#### Transcript Experiment 4 - Rensselaer Polytechnic Institute

```Agenda and Notes



Today, during class! 9:30 a.m. Boeing
Space and Intelligence Systems (Matt
and Matt)
4 extra credit assignments available at
the bottom of
http://hibp.ecse.rpi.edu/~connor/educat
Friday, Oct. 3 (EMPAC!), Open shop
2:00-5:00 p.m
Electronic Instrumentation
Experiment 4 (continued)
Part A. Op Amp Basics Review
Part B. Adder and Differential Op Amp
Part C. Op Amp Limitations
What is an op amp?


An inexpensive, versatile, integrated
circuit that is another basic building
block to electronics (made of resistors
and transistors)
Amplifier that has
• Large open loop gain (intrinsic)
• Differential input stage, inverting input (-)
and non-inverting input (+)
• One output
• Uses components in the feedback network
to control the relationship between the
input and output
What does an Op-Amp do?

Performs “operations” on an input
signal
•
•
•
•
Amplification
Buffering
Integration/Differentiation
Open Loop/Closed Loop and
Feedback


Open loop
• Very high gain (intrinsic gain)
• Poor stability
• Open loop gain assumed to be infinite for ideal op amps
Closed loop
• Uses feedback to add stability
• Reduces gain of the amplifier
• Output is applied back into the inverting (-) input
• Most amplifiers are used in this configuration
Feedback
Open loop
Vout
Vin
Σ
+
gain
Golden Rules of Op-Amp Analysis

Rule 1: VA = VB
• The output attempts to do whatever is necessary to
make the voltage difference between the inputs zero.
• The op-amp “looks” at its input terminals and swings
its output terminal around so that the external
feedback network brings the input differential to zero.

Rule 2: IA = IB = 0
• The inputs draw no current
• The inputs are connected to what is essentially an
open circuit
Steps in Analyzing Op-Amp Circuits
1) Remove the op-amp from the circuit and
draw two circuits (one for the + and one for
the – input terminals of the op amp).
2) Write equations for the two circuits.
3) Simplify the equations using the rules for op
amp analysis and solve for Vout/Vin
Why can the op-amp be removed from the circuit?
• There is no input current, so the connections at the
inputs are open circuits.
• The output acts like a new source. We can replace it
by a source with a voltage equal to Vout.
The Inverting Amplifier
Vout  
Rf
Rin
Vin
A
Rf
Rin
The Non-Inverting Amplifier
 Rf
Vout  1 
 R
g

Rf
A  1
Rg

Vin


The Voltage Follower
High input impedance
Low output impedance
Buffer circuit
Vout
A
1
Vin
Ideal Differentiator
Time domain (like oscilloscope)
Vout
dVin
  R f Cin
dt
Amplitude
changes by a
factor of
RfCin
analysis:
Zf
Rf
Vout


  j R f Cin
1
Vin
Z in
j Cin
Frequency domain
(like AC sweep)
Comparison of ideal and non-ideal
Both differentiate in sloped region.
Both curves are idealized, real output is less well behaved.
A real differentiator works at frequencies below c=1/RinCin
Ideal Integrator
Time domain (like oscilloscope)
Vout
1

Vindt ( VDC )

RinC f
What happens to a
capacitor at DC?
analysis:
Zf
Vout


Vin
Z in
1
j C f
Rin
Amplitude
changes by a
factor of
1/RinCf
1

j RinC f
Frequency domain (like AC sweep)
Miller (non-ideal) Integrator

If we add a resistor to the feedback path,
we get a device that behaves better, but
does not integrate at all frequencies.
Comparison of ideal and non-ideal
Both integrate in sloped region.
Both curves are idealized, real output is less well behaved.
A real integrator works at frequencies above c=1/RfCf
Comparison
original signal
Differentiaion
v(t)=Asin(t)
Integration
v(t)=Asin(t)
mathematically
dv(t)/dt = Acos(t)
v(t)dt = -(A/cos(t)
mathematical
phase shift
mathematical
amplitude change
H(j
electronic phase
shift
electronic
amplitude change
+90 (sine to cosine)
-90 (sine to –cosine)

1/
H(jjRC
-90 (-j)
H(jjRC = j/RC
+90 (+j)
RC
RC

The op amp circuit will invert the signal and multiply
the mathematical amplitude by RC (differentiator) or
1/RC (integrator)
In Class Problem
1.
Which op amp below has a gain of
+5?
a)
2.
b)
Op amp Analysis
1.
2.
3.
What are the golden rules for op amp
analysis?
For the circuit to the right draw two
circuits (one for – input and one for +
input)
Write the equation for each circuit
c)
In Class Problem
1.
Which op amp below has a gain of
+5? All of them! Topology may look
different but the functionality is the
same!
2.
Op amp Analysis
1.
2.
3.
What are the golden rules for op amp
analysis?
For the circuit to the right draw two
circuits (one for – input and one for +
input)
Write the equation for each circuit
Z3
+
V
Z2Z3
V 2
V1V
V  V out
Z1
Zf
--
--
Op Amps to know







Inverting
Non-inverting
Voltage Follower
Differentiator
Integrator
Differential (Subtracting)
Output signal is the
sum of the input
signals (V1 and V2).
 V1 V2 
Vout   R f   
 R1 R2 
Rf
Vout
if R1  R2 then

V1  V2
R1




also useful when R1 ≠ R2.
This is called a “Weighted Adder”
A weighted adder allows you to combine
several different signals with a different
gain on each input.
You can use weighted adders to build audio
mixers and digital-to-analog converters.
I1
If
I2
I f  I1  I 2
V1  VA
I1 
R1
V2  VA
I2 
R2
VA  Vout V1  VA V2  VA


Rf
R1
R2
 Vout V1 V2
 
Rf
R1 R2
Vout
VA  Vout
If 
Rf
VA  VB  0
 V1 V2 
  R f   
 R1 R2 
Differential (or Difference) Amplifier
Vout
 Rf 
 (V2  V1 )
 
 Rin 
A
Rf
Rin
Analysis of Difference Amplifier(1)
1)  :
:
Analysis of Difference Amplifier(2)
2)  : i 
V1  VB VB  Vout

Rin
Rf
 : VA 
Rf
Rin  R f
Note that step 2(-) here is very much
like step 2(-) for the inverting amplifier
and step 2(+) uses a voltage divider.
V2
V1 Vout

Rin R f
3) solve for VB : VB 
1
1

Rin R f
VA  VB :
Rf
Rin  R f
Rf
Vout

V2  V1 Rin
V2 
Rf
R f  Rin
R f V1  RinVout
VB 
Rin R f
R f  Rin
VB 
Rf
R f  Rin
V1 
Rin R f
V1 
Rin
Vout
R f  Rin
R f V2  R f V1  RinVout
Rin
Vout
R f  Rin
Op-Amp Limitations

Model of a Real Op-Amp
 Saturation
 Current Limitations
 Slew Rate
Internal Model of a Real Op-amp
+V
V2
Zin
Vin = V1 - V2
V1
Zout
+
AolVin
Vout
+
-
-V
• Zin is the input impedance (very large ≈ 2 MΩ)
• Zout is the output impedance (very small ≈ 75 Ω)
• Aol is the open-loop gain
Saturation

Even with feedback,
• any time the output tries to go above V+ the op-amp
will saturate positive.
• Any time the output tries to go below V- the op-amp
will saturate negative.

Ideally, the saturation points for an op-amp are
equal to the power voltages, in reality they are
1-2 volts less.
 V  Vout  V
Ideal: -9V < Vout < +9V
Real: -8V < Vout < +8V

Current Limits  If the load on the op-amp is
very small,
•
•
•
•

Most of the current goes through the load
Less current goes through the feedback path
Op-amp cannot supply current fast enough
Slew Rate
• The op-amp has internal current limits and internal
capacitance.
• There is a maximum rate that the internal capacitance
can charge, this results in a maximum rate of change
of the output voltage.
• This is called the slew rate.
Analog Computers (circa. 1970)
Analog computers use op-amp circuits to do real-time
mathematical operations (solve differential equations).
Using an Analog Computer
Users would hard wire adders, differentiators, etc.
using the internal circuits in the computer to perform
whatever task they wanted in real time.
Analog vs. Digital Computers




In the 60’s and 70’s analog and digital computers competed.
Analog