Download FIGURE 1-1 Analog versus digital: (a) analog - e

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
Operational Amplifiers
The operational amplifier (op-amp).
Note the op-amp has two inputs and one output.
Thomas L. Floyd
Digital Fundamentals, 8e
Copyright ©2003 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Electrical Characteristics
Note these ratings are for specific circuit conditions, and they often include minimum,
maximum and typical values.
Robert Boylestad
Digital Electronics
Copyright ©2002 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
CMRR
•
One rating worth mentioning that is unique to op-amps is
CMRR or Common-Mode Rejection Ratio.
•
Because the op-amp has two inputs that are opposite in phase
(inverting input and the non-inverting input) any signal that is
common to both inputs will be cancelled.
•
A measure of the ability to cancel out common signals is called
CMRR.
Robert Boylestad
Digital Electronics
Copyright ©2002 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Differential and common-mode operation: (a) differential-mode; (b) common-mode.
VC = ½ ( Vi1 – Vi2 )
EXAMPLE
Calculate the CMRR for the circuit measurements shown in Fig.
+
Vd
Vi1
=0.5mV
+
Vo
=8V
Vo
=8V
Vd
=1mV
-
-
Vi2
=0.5mV
(a)
+
+
Vo
=12V
Vi1
=1mV
-
-
V2
=1mV
Vc
=1mV
(b)
Vo
=12V
Solution
From the measurement shown in Figure, using the procedure in step 1
above, we obtain
Vo
8V
Ad 

 8000
Vd 1mV
The measurement shown in Fig. using the procedure in step 2 above, gives us
Ac 
Vo 12mV

 12
Vc
1mV
The value of CMRR is
Ad 8000
CMRR  
 666.7
Ac 12
Which can be expressed as
Ad
CMRR  20 log 1 0 
 20 log 1 0 666.7  56.48dB
Ac
Ac equivalent of op-amp circuit: (a) practical; (b) ideal.
Robert L. Boylestad and Louis Nashelsky
Electronic Devices and Circuit Theory, 8e
Copyright ©2002 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Operation of op-amp as constant-gain multiplier:
op-amp ac equivalent circuit.
Robert L. Boylestad and Louis Nashelsky
Electronic Devices and Circuit Theory, 8e
Copyright ©2002 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Operation of op-amp as constant-gain multiplier:
ideal op-amp equivalent circuit.
Robert L. Boylestad and Louis Nashelsky
Electronic Devices and Circuit Theory, 8e
Copyright ©2002 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Operation of op-amp as constant-gain multiplier:
redrawn equivalent circuit.
Robert L. Boylestad and Louis Nashelsky
Electronic Devices and Circuit Theory, 8e
Copyright ©2002 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Virtual ground in an op-amp.
Robert L. Boylestad and Louis Nashelsky
Electronic Devices and Circuit Theory, 8e
Copyright ©2002 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Inverting constant-gain multiplier.
Robert L. Boylestad and Louis Nashelsky
Electronic Devices and Circuit Theory, 8e
Copyright ©2002 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Noninverting constant-gain multiplier.
Robert L. Boylestad and Louis Nashelsky
Electronic Devices and Circuit Theory, 8e
Copyright ©2002 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
A Constant–Gain Amplifier
Rf
A
R1
Rf
A 1
R1
Robert Boylestad
Digital Electronics
Copyright ©2002 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Voltage Buffer
Any amplifier with no gain (or loss) is called a unity gain amplifier.
The advantages of using a unity gain amplifier:
• very high input impedance
• very low output impedance
Realistically these circuits will be designed using resistors that are equal
(R1 = Rf) to void out problems with offset voltages.
Robert Boylestad
Digital Electronics
Copyright ©2002 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Summing amplifier
Rf
Rf 
 Rf
Vo   V1 
V2 
V3 
R2
R3 
 R1
Robert L. Boylestad and Louis Nashelsky
Electronic Devices and Circuit Theory, 8e
Copyright ©2002 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
virtual-ground equivalent circuit.
Robert L. Boylestad and Louis Nashelsky
Electronic Devices and Circuit Theory, 8e
Copyright ©2002 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
A 4-bit DAC with binary-weighted inputs.
Thomas L. Floyd
Digital Fundamentals, 8e
Copyright ©2003 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Example : A 4-bit DAC with binary-weighted inputs.
Thomas L. Floyd
Digital Fundamentals, 8e
Copyright ©2003 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Active Filters
Adding capacitors to op-amp circuits provides an
external control for the cutoff frequencies. The
op-amp active filter provides controllable cutoff
frequencies and controllable gain.
• Low-Pass Filter
• High-Pass Filter
• Band pass Filter
Low-Pass Filter
First-order low-pass filter.
Low-Pass Filter (Cont’d)
Second-order low-pass filter:
By adding more RC networks the roll-off can be made steeper.
High-Pass Filter
The lower cutoff frequency:
1
fOL 
2πR1C1
Bandpass Filter
There are two cutoff frequencies: upper and lower. They can be calculated using
the same low-pass cutoff and high-pass cutoff frequency formulas in the
appropriate sections.
Questions
1. Calculate CMRR (dB) for the following Op Amp values:
Vd = 1 mV, Vo = 200 mV dan Vc = 1 mV, Vo = 30uV.
Calculate the output voltage if the input voltage is 150mV pmkd
750 k
+ 10 V
50 k
Vo
+
- 10 V
V1
Calculate the output voltage, Vo
10 k
500 k
750 k
850 k
+ 22 V
+ 22 V
+ 22 V
25 k
15 k
+
+
+
- 22 V
V1
Vo
- 22 V
- 22 V
Calculate the output voltage, Vo
150 k
780 k
15 k
78 k
V1
15 mV
+
Vo
78 k
+
V2
20 mV
Calculate the output voltage, Vo
15 
V1 = 10 V
V2 = 20 V
650 k


V3 = 30 V
Vo
+
Determine the cut-off frequency
k
k
Vo
0.02uF
+
0.02uF
V1
k
k
Determine the Low and High cut-off frequency
20 k
5 k
20 k
5 k
Vo
0.05uF
+
+
Vi
20 k
0.02uF
Calculate the Value of V1 & V2
Robert L. Boylestad and Louis Nashelsky
Electronic Devices and Circuit Theory, 8e
Copyright ©2002 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Vo = ?
Robert L. Boylestad and Louis Nashelsky
Electronic Devices and Circuit Theory, 8e
Copyright ©2002 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
Vo = ?
Related documents