2022 A/L Physics Past Paper · Question 46 (English Medium)

Answer: (2) $V_A = V_B < V_C$

Given information

What we know:

Three circuits $A$, $B$ and $C$ use identical op-amps. We treat each op-amp as an ideal one, as the book does.

The same input voltage $V_{in}$ goes into all three circuits.

In every circuit, a $10\ \text{k}\Omega$ resistor joins the $-$ input to the ground, and a feedback resistor joins the output to the $-$ input.

In $A$ and $B$ the feedback resistor is $40\ \text{k}\Omega$. In $C$ there are two, $40\ \text{k}\Omega$ and $10\ \text{k}\Omega$, one after the other.

In $B$ there is also a $10\ \text{k}\Omega$ resistor between $V_{in}$ and the $+$ input.

What the question asks

We want: how the sizes of $V_A$, $V_B$ and $V_C$ compare.

What each letter means:

$V_{in}$ ($\text{V}$): the input voltage, the same for all three circuits.

$V_A$, $V_B$, $V_C$ ($\text{V}$): the output voltages of $A$, $B$ and $C$.

$R$ ($\text{k}\Omega$): the resistor from the $-$ input to the ground, $10\ \text{k}\Omega$.

$R_f$ ($\text{k}\Omega$): the feedback resistor from the output to the $-$ input.

$G$: the voltage gain, the output voltage divided by the input voltage. It has no unit.

Relevant theory

What is happening:

The op-amp looks at its two inputs. It changes its output until the two inputs are at the same voltage.

Part of the output comes back to the $-$ input through the feedback resistor. The $10\ \text{k}\Omega$ resistor to the ground sets how much comes back.

So the op-amp has to push the output higher than $V_{in}$ to keep the $-$ input at $V_{in}$. How much higher depends on the two resistors.

We find the gain of each circuit and compare.

According to " Golden rule II", no current is flown into the input terminal.

is the voltage gain of the close-loop non-inverting amplifier.

If a number of resistors are connected so that the same current flows through them, then it is a series connection.

The first quote says that no current goes into an input terminal. So a resistor in front of the $+$ input has no current in it, and there is no voltage drop across it.

The second quote names the gain of this kind of circuit, the output voltage divided by the input voltage.

The third quote is about two resistors that carry the same current. In $C$ the two feedback resistors do, so they are in series and their resistances add.

For example, a person riding a bicycle along a painted line. They look at how far they are from the line and steer back. They keep correcting until they are on the line. The op-amp does the same with its two inputs.

Required equations

$V_- = V_+ = V_{in}$

$I = \dfrac{V_{in}}{R}$

$V_{out} - V_{in} = I R_f$

$G = \dfrac{V_{out}}{V_{in}} = 1 + \dfrac{R_f}{R}$

Applying it to this question

The gain $1 + \dfrac{R_f}{R}$ has the unit $\dfrac{\text{k}\Omega}{\text{k}\Omega} = 1$, so the gain has no unit and $V_{out}$ has the unit of $V_{in}$.

We use this for an ideal op-amp working in its linear region. The output cannot go beyond the supply voltage.

$5$

$G_A = 1 + \dfrac{40}{10}$

$G_B = 1 + \dfrac{40}{10}$

So $V_A = 5\,V_{in}$ and $V_B = 5\,V_{in}$.

$50$

$R_f = 40 + 10$

$6$

$G_C = 1 + \dfrac{50}{10}$

Reasoning and calculation

$V_A = 5\,V_{in}$, $V_B = 5\,V_{in}$ and $V_C = 6\,V_{in}$.

So $V_A$ and $V_B$ are equal, and $V_C$ is bigger. That is $V_A = V_B < V_C$, option (2).

Checks

$1 + rf / r$

Does it make sense?

A bigger feedback resistor sends less of the output back to the $-$ input, so the op-amp has to push the output higher. That gives a bigger gain. $C$ has the bigger feedback resistor and the bigger gain.

The extra resistor in $B$ is in front of an input that takes no current, so it can change nothing.

This holds when the op-amps stay in their linear region. An op-amp that reaches its supply limit would stop growing, and the paper treats the op-amps as ideal.

Conclusion

$V_A = V_B < V_C$

The gains are $5$, $5$ and $6$. So $V_A = V_B < V_C$.

The answer is (2).

For a circuit with the input on the $+$ terminal, find the gain $G = 1 + \dfrac{R_f}{R}$.

For an ideal op-amp, a resistor in front of the $+$ input carries no current, so it does not change the gain. Resistors in series in the feedback path add.

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