Specification: AQA A-level Physics (7408) 3.13.4.1 Electronics option | Paper: 3 Section B | Data and formulae booklet: given in the exam
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Ideal op-amp: an op-amp with infinite open-loop gain, infinite input resistance (no current flows into either input) and zero output resistance. With negative feedback this keeps V₊ and V₋ almost equal.
Virtual earth: in the inverting amplifier the non-inverting input is at 0 V, so the inverting input is held at very nearly 0 V even though it is not connected to 0 V. The current through R_in therefore equals the current through R_f.
Saturation: the output of an op-amp cannot go beyond (about) the supply voltages. If the calculated output is larger than the saturation voltage, the output simply stays at the saturation value and the signal is clipped.
The equation
voltage gain = − feedback resistance ÷ input resistance V_out ÷ V_in = −R_f ÷ R_in (input through R_in to the inverting input, R_f from the output back to the inverting input, non-inverting input at 0 V)
| Quantity | Symbol | Unit |
|---|---|---|
| output voltage | V_out | volts (V) |
| input voltage | V_in | volts (V) |
| feedback resistance | R_f | ohms (Ω) |
| input resistance | R_in | ohms (Ω) |
| voltage gain | V_out ÷ V_in | no unit (a ratio) |
Rearranged: V_out = −(R_f ÷ R_in) × V_in | R_f = R_in × (size of gain) | R_in = R_f ÷ (size of gain) | The minus sign means the output is inverted (180° out of phase with the input). Always compare your answer with the saturation voltage.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) An inverting amplifier (input resistor R_in from the input to the inverting (−) input, feedback resistor R_f from the output to the inverting input, non-inverting (+) input connected to 0 V) has R_in = 10 kΩ and R_f = 47 kΩ. Calculate its voltage gain.
Show solution
F: V_out ÷ V_in = −R_f ÷ R_in
I: gain = −47 ÷ 10 (both in kΩ, so the units cancel)
F: gain is already the subject
A: −4.7 (no unit)
Q2 (Core) An inverting amplifier has R_in = 2.2 kΩ and R_f = 22 kΩ. The input voltage is +0.35 V. Calculate the output voltage. Assume the op-amp is ideal and saturates at ±13.5 V (±15 V supply).
Show solution
F: V_out ÷ V_in = −R_f ÷ R_in
I: V_out ÷ 0.35 = −22 ÷ 2.2 = −10
F: V_out = −10 × 0.35
A: −3.5 V. This is inside ±13.5 V, so the op-amp does not saturate.
Q3 (Core) An inverting amplifier with R_f = 100 kΩ gives an output of −1.2 V when the input is +25 mV. Calculate R_in, to 2 significant figures.
Show solution
Convert first: 25 mV = 0.025 V
F: V_out ÷ V_in = −R_f ÷ R_in
I: −1.2 ÷ 0.025 = −100 000 ÷ R_in, so −48 = −100 000 ÷ R_in
F: R_in = 100 000 ÷ 48 = 2083.3… Ω
A: 2.1 kΩ (2 s.f.)
Q4 (Core) A microphone produces a signal of amplitude 12 mV. It feeds an inverting amplifier with R_in = 1.0 kΩ. Calculate the value of R_f needed to give an output amplitude of 3.0 V.
Show solution
Convert first: 12 mV = 0.012 V; 1.0 kΩ = 1000 Ω
F: V_out ÷ V_in = −R_f ÷ R_in
I: size of gain = 3.0 ÷ 0.012 = 250
F: R_f = 250 × R_in = 250 × 1000
A: 250 000 Ω = 250 kΩ
Watch out: using 12 instead of 0.012 V gives a gain of 0.25. Convert mV to V (or keep both voltages in the same unit) before dividing.
Q5 (Exam) An inverting amplifier has R_in = 15 kΩ and R_f = 120 kΩ. The input voltage is −0.80 V. Assume the op-amp is ideal and saturates at ±13.5 V (±15 V supply). (a) Calculate the output voltage. (b) Calculate the current in R_f, to 2 significant figures.
Show solution
Step 1 – (a) output voltage
F: V_out ÷ V_in = −R_f ÷ R_in
I: V_out ÷ (−0.80) = −120 ÷ 15 = −8.0
F: V_out = −8.0 × (−0.80)
A: +6.4 V (a negative input gives a positive output; inside ±13.5 V)
Step 2 – (b) current
the inverting input is a virtual earth at 0 V, and no current enters the op-amp, so the current in R_f equals the current in R_in
I = V_in ÷ R_in = 0.80 ÷ 15 000 = 5.33… × 10−5 A
A: 5.3 × 10−5 A (53 µA, 2 s.f.)
Check: I × R_f = 5.33… × 10−5 × 120 000 = 6.4 V across R_f, from 0 V to the output
Q6 (Exam) An inverting amplifier has R_in = 10 kΩ and R_f = 220 kΩ. Assume the op-amp is ideal and saturates at ±13.5 V (±15 V supply). (a) Find the output voltage when the input is +0.75 V. (b) Calculate the largest input voltage that can be amplified without saturation, to 2 significant figures.
Show solution
Step 1 – (a) output
F: V_out ÷ V_in = −R_f ÷ R_in
I: gain = −220 ÷ 10 = −22, so V_out = −22 × 0.75 = −16.5 V
F: −16.5 V is beyond the −13.5 V saturation voltage
A: the output saturates at −13.5 V
Step 2 – (b) largest input
V_in = 13.5 ÷ 22 = 0.613… V
A: 0.61 V (2 s.f.)
Watch out: giving −16.5 V as the answer. An op-amp output can never go beyond its saturation voltage, so always check this last.
Q7 (Exam) A sinusoidal signal of r.m.s. voltage 0.15 V is applied to an inverting amplifier with R_in = 2.0 kΩ and R_f = 100 kΩ. Assume the op-amp is ideal and saturates at ±13.5 V (±15 V supply). (a) Calculate the r.m.s. output voltage. (b) Calculate the peak output voltage, to 2 significant figures, and state whether the output is clipped.
Show solution
Step 1 – (a) r.m.s. output
F: V_out ÷ V_in = −R_f ÷ R_in
I: gain = −100 ÷ 2.0 = −50
F: V_out = −50 × 0.15
A: size of r.m.s. output = 7.5 V, inverted (180° out of phase with the input)
Step 2 – (b) peak output
V₀ = √2 × V_rms = √2 × 7.5 = 10.6… V
A: 11 V (2 s.f.). This is less than 13.5 V, so the output is not clipped.
Watch out: comparing the r.m.s. value with the saturation voltage. Clipping depends on the peak value, which is √2 times larger.
Q8 (Exam) Two inverting amplifiers are connected one after the other: the output of the first is the input of the second. Stage 1 has R_in = 10 kΩ and R_f = 100 kΩ. Stage 2 has R_in = 10 kΩ and R_f = 47 kΩ. The input to stage 1 is +0.12 V. Assume the op-amp is ideal and saturates at ±13.5 V (±15 V supply). Calculate the final output voltage, to 2 significant figures, and the overall voltage gain.
Show solution
Step 1 – stage 1
F: V_out ÷ V_in = −R_f ÷ R_in
I: gain₁ = −100 ÷ 10 = −10, so V₁ = −10 × 0.12 = −1.2 V
Step 2 – stage 2
I: gain₂ = −47 ÷ 10 = −4.7, so V₂ = −4.7 × (−1.2) = 5.64 V
F: overall gain = gain₁ × gain₂ = (−10) × (−4.7) = +47
A: +5.6 V (2 s.f.), overall gain +47. Neither stage reaches ±13.5 V, and two inversions put the output back in phase with the input.
Q9 (Stretch) A sinusoidal signal of peak voltage 0.40 V is applied to an inverting amplifier with R_in = 2.2 kΩ and R_f = 100 kΩ. Assume the op-amp is ideal and saturates at ±13.5 V (±15 V supply). (a) Show that the output is clipped. (b) Calculate the fraction of each cycle for which the output is clipped, to 2 significant figures. (c) Calculate the largest value of R_f that would avoid clipping, to 2 significant figures.
Show solution
Step 1 – (a) unclipped peak
F: V_out ÷ V_in = −R_f ÷ R_in
I: gain = −100 ÷ 2.2 = −45.45…
F: peak output = 45.45… × 0.40 = 18.18… V
A: 18.18… V is greater than 13.5 V, so the output is clipped
Step 2 – (b) fraction clipped
the output is clipped when |sin θ| is greater than 13.5 ÷ 18.18… = 0.7425, which starts at θ = sin⁻¹ 0.7425 = 47.94…°
in each half cycle it is clipped from 47.94…° to 180° − 47.94…°, which is 84.11…° out of 180°
A: fraction = 84.11… ÷ 180 = 0.467…, so the output is clipped for 0.47 (47%) of each cycle (2 s.f.)
Step 3 – (c) largest R_f
the largest size of gain = 13.5 ÷ 0.40 = 33.75, so R_f = 33.75 × 2.2 kΩ = 74.25 kΩ
A: 74 kΩ (2 s.f.)
Q10 (Stretch) An op-amp has a gain × bandwidth product of 1.0 MHz and saturates at ±13.5 V (±15 V supply). It is used in an inverting amplifier with R_in = 4.0 kΩ and R_f = 100 kΩ. Use gain × bandwidth = constant, with the size of the closed-loop gain. (a) Calculate the bandwidth. (b) A designer needs a bandwidth of at least 20 kHz for audio. Calculate the largest R_f that could be used with R_in = 4.0 kΩ. (c) For this new circuit, calculate the largest input amplitude that avoids saturation.
Show solution
Step 1 – (a) bandwidth
size of gain = R_f ÷ R_in = 100 ÷ 4.0 = 25
F: gain × bandwidth = 1.0 × 106 Hz
I: 25 × bandwidth = 1.0 × 106
F: bandwidth = 1.0 × 106 ÷ 25
A: 40 000 Hz = 40 kHz
Step 2 – (b) largest R_f
largest size of gain = 1.0 × 106 ÷ 20 000 = 50, so R_f = 50 × 4.0 kΩ
A: 200 kΩ
Step 3 – (c) largest input
V_in = 13.5 ÷ 50 = 0.27 V
A: 0.27 V
Watch out: increasing the gain always reduces the bandwidth. A high-gain stage may amplify loudly but cut off the higher audio frequencies.
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