Transcript Experiment 1 - Rensselaer Polytechnic Institute

Electronic Instrumentation
Quiz 1 Review
Op Amps (Continued)
Notes

Test is TOMORROW 6:00 p.m.-8:00 p.m. DCC
308
 If you’ve just started Project 1 you are a week
behind! (Get to open shop this weekend)
 Make sure to find the studio attendance sheet if it
doesn’t make it around to you
Review Topics
Quiz Review Tally
80
70
60
Arb. Units
50
40
30
20
10
0
Transfer
Functions
Resonant
Frequency
(Filters)
Transformer
Topics
Circuit Analysis
Pspice
Transfer Functions and Phasors

Apply the voltage divider equation and parallel
and series combination rules to find transfer
functions using complex impedance expressions
 Simplify the transfer function to find a function
which governs behavior at low and high
frequencies.
 Find an expression (or value) for the magnitude
and phase of the simplified transfer function at the
corner or resonant frequency
 Find Vout or Vin from the transfer function
(magnitude and phase)
Crib Sheet Highlighter
Crib Sheet Highlighter
Crib Sheet Highlighter
Transfer Functions and Phasors
1) Find the transfer function for the above
circuit.
Write in terms of Z impedance first
Transfer Functions and Phasors
Z
ZR  R
Z
ZC 
Z
1) Transfer function for the above circuit.
a) Combine Impedances: Find ZR2C1
b) Use voltage divider to find H=Vout/V1
c) Substitute component values
d) Simplify
j R 2 C 1  1
H  j 


j C 1 R 1  R 2  1
1
j C
Transfer Functions and Phasors
2) Assume R1=R2=1KΩ and
C1=1μF, evaluate the
magnitude of the transfer
function at ω=0 and ω=∞.
a) Magnitude of the transfer function at ω=0
For low frequencies, the lowest power dominates
H  j 
j R 2 C 1  1


j C 1 R 1  R 2  1
lim H  j 
 
0
1
1
Transfer Functions and Phasors
2) Assume R1=R2=1KΩ and
C1=1μF, evaluate the
magnitude of the transfer
function at ω=0 and ω=∞.
b) Magnitude of the transfer function at ω= ∞
For high frequencies, the highest power dominates
H  j 
j R 2 C 1  1


j C 1 R 1  R 2  1
lim
  
H  j 
1
2
Transfer Functions and Phasors
3)


V1


V1


V1
A e
j  t  
 
j 
4 

5e
5e
j ( 0.79)
See crib sheet
When doing analysis ωt
will eventually drop out so
just θ is used
Transfer Functions and Phasors
4 Given R1=1K , R2=1K , and C1=1F, what is the output phasor



Vout
2  1 KHz
H  j 
H  j 
Need both Magnitude
and Phase from the
transfer function.
j R2 C1  1


j C1 R1  R2  1



2
 R2 C1  12


 C1 R1  R2
6.283 2
12.566 2
 12
 12

2
 12
0.505
Magnitude
Transfer Functions and Phasors

2  1 KHz
H  j 
PhaseH
j R2  C1 

Need both Magnitude
and Phase from the
transfer function.
1

j C1 R1  R2 
1
  R2 C1 
  C1  R1  R2 
 1


tan
 tan
1
1




 1
PhaseH  atan  2    1  103  1  103  1  10 6  atan  2    1  103  2  103  1  10 6
PhaseH  0.078


Vout


H  j   V1


Vout


H  j   V1
H A e
.505  5  e

j   phaseH
j ( 0.790.078)
Phase

2.5  e
j  0.71
Transfer Functions and Phasors
Using Transfer Functions (on crib sheet!)


Vout


H  j  V1


Vout


H  j  V1

A e
j   phaseH
H
.505 5  e
j ( 0.790.078)

2.5 e
In the form Vout(t)=ACos( t+), i.e. find A, , and 
V out ( t)

2.5Cos 2 1KHz t
 0.71
j  0.71
Filters





Understand how capacitors and inductors work at
very low and high frequencies
Redraw a given RL, RC or RLC circuit at low
and/or high frequencies and identify low pass,
high pass, band pass and band reject filters
Find resonant frequency of RLC circuits
Find corner frequency of RC and RL circuits
Identify whether a signal of a certain frequency be
passed rejected or something in between by a filter
Crib Sheet Highlighter
Crib Sheet Highlighter
Filters
RS
1k
R2
4k
Va
Vb
L1
100uH
R3
20k
VS
8Vac
C1
10uF
2
1
0
1) Let Rs=0Ω Redraw the circuit above for
very low frequencies
2) What is Va and Vb at very low frequencies?
Filters
RS
1k
R2
4k
VS
8Vac
Va
Vb
R3
20k
0
2) Va=0V, Vb=0V
Filters
RS
1k
R2
4k
Va
Vb
L1
100uH
R3
20k
VS
8Vac
C1
10uF
2
1
0
1) Let Rs=0Ω Redraw the circuit above for
very high frequencies
2) What is Va at very high frequencies?
Filters
RS
1k
R2
4k
VS
8Vac
Va
Vb
R3
20k
0
2) Va=8V
If Va is considered the output, what type of
filter is this?
In Class Problem:
Filters: Resonant Frequency
1)
2)
3)
4)
Redraw this circuit at low frequencies
Redraw this circuit at high frequencies
What type of filter is this?
If L1=2mH, L2=2mH, C1=0.5uF and R13K,
what is the resonant frequency in Hertz?
In Class Answer:
Filters: Resonant Frequency
Low Frequencies
Inductors: short
Capacitor: open
Vout=0V
High Frequencies
Inductors: open
Capacitor: short
Vout=V1
High Pass Filter
In Class Answer:
Filters: Resonant Frequency
Resonant frequency
L1  2mH
L2  2mH
Ltotal  L1  L2
1
 0 
0 
f0 
Ltotal C1
2.236 
0
2
104
rad
s
f0 
3.559  103 Hz
C1 
0.5 F
Transformers and Inductors





How to apply transformer equations
Basic characteristics of transformers
Calculate unknown inductance given the
capacitance or visa versa
Calculate resonant frequency given
inductance or capacitance or visa versa
Estimate inductance of a coil given some
dimensions
Crib Sheet Highlighter
Crib Sheet Highlighter
Transformers (Homework 3, #9)
You are given the transformer pictured below. What is the
voltage across the load resistor if “a” is 9, R2 is 5K ohms,
R1 is 3K ohms, and V1 is 62 Volts?
Note that R1 is large so you CANNOT assume that it has
negligible resistance. Give your answer in volts.
Transformers (Homework 3, #9)
You are given the transformer pictured below. What is the
voltage across the load resistor if “a” is 9, R2 is 5K ohms,
R1 is 3K ohms, and V1 is 62 Volts?
Note that R1 is large so you CANNOT assume that it has
negligible resistance. Give your answer in volts.
N L VL
LL I S
a



N S VS
LS I L
RL
Z in  2
a
Transformers (Homework 3, #9)
1. Find Zin
2. Use the voltage divider to find Vs (due to nonnegligible voltage drop across R1
3. Use ratio relationship below
N L VL
LL I S
a



N S VS
LS I L
RL
Z in  2
a
Transformers (Homework 3, #9)
Zin 
Zin 
5k
Vs 
1.25
92
61.728 
62  Z
in
Vs 
3 K  Z
in
a
VL
Vs
VL  Vs 9
VL 
11.25
Circuit Analysis






Handle combinations of parallel and/or series
resistors
Give resistance expressions in equation form,
rather than as a number.
Find voltages or currents through any resistor
Find the total resistance or current
Know the voltage divider equation
Find the voltage across a resistor in a voltage
divider configuration.
Crib Sheet Highlighter
Crib Sheet Highlighter
Circuit Analysis
R1
R3
V1
R2
R4
0
1) Find the total resistance of the circuit, seen
from the voltage source
Write equation first then put in the numbers!
Given V1=5V, R1=2000Ω, R2=3000Ω,
R3=200Ω, R4=800Ω
Circuit Analysis
R1
R3
V1
R2
R4
0
2) Find the voltage across R1
3) Find the current through R3
Given V1=5V, R1=2000Ω, R2=3000Ω,
R3=200Ω, R4=800Ω
PSpice Instrumentation and
Components




Know which trace corresponds to which
voltage point on a simple circuit
Describe specific steps you’d follow to
obtain a certain output for AC sweep, DC
sweep or Transient Analysis
Understand how to set parameters for
function generator
Understand how to use the oscilloscope
Op-Amp Circuits: Quick Review





Op-Amps are most commonly used to
________ a signal.
Inputs to the op-amp are called the
_______ and _______ inputs.
Unpredictable high gain that is multiplied
by the input signal is called ____-____ ____
or ______ ______.
Extreme gain causes __________.
What day, time, and location is Quiz 1?
Op-Amp Circuits use Negative Feedback

A balancing act between gain and negative
feedback for a stable circuit
How do you “design” negative feedback in the circuit?
Op-Amp Analysis

We assume we have an ideal op-amp:
•
•
•
•
infinite input impedance (no current at inputs)
zero output impedance (no internal voltage losses)
infinite intrinsic gain
instantaneous time response
The Inverting Amplifier
Vout  
Rf
Rin
Vin
A
Rf
Rin
Is this the same as intrinsic gain?
Inverting Amplifier Analysis
Step 0: Understand the Golden Rules!
Rule
1: VA = VB (feedback network brings the
input differential to zero)
Rule 2: IA = IB = 0 (inputs draw no current)
Inverting Amplifier Analysis
inverting input (-):
non-inverting input (+):
Step 1: Re-draw the circuit
Remove the op-amp from the circuit and draw two
circuits (one for the + and one for the – input terminals
of the op amp).
Inverting Amplifier Analysis
Step 2: Write equations for the two circuits
inverting input (-):
non-inverting input (+):
inverting input (-):
i
non-inverting input (+):
V
V in  V B
V B  V out
R
R in
Rf
VA=0
Inverting Amplifier Analysis
Step 3: Simplify using Golden Rules and solve for
Vout/Vin
V in  0 0  V out
VA=VB=0
therefore
R in
Rf
Golden
Rule!
V in
V out
R in
Rf
V out
R f
V in
R in
What is this saying about how you can design your gain?
PSpice Inverting Amplifier
Use the uA741 op amp to model your circuits
Can’t find it?
It is in the EVAL library
Add library “Eval”
Inverting Amplifer
R3
Input amplitude: 200mV
Rf=10k Ω
Rin=1k Ω
10k
V2
U1 7
3
V+
OS2
V
+
0
R1
2
V3
1k
OUT
- 4
uA741
VOFF = 0
VAMPL = 200mV
FREQ = 1kHz
OS1
V-
5
V
9V
6
1
What should the
simulated output look like?
V1
-9V
R2
1k
0
The Non-Inverting Amplifier
 Rf
Vout  1 
 R
g

Rf
A  1
Rg

Vin


Non-inverting Amplifier Analysis
inverting input (-):
non-inverting input (+):
Step 1: Re-draw the circuit
Remove the op-amp from the circuit and draw two
circuits (one for the + and one for the – input terminals
of the op amp).
Non-inverting Amplifier Analysis
Step 2: Write equations for the two circuits
inverting input (-):
non-inverting input (+):
Voltage Divider
inverting input (-):
non-inverting input (+):
VA=Vin
VB
Rg
Rf  Rg
 V out
Non-inverting Amplifier Analysis
Step 3: Simplify using Golden Rules and solve for
Vout/Vin
VA=VB=Vin
Golden
Rule!
therefore
V out
V in
V in
Rg
Rf  Rg
1
Rf
Rg
 V out
PSpice Non-inverting Amplifier
Non-Inverting Amplifer
0
0
V5
9V
V
7
3
V+
OS2
+
OUT
2
-
OS1
V-
4
U2
uA741
5
6
V
V6
VOFF = 0
VAMP L = 200mV
FREQ = 1kHz
Using the uA741 op amp
Input amplitude: 200mV
Rf=1k Ω
Rg=1k Ω
1
V4
-9V
R40
1k
R5
1k
0
What should the simulated
output look like?
The Voltage Follower
Vout
1
Vin
Unity gain amplifier
analysis :
1] VA  Vout
VA  VB
2] VB  Vin
therefore, Vout  Vin
Why is it useful?

In this voltage divider, we get a
different output depending upon
the load we put on the circuit.
 Why?

We can use a voltage follower to convert this real
voltage source into an ideal voltage source.
 The power now comes from the +/- 15 volts to the
op amp and the load will not affect the output.
Integrators and Differentiators

General Op-Amp Analysis
 Differentiators
 Integrators
 Comparison
General Analysis Example(1)
into inverting input

Assume we have the circuit above,
where Zf and Zin represent any
combination of resistors, capacitors and
inductors.
General Analysis Example(2)

We remove the op amp from the circuit
and write an equation for each input
voltage.

Note that the current through Zin and Zf is
the same, because equation 1] is a series
circuit.
General Analysis Example(3)
I

Since I=V/Z, we can write the following:
Vin  VA VA  Vout
I

Zin
Zf

But VA = VB = 0, therefore:
Vin
Z in
 Vout

Zf
Zf
Vout

Vin
Z in
General Analysis Conclusion

For any op amp circuit where the positive input
is grounded, as pictured above, the equation for
the behavior is given by:
Zf
Vout

Vin
Z in
Ideal Differentiator
Phase shift
j/2
-  ±
Net-/2
analysis :
Zf
Rf
Vout


  j R f Cin
1
Vin
Z in
j Cin
Amplitude
changes by a
factor of
RfCin
Analysis in time domain
I
dVCin
I Cin  Cin
VRf  I Rf R f I Cin  I Rf  I
dt
d (Vin  VA ) VA  Vout
I  Cin

VA  VB  0
dt
Rf
therefore, Vout
dVin
  R f Cin
dt
Problem with ideal differentiator
Ideal
Real
Circuits will always have some kind of input resistance,
even if it is just the 50 ohms or less from the function
generator.
Analysis of real differentiator
I
1
Z in  Rin 
j Cin
Zf
Rf
j R f Cin
Vout



1
Vin
Z in
j RinCin  1
Rin 
j Cin
Low Frequencies
Vout
  j R f Cin
Vin
ideal differentiator
High Frequencies
Rf
Vout

Vin
Rin
inverting amplifier
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
Phase shift
1/j-/2
-  ±
Net/2
Amplitude
changes by a
factor of
1/RinCf
analysis :
Zf
Vout


Vin
Z in
1
j C f
Rin
1

j RinC f
Analysis in time domain
I
VRin  I Rin Rin
I Cf  C f
dVCf
I Cf  I Rin  I
dt
Vin  VA
d (VA  Vout )
I
 Cf
VA  VB  0
Rin
dt
dVout
1
1

Vin Vout  
Vin dt ( VDC )

dt
RinC f
RinC f
Analysis in time domain
I
VRin  I Rin Rin
I Cf  C f
dVCf
I Cf  I Rin  I
dt
Vin  VA
d (VA  Vout )
I
 Cf
VA  VB  0
Rin
dt
dVout
1
1

Vin Vout  
Vin dt ( VDC )

dt
RinC f
RinC f
Problem with ideal integrator (2)
With DC offset.
Saturates immediately.
What is the integration of a constant?
Problem with ideal integrator (2)
With DC offset.
Saturates immediately.
What is the integration of a constant?
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.
Behavior of Miller integrator
Low Frequencies
High Frequencies
Zf
Rf
Vout


Vin
Z in
Rin
Zf
Vout
1


Vin
Z in jRinCf
inverting amplifier
ideal integrator
The influence of the capacitor dominates at higher frequencies.
Therefore, it acts as an integrator at higher frequencies, where
it also tends to attenuate (make less) the signal.
Analysis of Miller integrator
I
Rf
1
Rf 
j C f
Rf
Zf 

1
j R f C f  1
Rf 
j C f
Zf
j R f C f  1
Rf
Vout



Vin
Z in
Rin
j Rin R f C f  Rin
Low Frequencies
Rf
Vout

Vin
Rin
inverting amplifier
High Frequencies
Vout
1

Vin
j RinC f
ideal integrator
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
Problem solved with Miller integrator
With DC offset.
Still integrates fine.
Why use a Miller integrator?

Would the ideal integrator work on a signal with
no DC offset?
 Is there such a thing as a perfect signal in real
life?
• noise will always be present
• ideal integrator will integrate the noise

Therefore, we use the Miller integrator for real
circuits.
 Miller integrators work as integrators at > c
where 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)