Download Lecture 2 - Operational Amplifier

CHAPTER 8
IDEAL OPERATIONAL AMPLIFIER AND
OP-AMP CIRCUITS
inverting
output
non-inverting
Op-amp circuit symbol
• Open loop mode
• Vo = Aod ( v2 – v1)
– Aod is referred to as the
open loop gain.
– Notice that is v2 = v1, the
open loop gain equals to 
• Two main characteristics:
• We want the open loop gain to be equal to  which means
that v2 = v1
0
0
• the input resistance to be equal to  , hence there is no
current going into the op-amp
• Op amp can be configured to be used for
different type of circuit applications:
– Inverting Amplifier
– Non – inverting Amplifier
– Summing Amplifier
– Integrator
– Differentiator
Inverting Amplifier
Op-amp as an inverting amplifier
Voltage at node 1 (inverting) = voltage at node 2 (non-inverting )
KCL at node 1:
I1 – I2 – Iin = 0
(Vi – 0) / R1 = (0 – Vo) / R2
Vi / R1 = - Vo / R2
Vo = - R2
Vi
R1
Example 1
Gain = - (R2 / R1) = -(150/12) = -12.5
Example 2
A voltage source 𝑉𝑆 with source resistance 𝑅𝑆 = 1.5 𝑘Ω is connected to the input
of an op-amp inverting amplifier circuit
(a) If the 𝑅1 = 1.0 𝑘Ω and 𝑅2 = 15.0 𝑘Ω then calculate the voltage gain, VO / VS
(b) Determine the output voltage 𝑉𝑜 for the source voltage 𝑉𝑆 = 45 mV
Answers:
(a) - 6
(b) - 0.27 V
Non - Inverting Amplifier
Voltage at node 1 (inverting) = voltage at node 2 (non-inverting )
KCL at node 1:
i1 – i2 = 0
(0– Vi) / R1 = (Vi – Vo) / R2
-(Vi / R1) = (Vi / R2) – (Vo / R2)
Vo / R2 = (Vi / R2) + (Vi / R1) = Vi 1 + 1
R2
R1
Vo / Vi = R2 1 + 1
R2
R1
Noninverting amplifier
Voltage Follower / Buffer Amplifier
vo = vI
Hence, gain = 1
Example 1
A voltage source 𝑉𝑆 with source resistance 𝑅𝑆 = 5 𝑘Ω is connected to a load
resistance input of 𝑅𝐿 = 5 𝑘Ω directly and through a buffer amplifier as shown in
the figures below.
Determine the output voltage across the load and the current in the resistance RL
for each circuit configuration
Answers:
Vo = 5 V, Current = 1 mA
Vo = 10 V, current = 2 mA
Summing Amplifier
Summing
Similarly,Amplifier
i1 + i2 + i3 – i4 – 0 = 0
Output voltage
Example 8.2
Design a summing amplifier as shown in figure to produce a specific output signal, such
that vo = 1.25 – 2.5 cos t volt. Assume the input signals are vI1 = -1.0 V, vI2 = 0.5 cos t
volt. Assume the feedback resistance RF = 10 k
Solution: output voltage
Other Op-Amp Applications
Integrator
When the feedback resistor of an inverter circuit is replaced by a capacitor the circuit is
worked as an integrator circuit -cause the output to respond to changes in the input voltage
over time
Integrator circuit
Example 1
The integrator circuit as shown in figure has an initial voltage 𝑉𝑥 = −1.4 V across the capacitor
at time 𝑡 = 0. A step input voltage 𝑉𝑆 = −2V is applied at time 𝑡 = 0. Determine the RC time
constant necessary such that the output voltage reaches +10.2 V at time 𝑡 = 5.0 ms.
Solution: The output voltage
𝑉𝑜 = 𝑉𝑥 −
1
𝑅1 𝐶
1
= 𝑉𝑥 −
𝑅1 𝐶
−2
10.2 = −1.4 −
𝑅1 𝐶
𝑅1 𝐶 = 0.862 𝑚𝑠
𝑉𝑆 𝑑𝑡
5
𝑉𝑆 𝑑𝑡
0
5
0
2
𝑑𝑡 = −1.4 +
5
𝑅1 𝐶
Differentiator
When the inverting input terminal resistor of an op-amp inverter circuit is replaced by a
capacitor the circuit is worked as a differentiator circuit.
Differentiator circuit
Because Q = CVS
Example 1
Solution:
The output voltage
𝑣𝑂 = −𝑅2 𝐶1
𝑑𝑣𝐼
𝑑𝑡
= − 1.5 × 10−3
𝑑 3.5 𝑐𝑜𝑠 100𝜋𝑡
𝑑𝑡
𝑣𝑂 = − 1.5 × 10−3 −3.5 × 100𝜋 × 𝑠𝑖𝑛 100𝜋𝑡
𝑣𝑂 = 1.65 𝑠𝑖𝑛 100𝜋𝑡 volt
Calculating Gain and Design Questions
NON - INVERTING
INVERTING
Calculating Output and Design Questions
SUMMING AMPLIFIER
INTEGRATOR AMPLIFIER
DIFFERENTIATOR AMPLIFIER
Va
NON - INVERTING
Vb
INVERTING
INVERTING
Calculate the input voltage if the final output, VO is 10.08 V.
Finally:
Va = (1 + 10/5) V1
0.504 = 3V1
V1 = 0.168 V
Then:
Vb = -(5/5) Va
-0.504 = - Va
Va = 0.504 V
Have to work backwards:
Vo = -(100/5) Vb
10.08 = -20 Vb
Vb = -0.504 V
Va
INVERTING
SUMMING
Calculate the output voltage, VO if V1 = V2 = 700 mV
Va = -(500/250) 0.7
Va = -1.4 V
Then:
Vo = - 500 [ Va / 100 + V2 / 50 ]
Vo = - 500 [ -1.4 / 100 + 0.7 / 50 ]
Vo = 0 V
Calculate the output voltage VO of the operational amplifier
circuit as shown in the figure.
Answer: -3 V