Op-amps: introduction * The Operational Amplifier (Op-Amp) is a - - PowerPoint PPT Presentation

Op-amps: introduction

* The Operational Amplifier (Op-Amp) is a versatile building block that can be used for realizing several electronic circuits.

• M. B. Patil, IIT Bombay

Op-amps: introduction

* The Operational Amplifier (Op-Amp) is a versatile building block that can be used for realizing several electronic circuits. * The characteristics of an op-amp are nearly ideal → op-amp circuits can be expected to perform as per theoretical design in most cases.

• M. B. Patil, IIT Bombay

Op-amps: introduction

* The Operational Amplifier (Op-Amp) is a versatile building block that can be used for realizing several electronic circuits. * The characteristics of an op-amp are nearly ideal → op-amp circuits can be expected to perform as per theoretical design in most cases. * Amplifiers built with op-amps work with DC input voltages as well → useful in sensor applications (e.g., temperature, pressure)

• M. B. Patil, IIT Bombay

Op-amps: introduction

* The Operational Amplifier (Op-Amp) is a versatile building block that can be used for realizing several electronic circuits. * The characteristics of an op-amp are nearly ideal → op-amp circuits can be expected to perform as per theoretical design in most cases. * Amplifiers built with op-amps work with DC input voltages as well → useful in sensor applications (e.g., temperature, pressure) * The user can generally carry out circuit design without a thorough knowledge

• f the intricate details of an op-amp. This makes the design process simple.
• M. B. Patil, IIT Bombay

Op-Amp 741

Q23 Q2 Q1 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q24 R1 R2 R3 R4 R6 R7 R8 R9 R10 R5 CC

Symbol

• ffset adjust

OUT

OUT

−VEE VCC VCC −VEE

• M. B. Patil, IIT Bombay

Op-amp: equivalent circuit

OUT OUT OUT

Vo Vo Vi Vi AV Vi AV Vi Ro −VEE VCC Ri

• M. B. Patil, IIT Bombay

Op-amp: equivalent circuit

OUT OUT OUT

Vo Vo Vi Vi AV Vi AV Vi Ro −VEE VCC Ri

* The external resistances (∼ a few kΩ) are generally much larger than Ro and much smaller than Ri → we can assume Ri → ∞, Ro → 0 without significantly affecting the analysis.

• M. B. Patil, IIT Bombay

Op-amp: equivalent circuit

OUT OUT OUT

Vo Vo Vi Vi AV Vi AV Vi Ro −VEE VCC Ri

* The external resistances (∼ a few kΩ) are generally much larger than Ro and much smaller than Ri → we can assume Ri → ∞, Ro → 0 without significantly affecting the analysis. * VCC and −VEE (∼ ±5 V to ±15 V ) must be supplied; an op-amp will not work without them!

• M. B. Patil, IIT Bombay

Op-amp: equivalent circuit

OUT OUT OUT

Vo Vo Vi Vi AV Vi AV Vi Ro −VEE VCC Ri

* The external resistances (∼ a few kΩ) are generally much larger than Ro and much smaller than Ri → we can assume Ri → ∞, Ro → 0 without significantly affecting the analysis. * VCC and −VEE (∼ ±5 V to ±15 V ) must be supplied; an op-amp will not work without them! In op-amp circuits, the supply voltages are often not shown explicitly.

• M. B. Patil, IIT Bombay

Op-amp: equivalent circuit

OUT OUT OUT

Vo Vo Vi Vi AV Vi AV Vi Ro −VEE VCC Ri

* The external resistances (∼ a few kΩ) are generally much larger than Ro and much smaller than Ri → we can assume Ri → ∞, Ro → 0 without significantly affecting the analysis. * VCC and −VEE (∼ ±5 V to ±15 V ) must be supplied; an op-amp will not work without them! In op-amp circuits, the supply voltages are often not shown explicitly. * Parameter Ideal Op-Amp 741 AV ∞ 105 (100 dB) Ri ∞ 2 MΩ Ro 75 Ω

• M. B. Patil, IIT Bombay

Op-Amp: equivalent circuit

linear saturation saturation 10 5 −5 −10 −10 5 −5 10 OUT OUT OUT saturation linear saturation −5 5 −0.2 −0.1 0.1 0.2

Vo Vo Vi Vi AV Vi AV Vi Ro −VEE VCC Ri

−Vsat Vsat

slope = AV Vi (V) Vo (V) Vo (V) Vi (mV)

• M. B. Patil, IIT Bombay

Op-Amp: equivalent circuit

linear saturation saturation 10 5 −5 −10 −10 5 −5 10 OUT OUT OUT saturation linear saturation −5 5 −0.2 −0.1 0.1 0.2

Vo Vo Vi Vi AV Vi AV Vi Ro −VEE VCC Ri

−Vsat Vsat

slope = AV Vi (V) Vo (V) Vo (V) Vi (mV)

* The output voltage Vo is limited to ±Vsat, where Vsat ∼ 1.5 V less than VCC .

• M. B. Patil, IIT Bombay

Op-Amp: equivalent circuit

linear saturation saturation 10 5 −5 −10 −10 5 −5 10 OUT OUT OUT saturation linear saturation −5 5 −0.2 −0.1 0.1 0.2

Vo Vo Vi Vi AV Vi AV Vi Ro −VEE VCC Ri

−Vsat Vsat

slope = AV Vi (V) Vo (V) Vo (V) Vi (mV)

* The output voltage Vo is limited to ±Vsat, where Vsat ∼ 1.5 V less than VCC . * For −Vsat < Vo < Vsat, Vi = V+ − V− = Vo/AV , which is very small → V+ and V− are virtually the same.

• M. B. Patil, IIT Bombay

Op-amp circuits

10 −10 −5 5 saturation linear saturation OUT OUT

−5 5

Vo Vi AV Vi Ro −VEE VCC Ri Vsat −Vsat Vi (V) Vo (V)

• M. B. Patil, IIT Bombay

Op-amp circuits

10 −10 −5 5 saturation linear saturation OUT OUT

−5 5

Vo Vi AV Vi Ro −VEE VCC Ri Vsat −Vsat Vi (V) Vo (V)

* Broadly, op-amp circuits can be divided into two categories:

• M. B. Patil, IIT Bombay

Op-amp circuits

10 −10 −5 5 saturation linear saturation OUT OUT

−5 5

Vo Vi AV Vi Ro −VEE VCC Ri Vsat −Vsat Vi (V) Vo (V)

* Broadly, op-amp circuits can be divided into two categories:

• op-amp operating in the linear region
• M. B. Patil, IIT Bombay

Op-amp circuits

10 −10 −5 5 saturation linear saturation OUT OUT

−5 5

Vo Vi AV Vi Ro −VEE VCC Ri Vsat −Vsat Vi (V) Vo (V)

* Broadly, op-amp circuits can be divided into two categories:

• op-amp operating in the linear region
• op-amp operating in the saturation region
• M. B. Patil, IIT Bombay

Op-amp circuits

10 −10 −5 5 saturation linear saturation OUT OUT

−5 5

Vo Vi AV Vi Ro −VEE VCC Ri Vsat −Vsat Vi (V) Vo (V)

* Broadly, op-amp circuits can be divided into two categories:

• op-amp operating in the linear region
• op-amp operating in the saturation region

* Whether an op-amp in a given circuit will operate in linear or saturation region depends on

• M. B. Patil, IIT Bombay

Op-amp circuits

10 −10 −5 5 saturation linear saturation OUT OUT

−5 5

Vo Vi AV Vi Ro −VEE VCC Ri Vsat −Vsat Vi (V) Vo (V)

* Broadly, op-amp circuits can be divided into two categories:

• op-amp operating in the linear region
• op-amp operating in the saturation region

* Whether an op-amp in a given circuit will operate in linear or saturation region depends on

• input voltage magnitude
• M. B. Patil, IIT Bombay

Op-amp circuits

10 −10 −5 5 saturation linear saturation OUT OUT

−5 5

Vo Vi AV Vi Ro −VEE VCC Ri Vsat −Vsat Vi (V) Vo (V)

* Broadly, op-amp circuits can be divided into two categories:

• op-amp operating in the linear region
• op-amp operating in the saturation region

* Whether an op-amp in a given circuit will operate in linear or saturation region depends on

• input voltage magnitude
• type of feedback (negative or positive)

(We will take a qualitative look at feedback later.)

• M. B. Patil, IIT Bombay

Op-amp circuits (linear region)

10 −10 −5 5 saturation linear saturation OUT OUT

−5 5

Vo Vi AV Vi Ro −VEE VCC Ri iin Vsat −Vsat Vi (V) Vo (V)

• M. B. Patil, IIT Bombay

Op-amp circuits (linear region)

10 −10 −5 5 saturation linear saturation OUT OUT

−5 5

Vo Vi AV Vi Ro −VEE VCC Ri iin Vsat −Vsat Vi (V) Vo (V)

In the linear region, * Vo = AV (V+ − V−), i.e., V+ − V− = Vo/AV , which is very small → V+ ≈ V−

• M. B. Patil, IIT Bombay

Op-amp circuits (linear region)

10 −10 −5 5 saturation linear saturation OUT OUT

−5 5

Vo Vi AV Vi Ro −VEE VCC Ri iin Vsat −Vsat Vi (V) Vo (V)

In the linear region, * Vo = AV (V+ − V−), i.e., V+ − V− = Vo/AV , which is very small → V+ ≈ V− * Since Ri is typically much larger than other resistances in the circuit, we can assume Ri → ∞ . → iin ≈ 0

• M. B. Patil, IIT Bombay

Op-amp circuits (linear region)

10 −10 −5 5 saturation linear saturation OUT OUT

−5 5

Vo Vi AV Vi Ro −VEE VCC Ri iin Vsat −Vsat Vi (V) Vo (V)

In the linear region, * Vo = AV (V+ − V−), i.e., V+ − V− = Vo/AV , which is very small → V+ ≈ V− * Since Ri is typically much larger than other resistances in the circuit, we can assume Ri → ∞ . → iin ≈ 0 These two “golden rules” enable us to understand several op-amp circuits.

• M. B. Patil, IIT Bombay

Op-amp circuits (linear region)

ii RL R2 R1 Vi Vo

Op-amp circuits (linear region)

ii RL R2 R1 Vi Vo i1 Since V+ ≈ V−, V− ≈ 0 V → i1 = (Vi − 0)/R1 = Vi/R1 . (The non-inverting input is at real ground here, and the inverting input is at virtual ground.)

Op-amp circuits (linear region)

ii RL R2 R1 Vi Vo i1 Since V+ ≈ V−, V− ≈ 0 V → i1 = (Vi − 0)/R1 = Vi/R1 . (The non-inverting input is at real ground here, and the inverting input is at virtual ground.) Since ii (current entering the op-amp) is zero, i1 goes through R2 .

Op-amp circuits (linear region)

ii RL R2 R1 Vi Vo i1 Since V+ ≈ V−, V− ≈ 0 V → i1 = (Vi − 0)/R1 = Vi/R1 . (The non-inverting input is at real ground here, and the inverting input is at virtual ground.) Since ii (current entering the op-amp) is zero, i1 goes through R2 .

Op-amp circuits (linear region)

ii RL R2 R1 Vi Vo i1 Since V+ ≈ V−, V− ≈ 0 V → i1 = (Vi − 0)/R1 = Vi/R1 . (The non-inverting input is at real ground here, and the inverting input is at virtual ground.) Since ii (current entering the op-amp) is zero, i1 goes through R2 . → Vo = V− − i1 R2 = 0 − Vi R1

• R2 = −

R2 R1

• Vi .

Op-amp circuits (linear region)

ii RL R2 R1 Vi Vo i1 Since V+ ≈ V−, V− ≈ 0 V → i1 = (Vi − 0)/R1 = Vi/R1 . (The non-inverting input is at real ground here, and the inverting input is at virtual ground.) Since ii (current entering the op-amp) is zero, i1 goes through R2 . → Vo = V− − i1 R2 = 0 − Vi R1

• R2 = −

R2 R1

• Vi .

The circuit is called an “inverting amplifier.”

Op-amp circuits (linear region)

ii RL R2 R1 Vi Vo i1 Since V+ ≈ V−, V− ≈ 0 V → i1 = (Vi − 0)/R1 = Vi/R1 . (The non-inverting input is at real ground here, and the inverting input is at virtual ground.) Since ii (current entering the op-amp) is zero, i1 goes through R2 . → Vo = V− − i1 R2 = 0 − Vi R1

• R2 = −

R2 R1

• Vi .

The circuit is called an “inverting amplifier.” Where does the current go?

Op-amp circuits (linear region)

ii RL R2 R1 Vi Vo i1 RL R2 R1 i1 Vi Vo 0.1 V −1 V Since V+ ≈ V−, V− ≈ 0 V → i1 = (Vi − 0)/R1 = Vi/R1 . (The non-inverting input is at real ground here, and the inverting input is at virtual ground.) Since ii (current entering the op-amp) is zero, i1 goes through R2 . → Vo = V− − i1 R2 = 0 − Vi R1

• R2 = −

R2 R1

• Vi .

The circuit is called an “inverting amplifier.” Where does the current go?

Op-amp circuits (linear region)

ii RL R2 R1 Vi Vo i1 RL R2 R1 i1 Vi Vo 0.1 V −1 V Since V+ ≈ V−, V− ≈ 0 V → i1 = (Vi − 0)/R1 = Vi/R1 . (The non-inverting input is at real ground here, and the inverting input is at virtual ground.) Since ii (current entering the op-amp) is zero, i1 goes through R2 . → Vo = V− − i1 R2 = 0 − Vi R1

• R2 = −

R2 R1

• Vi .

The circuit is called an “inverting amplifier.” Where does the current go? (Op-amp 741 can source or sink about 25 mA.)

• M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier

5 −5 0.5 1 1.5 2 t (msec)

RL R2 R1 Vi Vm = 0.5 V f = 1 kHz Vo 10 k 1 k Vi Vi , Vo (Volts) Vo

• M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier

5 −5 0.5 1 1.5 2 t (msec)

RL R2 R1 Vi Vm = 0.5 V f = 1 kHz Vo 10 k 1 k Vi Vi , Vo (Volts) Vo

* The gain of the inverting amplifier is −R2/R1. It is called the “closed-loop gain” (to distinguish it from the “open-loop gain” of the op-amp which is ∼ 105).

• M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier

5 −5 0.5 1 1.5 2 t (msec)

RL R2 R1 Vi Vm = 0.5 V f = 1 kHz Vo 10 k 1 k Vi Vi , Vo (Volts) Vo

* The gain of the inverting amplifier is −R2/R1. It is called the “closed-loop gain” (to distinguish it from the “open-loop gain” of the op-amp which is ∼ 105). * The gain can be adjusted simply by changing R1 or R2 !

• M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier

5 −5 0.5 1 1.5 2 t (msec)

RL R2 R1 Vi Vm = 0.5 V f = 1 kHz Vo 10 k 1 k Vi Vi , Vo (Volts) Vo

* The gain of the inverting amplifier is −R2/R1. It is called the “closed-loop gain” (to distinguish it from the “open-loop gain” of the op-amp which is ∼ 105). * The gain can be adjusted simply by changing R1 or R2 ! * For the common-emitter amplifier, on the other hand, the gain −gm (RC RL) depends on how the BJT is biased (since gm depends on IC ).

• M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier

5 −5 0.5 1 1.5 2 t (msec)

RL R2 R1 Vi Vm = 0.5 V f = 1 kHz Vo 10 k 1 k Vi Vi , Vo (Volts) Vo

* The gain of the inverting amplifier is −R2/R1. It is called the “closed-loop gain” (to distinguish it from the “open-loop gain” of the op-amp which is ∼ 105). * The gain can be adjusted simply by changing R1 or R2 ! * For the common-emitter amplifier, on the other hand, the gain −gm (RC RL) depends on how the BJT is biased (since gm depends on IC ).

(SEQUEL file: ee101 inv amp 1.sqproj)

• M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier

15 −15 0.5 1 1.5 2 t (msec)

RL R2 R1 Vi Vm = 2 V f = 1 kHz Vo 10 k 1 k Vo Vi Vi , Vo (Volts)

• M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier

15 −15 0.5 1 1.5 2 t (msec)

RL R2 R1 Vi Vm = 2 V f = 1 kHz Vo 10 k 1 k Vo Vi Vi , Vo (Volts)

* The output voltage is limited to ±Vsat.

• M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier

15 −15 0.5 1 1.5 2 t (msec)

RL R2 R1 Vi Vm = 2 V f = 1 kHz Vo 10 k 1 k Vo Vi Vi , Vo (Volts)

* The output voltage is limited to ±Vsat. * Vsat is ∼ 1.5 V less than the supply voltage VCC .

• M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier

10 20 40 60 80 −10

RL R2 R1 Vi Vm = 1 V f = 25 kHz Vo 10 k 1 k Vi , Vo (Volts) Vo (expected) Vo Vi t (µsec)

• M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier

10 20 40 60 80 −10

RL R2 R1 Vi Vm = 1 V f = 25 kHz Vo 10 k 1 k Vi , Vo (Volts) Vo (expected) Vo Vi t (µsec)

* If the signal frequency is too high, a practical op-amp cannot keep up with the input due to its “slew rate” limitation.

• M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier

10 20 40 60 80 −10

RL R2 R1 Vi Vm = 1 V f = 25 kHz Vo 10 k 1 k Vi , Vo (Volts) Vo (expected) Vo Vi t (µsec)

* If the signal frequency is too high, a practical op-amp cannot keep up with the input due to its “slew rate” limitation. * The slew rate of an op-amp is the maximum rate at which the op-amp output can rise (or fall).

• M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier

10 20 40 60 80 −10

RL R2 R1 Vi Vm = 1 V f = 25 kHz Vo 10 k 1 k Vi , Vo (Volts) Vo (expected) Vo Vi t (µsec)

* If the signal frequency is too high, a practical op-amp cannot keep up with the input due to its “slew rate” limitation. * The slew rate of an op-amp is the maximum rate at which the op-amp output can rise (or fall). * For the 741, the slew rate is 0.5 V /µsec.

• M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier

10 20 40 60 80 −10

RL R2 R1 Vi Vm = 1 V f = 25 kHz Vo 10 k 1 k Vi , Vo (Volts) Vo (expected) Vo Vi t (µsec)

* If the signal frequency is too high, a practical op-amp cannot keep up with the input due to its “slew rate” limitation. * The slew rate of an op-amp is the maximum rate at which the op-amp output can rise (or fall). * For the 741, the slew rate is 0.5 V /µsec.

(SEQUEL file: ee101 inv amp 2.sqproj)

• M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier

RL RL R2 R2 R1 R1 Vi Vi Vo Vo Circuit 1 Circuit 2

What if the + (non-inverting) and − (inverting) inputs of the op-amp are interchanged?

• M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier

RL RL R2 R2 R1 R1 Vi Vi Vo Vo Circuit 1 Circuit 2

What if the + (non-inverting) and − (inverting) inputs of the op-amp are interchanged? Our previous analysis would once again give us Vo = − R2 R1 Vi .

• M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier

RL RL R2 R2 R1 R1 Vi Vi Vo Vo Circuit 1 Circuit 2

What if the + (non-inverting) and − (inverting) inputs of the op-amp are interchanged? Our previous analysis would once again give us Vo = − R2 R1 Vi . However, from Circuit 1 to Circuit 2, the nature of the feedback changes from negative to positive. → Our assumption that the op-amp is working in the linear region does not hold for Circuit 2, and Vo = − R2 R1 Vi does not apply any more.

• M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier

RL RL R2 R2 R1 R1 Vi Vi Vo Vo Circuit 1 Circuit 2

What if the + (non-inverting) and − (inverting) inputs of the op-amp are interchanged? Our previous analysis would once again give us Vo = − R2 R1 Vi . However, from Circuit 1 to Circuit 2, the nature of the feedback changes from negative to positive. → Our assumption that the op-amp is working in the linear region does not hold for Circuit 2, and Vo = − R2 R1 Vi does not apply any more. (Circuit 2 is also useful, and we will discuss it later.)

• M. B. Patil, IIT Bombay

Op-amp circuits (linear region)

ii i1 i2 RL R2 R1 Vi Vo

* V+ ≈ V− = Vi

• M. B. Patil, IIT Bombay

Op-amp circuits (linear region)

ii i1 i2 RL R2 R1 Vi Vo

* V+ ≈ V− = Vi → i1 = (0 − Vi)/R1 = −Vi/R1 .

• M. B. Patil, IIT Bombay

Op-amp circuits (linear region)

ii i1 i2 RL R2 R1 Vi Vo

* V+ ≈ V− = Vi → i1 = (0 − Vi)/R1 = −Vi/R1 . * Since ii = 0, i2 = i1 → Vo = V− − i2 R2 = V+ − i1 R2 = Vi −

• − Vi

R1

• R2 = Vi
• 1 + R2

R1

• .
• M. B. Patil, IIT Bombay

Op-amp circuits (linear region)

ii i1 i2 RL R2 R1 Vi Vo

* V+ ≈ V− = Vi → i1 = (0 − Vi)/R1 = −Vi/R1 . * Since ii = 0, i2 = i1 → Vo = V− − i2 R2 = V+ − i1 R2 = Vi −

• − Vi

R1

• R2 = Vi
• 1 + R2

R1

• .

* This circuit is known as the “non-inverting amplifier.”

• M. B. Patil, IIT Bombay

Op-amp circuits (linear region)

ii i1 i2 RL R2 R1 Vi Vo

* V+ ≈ V− = Vi → i1 = (0 − Vi)/R1 = −Vi/R1 . * Since ii = 0, i2 = i1 → Vo = V− − i2 R2 = V+ − i1 R2 = Vi −

• − Vi

R1

• R2 = Vi
• 1 + R2

R1

• .

* This circuit is known as the “non-inverting amplifier.” * Again, interchanging + and − changes the nature of the feedback from negative to positive, and the circuit operation becomes completely different.

• M. B. Patil, IIT Bombay

Inverting or non-inverting?

Inverting amplifier Non−inverting amplifier RL RL R2 R2 R1 R1 Vs Vs Vo = −R2 R1 Vs Vo =

• 1 + R2

R1

• Vs

* If the sign of the output voltage is not a concern, which configuration should be preferred?

Inverting or non-inverting?

Inverting amplifier Non−inverting amplifier RL RL R2 R2 R1 R1 Vs Vs Vo = −R2 R1 Vs Vo =

• 1 + R2

R1

• Vs

i1 Vs RL R1 R2 Vo Vi AV Vi Ro Ri

* If the sign of the output voltage is not a concern, which configuration should be preferred?

Inverting or non-inverting?

Inverting amplifier Non−inverting amplifier RL RL R2 R2 R1 R1 Vs Vs Vo = −R2 R1 Vs Vo =

• 1 + R2

R1

• Vs

i1 Vs RL R1 R2 Vo Vi AV Vi Ro Ri

* If the sign of the output voltage is not a concern, which configuration should be preferred? * For the inverting amplifier, since V− ≈ 0 V , i1 = Vs/R1 → Rin = Vs/i1 = R1 .

Inverting or non-inverting?

Inverting amplifier Non−inverting amplifier RL RL R2 R2 R1 R1 Vs Vs Vo = −R2 R1 Vs Vo =

• 1 + R2

R1

• Vs

i1 Vs RL R1 R2 Vo Vi AV Vi Ro Ri Vs RL R1 R2 Vo Vi AV Vi Ro Ri

* If the sign of the output voltage is not a concern, which configuration should be preferred? * For the inverting amplifier, since V− ≈ 0 V , i1 = Vs/R1 → Rin = Vs/i1 = R1 .

• M. B. Patil, IIT Bombay

Inverting or non-inverting?

Inverting amplifier Non−inverting amplifier RL RL R2 R2 R1 R1 Vs Vs Vo = −R2 R1 Vs Vo =

• 1 + R2

R1

• Vs

i1 Vs RL R1 R2 Vo Vi AV Vi Ro Ri Vs RL R1 R2 Vo Vi AV Vi Ro Ri

* If the sign of the output voltage is not a concern, which configuration should be preferred? * For the inverting amplifier, since V− ≈ 0 V , i1 = Vs/R1 → Rin = Vs/i1 = R1 . * For the non-inverting amplifier, Rin ∼ Ri AV R1 R1 + R2 . Huge!

• M. B. Patil, IIT Bombay

Inverting and non-inverting amplifiers: summary

Inverting amplifier Non−inverting amplifier RL RL R2 R2 R1 R1 Vs Vs Vo = −R2 R1 Vs Vo =

• 1 + R2

R1

• Vs
• M. B. Patil, IIT Bombay

Non-inverting amplifier

R2 R1 Vi Vo

Non-inverting amplifier

R2 R1 Vi Vo R2 R1 Vi Vo

Non-inverting amplifier

R2 R1 Vi Vo R2 R1 Vi Vo R2 R1 Vi Vo

Non-inverting amplifier

R2 R1 Vi Vo R2 R1 Vi Vo R2 R1 Vi Vo R1 R2 Vi Vo

• M. B. Patil, IIT Bombay

Non-inverting amplifier

RL RL R2 R1 Vo Vo Vi Vi

Consider R1 → ∞ , R2 → 0 .

• M. B. Patil, IIT Bombay

Non-inverting amplifier

RL RL R2 R1 Vo Vo Vi Vi

Consider R1 → ∞ , R2 → 0 . Vo Vi → 1 + R2 R1 → 1 , i.e., Vo = Vi .

• M. B. Patil, IIT Bombay

Non-inverting amplifier

RL RL R2 R1 Vo Vo Vi Vi

Consider R1 → ∞ , R2 → 0 . Vo Vi → 1 + R2 R1 → 1 , i.e., Vo = Vi . This circuit is known as unity-gain amplifier/voltage follower/buffer.

• M. B. Patil, IIT Bombay

Non-inverting amplifier

RL RL R2 R1 Vo Vo Vi Vi

Consider R1 → ∞ , R2 → 0 . Vo Vi → 1 + R2 R1 → 1 , i.e., Vo = Vi . This circuit is known as unity-gain amplifier/voltage follower/buffer. What has been achieved?

• M. B. Patil, IIT Bombay

Loading effects

Vs RL Rs Vi Vo AV Vi Ro Ri

Consider an amplifier of gain AV . We would like to have Vo = AV Vs .

• M. B. Patil, IIT Bombay

Loading effects

Vs RL Rs Vi Vo AV Vi Ro Ri

Consider an amplifier of gain AV . We would like to have Vo = AV Vs . However, the actual output voltage is, Vo = RL Ro + RL AV Vi = AV RL Ro + RL Ri Ri + Rs Vs .

• M. B. Patil, IIT Bombay

Loading effects

Vs RL Rs Vi Vo AV Vi Ro Ri

Consider an amplifier of gain AV . We would like to have Vo = AV Vs . However, the actual output voltage is, Vo = RL Ro + RL AV Vi = AV RL Ro + RL Ri Ri + Rs Vs . To obtain the desired Vo, we need Ri → ∞ and Ro → 0 .

• M. B. Patil, IIT Bombay

Loading effects

Vs RL Rs Vi Vo AV Vi Ro Ri

Consider an amplifier of gain AV . We would like to have Vo = AV Vs . However, the actual output voltage is, Vo = RL Ro + RL AV Vi = AV RL Ro + RL Ri Ri + Rs Vs . To obtain the desired Vo, we need Ri → ∞ and Ro → 0 . The buffer (voltage follower) provides these features.

• M. B. Patil, IIT Bombay

Op-amp buffer: input resistance

Non−inverting amplifier

A B

IS RL R2 R1 VS RL R1 R2 VS Vi AV Vi Ro Ri Vo =

• 1 + R2

R1

• Vs
• M. B. Patil, IIT Bombay

Op-amp buffer: input resistance

Non−inverting amplifier

A B

IS RL R2 R1 VS RL R1 R2 VS Vi AV Vi Ro Ri Vo =

• 1 + R2

R1

• Vs

KCL at B: VB RL + VB − AV Vi Ro + VB − VA R2 = 0.

• M. B. Patil, IIT Bombay

Op-amp buffer: input resistance

Non−inverting amplifier

A B

IS RL R2 R1 VS RL R1 R2 VS Vi AV Vi Ro Ri Vo =

• 1 + R2

R1

• Vs

KCL at B: VB RL + VB − AV Vi Ro + VB − VA R2 = 0. Source current: IS = VA R1 + VA − VB R2 .

• M. B. Patil, IIT Bombay

Op-amp buffer: input resistance

Non−inverting amplifier

A B

IS RL R2 R1 VS RL R1 R2 VS Vi AV Vi Ro Ri Vo =

• 1 + R2

R1

• Vs

KCL at B: VB RL + VB − AV Vi Ro + VB − VA R2 = 0. Source current: IS = VA R1 + VA − VB R2 . Using Vi = ISRi, VA = VS − Vi, and after some algebra, we get Rin = VS IS =

• 1 + Ro

RL + Ro R2

• + Ri

1 R1 + 1 R2 1 + Ro RL + Ro R2

• − Ro

R2

2

+ AV R2

• 1

R1 + 1 R2 1 + Ro RL + Ro R2

• − Ro

R2

2

.

• M. B. Patil, IIT Bombay

Op-amp buffer: input resistance

Non−inverting amplifier

A B

IS RL R2 R1 VS RL R1 R2 VS Vi AV Vi Ro Ri Vo =

• 1 + R2

R1

• Vs

KCL at B: VB RL + VB − AV Vi Ro + VB − VA R2 = 0. Source current: IS = VA R1 + VA − VB R2 . Using Vi = ISRi, VA = VS − Vi, and after some algebra, we get Rin = VS IS =

• 1 + Ro

RL + Ro R2

• + Ri

1 R1 + 1 R2 1 + Ro RL + Ro R2

• − Ro

R2

2

+ AV R2

• 1

R1 + 1 R2 1 + Ro RL + Ro R2

• − Ro

R2

2

.

STOP

• M. B. Patil, IIT Bombay

Non-inverting amplifier: input resistance (continued)

Non−inverting amplifier

A B

IS RL R2 R1 VS RL R1 R2 VS Vi AV Vi Ro Ri Vo =

• 1 + R2

R1

• Vs

Rin = VS IS =

• 1 + Ro

RL + Ro R2

• + Ri

1 R1 + 1 R2 1 + Ro RL + Ro R2

• − Ro

R2

2

+ AV R2

• 1

R1 + 1 R2 1 + Ro RL + Ro R2

• − Ro

R2

2

.

• M. B. Patil, IIT Bombay

Non-inverting amplifier: input resistance (continued)

Non−inverting amplifier

A B

IS RL R2 R1 VS RL R1 R2 VS Vi AV Vi Ro Ri Vo =

• 1 + R2

R1

• Vs

Rin = VS IS =

• 1 + Ro

RL + Ro R2

• + Ri

1 R1 + 1 R2 1 + Ro RL + Ro R2

• − Ro

R2

2

+ AV R2

• 1

R1 + 1 R2 1 + Ro RL + Ro R2

• − Ro

R2

2

. Since Ro is much smaller than R1, R2, RL, or Ri, Rin ≈ 1 + Ri 1 R1 + 1 R2

• + AV

R2

• 1

R1 + 1 R2

• ≈

Ri R1 + R2 R1R2 + AV R2

• R1 + R2

R1R2 ≈ AV Ri R1 R1 + R2 .

• M. B. Patil, IIT Bombay

Op-amp buffer: input resistance

Buffer IS RL VS RL Vs Vi AV Vi Ro Ri Vo = Vs

• M. B. Patil, IIT Bombay

Op-amp buffer: input resistance

Buffer IS RL VS RL Vs Vi AV Vi Ro Ri Vo = Vs

Let Ro → 0.

• M. B. Patil, IIT Bombay

Op-amp buffer: input resistance

Buffer IS RL VS RL Vs Vi AV Vi Ro Ri Vo = Vs

Let Ro → 0. VS = Vi + AV Vi = Vi(1 + AV ).

• M. B. Patil, IIT Bombay

Op-amp buffer: input resistance

Buffer IS RL VS RL Vs Vi AV Vi Ro Ri Vo = Vs

Let Ro → 0. VS = Vi + AV Vi = Vi(1 + AV ). IS = Vi Ri .

• M. B. Patil, IIT Bombay

Op-amp buffer: input resistance

Buffer IS RL VS RL Vs Vi AV Vi Ro Ri Vo = Vs

Let Ro → 0. VS = Vi + AV Vi = Vi(1 + AV ). IS = Vi Ri . → Rin = VS IS = Ri(AV + 1)

• M. B. Patil, IIT Bombay

Op-amp buffer: output resistance

Non−inverting amplifier RL Vo R2 R1 Vs RL Vs R1 R2 Vi AV Vi Ro Ri Rout

To find Rout, * Deactivate the input source.

• M. B. Patil, IIT Bombay

Op-amp buffer: output resistance

Non−inverting amplifier RL Vo R2 R1 Vs RL Vs R1 R2 Vi AV Vi Ro Ri Rout

To find Rout, * Deactivate the input source. * Replace RL with a test source V ′.

• M. B. Patil, IIT Bombay

Op-amp buffer: output resistance

Non−inverting amplifier RL Vo R2 R1 Vs RL Vs R1 R2 Vi AV Vi Ro Ri Rout

To find Rout, * Deactivate the input source. * Replace RL with a test source V ′. * Find the current (I ′) through V ′.

• M. B. Patil, IIT Bombay

Op-amp buffer: output resistance

Non−inverting amplifier RL Vo R2 R1 Vs RL Vs R1 R2 Vi AV Vi Ro Ri Rout

To find Rout, * Deactivate the input source. * Replace RL with a test source V ′. * Find the current (I ′) through V ′. * Rout = V ′ I ′ .

• M. B. Patil, IIT Bombay

Op-amp buffer: output resistance (continued)

Non−inverting amplifier I′ I2 I1 RL Vo R2 V′ V′ −Vi R1 R1 R2 Vs Vi AV Vi Ro Ri AV Vi

• M. B. Patil, IIT Bombay

Op-amp buffer: output resistance (continued)

Non−inverting amplifier I′ I2 I1 RL Vo R2 V′ V′ −Vi R1 R1 R2 Vs Vi AV Vi Ro Ri AV Vi

Vi = − (Ri R1) R2 + (Ri R1) V ′ ≡ −kV ′.

• M. B. Patil, IIT Bombay

Op-amp buffer: output resistance (continued)

Non−inverting amplifier I′ I2 I1 RL Vo R2 V′ V′ −Vi R1 R1 R2 Vs Vi AV Vi Ro Ri AV Vi

Vi = − (Ri R1) R2 + (Ri R1) V ′ ≡ −kV ′. I ′ = I1 + I2 = V ′ − AV Vi Ro + V ′ − (−Vi) R2 = 1 Ro

• V ′ + kAV V ′

+ 1 R2

• V ′ − kV ′

.

• M. B. Patil, IIT Bombay

Op-amp buffer: output resistance (continued)

Non−inverting amplifier I′ I2 I1 RL Vo R2 V′ V′ −Vi R1 R1 R2 Vs Vi AV Vi Ro Ri AV Vi

Vi = − (Ri R1) R2 + (Ri R1) V ′ ≡ −kV ′. I ′ = I1 + I2 = V ′ − AV Vi Ro + V ′ − (−Vi) R2 = 1 Ro

• V ′ + kAV V ′

+ 1 R2

• V ′ − kV ′

. I ′ V ′ = 1 Ro (1 + kAV ) + 1 R2 (1 − k) → Rout = V ′ I ′ = Ro (1 + kAV ) R2 (1 − k) ≈ Ro (1 + kAV )

• M. B. Patil, IIT Bombay

Op-amp buffer: output resistance (continued)

Non−inverting amplifier I′ I2 I1 RL Vo R2 V′ V′ −Vi R1 R1 R2 Vs Vi AV Vi Ro Ri AV Vi

Vi = − (Ri R1) R2 + (Ri R1) V ′ ≡ −kV ′. I ′ = I1 + I2 = V ′ − AV Vi Ro + V ′ − (−Vi) R2 = 1 Ro

• V ′ + kAV V ′

+ 1 R2

• V ′ − kV ′

. I ′ V ′ = 1 Ro (1 + kAV ) + 1 R2 (1 − k) → Rout = V ′ I ′ = Ro (1 + kAV ) R2 (1 − k) ≈ Ro (1 + kAV ) Special case: Op-amp buffer k = (Ri R1) R2 + (Ri R1) → 1 ⇒ Rout ≈ Ro 1 + AV

• M. B. Patil, IIT Bombay

Op-amp buffer

RL Vs RL Vs Vs Rin Rout

In summary, the buffer (voltage follower) provides

• M. B. Patil, IIT Bombay

Op-amp buffer

RL Vs RL Vs Vs Rin Rout

In summary, the buffer (voltage follower) provides * a large input resistance Rin as seen from the source.

• M. B. Patil, IIT Bombay

Op-amp buffer

RL Vs RL Vs Vs Rin Rout

In summary, the buffer (voltage follower) provides * a large input resistance Rin as seen from the source. * a small output resistance Rout as seen from the load.

• M. B. Patil, IIT Bombay

Op-amp buffer

RL Vs RL Vs Vs Rin Rout

In summary, the buffer (voltage follower) provides * a large input resistance Rin as seen from the source. * a small output resistance Rout as seen from the load. * a gain of 1, i.e., the output voltage simply follows the input voltage.

• M. B. Patil, IIT Bombay

Loading effects (revisited)

Vs RL Rs Vi Vo AV Vi Ro Ri

Problem: We would like to have Vo = AV Vs .

• M. B. Patil, IIT Bombay

Loading effects (revisited)

Vs RL Rs Vi Vo AV Vi Ro Ri

Problem: We would like to have Vo = AV Vs . But the actual output voltage is, Vo = RL Ro + RL AV Vi = AV RL Ro + RL Ri Ri + Rs Vs.

• M. B. Patil, IIT Bombay

Op-amp buffer

buffer 2 load amplifier buffer 1 source

RL Vs Rs Vi AV Vi Ro Ri Vo i1 i2 Vo1 Vo2

• M. B. Patil, IIT Bombay

Op-amp buffer

buffer 2 load amplifier buffer 1 source

RL Vs Rs Vi AV Vi Ro Ri Vo i1 i2 Vo1 Vo2

Since the buffer has a large input resistance, i1 ≈ 0 A, and V+ (on the source side) = Vs → Vo1 = Vs .

• M. B. Patil, IIT Bombay

Op-amp buffer

buffer 2 load amplifier buffer 1 source

RL Vs Rs Vi AV Vi Ro Ri Vo i1 i2 Vo1 Vo2

Since the buffer has a large input resistance, i1 ≈ 0 A, and V+ (on the source side) = Vs → Vo1 = Vs . Similarly, i2 ≈ 0 A, and Vo2 = AV Vi = AV Vs .

• M. B. Patil, IIT Bombay

Op-amp buffer

buffer 2 load amplifier buffer 1 source

RL Vs Rs Vi AV Vi Ro Ri Vo i1 i2 Vo1 Vo2

Since the buffer has a large input resistance, i1 ≈ 0 A, and V+ (on the source side) = Vs → Vo1 = Vs . Similarly, i2 ≈ 0 A, and Vo2 = AV Vi = AV Vs . Finally, Vo = Vo2 = AV Vs , as desired, irrespective of RS and RL.

• M. B. Patil, IIT Bombay

Op-amp buffer

buffer 2 load amplifier buffer 1 source

RL Vs Rs Vi AV Vi Ro Ri Vo i1 i2 Vo1 Vo2

Since the buffer has a large input resistance, i1 ≈ 0 A, and V+ (on the source side) = Vs → Vo1 = Vs . Similarly, i2 ≈ 0 A, and Vo2 = AV Vi = AV Vs . Finally, Vo = Vo2 = AV Vs , as desired, irrespective of RS and RL. Note that the load current is supplied by the second buffer which acts as a voltage source (= AV Vs) with zero source resistance.

• M. B. Patil, IIT Bombay

Op-amp circuits (linear region)

Vi3 Vi2 Vi1 RL Vo Rf R3 R2 R1 i3 i2 i1 i ii if

• M. B. Patil, IIT Bombay

Op-amp circuits (linear region)

Vi3 Vi2 Vi1 RL Vo Rf R3 R2 R1 i3 i2 i1 i ii if

V− ≈ V+ = 0 V → i1 = Vi1/R1, i2 = Vi2/R2, i3 = Vi3/R3 .

• M. B. Patil, IIT Bombay

Op-amp circuits (linear region)

Vi3 Vi2 Vi1 RL Vo Rf R3 R2 R1 i3 i2 i1 i ii if

V− ≈ V+ = 0 V → i1 = Vi1/R1, i2 = Vi2/R2, i3 = Vi3/R3 . i = i1 + i2 + i3 = Vi1 R1 + Vi2 R2 + Vi3 R3

• .
• M. B. Patil, IIT Bombay

Op-amp circuits (linear region)

Vi3 Vi2 Vi1 RL Vo Rf R3 R2 R1 i3 i2 i1 i ii if

V− ≈ V+ = 0 V → i1 = Vi1/R1, i2 = Vi2/R2, i3 = Vi3/R3 . i = i1 + i2 + i3 = Vi1 R1 + Vi2 R2 + Vi3 R3

• .

Because of the large input resistance of the op-amp, ii ≈ 0 → if = i, which gives

• M. B. Patil, IIT Bombay

Op-amp circuits (linear region)

Vi3 Vi2 Vi1 RL Vo Rf R3 R2 R1 i3 i2 i1 i ii if

V− ≈ V+ = 0 V → i1 = Vi1/R1, i2 = Vi2/R2, i3 = Vi3/R3 . i = i1 + i2 + i3 = Vi1 R1 + Vi2 R2 + Vi3 R3

• .

Because of the large input resistance of the op-amp, ii ≈ 0 → if = i, which gives Vo = V− − if Rf = 0 − Vi1 R1 + Vi2 R2 + Vi3 R3

• Rf = −

Rf R1 Vi1 + Rf R2 Vi2 + Rf R3 Vi3

• ,

i.e., Vo is a weighted sum of Vi1, Vi2, Vi3.

• M. B. Patil, IIT Bombay

Op-amp circuits (linear region)

Vi3 Vi2 Vi1 RL Vo Rf R3 R2 R1 i3 i2 i1 i ii if

V− ≈ V+ = 0 V → i1 = Vi1/R1, i2 = Vi2/R2, i3 = Vi3/R3 . i = i1 + i2 + i3 = Vi1 R1 + Vi2 R2 + Vi3 R3

• .

Because of the large input resistance of the op-amp, ii ≈ 0 → if = i, which gives Vo = V− − if Rf = 0 − Vi1 R1 + Vi2 R2 + Vi3 R3

• Rf = −

Rf R1 Vi1 + Rf R2 Vi2 + Rf R3 Vi3

• ,

i.e., Vo is a weighted sum of Vi1, Vi2, Vi3. If R1 = R2 = R3 = R , the circuit acts as a summer, giving Vo = −K (Vi1 + Vi2 + Vi3) with K = Rf /R .

• M. B. Patil, IIT Bombay

Summer example

1.2 0.6 −0.6 −1 −2 −3 1 2 3 4 t (msec)

Vi3 Vi2 Vi1 RL Vo Rf R3 R2 R1 SEQUEL file: ee101 summer.sqproj R1 = R2 = R3 = 1 kΩ Rf = 2 kΩ → Vo = −2 (Vi1 + Vi2 + Vi3) i3 i2 i1 i ii if Vi2 Vi1 Vi3 Vo

• M. B. Patil, IIT Bombay

Summer example

1.2 0.6 −0.6 −1 −2 −3 1 2 3 4 t (msec)

Vi3 Vi2 Vi1 RL Vo Rf R3 R2 R1 SEQUEL file: ee101 summer.sqproj R1 = R2 = R3 = 1 kΩ Rf = 2 kΩ → Vo = −2 (Vi1 + Vi2 + Vi3) i3 i2 i1 i ii if Vi2 Vi1 Vi3 Vo

* Note that the summer also works with DC inputs (so do inverting and non-inverting amplifiers).

• M. B. Patil, IIT Bombay

Summer example

1.2 0.6 −0.6 −1 −2 −3 1 2 3 4 t (msec)

Vi3 Vi2 Vi1 RL Vo Rf R3 R2 R1 SEQUEL file: ee101 summer.sqproj R1 = R2 = R3 = 1 kΩ Rf = 2 kΩ → Vo = −2 (Vi1 + Vi2 + Vi3) i3 i2 i1 i ii if Vi2 Vi1 Vi3 Vo

* Note that the summer also works with DC inputs (so do inverting and non-inverting amplifiers). * Op-amps make life simpler! Think of adding voltages in any other way.

• M. B. Patil, IIT Bombay

Choice of resistance values

* If resistances are too small, they draw larger currents → increased power dissipation

• M. B. Patil, IIT Bombay

Choice of resistance values

* If resistances are too small, they draw larger currents → increased power dissipation * If resistances are too large,

• M. B. Patil, IIT Bombay

Choice of resistance values

* If resistances are too small, they draw larger currents → increased power dissipation * If resistances are too large,

• The effect of offset voltage and input bias currents becomes more

pronounced (to be discussed).

• M. B. Patil, IIT Bombay

Choice of resistance values

* If resistances are too small, they draw larger currents → increased power dissipation * If resistances are too large,

• The effect of offset voltage and input bias currents becomes more

pronounced (to be discussed).

• Combined with parasitic (wiring) capacitances, large resistances can

affect the frequency response and stability of the circuit.

• M. B. Patil, IIT Bombay

Choice of resistance values

* If resistances are too small, they draw larger currents → increased power dissipation * If resistances are too large,

• The effect of offset voltage and input bias currents becomes more

pronounced (to be discussed).

• Combined with parasitic (wiring) capacitances, large resistances can

affect the frequency response and stability of the circuit.

• Thermal noise increases as R increases, and it may not be desirable in

some applications.

• M. B. Patil, IIT Bombay

Choice of resistance values

* If resistances are too small, they draw larger currents → increased power dissipation * If resistances are too large,

• The effect of offset voltage and input bias currents becomes more

pronounced (to be discussed).

• Combined with parasitic (wiring) capacitances, large resistances can

affect the frequency response and stability of the circuit.

• Thermal noise increases as R increases, and it may not be desirable in

some applications. * Typical resistance values: 0.1 k to 100 k.

• M. B. Patil, IIT Bombay

Design an amplifier with Rin = 10 k and AV = −100.

Design an amplifier with Rin = 10 k and AV = −100. Vo Vi Rin R′

2

R′

1

Design an amplifier with Rin = 10 k and AV = −100. Vo Vi Rin R′

2

R′

1

Rin = R′

1 = 10 k.

Design an amplifier with Rin = 10 k and AV = −100. Vo Vi Rin R′

2

R′

1

Rin = R′

1 = 10 k.

AV = − R′

2

R′

1

= −100 → R′

2 = 100 × 10 k = 1 MΩ

Design an amplifier with Rin = 10 k and AV = −100. Vo Vi Rin R′

2

R′

1

Rin = R′

1 = 10 k.

AV = − R′

2

R′

1

= −100 → R′

2 = 100 × 10 k = 1 MΩ

R′

2 may be unacceptable from practical considerations.

Design an amplifier with Rin = 10 k and AV = −100. Vo Vi Rin R′

2

R′

1

Rin = R′

1 = 10 k.

AV = − R′

2

R′

1

= −100 → R′

2 = 100 × 10 k = 1 MΩ

R′

2 may be unacceptable from practical considerations.

→ need a design with smaller resistances.

Design an amplifier with Rin = 10 k and AV = −100. Vo Vi Rin R′

2

R′

1

Vo Vi R′

1

I1 V1 0 V Rin = R′

1 = 10 k.

AV = − R′

2

R′

1

= −100 → R′

2 = 100 × 10 k = 1 MΩ

R′

2 may be unacceptable from practical considerations.

→ need a design with smaller resistances.

• M. B. Patil, IIT Bombay

Design an amplifier with Rin = 10 k and AV = −100. Vo Vi Rin R′

2

R′

1

Vo Vi R′

1

I1 V1 0 V Rin = R′

1 = 10 k.

AV = − R′

2

R′

1

= −100 → R′

2 = 100 × 10 k = 1 MΩ

R′

2 may be unacceptable from practical considerations.

→ need a design with smaller resistances. If we ensure V1 I1 = R′

2, we will satisfy the gain condition.

• M. B. Patil, IIT Bombay

I1 V1 I2 R1 R2 R3

• M. B. Patil, IIT Bombay

I1 V1 I2 R1 R2 R3 I2 = V1 R3 + (R1 R2)

• M. B. Patil, IIT Bombay

I1 V1 I2 R1 R2 R3 I2 = V1 R3 + (R1 R2) I1 = R2 R1 + R2 I2 = R2 R1 + R2 × R1 + R2 R3(R1 + R2) + R1R2 V1

• M. B. Patil, IIT Bombay

I1 V1 I2 R1 R2 R3 I2 = V1 R3 + (R1 R2) I1 = R2 R1 + R2 I2 = R2 R1 + R2 × R1 + R2 R3(R1 + R2) + R1R2 V1 Reff ≡ V1 I1 = R1R2 + R2R3 + R3R1 R2

• M. B. Patil, IIT Bombay

I1 V1 I2 R1 R2 R3 I2 = V1 R3 + (R1 R2) I1 = R2 R1 + R2 I2 = R2 R1 + R2 × R1 + R2 R3(R1 + R2) + R1R2 V1 Reff ≡ V1 I1 = R1R2 + R2R3 + R3R1 R2 → Choose R1, R2, R3 such that Reff = R′

2 = 1 MΩ.

• M. B. Patil, IIT Bombay

I1 V1 I2 R1 R2 R3 I2 = V1 R3 + (R1 R2) I1 = R2 R1 + R2 I2 = R2 R1 + R2 × R1 + R2 R3(R1 + R2) + R1R2 V1 Reff ≡ V1 I1 = R1R2 + R2R3 + R3R1 R2 → Choose R1, R2, R3 such that Reff = R′

2 = 1 MΩ.

Vo Vi R′

2

Vo Vi R2 R′

1

R′

1

R1 R3

• M. B. Patil, IIT Bombay

Vo Vi R′

2

Vo Vi R2 R′

1

R′

1

R1 R3 Reff = R1R2 + R2R3 + R3R1 R2 We want Reff = R′

2 = 1 MΩ.

• M. B. Patil, IIT Bombay

Vo Vi R′

2

Vo Vi R2 R′

1

R′

1

R1 R3 Reff = R1R2 + R2R3 + R3R1 R2 We want Reff = R′

2 = 1 MΩ.

Let R1 = R3 ≡ R

• M. B. Patil, IIT Bombay

Vo Vi R′

2

Vo Vi R2 R′

1

R′

1

R1 R3 Reff = R1R2 + R2R3 + R3R1 R2 We want Reff = R′

2 = 1 MΩ.

Let R1 = R3 ≡ R → Reff = R2 + 2 R R2 R2 = R R R2 + 2

• M. B. Patil, IIT Bombay

Vo Vi R′

2

Vo Vi R2 R′

1

R′

1

R1 R3 Reff = R1R2 + R2R3 + R3R1 R2 We want Reff = R′

2 = 1 MΩ.

Let R1 = R3 ≡ R → Reff = R2 + 2 R R2 R2 = R R R2 + 2

• → R2 =

R Reff R − 2

• M. B. Patil, IIT Bombay

Vo Vi R′

2

Vo Vi R2 R′

1

R′

1

R1 R3 Reff = R1R2 + R2R3 + R3R1 R2 We want Reff = R′

2 = 1 MΩ.

Let R1 = R3 ≡ R → Reff = R2 + 2 R R2 R2 = R R R2 + 2

• → R2 =

R Reff R − 2 For R = 10 k, R2 = 10 k 100 − 2 ≈ 102 Ω.

• M. B. Patil, IIT Bombay

Vo Vi R′

2

Vo Vi R2 R′

1

R′

1

R1 R3 Reff = R1R2 + R2R3 + R3R1 R2 We want Reff = R′

2 = 1 MΩ.

Let R1 = R3 ≡ R → Reff = R2 + 2 R R2 R2 = R R R2 + 2

• → R2 =

R Reff R − 2 For R = 10 k, R2 = 10 k 100 − 2 ≈ 102 Ω.

Ref: Wait et al, Introduction to op-amp theory and applications, McGraw-Hill, 1992.

• M. B. Patil, IIT Bombay

Vo Vi R′

2

Vo Vi R2 R′

1

R′

1

R1 R3 Reff = R1R2 + R2R3 + R3R1 R2 We want Reff = R′

2 = 1 MΩ.

Let R1 = R3 ≡ R → Reff = R2 + 2 R R2 R2 = R R R2 + 2

• → R2 =

R Reff R − 2 For R = 10 k, R2 = 10 k 100 − 2 ≈ 102 Ω.

Ref: Wait et al, Introduction to op-amp theory and applications, McGraw-Hill, 1992.

Vo Vi 10 k 10 k 102 Ω 10 k

• M. B. Patil, IIT Bombay