EE 205
Coifman
Homework set #5, due 5/2/05
4-2, 4-5, 4-8, 4-9, 4-13, 4-16, 4-20, 4-21, 4-23, 4-24
Meet the Operational Amplifier, its friends call it "the Op Amp"
Positive
power supply
+V CC vO
8
Noninverting
input
+
7
6
5
Output
+
-
Top
Negative
power supply
Inverting
input
1
2
vN
3
vP
4
V CC
This is a real device that you can buy... made up of lots of
transistors and stuff. We will be working with a few simple
models of the Op Amp...
I C+
iP
iN
+
iO
+
-
+
+
vP
-
-
IC vN
-
- V CC
+
vP
+V CC
-
vO
+
-
vN
+
+
iP
iN
iO
+
-
iO≠iP+iN
iO=IC++IC-+iP+iN
13-1
+
vO
EE 205
Coifman
Transfer characteristic for an Op Amp
vO
+V CC
Output voltage
swing
1
vP -vN
µ
- V CC
- Saturation
noninverting input
inverting input
output voltage
voltage gain
Linear +Saturation
vP
vN
vO
µ
vO=µ(vP-vN)
3 modes of operation:
- saturation
linear
+ saturation
13-2
EE 205
Coifman
Model of an Ideal Op Amp operating in the linear region
vP
+
iP
+
RO
RI
vN
+
+
-
iN
iO
+
vO
µ(vP - vN )
-
R1∈[106,1012]Ω
RO∈[10,100]Ω
µ∈[105,108]
-VCC < vO < +VCC
-VCC/µ< (vP-vN) < +VCC/µ
Since VCC ~ 15V; µ >> 0,
(vP-vN) ≈ 0
Next, we assume µ → ∞ and arrive at the "ideal model" of
the Op Amp:
vP=vN
iP = iN = 0
In other words (pictures?)
13-3
EE 205
Coifman
i =0
+P
iO
Voltages vP
i =0 +
are
+N
equal vN
-
+
vO
-
Now for the final piece of the puzzle: feedback.
Feedback ensures (vP-vN) ≈ 0 and thus, keeps the Op
Amp in the linear region.
13-4
EE 205
Coifman
My friends call me an example- I’m a non-inverting Op
Amp circuit
vP
vS
+
-
vN
iP
+
-
iN
vO
+
R1
R2
Feedback path
K= closed loop gain (of the circuit)
13-5
vS
K
vO
EE 205
Coifman
Placing the non-inverting Op Amp circuit into a larger circuit...
R1
+
-
vP
vN
vS
iP
iN
vO
+
-
R3
R2
RL
R4
Source
vS
KS
Noninverting amplifier
(Active circuit)
vP
K AMP
13-6
vO
Load
EE 205
Coifman
Now for a voltage follower
RS
vP i P
vN
iN
+
-
vO i O
vS +
-
RS
RL
13-7
vS +
-
RL
EE 205
Coifman
And an inverting amplifier
A
R1
i1
+
- vS
iN
vN
vP
Feedback path
R2
vO
-
vS
i2
+
13-8
K
vO