Upload
cloris
View
29
Download
0
Embed Size (px)
DESCRIPTION
Loop Analysis (3.2) Circuits with Op-Amps (3.3). Prof. Phillips February 19, 2003. Op Amps. Op Amp is short for operational amplifier. An operational amplifier is modeled as a voltage controlled voltage source. An operational amplifier has a very high input impedance and a very high gain. - PowerPoint PPT Presentation
Citation preview
lecture 9 1
Loop Analysis (3.2)Circuits with Op-Amps (3.3)
Prof. Phillips
February 19, 2003
lecture 9 2
Op Amps
• Op Amp is short for operational amplifier.
• An operational amplifier is modeled as a voltage controlled voltage source.
• An operational amplifier has a very high input impedance and a very high gain.
lecture 9 3
Use of Op Amps
• Op amps can be configured in many different ways using resistors and other components.
• Most configurations use feedback.
lecture 9 4
Applications of Op Amps
• Amplifiers provide gains in voltage or current.
• Op amps can convert current to voltage.
• Op amps can provide a buffer between two circuits.
• Op amps can be used to implement integrators and differentiators.
• Lowpass and bandpass filters.
lecture 9 5
The Op Amp Symbol
+
-
Non-inverting input
Inverting input
Ground
High Supply
Low Supply
Output
lecture 9 6
The Op Amp Model
+
–Inverting input
Non-inverting input
Rin
v+
v-
+–
A(v+ -v- )
vo
lecture 9 7
Typical Op Amp
• The input resistance Rin is very large (practically infinite).
• The voltage gain A is very large (practically infinite).
lecture 9 8
“Ideal” Op Amp
• The input resistance is infinite.
• The gain is infinite.
• The op amp is in a negative feedback configuration.
lecture 9 9
The Basic Inverting Amplifier
–
+Vin+
–Vout
R1
R2
+–
lecture 9 10
Consequences of the Ideal
• Infinite input resistance means the current into the inverting input is zero:
i- = 0
• Infinite gain means the difference between v+ and v- is zero:
v+ - v- = 0
lecture 9 11
Solving the Amplifier Circuit
Apply KCL at the inverting input:
i1 + i2 + i-=0
–R1
R2
i1 i-
i2
lecture 9 12
KCL
0i
111 R
v
R
vvi inin
222 R
v
R
vvi outout
lecture 9 13
Solve for vout
Amplifier gain:
21 R
v
R
v outin
1
2
R
R
v
v
in
out
lecture 9 14
Recap
• The ideal op-amp model leads to the following conditions:
i- = 0 = i+
v+ = v-
• These conditions are used, along with KCL and other analysis techniques, to solve for the output voltage in terms of the input(s).
lecture 9 15
Where is the Feedback?
–
+Vin+
–Vout
R1
R2
+–
lecture 9 16
Review
• To solve an op-amp circuit, we usually apply KCL at one or both of the inputs.
• We then invoke the consequences of the ideal model.
– The op amp will provide whatever output voltage is necessary to make both input voltages equal.
• We solve for the op-amp output voltage.
lecture 9 17
The Non-Inverting Amplifier
+
–
vin
+
–
vout
R1
R2
+–
lecture 9 18
KCL at the Inverting Input
+
–
vin
+
–
vout
R1
R2
i-
i1 i2
+–
lecture 9 19
KCL
0i
111 R
v
R
vi in
222 R
vv
R
vvi inoutout
lecture 9 20
Solve for Vout
021
R
vv
R
v inoutin
1
21R
Rvv inout
lecture 9 21
A Mixer Circuit
–
+v2+
–vout
R2
RfR1
v1
+–
+–
lecture 9 22
KCL at the Inverting Input
–
+v2+
–vout
R2
RfR1
v1
i1
i2
if
i-
+–
+–
lecture 9 23
KCL
1
1
1
11 R
v
R
vvi
2
2
2
22 R
v
R
vvi
lecture 9 24
KCL
0i
f
out
f
outf R
v
R
vvi
lecture 9 25
Solve for Vout
02
2
1
1 f
out
R
v
R
v
R
v
22
11
vR
Rv
R
Rv ff
out
lecture 9 26
Class Example