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The Age of Silicon: 6. Amplifiers Answers

1. Q32 (c) (i), (ii) and (iii) from the 2002 HSC Physics exam

You will have to read the table and pick up relevant data (over what range are the output/input voltages in a linear relationship? If in doubt, graph the table but watch your time!

(-200 V to +200 V)
You will need to recognize that the gain is equal to the ratio of output to input voltage (see formula sheet)

Gain = 6/150 x 10^-6 = 40 000
From the table, we can see that the output voltage peaks at 8 volts for an input ≥ 200 V which means that the output signal is distorted (in this case, clipped) for inputs over that value.

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2. Q32(d) part (i) and (ii) from the 2003 HSC Physics exam

V_A = {R_T / (RT + 22 kΩ)} x 12 V

At 15⁰ C, R_T = 1.4 kΩ

VA = {1.4/23.4} x 12 V

= 0.72 V
At and below 15⁰ C, the voltage at A will be ≤0.72 volts

That means the amplifier will need to produce a voltage of - 6 volts (inverting amplifier) to activate the relay at or below an input voltage of 0.72 V.

V₀ / V _in = - R_f / R_i

- 6 / 0.72 = R_f / 100 kΩ

Rf e" 6 / .72 x 100 kΩ

≥833 kΩ

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3. Q32(d) (i), (ii) and (iii) from the 2004 HSC Physics Exam

V_out / V_in =" -" R_f / R_i

V_out /0.2 = - 300 kΩ / 10 kΩ

V_out = - 6 volts
For the summing amp shown, the total output voltage would be:

   = V_out ₁ + V_out ₂

   ="(" - 6 ) + ( - 2)

   = - 8 volts

Using the formula provided to verify the result

V_out = - R 3 ( V 1 / R 1 + V 2 / R 2 )

   = -300kΩ (0.2/10kΩ +0.1/15kΩ)

   = - 8 volts
The first two resistors provided a summed input current

The 300 kΩ resistor is the negative feedback resistor

The amplifier output provides an exactly opposing current through the feedback resistor to cancel the input current.

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