Revision Amplifier Configuration Electronics Theory

Electronic Circuits
Revision on
Basic Transistor Amplifiers
Contents
• Biasing
• Amplification principles
• Small-signal model development for BJT

2
Prof. C.K. Tse: Revision on Amplifier Configurations
Aim of this chapter
To show how transistors can
be used to amplify a signal.
amplifier

3
Basic idea
amplifier
Step 1: Set the transistor at a certain DC level
0.6V
7V
Step 2: Inject a small signal to the input and get a bigger output
— biasing
— coupling

4
Biasing the transistor
To set the transistor to a certain DC level = To set VCE and IC
RL
RB
VBE
+
–
VCE
+
–
IC
IB
Transistor:
β = 100
VCC=10V Suppose we want the following biasing condition:
IC = 10 mA and VCE = 5 V
Find RB and RL
Start with VBE ≈ 0.7 V.
Then, IB = (10 – VBE )/ RB = (10 – 0.7)/ RB
IC = βIB = 100 (10 – 0.7)/ RB = 10 mA
So, RB = 94kΩ
Also, VCE = 10 – RL IC
Hence, 5 = 10 – 10RL
So, RL = 0.5kΩ

5
β dependent biasing — bad biasing
RL
RB
VBE
+
–
VCE
+
–
IC
IB
Transistor:
β = 100
VCC=10V
This is a bad biasing circuit!
because it relies on the accuracy of β,
but β can be ±50% different from what is
given in the databook.
Now, let’s go to the lab and try using RB = 94kΩ and
RL = 0.5kΩ, and see if we get what we want.
…totally wrong! We don’t get IC = 10mA and VCE = 5V

6
A slightly better biasing method
RL
RB1
VBE
+
–
VCE
+
–
IC
IB
VCC=10V
Again, our objective is to find the resistors such
that IC = 10mA and VCE = 5V.
RB2
0.6 = 10 ×
RB 2
RB1 + RB2
First, if IB is small, we can approximately write
Suppose we get IC = 10mA. Then RL = 0.5kΩ.
RB1
RB 2
=
94
6
⇒
We can start with RB1 = 940Ω and RB2 = 60Ω.
Such resistors will make sure IB is much smaller
than the current flowing down RB1 and RB2, which
is consistent with the assumption.
What we need in practice is to fine tune RB1 or RB2
such that VCE is exactly 5V.

7
A much better biasing method —
emitter degeneration
RL
RB1
VCE
+
–
IC
IB
VCC=10V
Again, our objective is to find the resistors such
that IC = 10mA and VCE = 5V.
RB2
RE
Set VE = 2V, say. Then, RE = 2V/10mA = 0.2kΩ.
Surely, RL = 0.5kΩ in order to get VCE = 5V.
Finally, we have VB = VE + 0.6. Therefore, if IB is
small compared to IRB1 and IRB2, we have
+
VE
–
RB1
RB 2
=
74
26
Hence, RB1 = 740Ω and RB1 = 260Ω.
NOTE: β is never used in calculation!!

8
Stable (good) biasing
RL
RB1
VCE
+
–
IC
IB
VCC=10V
RB2
RE
+
VE
–
Summary of biasing with emitter degeneration:
Choose VE , IC and VCE .
RE RL
Use VBE ≈ 0.6 to get VB.
Then use
to choose RB1 and RB2 such that IB is much smaller the
current flowing in RB1 and RB2.
VB
RB1
RB 2
=
10−VB
VB

9
Terminology
The following are the same:
Biasing point
Quiescent point
Operating point (OP)
DC point

10
Alternative view of biasing
RL
VBE
+
–
VCE
+
–
IC
VCE
+
–
VR
+
–
+
–
VCC
VCC
IC
VCE
IC
VCC
RL Load line
Slope=–1/RL
operating point

11
What controls the operating point?
RL
VBE
+
–
VCE
+
–
IC
VCC
VCE
IC
Load line
Slope=–1/RL
VCC
operating point
a bigger VBE
a smaller RL
VBE or IB controls the OP
RL also controls the OP
CONCLUSION:

12
What happens if VBE dances up and
down?
RL
VBE
+
–
VCE
+
–
IC
VCC
VCE
IC
Load line
Slope=–1/RL
VCC
a bigger VBE = 0.65
a smaller VBE = 0.6
The OP also
dances up and
down along the
load line.
VCE also moves up
and down.
Typically, when VBE
moves a little bit, VCE
moves a lot!
THIS IS CALLED
AMPLICATION.

13
Animation to show amplifier action

14
Derivation of voltage gain
Question: what is ?
∆Vo
∆Vin
=
∆VCE
∆VBE
RL
VBE
+
–
VCE
+
–
IC
VCC
Then, what relates ∆IC and ∆VBE ?
Clearly, Ohm’s law says that
VCE = VCC − IC RL ⇒ ∆VCE = −RL .∆IC
Last lecture: transconductance gm =
∆IC
∆VBE
Hence, ∆VCE
∆VBE
= −gmRL

15
Common-emitter amplifier
RL
vBE
+
–
vCE
+
–
IC
VCC
RB1
RB2
The one we have just studied is called COMMON-EMITTER amplifier.
Small-signal voltage gain
= –gmRL
That means we can
increase the gain by
increasing gm and/or RL.
Output waveform is anti-
phase.
SUMMARY:

16
How do we inject signal into the amplifier?
RL
VBE
+
–
vCE
+
–
IC
VCC
RB1
RB2
?
vin
±20mV
= VCE + vCE
~
~
∆vCE
or
∆vin

17
Note on symbols
vCE = VCE + vCE
~
total signal
(large signal) operating point
or
DC value
or
quiescent point
small signal
or
ac signal
= + Total signal
a
DC point
A
Small signal
a or ∆a
~

18
Solution: Add the same biasing DC level
RL
vBE
+
–
vCE
+
–
IC
VCC
RB1
RB2
But, it is impossible to find a
voltage source which is
equal to the exact biasing
voltage across B-E.
VBE could actually be
0.621234V, which is
determined by the network
RB1, RB2 and the transistor
characteristic!!
How to apply the exact VBE?
vin
±20mV
+
–
Exactly the
same biasing
VBE
~

19
The wonderful voltage source: capacitor
RL
VBE
+
–
VCE
+
–
IC
VCC
RB1
RB2
VC
+
–
The capacitor voltage is
exactly equal to VBE
because DC current
must be zero
0A

20
Solution — insert coupling capacitor
RL
vBE
+
–
vCE
+
–
IC
VCC
RB1
RB2
vin
±20mV
– +
DC voltage
equal to exactly
the same
biasing VBE
This is called a
coupling capacitor
~

21
Complete common emitter amplifier
RL
vBE
+
–
vCE
+
–
IC
VCC
RB1
RB2
vin
vo
+
–
+
–
– + + –
coupling capacitors
(large enough so that they become
short-circuit at signal frequencies)
~
~

22
Can we simplify the analysis?
vin vo
+
–
+
–
common emitter
amplifier
We are mainly interested in the ac signals.
The DC bias does not matter!
Can we create a simple circuit just to look at ac signals?
~ ~

23
Small-signal model
vin vo
+
–
+
–
common emitter
amplifier
? ?
What is the loading
(resistance) seen
here?
What is the
Thévenin or Norton
equivalent circuit
seen here?
1 2
Two basic questions:
~ ~

24
Small-signal model of BJT: objectives
rin
Ro
To find:
rin
Ro
Gm
rin
Ro
rin
Ro
Am
+
–
Gmvin Amvin
or
Norton form Thévenin form

25
Derivation of the small-signal model
rπ
Input side:
iB
vBE
+
–
rin =
vB E
iB
=
vB E
iC / β
For small-signal,
rin =
∆vBE
∆iB
=
∆vBE
∆iC / β
=
β
(∆iC / ∆vBE )
=
β
gm
where gm is the BJT’s transconductance
rπ = β/gm

26
Output side:
∆vCE = −∆iC × RL
= −gm∆vBE ×RL
where gm is the BJT’s transconductance
RL
vCE
+
–
IC
RL
vCE = VCC – ICRL
VCC
For small-signal,
gmvBE
~
vCE
~
+
–

27
Output side:
∆vCE
RL
+
∆vCE
ro
= −∆iC
where ro is the Early resistor of the BJT.
RL
vCE
+
–
IC
RL
VCC
gmvBE
~
vCE
~
+
–
Including BJT’s Early effect
∆vCE = −∆iC (RL ro )
= −gm∆vBE (RL ro)
ro
Recall: ro = VA/IC , where VA is typically about 100V.
A very rough approx. is ro = ∞.

28
Initial small-signal model for BJT
gmvBE
~
vCE
~
+
–
ro
rπ
B
E
C
vBE
~
+
–
“MUST” REMEMBER Small-signal
BJT parameters:
gm =
IC
(kT /q)
=
IC
VT
rπ =
β
gm
ro =
VA
IC
VT is thermal voltage
≈ 25mV
VA is Early voltage
typically ~ 100V
BJT model

29
Initial small-signal model for FET
gmvGS
~
vDS
~
+
–
ro
G
S
D
vGS
~
+
–
Similar to BJT, but input resistance is ∞.
Small-signal
FET parameters:
gm = 2 K ID
ro =
1
λ
λ is the channel length modulation
parameter
K is a semiconductor parameter
FET model
All amplifier configurations using BJT can be likewise constructed
using FET.

30
Example: common-emitter amplifier
RL
vBE
+
–
VCC
RB1
RB2
+
vin
–
+
vo
–
Assume the coupling caps are large
enough to be considered as short-
circuit at signal frequency
B C
E E
gmvBE
~
rπ
ro
RL
VCC is ac 0V.
RB1 ||RB1

31
Complete model for common-emitter amplifier
Complete model:
gmvBE
~
rπ ro RL
RB1 ||RB1
+
vin
–
+
vo
–
+
vBE
–
~
gmvBE
~
RL||ro
RB1 ||RB1 || rπ
+
vin
–
+
vo
–
+
vBE
–
~
Simplified model:
Total input resistance
Rin = RB1 ||RB1 || rπ
Total output resistance
Ro = RL||ro
Voltage gain
vo
vin
= −gm (RL || ro )
≈ −gmRL

32
Alternative model for common-emitter amplifier
gm(RL||ro) vBE
RL||ro
RB1 ||RB1 || rπ
+
vin
–
+
vo
–
+
vBE
–
~
Total input resistance
Rin = RB1 ||RB1 || rπ
Total output resistance
Ro = RL||ro
Voltage gain
vo
vin
= −gm (RL || ro )
≈ −gmRL
Output in Thévenin form:
–
+
~

33
More about common-emitter amplifier
gm(RL||ro) vBE
RL||ro
RB1 ||RB1 || rπ
+
vin
–
+
vo
–
+
vBE
–
~ –
+
~
Because the output resistance is quite large (equal to RL||ro ≈ RL), the
common-emitter amplifier is a POOR voltage driver. That means, it is not a
good idea to use such an amplifier for loads which are smaller than RL. This
makes it not suitable to deliver current to load.
1kΩ, for example
10Ω
practically no
output!!

34
Bad idea — wrong use of common-emitter amplifier
1kΩ
vBE
+
–
+10V
RB1
RB2
+
vin
–
+
vo
–
speaker
10Ω
5mA
Transconductance gm = IC /(25mV) = 5/25 = 0.2 A/V
Expected gain = gmRL = (0.2)(1k) = 200 or 46dB
But the output circuit is:
–
+
200vin
1kΩ
10Ω
+
vo
–
The effective gain drops to
200×
10
1000+10
= 1.98

35
Proper use of common-emitter amplifier
1kΩ
vBE
+
–
+10V
RB1
RB2
+
vin
–
+
vo
–
10MΩ
5mA
Now the output circuit is:
–
+
200vin
1kΩ
10MΩ
+
vo
–
The effective gain is
200×
107
1000+107
≈ 200
The load must be much larger than RL.
nearly open
circuit

36
How can we use the amplifier in practice?
1kΩ
vBE
+
–
+10V
RB1
RB2
+
vin
–
5mA
How to connect the output to load?
+
vo
–
speaker
10Ω
?

37
Emitter follower
+10V
RB1
RB2
+
vin
–
Biasing conditions:
RE
+
vo
–
IC biased to 10mA
VCE biased to 5V
Base voltage ≈ 5.6V
Emitter voltage ≈ 5V
Collector current ≈ 10mA
RE = 500Ω
RB1:RB2 ≈ 44:56
Say, RB1 = 440kΩ
RB2 = 560kΩ
Thus, for small signal,
∆VE = ∆VB
or vo = vin
VE = VB – 0.6
Gain = vo / vin = 1

38
Small-signal model of emitter follower
+10V
RB1
RB2
+
vin
–
RE
+
vo
–
B C
E E
gmvBE
~
rπ
ro
RE
RB1 || RB2
+
vo
–

39
B C
E E
gmvBE
~
rπ
ro
RE
RB1 || RB2
+
vo
–
rin
vin
iB
rin =
vin
iB
=
vBE + vE
iB
=
vB E
iB
+
vE
iB
= rπ +
vE
iB
= rπ +
vE
iE /(1 + β)
= rπ + (1 + β)RE
which is quite large (good)!!
Input resistance is

40
B C
E E
gmvBE
~
rπ
ro
RE
rout
which is quite small (good)!!
Output resistance is
rout =
vm
im
=
−vBE
im
=
−vBE
iE −iB − gmvBE
=
vE
iE +
vE
rπ
+ gmvE
=
1
1
RE
+
1
rπ
+ gm
=
1
1
RE
+
gm
β
+ gm
≈
1
1
RE
+ gm
= RE ||
1
gm
+
–
vm
im

41
1 vin
RE||(1/gm)
rπ + (1+β)RE
+
vin
–
+
vo
–
+
–
very large very small
Large input resistance
Small output resistance
Voltage gain = 1
Good for any load
Draw no current from previous stage
Thevenin form:

42
A better “emitter follower”
+10V
RB1
RB2 +
vo
–
IE
+
vin
–
Input resistance is very LARGE
because RE = ∞.
Output resistance is 1/gm.
Gain = 1.
This circuit is also called CLASS A
output stage. Details to be studied
in second year EC2.
∞
1/gm

43
Common-emitter amplifier
with emitter follower as buffer
+10V
vBE
+
–
RB1
RB2
+
vin
–
RL
+
vo
–
speaker
10Ω
common-emitter amplifier
(high gain)
emitter follower
(unit gain)
∞
1
gm
IE

44
FET amplifiers (similar to BJT amplifiers)
+10V
vGS
+
–
RG1
RG2
+
vin
–
RL
+
vo
–
speaker
10Ω
common-source amplifier
(high gain = –gmRL)
source follower
(unit gain)
∞
1
gm
IS

45
Further thoughts
Will the biasing resistors affect the gain?
RL
Rbias
+
vin
–
+
vo
–
Seems not, because
Gain = –gmRL
which does not depend on Rbias .
However, a realistic voltage source has finite internal
resistance. This will affect the gain.

46
Input source with finite resistance
RL
Rbias
+
vin
–
+
vo
–
The input has a voltage divider network.
Therefore, the gain decreases to
assuming ro very large.
Rs
gmvBE ro
rπ
+
vBE
–
+
vin
–
Rbias RL
+
vo
–
Rs
vBE = vin
Rbias ||rπ
Rbias ||rπ + Rs
vo
vin
=
Rbias ||rπ
Rbias ||rπ + Rs
(−gmRL )

47
Example
1kΩ
94kΩ
+
vin
–
+
vo
–
50Ω
600Ω
10V
By how much does the gain drop?
rπ
+
vBE
–
+
vin
–
Rbias
50Ω
94k||600 = 596Ω
5mA
gm = 5mA/25mV = 0.2A/V
rπ = β/gm = 100/0.2 = 500Ω
Voltage divider attenuation =
Hence, the gain is reduced to 0.845(gmRL) = 169
Rbias ||rπ
50 + Rbias ||rπ
=
596|| 500
50 + 596||500
= 0.845 or −1.463dB

48
Further thoughts
1kΩ
84kΩ
+
vin
–
+
vo
–
16kΩ
10V
Recall that the best biasing scheme should be β independent.
5mA
200Ω
One good scheme is emitter degeneration, i.e.,
using RE to fix biasing current directly. Here,
since VB is about 1.6V, as fixed by the base
resistor divider, VE is about 1V.
Therefore, IC ≈ VE/RE = 5mA (no β needed!)
Question:
Will this biasing scheme affect the
gain?

49
Common-emitter amplifier with emitter
degeneration
RL
RB1
+
vin
–
+
vo
–
RB2
VCC
Exercise: Find the small-signal gain of this
amplifier.
RE
Answer:
The gain is MUCH smaller.
vo
vin
=
−gmRL
1+ 1+
1
β
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟gmRE
≈
−gmRL
1 + gmRE
≈
−RL
RE
We have a good biasing, but a poor gain! Can we improve the gain?

50
Common-emitter amplifier with emitter by-pass
RL
RB1
+
vin
–
+
vo
–
RB2
VCC
RE CE
Add CE such that the effective emitter
resistance becomes zero at signal
frequency.
So, this circuit has good biasing, and the
gain is still very high!
Gain = – gmRL
which is unaffected by RE because
effectively RE is shorted at signal
frequency.
CE is called bypass capacitor.

51
Summary
Basic BJT model
(small-signal ac model):
gmvBE ro
rπ
B
E
C
E
+
vBE
–
gm =
IC
(kT /q)
=
IC
VT
rπ =
β
gm
ro =
VA
IC
VT is thermal voltage
≈ 25mV
VA is Early voltage
typically ~ 100V

52
Summary
Basic FET model
(small-signal ac model):
Similar to the BJT model,
but with infinite input
resistance.
Therefore, the FET can be
used in the same way as
amplifiers.
gmvGS ro
G
S
D
S
+
vGS
–
∞

53
Summary
Common-emitter (CE) amplifier
small-signal ac model:
gmvBE ro
rπ
+
vBE
–
+
vin
–
Rbias RL
+
vo
–
Gain = –gmRL
Input resistance = Rbias || rπ (quite large — desirable)
Output resistance = RL ||ro ≈ RL (large — undesirable)
RL
Rbias
+
vin
–
+
vo
–

54
Summary
Emitter follower (EF)
small-signal ac model:
gmvBE
rπ
+
vBE
–
+
vin
–
Rbias
+
vo
–
Gain = 1
Input resistance = Rbias || [rπ +(1+β)RE ] (quite large — desirable)
Output resistance = RE || (1/gm) (small — desirable)
RE
Rbias
RE
+
vin
–
+
vo
–

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