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

Op-amp circuits (linear region) R 2 i 1 V i R 1 i i V o R L Since V + ≈ V − , V − ≈ 0 V → i 1 = ( V i − 0) / R 1 = V i / R 1 . (The non-inverting input is at real ground here, and the inverting input is at virtual ground.) Since i i (current entering the op-amp) is zero, i 1 goes through R 2 .

Op-amp circuits (linear region) R 2 i 1 V i R 1 i i V o R L Since V + ≈ V − , V − ≈ 0 V → i 1 = ( V i − 0) / R 1 = V i / R 1 . (The non-inverting input is at real ground here, and the inverting input is at virtual ground.) Since i i (current entering the op-amp) is zero, i 1 goes through R 2 .

Op-amp circuits (linear region) R 2 i 1 V i R 1 i i V o R L Since V + ≈ V − , V − ≈ 0 V → i 1 = ( V i − 0) / R 1 = V i / R 1 . (The non-inverting input is at real ground here, and the inverting input is at virtual ground.) Since i i (current entering the op-amp) is zero, i 1 goes through R 2 . � V i � R 2 � � → V o = V − − i 1 R 2 = 0 − R 2 = − V i . R 1 R 1

Op-amp circuits (linear region) R 2 i 1 V i R 1 i i V o R L Since V + ≈ V − , V − ≈ 0 V → i 1 = ( V i − 0) / R 1 = V i / R 1 . (The non-inverting input is at real ground here, and the inverting input is at virtual ground.) Since i i (current entering the op-amp) is zero, i 1 goes through R 2 . � V i � R 2 � � → V o = V − − i 1 R 2 = 0 − R 2 = − V i . R 1 R 1 The circuit is called an “inverting amplifier.”

Op-amp circuits (linear region) R 2 i 1 V i R 1 i i V o R L Since V + ≈ V − , V − ≈ 0 V → i 1 = ( V i − 0) / R 1 = V i / R 1 . (The non-inverting input is at real ground here, and the inverting input is at virtual ground.) Since i i (current entering the op-amp) is zero, i 1 goes through R 2 . � V i � R 2 � � → V o = V − − i 1 R 2 = 0 − R 2 = − V i . R 1 R 1 The circuit is called an “inverting amplifier.” Where does the current go?

Op-amp circuits (linear region) R 2 R 2 i 1 i 1 V i V i − 1 V R 1 i i R 1 V o 0.1 V V o R L R L Since V + ≈ V − , V − ≈ 0 V → i 1 = ( V i − 0) / R 1 = V i / R 1 . (The non-inverting input is at real ground here, and the inverting input is at virtual ground.) Since i i (current entering the op-amp) is zero, i 1 goes through R 2 . � V i � R 2 � � → V o = V − − i 1 R 2 = 0 − R 2 = − V i . R 1 R 1 The circuit is called an “inverting amplifier.” Where does the current go?

Op-amp circuits (linear region) R 2 R 2 i 1 i 1 V i V i − 1 V R 1 i i R 1 V o 0.1 V V o R L R L Since V + ≈ V − , V − ≈ 0 V → i 1 = ( V i − 0) / R 1 = V i / R 1 . (The non-inverting input is at real ground here, and the inverting input is at virtual ground.) Since i i (current entering the op-amp) is zero, i 1 goes through R 2 . � V i � R 2 � � → V o = V − − i 1 R 2 = 0 − R 2 = − V i . R 1 R 1 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 V m = 0.5 V 10 k f = 1 kHz V o V i , V o (Volts) R 2 1 k V i 0 R 1 V o V i R L −5 0 0.5 1 1.5 2 t (msec) M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier 5 V m = 0.5 V 10 k f = 1 kHz V o V i , V o (Volts) R 2 1 k V i 0 R 1 V o V i R L −5 0 0.5 1 1.5 2 t (msec) * The gain of the inverting amplifier is − R 2 / R 1 . It is called the “closed-loop gain” (to distinguish it from the “open-loop gain” of the op-amp which is ∼ 10 5 ). M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier 5 V m = 0.5 V 10 k f = 1 kHz V o V i , V o (Volts) R 2 1 k V i 0 R 1 V o V i R L −5 0 0.5 1 1.5 2 t (msec) * The gain of the inverting amplifier is − R 2 / R 1 . It is called the “closed-loop gain” (to distinguish it from the “open-loop gain” of the op-amp which is ∼ 10 5 ). * The gain can be adjusted simply by changing R 1 or R 2 ! M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier 5 V m = 0.5 V 10 k f = 1 kHz V o V i , V o (Volts) R 2 1 k V i 0 R 1 V o V i R L −5 0 0.5 1 1.5 2 t (msec) * The gain of the inverting amplifier is − R 2 / R 1 . It is called the “closed-loop gain” (to distinguish it from the “open-loop gain” of the op-amp which is ∼ 10 5 ). * The gain can be adjusted simply by changing R 1 or R 2 ! * For the common-emitter amplifier, on the other hand, the gain − g m ( R C � R L ) depends on how the BJT is biased (since g m depends on I C ). M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier 5 V m = 0.5 V 10 k f = 1 kHz V o V i , V o (Volts) R 2 1 k V i 0 R 1 V o V i R L −5 0 0.5 1 1.5 2 t (msec) * The gain of the inverting amplifier is − R 2 / R 1 . It is called the “closed-loop gain” (to distinguish it from the “open-loop gain” of the op-amp which is ∼ 10 5 ). * The gain can be adjusted simply by changing R 1 or R 2 ! * For the common-emitter amplifier, on the other hand, the gain − g m ( R C � R L ) depends on how the BJT is biased (since g m depends on I C ). (SEQUEL file: ee101 inv amp 1.sqproj ) M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier 15 V m = 2 V 10 k f = 1 kHz V o V i , V o (Volts) R 2 1 k V i R 1 0 V o V i R L −15 0 0.5 1 1.5 2 t (msec) M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier 15 V m = 2 V 10 k f = 1 kHz V o V i , V o (Volts) R 2 1 k V i R 1 0 V o V i R L −15 0 0.5 1 1.5 2 t (msec) * The output voltage is limited to ± V sat . M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier 15 V m = 2 V 10 k f = 1 kHz V o V i , V o (Volts) R 2 1 k V i R 1 0 V o V i R L −15 0 0.5 1 1.5 2 t (msec) * The output voltage is limited to ± V sat . * V sat is ∼ 1 . 5 V less than the supply voltage V CC . M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier 10 V m = 1 V V o (expected) 10 k f = 25 kHz R 2 V o V i , V o (Volts) 1 k V i 0 R 1 V o V i R L −10 0 20 40 60 80 t ( µ sec) M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier 10 V m = 1 V V o (expected) 10 k f = 25 kHz R 2 V o V i , V o (Volts) 1 k V i 0 R 1 V o V i R L −10 0 20 40 60 80 t ( µ sec) * 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 V m = 1 V V o (expected) 10 k f = 25 kHz R 2 V o V i , V o (Volts) 1 k V i 0 R 1 V o V i R L −10 0 20 40 60 80 t ( µ sec) * 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 V m = 1 V V o (expected) 10 k f = 25 kHz R 2 V o V i , V o (Volts) 1 k V i 0 R 1 V o V i R L −10 0 20 40 60 80 t ( µ sec) * 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 /µ sec. M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier 10 V m = 1 V V o (expected) 10 k f = 25 kHz R 2 V o V i , V o (Volts) 1 k V i 0 R 1 V o V i R L −10 0 20 40 60 80 t ( µ sec) * 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 /µ sec. (SEQUEL file: ee101 inv amp 2.sqproj ) M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier R 2 R 2 V i V i R 1 R 1 V o V o R L R L 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 R 2 R 2 V i V i R 1 R 1 V o V o R L R L 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 V o = − R 2 V i . R 1 M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier R 2 R 2 V i V i R 1 R 1 V o V o R L R L 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 V o = − R 2 V i . R 1 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 V o = − R 2 V i does not apply any more. R 1 M. B. Patil, IIT Bombay

Op-amp circuits: inverting amplifier R 2 R 2 V i V i R 1 R 1 V o V o R L R L 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 V o = − R 2 V i . R 1 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 V o = − R 2 V i does not apply any more. R 1 (Circuit 2 is also useful, and we will discuss it later.) M. B. Patil, IIT Bombay

Op-amp circuits (linear region) i 2 R 2 i 1 R 1 i i V o V i R L * V + ≈ V − = V i M. B. Patil, IIT Bombay

Op-amp circuits (linear region) i 2 R 2 i 1 R 1 i i V o V i R L * V + ≈ V − = V i → i 1 = (0 − V i ) / R 1 = − V i / R 1 . M. B. Patil, IIT Bombay

Op-amp circuits (linear region) i 2 R 2 i 1 R 1 i i V o V i R L * V + ≈ V − = V i → i 1 = (0 − V i ) / R 1 = − V i / R 1 . � − V i � � 1 + R 2 � * Since i i = 0, i 2 = i 1 → V o = V − − i 2 R 2 = V + − i 1 R 2 = V i − R 2 = V i . R 1 R 1 M. B. Patil, IIT Bombay

Op-amp circuits (linear region) i 2 R 2 i 1 R 1 i i V o V i R L * V + ≈ V − = V i → i 1 = (0 − V i ) / R 1 = − V i / R 1 . � − V i � � 1 + R 2 � * Since i i = 0, i 2 = i 1 → V o = V − − i 2 R 2 = V + − i 1 R 2 = V i − R 2 = V i . R 1 R 1 * This circuit is known as the “non-inverting amplifier.” M. B. Patil, IIT Bombay

Op-amp circuits (linear region) i 2 R 2 i 1 R 1 i i V o V i R L * V + ≈ V − = V i → i 1 = (0 − V i ) / R 1 = − V i / R 1 . � − V i � � 1 + R 2 � * Since i i = 0, i 2 = i 1 → V o = V − − i 2 R 2 = V + − i 1 R 2 = V i − R 2 = V i . R 1 R 1 * 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? R 2 V s R 1 V o = − R 2 V s R 1 R L Inverting amplifier R 2 R 1 � 1 + R 2 � V o = V s V s R 1 R L Non−inverting amplifier * If the sign of the output voltage is not a concern, which configuration should be preferred?

Inverting or non-inverting? R 2 i 1 R 2 V s R o R 1 V s R 1 V i V o V o = − R 2 R L V s A V V i R i R 1 R L Inverting amplifier R 2 R 1 � 1 + R 2 � V o = V s V s R 1 R L Non−inverting amplifier * If the sign of the output voltage is not a concern, which configuration should be preferred?

Inverting or non-inverting? R 2 i 1 R 2 V s R o R 1 V s R 1 V i V o V o = − R 2 R L V s A V V i R i R 1 R L Inverting amplifier R 2 R 1 � 1 + R 2 � V o = V s V s R 1 R L Non−inverting amplifier * If the sign of the output voltage is not a concern, which configuration should be preferred? * For the inverting amplifier, since V − ≈ 0 V , i 1 = V s / R 1 → R in = V s / i 1 = R 1 .

Inverting or non-inverting? R 2 i 1 R 2 V s R o R 1 V s R 1 V i V o V o = − R 2 R L V s A V V i R i R 1 R L Inverting amplifier R 2 R 2 R o R 1 R 1 V i V o � 1 + R 2 � R L V o = V s A V V i V s R i R 1 R L V s Non−inverting amplifier * If the sign of the output voltage is not a concern, which configuration should be preferred? * For the inverting amplifier, since V − ≈ 0 V , i 1 = V s / R 1 → R in = V s / i 1 = R 1 . M. B. Patil, IIT Bombay

Inverting or non-inverting? R 2 i 1 R 2 V s R o R 1 V s R 1 V i V o V o = − R 2 R L V s A V V i R i R 1 R L Inverting amplifier R 2 R 2 R o R 1 R 1 V i V o � 1 + R 2 � R L V o = V s A V V i V s R i R 1 R L V s Non−inverting amplifier * If the sign of the output voltage is not a concern, which configuration should be preferred? * For the inverting amplifier, since V − ≈ 0 V , i 1 = V s / R 1 → R in = V s / i 1 = R 1 . R 1 * For the non-inverting amplifier, R in ∼ R i A V . Huge! R 1 + R 2 M. B. Patil, IIT Bombay

Inverting and non-inverting amplifiers: summary R 2 R 2 V s V o = − R 2 � 1 + R 2 � R 1 R 1 V s V o = V s R 1 R 1 V s R L R L Inverting amplifier Non−inverting amplifier M. B. Patil, IIT Bombay

Non-inverting amplifier R 2 R 1 V o V i

Non-inverting amplifier V i R 2 V o R 1 V o V i R 1 R 2

Non-inverting amplifier V i R 2 V o R 1 V o V i R 1 R 2 V i V o R 2 R 1

Non-inverting amplifier V i R 2 V o R 1 V o V i R 1 R 2 V i V i V o V o R 2 R 2 R 1 R 1 M. B. Patil, IIT Bombay

Non-inverting amplifier R 2 R 1 V o V o V i V i R L R L Consider R 1 → ∞ , R 2 → 0 . M. B. Patil, IIT Bombay

Non-inverting amplifier R 2 R 1 V o V o V i V i R L R L Consider R 1 → ∞ , R 2 → 0 . V o → 1 + R 2 → 1 , i.e., V o = V i . V i R 1 M. B. Patil, IIT Bombay

Non-inverting amplifier R 2 R 1 V o V o V i V i R L R L Consider R 1 → ∞ , R 2 → 0 . V o → 1 + R 2 → 1 , i.e., V o = V i . V i R 1 This circuit is known as unity-gain amplifier/voltage follower/buffer. M. B. Patil, IIT Bombay

Non-inverting amplifier R 2 R 1 V o V o V i V i R L R L Consider R 1 → ∞ , R 2 → 0 . V o → 1 + R 2 → 1 , i.e., V o = V i . V i R 1 This circuit is known as unity-gain amplifier/voltage follower/buffer. What has been achieved? M. B. Patil, IIT Bombay

Loading effects R o R s V i V o V s R L A V V i R i Consider an amplifier of gain A V . We would like to have V o = A V V s . M. B. Patil, IIT Bombay

Loading effects R o R s V i V o V s R L A V V i R i Consider an amplifier of gain A V . We would like to have V o = A V V s . However, the actual output voltage is, R L R L R i V o = A V V i = A V V s . R o + R L R o + R L R i + R s M. B. Patil, IIT Bombay

Loading effects R o R s V i V o V s R L A V V i R i Consider an amplifier of gain A V . We would like to have V o = A V V s . However, the actual output voltage is, R L R L R i V o = A V V i = A V V s . R o + R L R o + R L R i + R s To obtain the desired V o , we need R i → ∞ and R o → 0 . M. B. Patil, IIT Bombay

Loading effects R o R s V i V o V s R L A V V i R i Consider an amplifier of gain A V . We would like to have V o = A V V s . However, the actual output voltage is, R L R L R i V o = A V V i = A V V s . R o + R L R o + R L R i + R s To obtain the desired V o , we need R i → ∞ and R o → 0 . The buffer (voltage follower) provides these features. M. B. Patil, IIT Bombay

Op-amp buffer: input resistance R 2 R 2 B A R o R 1 R 1 V i � 1 + R 2 � R L V o = V s A V V i V S R i R 1 I S R L V S Non−inverting amplifier M. B. Patil, IIT Bombay

Op-amp buffer: input resistance R 2 R 2 B A R o R 1 R 1 V i � 1 + R 2 � R L V o = V s A V V i V S R i R 1 I S R L V S Non−inverting amplifier V B + V B − A V V i + V B − V A KCL at B: = 0. R L R o R 2 M. B. Patil, IIT Bombay

Op-amp buffer: input resistance R 2 R 2 B A R o R 1 R 1 V i � 1 + R 2 � R L V o = V s A V V i V S R i R 1 I S R L V S Non−inverting amplifier V B + V B − A V V i + V B − V A KCL at B: = 0. R L R o R 2 I S = V A + V A − V B Source current: . R 1 R 2 M. B. Patil, IIT Bombay

Op-amp buffer: input resistance R 2 R 2 B A R o R 1 R 1 V i � 1 + R 2 � R L V o = V s A V V i V S R i R 1 I S R L V S Non−inverting amplifier V B + V B − A V V i + V B − V A KCL at B: = 0. R L R o R 2 I S = V A + V A − V B Source current: . R 1 R 2 Using V i = I S R i , V A = V S − V i , and after some algebra, we get �� 1 � � + 1 � � � � 1 + R o + R o 1 + R o + R o − R o + A V + R i R 2 R L R 2 R 1 R 2 R L R 2 R 2 R in = V S 2 = . � 1 I S + 1 � � 1 + R o + R o � − R o R 2 R 1 R 2 R L R 2 2 M. B. Patil, IIT Bombay

Op-amp buffer: input resistance R 2 R 2 B A R o R 1 R 1 V i � 1 + R 2 � R L V o = V s A V V i V S R i R 1 I S R L V S Non−inverting amplifier V B + V B − A V V i + V B − V A KCL at B: = 0. R L R o R 2 I S = V A + V A − V B Source current: . R 1 R 2 Using V i = I S R i , V A = V S − V i , and after some algebra, we get �� 1 � � + 1 � � � � 1 + R o + R o 1 + R o + R o − R o + A V + R i R 2 R L R 2 R 1 R 2 R L R 2 R 2 R in = V S STOP 2 = . � 1 I S + 1 � � 1 + R o + R o � − R o R 2 R 1 R 2 R L R 2 2 M. B. Patil, IIT Bombay

Non-inverting amplifier: input resistance (continued) R 2 R 2 B A R o R 1 R 1 V i � � R L 1 + R 2 V o = V s A V V i V S R i R 1 I S R L Non−inverting amplifier V S �� 1 � 1 + R o + R o � + 1 � � 1 + R o + R o � − R o + A V � + R i R 2 R in = V S R L R 2 R 1 R 2 R L R 2 R 2 2 = . � 1 I S + 1 � � 1 + R o + R o � − R o R 2 R 1 R 2 R L R 2 2 M. B. Patil, IIT Bombay

Non-inverting amplifier: input resistance (continued) R 2 R 2 B A R o R 1 R 1 V i � � R L 1 + R 2 V o = V s A V V i V S R i R 1 I S R L Non−inverting amplifier V S �� 1 � 1 + R o + R o � + 1 � � 1 + R o + R o � − R o + A V � + R i R 2 R in = V S R L R 2 R 1 R 2 R L R 2 R 2 2 = . � 1 I S + 1 � � 1 + R o + R o � − R o R 2 R 1 R 2 R L R 2 2 Since R o is much smaller than R 1 , R 2 , R L , or R i , �� 1 � R 1 + R 2 + 1 � + A V � + A V � 1 + R i R i R 1 R 2 R 2 R 1 R 2 R 2 R 1 R in ≈ ≈ ≈ A V R i . � 1 R 1 + R 2 + 1 � R 1 + R 2 R 1 R 2 R 1 R 2 M. B. Patil, IIT Bombay

Op-amp buffer: input resistance R o V o = V s V i R L A V V i V s R i I S R L V S Buffer M. B. Patil, IIT Bombay

Op-amp buffer: input resistance R o V o = V s V i R L A V V i V s R i I S R L V S Buffer Let R o → 0. M. B. Patil, IIT Bombay

Op-amp buffer: input resistance R o V o = V s V i R L A V V i V s R i I S R L V S Buffer Let R o → 0. V S = V i + A V V i = V i (1 + A V ) . M. B. Patil, IIT Bombay

Op-amp buffer: input resistance R o V o = V s V i R L A V V i V s R i I S R L V S Buffer Let R o → 0. V S = V i + A V V i = V i (1 + A V ) . I S = V i . R i M. B. Patil, IIT Bombay

Op-amp buffer: input resistance R o V o = V s V i R L A V V i V s R i I S R L V S Buffer Let R o → 0. V S = V i + A V V i = V i (1 + A V ) . I S = V i . R i → R in = V S = R i ( A V + 1) I S M. B. Patil, IIT Bombay

Op-amp buffer: output resistance R 2 R 2 R o R 1 R 1 V o V i R L A V V i V s R i R L V s R out Non−inverting amplifier To find R out , * Deactivate the input source. M. B. Patil, IIT Bombay

Op-amp buffer: output resistance R 2 R 2 R o R 1 R 1 V o V i R L A V V i V s R i R L V s R out Non−inverting amplifier To find R out , * Deactivate the input source. * Replace R L with a test source V ′ . M. B. Patil, IIT Bombay

Op-amp buffer: output resistance R 2 R 2 R o R 1 R 1 V o V i R L A V V i V s R i R L V s R out Non−inverting amplifier To find R out , * Deactivate the input source. * Replace R L with a test source V ′ . * Find the current ( I ′ ) through V ′ . M. B. Patil, IIT Bombay

Op-amp buffer: output resistance R 2 R 2 R o R 1 R 1 V o V i R L A V V i V s R i R L V s R out Non−inverting amplifier To find R out , * Deactivate the input source. * Replace R L with a test source V ′ . * Find the current ( I ′ ) through V ′ . * R out = V ′ I ′ . M. B. Patil, IIT Bombay

Op-amp buffer: output resistance (continued) I 2 R 2 I ′ R 2 A V V i V ′ − V i R 1 R o I 1 R 1 V o V ′ V i A V V i V s R i R L Non−inverting amplifier M. B. Patil, IIT Bombay

Op-amp buffer: output resistance (continued) I 2 R 2 I ′ R 2 A V V i V ′ − V i R 1 R o I 1 R 1 V o V ′ V i A V V i V s R i R L Non−inverting amplifier ( R i � R 1 ) R 2 + ( R i � R 1 ) V ′ ≡ − kV ′ . V i = − M. B. Patil, IIT Bombay

Op-amp buffer: output resistance (continued) I 2 R 2 I ′ R 2 A V V i V ′ − V i R 1 R o I 1 R 1 V o V ′ V i A V V i V s R i R L Non−inverting amplifier ( R i � R 1 ) R 2 + ( R i � R 1 ) V ′ ≡ − kV ′ . V i = − I ′ = I 1 + I 2 = V ′ − A V V i + V ′ − ( − V i ) = 1 + 1 V ′ + kA V V ′ � V ′ − kV ′ � � � . R o R 2 R o R 2 M. B. Patil, IIT Bombay

Op-amp buffer: output resistance (continued) I 2 R 2 I ′ R 2 A V V i V ′ − V i R 1 R o I 1 R 1 V o V ′ V i A V V i V s R i R L Non−inverting amplifier ( R i � R 1 ) R 2 + ( R i � R 1 ) V ′ ≡ − kV ′ . V i = − I ′ = I 1 + I 2 = V ′ − A V V i + V ′ − ( − V i ) = 1 + 1 V ′ + kA V V ′ � V ′ − kV ′ � � � . R o R 2 R o R 2 I ′ (1 − k ) → R out = V ′ V ′ = 1 (1 + kA V ) + 1 R o R 2 R o I ′ = (1 + kA V ) � (1 − k ) ≈ R o R 2 (1 + kA V ) M. B. Patil, IIT Bombay

Op-amp buffer: output resistance (continued) I 2 R 2 I ′ R 2 A V V i V ′ − V i R 1 R o I 1 R 1 V o V ′ V i A V V i V s R i R L Non−inverting amplifier ( R i � R 1 ) R 2 + ( R i � R 1 ) V ′ ≡ − kV ′ . V i = − I ′ = I 1 + I 2 = V ′ − A V V i + V ′ − ( − V i ) = 1 + 1 V ′ + kA V V ′ � V ′ − kV ′ � � � . R o R 2 R o R 2 I ′ (1 − k ) → R out = V ′ V ′ = 1 (1 + kA V ) + 1 R o R 2 R o I ′ = (1 + kA V ) � (1 − k ) ≈ R o R 2 (1 + kA V ) Special case: Op-amp buffer ( R i � R 1 ) R o k = R 2 + ( R i � R 1 ) → 1 ⇒ R out ≈ 1 + A V M. B. Patil, IIT Bombay

Op-amp buffer V s R L V s V s R L R in R out In summary, the buffer (voltage follower) provides M. B. Patil, IIT Bombay

Op-amp buffer V s R L V s V s R L R in R out In summary, the buffer (voltage follower) provides * a large input resistance R in as seen from the source. M. B. Patil, IIT Bombay

Op-amp buffer V s R L V s V s R L R in R out In summary, the buffer (voltage follower) provides * a large input resistance R in as seen from the source. * a small output resistance R out as seen from the load. M. B. Patil, IIT Bombay

Op-amp buffer V s R L V s V s R L R in R out In summary, the buffer (voltage follower) provides * a large input resistance R in as seen from the source. * a small output resistance R out 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) R s R o V s V i V o R L R i A V V i Problem: We would like to have V o = A V V s . M. B. Patil, IIT Bombay

Loading effects (revisited) R s R o V s V i V o R L R i A V V i Problem: We would like to have V o = A V V s . But the actual output voltage is, R L R L R i V o = A V V i = A V V s . R o + R L R o + R L R i + R s M. B. Patil, IIT Bombay