / is
Small-Signal Voltage Gain: Av = Vo / Vs
ib
Common-Emitter Common-Emitter AmplifierAmplifier
Remember that for Common Emitter Amplifier, the output is measured at the collector terminal. the gain is a negative value
Three types of common emitter Emitter grounded With RE
With bypass capacitor CE
STEPSOUTPUT SIDE
1.Get the equivalent resistance at the output side, ROUT
2.Get the vo equation where vo = - ib ROUT
INPUT SIDE
3.Calculate Rib using KVL where Rib = vb / ib
4.Calculate Ri
5.Get vb in terms of vs – eg: using voltage divider.
6.Go back to vo equation and replace where necessary
VCC = 12 V
RC = 6 k93.7 k
6.3 k
0.5 kβ = 100VBE = 0.7VVA = 100 V
Emitter Grounded
Voltage Divider biasing:Change to Thevenin EquivalentRTH = 5.9 kVTH = 0.756 V
Perform DC analysis to obtain the value of IC
BE loop: 5.9IB + 0.7 – 0.756 = 0
IB = 0.00949
IC = βIB = 0.949 mA
Calculate the small-signal parameters
r = 2.74 k , ro = 105.37 k and gm = 36.5 mA/V
VCC = 12 V
RC = 6 k93.7 k
6.3 k
0.5 kβ = 100VBE = 0.7VVA = 100 V
Emitter Grounded
vb
Follow the steps1. Rout = ro || RC = 5.677 k
2. Equation of vo : vo = - ( ro || RC ) ib= - 567.7 ib
4. Calculate Ri RTH||r = 1.87 k
5. vb in terms of vs use voltage divider: vb = [ Ri / ( Ri + Rs )] * vs = 0.789 vs
3. Calculate Rib using KVL: ib r - vb = 0 hence Rib = r
so: vb = 0.789 vs replace in equation from step 1
6. Go back to equation of vo
vo = - 567.7 ib
but ib = vb / Rib (from step 3)
vo = - 567.7 [0.789 vs] / 2.74
vo = -163.5 vs
AV = vo / vs = - 163.5
vb
bring VS over
TYPE 2: Emitter terminal connected with RE – normally ro = in this type
VCC = 5 V
RC = 5.6 k250 k
75 k
0.5 k
RE = 0.6 k
β = 120VBE = 0.7VVA =
0.5 k
57.7 k
RC = 6 k
7.46 k
RE = 0.6 k
vb
1. Rout = RC = 6 k
2. Equation of vo : vo = - RC ib= - 720 ib
4. Calculate Ri RTH||Rib = 33.53 k
5. vb in terms of vs use voltage divider: vb = [ Ri / ( Ri + Rs )] * vs = 0.9853 vs
3. Calculate Rib using KVL: ib r + ie RE - vb = 0
but ie = (1+ ) ib = 121 ibso: ib [ 121(0.6) + 7.46 ] = vb Rib = 80.06 k
vb
so: vb = 0.9853 vs
6. Go back to equation of vo
vo = - 720 ib = - 720 [ vb / Rib ]
vo = - 720 [ 0.9853 vs / 80.06 ]
vo = - 8.86vs
AV = vo / vs = - 8.86bring VS over
vb
Circuit with Emitter Bypass Capacitor
● There may be times when the emitter resistor must be large for the purpose of DC design, but degrades the small-signal gain too severely.
● An emitter bypass capacitor can be used to effectively create a short circuit path during ac analysis hence avoiding the effect RE
TYPE 3: With Emitter Bypass Capacitor, CE
vb
CE becomes a short circuit path – bypass RE; hence similar to Type 1
VCC = 5 V
RC = 2.3 k20 k
20 k
0 k
RE = 5k
Bypass capacitor
β = 125VBE = 0.7VVA = 200 VIC = 0.84 mA
0 k
10 kRC = 2.3 k
238 k
3.87 k
vb
β = 125VBE = 0.7VVA = 200 VIC = 0.84 mA
Short-circuited (bypass) by the capacitor CE
vb
Follow the steps1. Rout = ro || RC = 2.278 k
2. Equation of vo : vo = - ( ro || RC ) ib= - 284.75 ib
4. Calculate Ri RTH||r = 2.79 k
5. vb in terms of vs use voltage divider: vb = [ Ri / ( Ri + Rs )] * vs = vs since RS = 0 k
3. Calculate Rib using KVL: ib r - vb = 0 hence Rib = r = 3.87 k
so: v b = vs
6. Go back to equation of vo
vo = - 284.75 ib = - 284.75 [ vb / Rib ]
vo = - 284.75 (vs) / 3.87
vo = -73.58 vs
AV = vo / vs = - 73.58bring VS over
vb