Download Week 3 - Chapter 2 (Part 1)

Analogue Electronics Circuit 2
EKT 214
Chapter 2
Op-Amp Applications and
Frequency Response
Semester 2 2010/11
By: Norizan Binti Mohamed Nawawi
1
Op-Amp Applications
 2.1 Voltage Follower
 2.2 Summing Amplifier
 2.3 Differencing Amplifier
 2.4 Integrator
 2.5 Differentiator
 2.6 Comparator
2
Voltage Follower / Buffer Amplifier

This “buffer” is used to control impedance levels in the circuit
– it isolates part of the overall (measurement) circuit from the
output (driver).
 The input impedance to the buffer is very high and its output
impedance is low.

3
The output voltage from a source with high output impedance
can, via the buffer (voltage follower), supply signal to one or
more loads that have a low impedance.
Voltage Follower / Buffer Amplifier
 High input impedance
 Low output impedance
 Voltage gain = 1
UNITY GAIN;
Vout
Av 
1
Vin
Vout  Vin
4
Summing Amplifiers
 The inverting amplifier can accept two or more inputs
and produce a weighted sum.
 Assuming that V- ≈ 0 (voltage drop at V- is approx. 0).
iin
The sum of the currents
through R1, R2,…,Rn is:
iin 
V
V1 V2

 ...  n
R1 R2
Rn
Summing Amplifiers
The op-amp adjusts itself to draw iin through Rf (iin = if)
if
if 
iin

V
V1 V2

 ...  n
R1 R2
Rn
Vout V1 V2
V


 ...  n
Rf
R1 R2
Rn
Rf
Rf 
 Rf


Vout  i f R f  V1
 V2
 ...  VN
R2
RN 
 R1
The output will thus be the sum of V1,V2,…,Vn, weighted by the
Rf Rf
Rf
,
,.....
different gain factors,
respectively.
R1 R2
R3
Summing Amplifiers
Special Cases for this Circuit:
1. If R1 = R2 =……= R then:
Vout  
Rf
R1
V1  V2  .....  VN 
i.e. the output voltage is proportional to the sum of the input
voltages (unity gain summing amplifier).
2. If R1 = R2 = … = R and v1, v2, … are either 0V (digital “0”) or
5V (digital “1”) then the output voltage is now proportional to
the number of (digital) 1’s input.
Example 1
 Calculate the output voltage for the circuit below.
 The inputs are V1 = 50 mV sin(1000t) and V2 = 10 mV sin(3000t).
330 k
33 k
_
V1
7
2
Vout
741
10 k
V2
3
+
6
4
Answer
Vout = - [0.5 sin(1000t) + 0.33 sin(3000t)]
8
Summing Amplifiers Applications
Digital to Analog Converter (DAC)
• Binary-weighted Resistor DAC (scaling adder)
Note:
Rf = 8R
Summing Amplifiers Applications
Digital to Analog Converter
• R/2R Ladder DAC
Differencing Amplifiers
Vout 
Rf
R1
V1  V2 
This circuit produces an output which is proportional to the
difference between the two inputs.
Differencing Amplifiers
The circuit is linear so we can look at the output due to each
input individually and then add them.
Output due to V2 is the same as the inverting amplifier, so:
Vout2  
Rf
R1
V2
The signal to the non-inverting output, is reduced by the voltage
divider:
Vin 
Rf
R1  R f
V1
Differencing Amplifiers
The output V1 is that for a non-inverting amplifier:
Vout1 
R1  R f
R1
Vin 
R1  R f
Rf
R1
R1  R f
V1 
Rf
R1
V1
Thus the output is:
Vout  Vout1  Vout2 
Vout 
Rf
R1
Rf
R1
V1 
Rf
R1
V2
(V1  V2 )
Thus the amplifier subtracts the inputs and amplifies their
difference.