Past exam of the physics 2 paper of the University of Oxford / 2013 / 8

Table of contents
- i 8
  - Solution i
- ii 8
  - Solution ii
- iii 8
  - Solution iii
- iv 8
  - Solution iv
- v 8
  - Solution v

i

Solution

Ground the non-inverting input and take the ideal operational amplifier in stable, unsaturated negative feedback. Its inverting input is a virtual ground and draws no current. Kirchhoff's current law there gives

\frac{V _{i}}{R _{i}} + \frac{V _{o}}{R _{f}} = 0, \frac{V _{o}}{V _{i}} = - \frac{R _{f}}{R _{i}}, ∣ A_{v} ∣ = \frac{R _{f}}{R _{i}} .

The minus sign is the phase inversion of the inverting amplifier.

ii

Solution

The input source supplies

I_{i} = V_{i} / R_{i}

into the virtual ground, so the complete circuit's input resistance is

R_{i}

. The ideal output is a zero-resistance controlled voltage source:

R_{in} = R_{i}, R_{out} = 0.

These are properties of the closed-loop circuit, distinct from the infinite input resistance of the amplifier terminals themselves.

iii

Solution

With

R_{f} = R_{i}

, the output and input voltage magnitudes are equal. The signal power supplied at the input is

P_{i} = V_{i, rms}^{2} / R_{i}

, while the load receives

P_{L} = V_{o, rms}^{2} / R_{L}

. Hence

\frac{P _{L}}{P _{i}} = (\frac{R _{f}}{R _{i}})^{2} \frac{R _{i}}{R _{L}} = \frac{1000}{100} = 10.

This is

10

dB of power gain into the specified load. Output energy comes from the amplifier's supply rails; unity voltage-gain magnitude does not imply unity power gain. Dissipation in the feedback resistor is additional to the specified load power.

iv

Solution

Replace the feedback resistor by a capacitor

C_{f}

, retaining the input resistor

R_{i}

. The virtual ground makes the input charge flow onto the feedback capacitor rather than into the ideal amplifier. Integrating its current gives

V_{o} (t) - V_{o} (0) = - \frac{1}{C _{f}} \int_{0}^{t} I_{i} (t^{'}) d t^{'}, \frac{Δ V _{o}}{Q _{in}} = - \frac{1}{C _{f}} .

If the calorimeter delivers

Q_{in} = κ E

for deposited energy

E

, then

Δ V_{o} = - κ E / C_{f}

. A reset establishes the initial output; a practical large resistor across

C_{f}

limits long-time drift, but is omitted from the ideal integration interval.

Figure 1.
Ideal inverting charge integrator for the calorimeter signal
. The original feedback resistor is replaced by C_f; the input resistor remains for the subsequent voltage-driven integrator calculation. Details.

v

Solution

For the modified circuit,

I_{i} = V_{i} / R_{i}

and

\dot{V}_{o} = - V_{i} / (R_{i} C_{f})

. With

V_{i} = V_{0} cos ω t

,

V_{o} (t) = V_{o} (0) - \frac{V _{0}}{ω R _{i} C _{f}} sin ω t .

The sinusoidal component has amplitude

V_{0} / (ω R_{i} C_{f})

and leads the cosine input by

9 0^{\circ}

; the constant is set by the initial capacitor charge. This assumes the output remains within the amplifier's linear range.

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