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EE 201 op amps – 1
Operational amplifiers (Op amps)
View it as an ideal amp. Take the properties to the extreme: Ri → ∞, Ro = 0, A → ∞. ?!?!?!?!
Ri–
+vi
+– Avi
Ro
v+
v––
+vo
–
+vi
+– Avi
v+
v––
+vo
A → ∞
Consequences: No voltage dividers at input or output. (That’s good.) No current flows into the input. (That’s good.) The gain is infinite. (Is that good?)
EE 201 op amps – 2
How do we handle this infinite gain business?
vo = Avi
In order to keep vo finite, then as A → ∞ , vi → 0. In other words, we must force the difference signal at the input to go to zero. How do we do that? With feedback, of course.
6G =�
�+ $�6L
Recall that the difference signal in a feedback arrangement must become very small if the gain is very big.
→ 0, as A → ∞
One input is connected to the source voltage in some fashion. The other input is connected to the feedback network. If the feedback is working properly, then vi = v+ – v_ = 0.
The condition of v+ = v_ is called a virtual short at the input. This should be the case, if negative feedback is working in the circuit.
EE 201 op amps – 3
–+
–vo+
v–
v+
v+ → non-inverting input v_ → inverting input
Ideal op amp
v+ ≈ v_ → virtual short (assuming a proper negative feedback configuration.)
i+ = i_ = 0 → due to infinite input resistance
No voltage divider effects at output. This means that we can connect anything to the output without worrying about loading effects.
When using an op amp in a circuit:
EE 201 op amps – 4
Non-inverting amplifier
–+
+–
+vo–
R1
R2
vS
Write a node equation at the inverting terminal.
YR � Y�5�
=Y�5�
+ L�
v– = v+ = vs (virtual short due to feedback) i– = 0 (infinite input resistance)
YR � Y65�
=Y65�
* =YRY6
=5� + 5�5�
= �+5�5�
(Worth memorizing.)
EE 201 op amps – 5
Inverting amplifier
–++
–
R1
R2
–vo+
vS
Write a node equation at the inverting terminal.
v– = v+ = 0 (virtual ground!) i– = 0 (infinite input resistance)
YV � Y�5�
=Y� � YR
5�+ L�
YV5�
=�YR5�
* =YRY6
= �5�5�
Note the negative sign!
(Also worth memorizing.)
EE 201 op amps – 6
Summing amp (weighted summer)
–++
–R1
RF
–vo+
vS1+–
vS2
R2
R3
+–
vS3
At the inverting terminal: v– = v+ = 0 (virtual ground). Write a node equation there. (Or use superposition.)
YR = ��5)5�
Y6� +5)5�
Y6� +5)5�
Y6��
Y6�5�
+Y6�5�
+Y6�5�
=�YR5)
Use another inverter if you don’t like the negative sign.
The “virtually grounded” inverting terminal becomes a summing node.
EE 201 op amps – 7
Difference ampWe would like to amplify only the difference between va and vb. Anything this applied in common to both, will not be amplified.
–+va
vbR1
R2
R3 R4
At the inverting input:
YE � Y�5�
=Y� � YR
5�
YR =
��+
5�5�
�Y� � 5�
5�YE
At the non-inverting input:YD � Y+
5�=
Y+
5�Y+ =
5�5� + 5�
YD
v+ = v–
YR =
��+
5�5�
� �5�
5� + 5�YD
�� 5�
5�YE
5�5�
=5�5�
if then YR =5�5�
(YD � YE)Difference only! (Check it.)
EE 201 op amps – 8
Unity gain buffer
–+vs vo
Non-inverting amp with R2 = 0 and R1 → ∞ . So G = 1, meaning vo = vs. What good is that?
Connecting two circuits. If Ro1> Ri2, then much of the voltage is lost in the voltage divider, vi2 << v1.
+–
v1Ro1
Ri2 +–
v1Ro1 –
+
Ri2
High input resistance of op amp makes v+ = v1. Zero output resistance of op amp makes vi2 = vo. Since vo = v+, then vi2 = v1. The op amp served as a buffer between the two circuits, eliminated the voltage divider problem.
EE 201 op amps – 9
Digital-to-analog converter (DAC)Data is stored as digital information on your phone or computer. Some of that data needs to be converted to analog form in order to use it, i.e. music.
4-bit DAC
The digital bits control the operation of the switches – one bit per switch. Si = 0, switch is open, Si = 1 switch is closed.
So only closed: vo = –VREF/8. S1 only closed: vo = –VREF/4.
S2 only closed: vo = –VREF/2. S3 only closed: vo = –VREF.
S vo (V)0000 00001 0.1250010 0.250011 0.3750100 0.50101 0.6250110 0.750111 0.8751000 11001 1.1251010 1.251011 1.3751100 1.51101 1.6251110 1.751111 2
R3 = RF R2 = RF / 2R1 = RF / 4 Ro = RF / 8
VREF = -1 V
–+
VREF
–vo+
Ro
R1
R2
R3S3
S2
S1
S0
RF
EE 201 op amps – 10
Cascading amps
The various types of circuits can serve as building blocks for more complicated circuits.
–+
R3
R4
vo1
–+
vo2
R1
R2
–+R6
R7
vo3
R5va
vb
inverting
non-inverting
summing
YR� = �5�5�
YD
YR = �5�5�
YR� � 5�5�
YR�
YR =5�5�
5�5�
YD � 5�5�
��+
5�5�
�YE
YR� =
��+
5�5�
�YE
EE 201 op amps – 11
10 k!
1 k!
20 k!
8 k!
2 k!
non-inverting
YR� =
��+
5�5�
�YR�
= �YR�
summing
YR� = �5�5�
Y6 � 5�5�
YR�
want vo2 / vS.
–+
vo2
R3
R4
–+
R1R2
vS
R5
vo1
= ���Y6 � �.�YR�
YR�� = ���Y6 � �.�YR�
�.�YR� = ���Y6YR�Y6
= ���.��
Feedback loop around feedback loops !!