2018 A/L Physics Past Paper · Question 43 (English Medium)

Circuit with an ideal diode — overall resistance against V_X

In the circuit shown, $X$ and $Y$ represent the terminals of a variable voltage source situated in the box with broken lines. $P$ is a variable resistor. $D$ is an ideal diode. As the value of the voltage $V_X$ at point $X$ is gradually increased from $0$ to $15$ V, which of the following graphs correctly represents the variation of the overall resistance $R$ of the section of the circuit to the right of $XY$ ?

(1) Figure (1)
(2) Figure (2)
(3) Figure (3)
(4) Figure (4)
(5) Figure (5)

Answer and explanation

Answer: (2) Figure (2)

What the question asks

We want: how the overall resistance $R$ to the right of $XY$ changes as $V_X$ rises from 0 to 15 V.

$R = \frac{V_{XY}}{I}$

Relevant theory

To make a p - n junction forward biased, an external potential difference (Es) should be connected so that the p - region gets the positive potential and n - region gets the negative potential.

Reverse current flows because of the minority carriers created by thermal agitation.

An ideal diode is like a wire when it is forward biased.

It is like a gap when it is reverse biased.

In the box, $P$ is across the 15 V cell. Its bottom is earthed. So $V_X$ goes from 0 to 15 V.

The $+$ end of the 5 V cell is at $Y$ . Its $-$ end is earthed. So $V_Y = +5$ V.

On the right, $X$ goes through 1 kΩ to a point $N$ . Then through 1 kΩ to $Y$ . The diode D is across the lower 1 kΩ. Its anode is on the $Y$ side.

Required equations

$R = \frac{V_X - V_Y}{I}$

$V_Y > V_N \iff V_X < V_Y = 5\ \text{V}$

Reasoning and calculation

The graph has $V_X$ (V) across and $R$ (kΩ) up the side.

At the start, $V_X = 0$ . That is lower than $V_Y = 5$ V.

So current flows up from $Y$ to $X$ . The anode of D (the $Y$ side) is higher than its cathode. D is forward biased.

There are two parts: $V_X < 5$ V and $V_X > 5$ V.

The change is at $V_X = V_Y = 5$ V.

For $V_X < 5$ V, D conducts. It bypasses the lower 1 kΩ. So $R = 1$ kΩ.

For $V_X > 5$ V, the current flows down from $X$ . D is reverse biased, so it is open. The two 1 kΩ are in series. So $R = 2$ kΩ.

At $V_X = 5$ V no current flows. There $R$ jumps from 1 kΩ to 2 kΩ. That is the broken line.

In each part, $R$ stays the same.

The graph is flat at 1 kΩ from 0 to 5 V.

It steps up at 5 V.

Then it is flat at 2 kΩ from 5 to 15 V.

Reading the options

Only figure (2) has this shape.

The others step the wrong way, or at the wrong place.

Why the other options fail

The diode is taken the wrong way round.

The step is at the wrong place and goes the wrong way.

The step is at 5 V, not 10 V.

This is as if the diode never conducts.

Conclusion

$R = 1$ kΩ for $V_X < 5$ V.

$R = 2$ kΩ for $V_X > 5$ V.

The answer is option (2).

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