Electrical · real student question

An ideal inverting op-amp amplifier has R1 = 1.5 kΩ in series with the input, R2 = 5.6 kΩ as the feedback resistor, and R3 = 1.2 kΩ from the non-inverting input to ground. Supplies are ±15 V and the input is 0.1 Vp at 1 kHz. (a) Find Vout. (b) Find the largest Vin that does not clip the output. (c) What value of R2 would give Vout = −5 Vp?

Question

An ideal op-amp is wired as an inverting amplifier with R1=1.5 kΩR_1=1.5\ \text{k}\Omega at the input, R2=5.6 kΩR_2=5.6\ \text{k}\Omega in the feedback path and R3=1.2 kΩR_3=1.2\ \text{k}\Omega from the non-inverting input to ground. The supplies are ±VCC=±15\pm V_{CC}=\pm 15 V and the input is Vin=0.1 VpV_{in}=0.1\ \text{V}_p at 11 kHz.

(a) Determine VoutV_{out}.
(b) Determine the largest VinV_{in} for which the output is not clipped.
(c) What value of R2R_2 gives Vout=−5 VpV_{out}=-5\ \text{V}_p?

Step-by-step solution

  1. Identify the topology and the role of R3. The signal enters through R1R_1 into the inverting input with R2R_2 feeding back, so this is the classic inverting amplifier. R3R_3 sits between the non-inverting input and ground purely for bias-current balancing; for an ideal op-amp it draws no current and therefore does not appear in any of the three answers.

  2. (a) Apply the inverting gain formula. Because the inverting input is a virtual earth,

    Av=−R2R1=−5.6 kΩ1.5 kΩ=−3.733A_v=-\frac{R_2}{R_1}=-\frac{5.6\ \text{k}\Omega}{1.5\ \text{k}\Omega}=-3.733

    Vout=AvVin=−3.733×0.1=−0.373 VpV_{out}=A_vV_{in}=-3.733\times 0.1=-0.373\ \text{V}_p

    The minus sign means the output is inverted, i.e. 180∘180^\circ out of phase with the input, not that it is "negative" for a sine wave.

  3. (b) Set the clipping condition. An ideal op-amp can swing to the rails, so the peak output is limited by ∣Vout∣≤VCC=15|V_{out}|\le V_{CC}=15 V. Hence

    ∣Vin,max⁡∣=VCC∣Av∣=153.733=4.018 Vp|V_{in,\max}|=\frac{V_{CC}}{|A_v|}=\frac{15}{3.733}=4.018\ \text{V}_p

    So inputs up to about 4.02 Vp4.02\ \text{V}_p pass undistorted; beyond that the peaks flatten. (A real op-amp saturates a volt or two short of the rails, which lowers this figure.)

  4. (c) Solve the gain equation for R2. Requiring Vout=−5 VpV_{out}=-5\ \text{V}_p with Vin=0.1 VpV_{in}=0.1\ \text{V}_p means ∣Av∣=50|A_v|=50, so

    R2R1=50⟹R2=50×1.5 kΩ=75 kΩ\frac{R_2}{R_1}=50\quad\Longrightarrow\quad R_2=50\times 1.5\ \text{k}\Omega=75\ \text{k}\Omega

  5. Collect the three answers.

    Vout=−0.373 Vp,Vin,max⁡≈4.02 Vp,R2=75 kΩ\boxed{V_{out}=-0.373\ \text{V}_p,\qquad V_{in,\max}\approx 4.02\ \text{V}_p,\qquad R_2=75\ \text{k}\Omega}

  6. Consistency check on part (c). With R2=75 kΩR_2=75\ \text{k}\Omega the peak output is 55 V, comfortably inside the ±15\pm 15 V rails, so the new design does not clip — worth verifying, because raising the gain also lowers the maximum usable input, here to 15/50=0.3 Vp15/50=0.3\ \text{V}_p.

Answer

Vout=−0.373 Vp;Vin,max⁡≈4.02 Vp;R2=75 kΩV_{out}=-0.373\ \text{V}_p;\quad V_{in,\max}\approx 4.02\ \text{V}_p;\quad R_2=75\ \text{k}\Omega

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