OPAMP
By: Sachin K Jain
Operational Amplifier (OP-AMP)
An op-amp amplifies the difference between its inverting and noninverting inputs .
Ideal OP-AMP
Ideal OP-AMP • First, the ideal op-amp has infinite voltage gain (Vout/Vin) and, •
Infinite bandwidth (have the same voltage gain for all input frequencies).
• Also, it has an infinite input impedance (Zin). Impedance is related to resistance hence its input draws 0 current. • Finally, it has a zero output impedance. (Its output has no resistance)
Practical OP-AMP
Practical OP-AMP • Although modern integrated circuit op-amps approach parameter values that can be treated as ideal in many cases, the ideal device cannot be made. • Any device has limitations, and the integrated circuit op-amp is no exception. Op-amps have both voltage and current limitations. Peak-to-peak output voltage, for example, is usually limited to slightly less than the supply voltages. Output current is also limited by internal restrictions such as power dissipation and component ratings. • Characteristics of a practical op-amp are very high voltage gain, very high input impedance, very low output impedance and finite bandwidth (different voltage gain for different frequencies).
Useful OP-AMP terminologies • Input Offset Voltage: – The ideal op-amp produces zero volts out for zero volts in. Practical op-amps produces a small dc voltage at the output when the input is zero.
• Input Offset Voltage drift: – Specifies how much the Input Offset Voltage changes with each degree change in temperature
Useful OP-AMP terminologies • Input Impedance: Has two definitions depending on how you are using the amplifier – (1) The resistance between the ‘+’ (non inverting) and ‘-’ (inverting) inputs. – (2) The resistance between each input and ground.
Useful OP-AMP terminologies •
Open-Loop voltage gain Aol: –
•
Gain without feedback (we discuss feedback as the next topic). Ideal op-amps have infinite gain. Practical op-amps range from 50,000 to 200,000.
Frequency response: – This has to do with how the gain changes with different input frequencies. For ideal op-amps the gain does not change with frequency. For practical op-amps, the gain decreases as the frequency f requency increases.
Negative Feedback
Negative feedback is one of the most useful concepts in electronic circuits, particularly ) op-amp applications. Negative feedback is the process whereby a portion of the out- voltage of an amplifier is returned to the input with a phase angle that opposes the input signal.
Why Negative Feedback? • An amplifier cannot give an output that has a voltage greater its supply voltage. • If the gain of the amp is 20,000, the supply is 12Volts and the input voltage is 2 Volts then the output is calculated as 20,000 * 2 = 40,000Volts. • Obviously 40,000Volts 40,000Volts is just not practical and the output will go as high as it can and stay at 12 volts. • Negative feedback reduces the gain and increases the bandwidth.
Noninverting Amplifier
Noninverting Amplifier • An op-amp connected in a closed loop configuration as a noninverting amplifier with a controlled amount of voltage gain is shown in the previous slide. • The input signal is applied to the t he noninverting input (‘+’). • The output is applied back to the inverting input (‘-’) through the feedback circuit formed by Ri and Rf • The voltage across Ri, Vi is calculated on the next slide.
Noninverting Amplifier •
Vi = Voltage across R i and It = Total current
• • • •
Vi = It*Ri We need to find I t. Total current = total voltage/total resistance. For a series circuit, the total resistance (R t) = sum of all resistors in the circuit. => Rt= Rf +Ri, Also Vt = Vout Finally, It = Vout / Rf +Ri Now we can find V i, Vi= (Vout / Rf +Ri)* Ri
• • • • •
Calculating the gain with feed back for noninverting configuration
Calculating the gain with feed back for noninverting configuration • Recall that AOL is the open loop gain i.e. gain without feed back. • VOUT= AOL(VIN - Vi) • Recall that Vi= (Vout / Rf +Ri)* Ri • Let R i /(R i + R f )= B, • Then apply basic algebra as follows: • V out = Aol V in -Aol BV out • V out + Aol BV out = Aol V in • V out (1 + Aol B) = Aol V in • Since the total voltage gain of the amplifier in the previous slide is V out /V in
Calculating the gain with feed back for noninverting configuration • The product The product of A Aol B is typically much greater than 1 hence,
Becomes,
The above formula gives the gain of a noninverting amplifier.
Inverting Amplifier
Inverting Amplifier •
•
• •
Let's say a current of 1 milliamp is caused to flow to the inverting input pin through a 1000 ohm input resistor, R i, the Op-Amp tries to maintain equilibrium, i.e., no current flow in that inverting input pin (high resistance). To do this marvelous feat, it generates an output voltage of the opposite polarity, which maintains that 1 milliamp to flow through a 10 K feedback resistor, R f . Because the feedback resistor is ten times the value of the input resistor, it will w ill require ten times the voltage to cause that same 1 mA to flow. The view from the input pin: there is a current of 1 milliamp coming down the input resistor, and at the same time AND there is a current of 1 milliamp coming from the feedback resistor. There is no current left over for the input pin; This satisfying the zero current requirement of the Op-Amp. This is because the output is the inverse of the input and both waves cancel. This the zero current results in zero voltage at the inverting input terminal and is referred to as virtual ground. This condition is illustrated in the next slide.
Inverting Amplifier Because Iin and If are the same but opposite, the voltage at the inverting input must be 0.
Since there is zero current flowing to the inverting input, the current through Ri is therefore equal but opposite to R f . => If = -Ii The voltage across Ri=-Vin because of the virtual ground on the other side of the resistor. i.e. 0 – Vin = - Vin I = - Vin /Ri
⇒ in
The voltage across Rf = Vout because of the virtual ground. i.e. Vout – 0= Vout
Inverting Amplifier • Recall that there is zero current flowing to the inverting input, the current through Ri is therefore equal but opposite to Rf . • => If = -Ii • Vout /Rf = - Vin /Ri • => Vout /Vin = - R /R f i • This final equation gives the gain for an inverting amplifier. The minus sign implies that the t he output is inverted. • Recall from previous slide that gain is V out /V in
Summing Amplifier (Variation of the inverting amp)
Now, since –(I1+I2) = It -(V1 /R1 + V2 /R2) = Vout /Rf Vout = - Rf (V1 /R1 + V2 /R2)
⇒
If all the resistors are the same then Vout = - (V1 + V2)
⇒
Hence we end up with a summing device.
⇒
Summing Amplifier (Variation of the inverting amp) • The formula is true for more inputs:
Exercise: When Rf is larger than the other resistors using the same analysis show that :
OP-AMP as an Integrator and Differentiator • Differentiator
• Now, the charge through the capacitor is given by qc =CVin …..(1) • If we differentiate equation (1) with respect to time we get:
dq
c
dt
= C
dV in dt
• Now recall that the rate at which charge flows is current.
=> I c =
C
dV in dt
………………(2)
Recall that because of the feed back, Ir I r = -Ic also because of the zero volts between the resistor and the capacitor, the voltage across the capacitor is – Vin and the voltage across the resistor is Vout •
• Now IT = V OUT / R
V out
•
R
Finally
V out
= -
c
dV in dt
= − RC
dV in
………….(3)
dt
From equation (3) we see that if the product of RC = 1
•
then Vout = - dVIN / dt
Integrator
Again, the charge through the capacitor is given by
qc=CVout
……………(4)
If we differentiate equation (1) with respect to time we get:
•
dq c dt
= C
dV out dt
Again recall that the rate at which charge flows is current.
•
=> I c = C Now IT = VIN / R
•
dV out dt
……………………. (3)
•
Ic = - Ir, hence C
dV out dt
1
=-
RC
dV out dt
∗ V in
=-
V in R
…………………(4)
Integrating both sides of equation (4) gives
Vout = -
1
t
∫ RC
V in dt + k
0
Where k is the Vout at start time (t=0)
• Finally if RC = 1, then t
• Vout = -
∫
V in dt + k
0
