#### Transcript 1st Order Op Amp Circuits

```Objective of Lecture
 Discuss analog computing and the application of 1st
order operational amplifier circuits.
 Derive the equations that relate the output voltage to
the input voltage for a differentiator and integrator.
 Explain the source of the phase shift between the
output and input voltages.
Mechanical Analog Computers
Designed by Vannevar Bush in 1930 and used to control position of
artillery through WWII. Replaced by electrical analog computers
towards the end of WWII, which performed the needed calculations
much faster.
http://www.science.uva.nl/museum/AnalogComputers.html
Why Use an Analog Computer?
 Calculations performed in real time without the use of a
‘real’ computer.
 Can be integrated into the instrumentation circuitry.


Commonly used in control circuits to rapidly monitor and change
conditions without the need to communicate back and forth with
a digital computer.
Power consumption is not high.
 Input can be any value between V+ and V-.
 Can be designed to handle large (or small) voltages.
 No digitizing errors.
 Analog computers are more robust
 Less susceptible to electromagnetic radiation damage (cosmic
rays), electrostatic discharge, electromagnetic interference
(pick-up of electric noise from the environment), etc.
Disadvantage
 Slow
 Maximum frequency is less than 10 MHz

Compare this to the clock speed of your digital computer.
 Voltage transfer function is nonlinear over entire range
of input voltages.
 Timing of inputs needs to be carefully considered.

Any time delays can cause errors in the calculations
performed.
Subsystems
 Multipliers
 Inverting and non-inverting amplifiers

Typically fixed number, which means fixed resistor values in
amplifiers
 Adders and Subtractors
 Summing and difference amplifiers
 Differentiators
 Integrators
1st order op amp circuits
Capacitors
dvC
iC (t )  C
dt
t1
1
vC (t )   iC (t )dt  vC (to )
C to
Differentiator
Ideal Op Amp Model
Virtual ground
Op Amp Model
Virtual ground
Analysis
 Since current is not
allowed to enter the
input terminals of an
ideal op amp.
iC (t )  iR (t )
vC (t )  vS (t )
dvC
dvS
iC (t )  C
C
dt
dt
vo
iR (t )  
R
vo
dvS
 C
R
dt
dvS (t )
vo (t )   RC
dt
Example #1
 Suppose vS(t) = 3V u(t-5s)
 The voltage source changes from 0V to 3V at t = 5s.


Initial condition of VC = 0V when t <5s.
Final condition of VC = 3V when t > 5RC.
Example #1 (con’t)
vC (t )  0V when t t o


vC (t )  VCinitial  VC final  VCinitial e t to   when t t o
vC (t )  0V  3V  0V  e t 5 s  0.8 s when t t o
vC (t )  3V e t 5 s  0.8 s when t 5s
Example #1 (con’t)
dvC (t )
vo (t )   RC
dt
vo (t )  0V when t 5s
vo (t )  0V
when t to  5 , where   RC
vo (t )  0V
when t 5s  5(20k)(40 F)  9s
1
vo (t ) 
(20 x103 )( 40 x10 6 F )(3V ) e t 5 s  0.8 s
0.8s
vo (t )  3V e t 5 s  0.8 s
Example #2
 Let R = 2 k, C = 0.1F, and vS(t) = 2V sin(500t) at t = 0s
Since vC (t )  vS (t )
dvS
vo (t )   RC
dt
d 2V sin(500t )
vo (t )  (2000)(10 F )
dt
vo (t )   0.2m s2V 500 cos(500t )
7
vo (t )  0.2V cos(500t ) when t 0s
vo (t )  0V when t  0s
Cosine to Sine Conversion
vo (t )  0.2V cos(500t )
vo (t )  0.2V sin(500t  90o )
vo (t )  0.2V sin(500t  90o  180o )
vo (t )  0.2V sin(500t  90o )
As vS (t )  2V sin(500t ) , the output voltage lags the
input voltage by 90 degrees.
vS(t) vo(t)
PSpice Simulation
Shows the 90 degree phase shift as well as the deamplification.
Integrator
Virtual ground
Op Amp Model
Integrator
vS (t )  v1 vS (t )
iR 

R
R
dvC
iC  C
dt
vC (t )  v1  vo (t )  vo (t )
iR  iC  0m A
vS (t )
d  vo (t )
C
0
R
dt
dvo (t ) vS (t )

0
dt
RC
1
vo (t 2 ) 
vS (t )dt  vo (t1 )

RC t1
t2
Example #3
rad 

 Let R = 25 k, C = 5nF, vS (t )  3V sin 6.24k
t  at t=0s
s 
t

1
Vo (t 2 ) 
2
V

RC
in
(t )dt  Vo (t1 )
t1
Vo (t 2 ) 
1
rad 

3
V
sin
6
.
24
k
t dt


25k(5nF ) t1
s 

t2
t2
rad 

Vo (t 2 )  3.85V cos 6.24k
t   Vo (t1 )
s  t1

rad


Vo (t 2 )  3.85V sin  6.24k
t 2  90o   3.85V when t1  0 s
s


since vo(t) = -vC(t) and the voltage across a capacitor can’t be discontinuous.
PSpice Simulation
Shows that the output voltage leads the input voltage by +90
degree and the voltage offset due to the Vo(t1) term.
Example #4
Initial Charge on Capacitor
Example #4 (con’t)
If there is an initial charge that produces a voltage on the
capacitor at some time, to:
The voltage on the negative input of the op amp is:
v1 = VC + VR1
v1= v2 = 0V
Example #4 (con’t)
The current flowing through R1 is the same current flowing
through C.
dvC (t )
iC (t )  C
dt
vC (t )
VR1 [v1  vC (t )] [0V  vC (t )]
iR1 (t ) 



R1
R1
R1
R1
vC (to )
at t  t o , iR (to )  
R1
as t  , vC (t )  0V , iC (t )  0m A
iC (t )  iR1 (t )  0
dvC (t ) vC (t )
C

0
dt
R1
dvC (t ) vC (t )

0
dt
R1C
vC (t )  vC (to )e

t to
R1C
dvC (t )
1
iC  C
  vC (to )e
dt
R1

R1C is the time constant, .
t to
R1C
iR 2  iC
iR 2
0V  vo (t )
vo (t )


R2
R2
R2
vo (t ) 
vC (to )e
R1

t to
R1C
Summary
 Differentiator and integrator circuits are 1st order op
amp circuits.
 When the C is connected to the input of the op amp, the
circuit is a differentiator.

If the input voltage is a sinusoid, the output voltage lags the
input voltage by 90 degrees.
 The output voltage may be discontinuous.
 When the C is connected between the input and output
of the op amp, the circuit is an integrator.

If the input voltage is a sinusoid, the output voltage leads the
input voltage by 90 degrees.
 The output voltage must be continuous.
```