Recitation 20
Amplifiers Review
6.012 Spring 2009
Recitation 20: Amplifiers Review
Yesterday, we introduced two more amplifier circuits: C-drain, C-base.
As we know, there is an analogy between MOS & BJT:
MOS
Common Source ←→
Common Drain ←→
Common Gate ←→
BJT
Common-Emitter
Common-Collector
Common-Base
Function
Voltage or Gm Amp.
Voltage Buffer
Current Buffer
Note: Buffer is an amplifier with gain 1, but input or output impedance changed
We have also learned that there are 4 types of amplifiers, their two port models are
Voltage
Amplifier
Current
Amplifier
Transconductance
Amplifier
Transresistance
Amplifier
For the above single stage amplifiers (i.e. CS, CD, CG, CE, CC, CB), as we identify their
particular function, e.g. current buffer is a type of current amplifier. We can use a two-port
model for current amplifier to model a CB or CG circuit. Their corresponding Rin , Rout , Aio
will depend on the circuit (or device parameter), which we can derive based on the small
signal circuit model of the circuit.
Yesterday, we looked at the example of CD & CG. Today we will look at CC & CB.
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Recitation 20
Amplifiers Review
6.012 Spring 2009
Common-Base Amplifier
Cast this into two port model
Need to find what is the corresponding Aio , Rin , Rout
Aio
Intrinsic current gain: ignore Rs , just consider iin = is ; RL short.
short RL at the output
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Recitation 20
Amplifiers Review
6.012 Spring 2009
iin
=⇒ vπ
vπ
vπ
vπ
= −
+ gm vπ +
, iout = gm vπ +
γπ
γo
γo
iin
iin
= − 1
1 = g +g +g
π
m
o
vπ + gm + γo
=⇒ Aio =
∵
1
gm
iout
=−
iin
(gm + go ) ·
iin
+ go
gm + go
gπ + gm
=−
−1
iin
gπ + gm + go
1 kΩ, γo ≈ 100 kΩ
gm go , γπ =
βF
gm
=⇒ gπ =
gπ gm
gm
βF
Rin
γπ , γo ∴
it
=
∴ Rin
=
1
as we just discussed
gm
transconductance generator gm dominates currents at the input node
vπ
vo
−
+ gm vπ +
−gm vπ = gm vt
γπ
γo
vt
vt
1
=
LOW! (good for getting current in)
it
gm vt
gm
Exact: see pp 150 Rin =
1
γπ
1
1 − gm (γco ||RL )
+ gm +
γo + (Voc ||RL )
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Recitation 20
Amplifiers Review
6.012 Spring 2009
Rout
Similarly
1. shut down all independent sources
2. load input with Rs
3. put test current source at output
vt
4. Rout =
it
vt + vπ
γo
= −it · (γπ ||Rs )
it = gm vπ +
vπ
voltage across γo is vt + vπ
(2)
=⇒ plug (2) into (1)
(3)
it =
=⇒
vt
it
(1)
vt /γo
γπ ||Rs
+ gm (γπ ||Rs )
1+
γo
= γo + (γπ ||Rs ) + gm γo (γπ ||Rs )
(4)
(5)
∴ Rout = γoc ||[γo + (γπ ||Rs ) + gm γo (γπ ||Rs )] γoc ||γo [1 + gm (γπ ||Rs )]
(6)
If Rs γπ , Rout γoc ||γo [1 + gm γπ ] = γoc ||γo · βF
βp
(7)
large
Excellentcurrent
buffer: can use current source with source resistance only slightly higher
1
than Rin
, and get same current with high Rout
gm
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Recitation 20
Amplifiers Review
6.012 Spring 2009
Common-Collector Amplifier
Rearrange,
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Recitation 20
Amplifiers Review
6.012 Spring 2009
Cast this into two port voltage amplifier model
Avo (RL = ∞, Rs = 0)
Vout
iin
vπ
= Avo Vin = gm vπ + gm
βF
1
= gm 1 +
vπ (γo ||γoc )
βF
vπ
gm
=
= vπ
γπ
βF
But vπ = vin − vout =⇒ vout = gm
Avo =
vout
=
vin
1+
1
gm
1
1+
1
βF
· (γo ||γoc )
1
1+
βF
(γo ||γoc )
(vin − vout )(γo ||γoc )
1
Rin
Leave RL in place, replace source with
vt = it · γπ + (it + gm vπ ) · (γo ||γoc ||RL )
1
= it vπ + gm 1 +
vπ (γo ||γoc ||RL )
βF
1
g
1
+
m
βF vπ (γo ||γoc ||RL )
vt
= γπ +
Rin =
vπ
it
gm
βF
= γπ + (βF + 1)(γo ||γoc ||RL ) much larger than γπ
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Recitation 20
Amplifiers Review
6.012 Spring 2009
Rout
vs = 0, leave Rs , apply vt , it at the output
than gm vπ
vπ
(it + gm vπ +
) · (γo ||γoc ) = vt
γ
π
γπ
· vt
γπ + R s
vt
−gm vπ +
γo ||γoc
1
gm γπ
vt
βF
· vt +
=
+
vt
γπ + Rs γo ||γoc
γπ + R s
γo ||γoc
βF
vt
γπ + Rs
vt
γπ + Rs
1
Rs
=
=
+
LOW! ∵ gm , βF large
βF
gm βF
it
voltage divider vπ = −
=⇒ it =
=⇒ it =
∴ it Rout =
In conclusion, see the summary sheet handout
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