Physics 111 – BSC Lecture 7
Jim Siegrist
Phone: 486-4397
Email: JLSiegrist@lbl.gov
Room (at LBL): 50-4055
Advice:
Today:
lec 7
lec 8
lec 9
lec 10
OP AMP I
TH
Feb 22
TH
Mar 8 – Note Dates!
TH
Mar 15 – Note Dates!
NB: problems 8.12 – 8.13 → supplementary
Report labs posted.
Concepts
Op-Amps
IC ' s with Ri > ~ 10 4 Ω
Ro < ~ 50Ω
gain ~ 10 4 or so
rugged, inexpensive building block to make actual circuits.
Special Symbol:
A(v(+) − v(−))
‘non-inverting’
MODEL:
Ri
Ro
‘inverting’
+ Power supply, trim voltage connections, hook up power and ground or nothing happens!!
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Physics 111 – BSC Lecture 7
Feedback
Op-amps usually connected in feedback circuits. Literally, a bit of the output signal is ‘fed back’
to the input:
Negative Feedback
decrease the output when it is too high
increase the output when it is too low
Positive Feedback
increase output when it is too high
decrease output when it is too low
⇒ oscillations, generally bad
Use of (negative) feedback elements external to the op-amp reduces sensitivity to device
variation (as we shall see) – very important.
Path that returns some of the output to the input is the feedback loop ⇒ A = ‘open loop’ gain
To avoid using the detailed model, consider ideal op-amp: A, Ri → ∞, Ro → 0
&
⇒ Assume potential difference between the two input terminals v(+) – v(–) = 0
the two input currents both = 0.
Why? |vout| is limited by supply voltage. From the model,
vout = A v(+ ) − v(− ) < Vsup ply
⇒ v( + ) − v( − ) <
& iin =
Vsup ply
A
→0 A→∞
v+ − v−
, Ri → ∞ , v+ − v− → 0
Ri
⇒ iin = 0
Circuit Analysis
Examples
1) Voltage follower:
vin
RS
RL
vout
Note, ideal model:
Ri
vout = v− = v+ = vin ⇒
A' =
v
vout
= `closed loo p ′ gain = 1.0 in
vin
RS
Ro
A(v(+ ) − v(−))
vout
Iout
RL
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Physics 111 – BSC Lecture 7
Exact value? – use model
node equation at + node:
vin − v + v + − vout
+
=0
RS
Ri
(1)
but vout = I out ⋅ RL
vout =
A(v + − v − )
⋅ RL
Ro
but v− = vout ⇒ vout = A(v+ − vout ) ⋅
RL
Ro
Ro
vout + vout = v +
AR L
R
R
vin − vout − o v out vout + o vout − vout
ARL
ARL
substitute into (1) ⇒
+
=0
RS
Ri
vin v out
Ro
Ro
−
−
vout +
v out = 0
RS RS AR L RS
AR L Ri
solve for v + ⇒
⎛
R
R R ⎞
vin = vout ⎜⎜1 + o − o S ⎟⎟
ARL ARL Ri ⎠
⎝
ARL Ri
⇒
vout = vin
ARL Ri + Ro Ri − Ro RS
v
R (R − R S )
≈1
A′ = out ≈ 1 − o i
vin
ARL Ri
Ro ~ 50 Ω
RS ~ 50 Ω
RL ~ 1 k
Ri ~ 10 k
A ~ 10 k
Note: Let RS → 0
Ro → 0
Ro
→ 1 as
RL → ∞
RL
1
, same as equation 8.3, page 311 S&S.
A
Should note: A′ = 1 did not depend on op-amp parameters (A, Ri, Ro), so it is very probably
accurately enough known by the ideal model.
then my expression reduces to A′ ≈ 1 −
Input & Output Impedance
Input
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Physics 111 – BSC Lecture 7
Apply Vtest at input terminals; get equivalent input resistance from
Vtest
I test
N.B. This will depend on A, Ri, etc. so we have to resort to the model:
(To see that, itest = 0 in ideal model ⇒ Ri = ∞)
Itest
To obtain a better value:
( ignore RS)
(+)
vtest
Ri
(−)
vout
Ro
itest
i2
RL
A(v+−v−)
Go around 1st loop:
vtest − itest Ri − Ro (itest − i2 ) − A(v(+ ) − v(− ) ) = 0
2nd loop:
A(v(+ ) − v(− ) ) − Ro (i2 − itest ) − i2 RL = 0
but v + = vtest , v(− ) = vtest − itest Ri ⇒
(1) vtest − itest Ri − Ro (itest − i2 ) − A(vtest − vtest + itest Ri ) = 0
(2) A(vtest − vtest + itest Ri ) − Ro (i2 − itest ) − i2 RL = 0
solve (2) for i2 ⇒ i2 =
itest ( ARi + Ro )
Ro + R L
put back in (1) ⇒
⎡ i ( ARi + Ro ) ⎤
vtest − itest (Ro + (1 + A)Ri ) + Ro ⎢ test
⎥=0
⎣ Ro + R L ⎦
⎡ [R + (1 + A)Ri ](Ro + RL ) − Ro [Ro + ARi ]⎤
vtest = itest ⎢ o
⎥
Ro + R L
⎣
⎦
Ri′ =
vtest
Ro
= Ro + (1 + A)Ri −
[Ro + ARi ]
itest
Ro + R L
⎡ R + ARi ⎤
= Ri (1 + A) + Ro ⎢1 − o
⎥
Ro + R L ⎦
⎣
Ri′ ≈ Ri A ~ 1010 Ω ! ≈ ∞
Output Impedance
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Physics 111 – BSC Lecture 7
Ro′ = −
voc
since v+ = v− , voc = vin (ideal model)
i sc
RS
(+)
isc is given by:
iSC
Ri
(−)
Ro
vS
iout
A(v+−v−)
Note v− is grounded ⇒
(v )
iout = −isc ≈ A +
Ro
v
= A in
Ro
v
vin
⇒ Ro′ = − oc =
v
isc
A[ in ]
Ro
ignore current from input
Ro
~ 10− 3 Ω ≈ 0
A
Very good emitter follower.
⇒ Buffer circuit to isolate one part of the circuit from another.
=
( 1 = follower)
Example 2
(gives voltage gain)
Non-Inverting Amplifier
R1
vin
1
RF
vout
Like follower, but a voltage divider in the feedback path.
v − = vin = v +
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Physics 111 – BSC Lecture 7
No current flows into the amp ⇒ at node (1)
vin vin − vout
+
=0
R1
RF
⇒ vout =
⇒ A′ =
R1 + RF
vin
R1
v out R1 + RF
=
vin
R1
follower _ + _ new _ term
One finds (like follower)
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8
⎛ R1 ⎞
⎟⎟
Ri′ = ARi ⎜⎜
⎝ R1 + RF ⎠
Ro′ =
Ro
A
⎛ R1 + RF
⎜⎜
⎝ R1
⎞
⎟⎟
⎠
Example 3
Inverting Amplifier
RF
R1
vin
vout
Voltage at (−) = 0
⇒ current through R1 goes through RF
⇒ (none goes into amplifier)
vin v out
+
=0
R1 RF
v
R
(inverting amp)
⇒ A′ = out = − F
vin
R1
↑
6444447
444448
RF
vin RF = R1 ⇒ unity gain inverter
vout = −
R1
Input & Output Impedance:
Input
v
itest = test ⇒ Ri′ = R1
R1
Output
R (R + R1 + RS )
turns out to be Ro′ = o F
(R1 + RS )
A
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