Ideal Operational Amplifier analysis and design

Ideal Operational Amplifier analysis and design
Ideal OP-Amps: an op-Amp is high gain dc
differential amplifier.
Ideal characteristics :
1- Gain infinity
Vo=0 when v1=v2
Rd=infinite (input impedance
Ro=0 (output impedance is zero)
Equivalent Op-Amp circuit
Ro
Rd
A(v2-v1)
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University
1
Two Basic Rules for design of OP-Amp
Rule1: When the Op-Amp output is in its linear range the
two inputs are the same voltage.
Rule 2: No current flows into either input terminal of the
Op-Amp.
Inverting Amplifiers
vi
Rf
i
Ri
v1
v2
-
vo
+
It is widely used in instrumentation
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University
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Equation : v2=0; v1=0 by using rule 1.
Ohm’s Law: Vi=I. Ri
Rule 2 : Vo=-I. Rf=-ViRf/Ri; imply that
Vo/Vi=-Rf/Ri;
Input output plot of inverting amplifier :
The input output plot shows an slope of –Rf/Ri.
In the central portion but the output saturate at about
±13 volts.
+13 V
Slope=-Rf/Ri
+VT
-VT
-13 V
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University
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SUMMING Amplifier
The Inverting Amplifier may be extended to for m a circuit
that yields sum of several input voltages. Each input voltage
V1, V2, V3, …Vk is connected to the negative input of the OpAmp by individual resistor Rk, the conductance 1/Rk is
proportional to the desired weighting.
if
v1
R1
v2
v3
If=I1 +I2 +I3
Rf
R2
-
vo
+
R3
If=-Vo/Rf ; I1=V1/R1;
I2=V2/R2;
I3=V3/R3
Vo=-Rf (V1/R1+V2/R2+V3/R3);
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University
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Non Inverting Amplifier
Ri
-
vi
i=vi/Ri ;
i
Rf
vo
+
V0=i(Rf+Ri); Vo/vi=i(Rf+Ri)/iRi = 1+ Rf/Ri;
vo +13 V
Input Output Plot
Slope=1+Rf/Ri
-VT
+VT
vi
-13 V
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University
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Differential Amplifiers
i
v1
R2
R1
v2
v3
+
R1
vo
R2
V3=V2 R2/(R1+R2); i=(V1-V3)/R1 = (V3-Vo)/R2;
Vo=(V2-V1)R2/R1; Differential Amplifiers
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Comparators
R1
Vi
Vref
+
Rf
R2
R3
Comparator is a circuit that compares the input voltage
with some reference voltage.
Vo
+13V
If R3 =0;
V0=+13V if Vi=-Vref
Vo flip when Vi+Vref=0;
Vi >Vref → Vo=-13V;
Vi< Vref → V0=+13V.
Vi
-Vref
-13V
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University
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Non Inverting Configuration (Real Op-Amp)
Negative Feed Back on voltage Gain
Vin
+
Vf
Vid
Aol
Vo
-
B
Vid=Vin-Bvo =Vin -Vf;
Vo=Aol (Vin-Bvo);
Av=Vo/Vin=Aol/(1+BAol);
Ideal Case Aol→∞ which imply Av=1/B;
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University
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Non Inverting
i
Rf
Ri
Av=1+Rf/R1
-
vi
+
vo
Effect of Negative Feed Back on Output Resistance
RoF =Ro/(1+B.Aol).
For inverting B=R1/Rf;
For Non-inverting amplifier B=R1/(R1+Rf);
Inverting Amplifier input resistance
Rinf=R1 (Without feed back)
With Feed Back R’F=Rinf/(Aol+1)≈Rinf/Aol.
If Aol very big → R’F≈ Very Small can be neglected.
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University
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Non Inverting Op-Amp input resistance
RinF=VinRin/Vid=Rin(1+BAol) → input resistance very High.
Voltage Follower:
Vid
+
Vo
vin
Vo=V1 ; the circuit operate like an emitter follower,
except that the gain is exactly 1.
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University
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Bode plot and Bandwidth limitation
100
AMP(dB)
80
60
40
20
10
100
1k
10k
100k
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University
1M
10M
11
fc: corner frequency, Break frequency or critical frequency.
fc : is defined as frequency at which the gain of the amplifier
reduced by 3dB from the zero frequency gain or mid frequency.
Midband ranges : from mid frequency to fc : [0,fc].
Example : A certain Op-Amp has Aol =120,000 fc= 100Hz;
Determine Amid (dB) and Ac (numerical).
Solution :
Amid(dB)=20logAol= 20x5,08=101,6dB.
Ac(dB)=101,6-3=98.6dB
Ac(num)=log-1(Ac(dB)/20)=85000.
Ac=0.707 Amid.
Unity Gain : When the bode plot cross zero, the frequency at that
position is equal to the unity gain. For the OP-Amp of the previous
slide funity=Gain-Band width=10Mhz. GBW=gain(numerical)x
frequency at any point of the curve (Bode plot).
GBW=Av (num) xBW;
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Example
Determine Av and BW?
vi
176K
1.76K
-
vo
+
The amplifier follow the bode plot of slide 11.
Solution : GBW=BWx Av
Av=-176/1.76=-100=40dB.
BW=fc=GBW/ Av
Funity=frequency at which Aol(num)=1;
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University
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Prepared by Dr. Ali Saad King Saud
University
14