ES309
Elements of Electrical Engineering
Lecture #12
Jeffrey Miller, Ph.D.
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Elements of Electrical Engineering: Op Amp Circuits and Amplification - Prof. Jeffrey A. M, Study notes of Engineering
A lecture note from the es309: elements of electrical engineering course, focusing on operational amplifiers (op ams) and their applications in various amplifier circuits such as inverting, summing, and non-inverting amplifiers, as well as difference amplifiers. The operating regions of op ams, voltage divider equations, and the calculation of output voltages.
Typology: Study notes
2009/2010
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ES
Elements of Electrical Engineering
Lecture
Jeffrey Miller, Ph.D.
Outline
Op Amp Review
Chapter 5.5-5.
Op Amp Review
The linear operating region is between v 0 is between –V cc and V cc
Positive saturation occurs when v 0
=V
cc
Negative saturation occurs when v 0
=-V
cc
Inverting-Amplifier Circuit Review
v
o
= -v
s
* R
f
/ R
s
Non-Inverting Amplifier Circuit
Non-Inverting Amplifier Circuit
We know i
n
= i
p
= 0, so we can calculate the voltage
drop across the resistor R
g
using Ohm’s Law (v=iR),
which we see will give us no voltage drop across R
g
So, v p = v g - And, v n = v p , which means that v n = v g
Non-Inverting Amplifier Circuit
To operate in the linear operating region, v
o
< V
cc
v
o
= v
g
* (R
s
+ R
f
) / R
s
| V
cc
/ v
g
| > (R
s
+ R
f
) / R
s
Difference-Amplifier Circuit
Difference-Amplifier Circuit
We already found -i n = (v n - v a ) / R a^ + (v n - v o ) / R b in = 0 - So then 0 = (v n - v a ) / R a^ + (v n - v o ) / R b - We already found v n = v p = v b * R d / (R c + R d ) - So then 0 = (v b^ * R d^ / (R c^ + R ) – vd ) / Ra a^ + (v b^ * R d^ / (R c^ + R ) – vd ) / Ro b 0 = v b^ * R d^ / (R a^ * (R c^ + R ) ) – vd a^ / R a^ + v b^ * R d^ / (R b^ * (R c^ + R ) ) – vd o^ / R b 0 = v b^ R b^ R d^ - v a^ R b^ (R c^ + R ) + vd b^ R a^ R d^ - v o^ R a^ (R c^ + R ) / (Rd a^ R b^ (R c^ + R ) )d 0 = v b^ R b^ R d^ - v a^ R b^ (R c^ + R ) + vd b^ R a^ R d^ - v o^ R a^ (R c^ + R )d v o^ R a^ (R c^ + R ) = vd b^ (R b^ R d^ + R a^ R ) – vd a^ (R b^ (R c^ + R ) )d v o^ = v b^ (R d^ (R a^ + R ) ) / (Rb a^ (R c^ + R ) ) – vd a^ (R b^ / R )a
Difference-Amplifier Circuit
v o = v b
(R
d
(R
a
+ R
b
) ) / (R
a
(R
c
+ R
d ) ) – v a
(R
b
/ R
a
The above equation shows that the output is a scaled difference ofthe input voltages
The scaling factor is not the same, but it can be if R a /R b = R /Rc d
Difference-Amplifier Example
What range of values for v
a
will result in linear
operation?
Difference-Amplifier Example vo
= v b^ (R d^ (R
- Ra ) ) / (Rb (Ra c^
- R ) ) – vd (Ra b^ / R )a vb = 4.0V R = 4kc Ω R = 10ka Ω R d^ = 20k Ω R b^ = 50k Ω v o^ = -10V and v o^ = 10V va = v b^ (R / Ra ) (Rb d^ (R a^
- R ) ) / (Rb (Ra c^
- R ) ) - (Rd / Ra ) vb o va = 4.0V * (10k Ω / 50k Ω ) (20k Ω (10k Ω
- 50k Ω )) / (10k Ω (4k Ω
- 20k Ω )) - (10k Ω / 50k Ω ) * -10V va = 0.8V * 1200 / 240 - -2V va = 4.0V + 2V va = 6.0V va = v b^ (R / Ra ) (Rb d^ (R a^
- R ) ) / (Rb (Ra c^
- R ) ) - (Rd / Ra ) vb o va = 4.0V * (10k Ω / 50k Ω ) (20k Ω (10k Ω
- 50k Ω )) / (10k Ω (4k Ω
- 20k Ω )) - (10k Ω / 50k Ω ) * 10V va = 0.8V * 1200 / 240 + -2V va = 4.0V - 2V va = 2.0V