2019 A/L Physics Past Paper · Question 03 (English Medium)

Answer: (5) not change the collector current.

What the question asks

We want:

What happens when the base current $I_B$ is raised even more.

The transistor is already in saturation.

We look at the collector current $I_C$ and at the switch state.

Relevant theory

The collector circuit:

$$V_{CC} = I_CR_C + V_{CE}$$

In the active region $I_C = \beta I_B$.

In saturation both junctions are forward biased.

$V_{CE}$ is then at its lowest, $V_{CE(sat)} \approx 0.2$ V.

In the saturated region, that is where IC is no longer increasing with IB, IC is less than β IB

Applying it to this question

What stays fixed:

$V_{CC}$, $R_C$ and $V_{CE(sat)}$.

So the biggest current the collector circuit can carry is fixed too.

$$I_{C(sat)} = \dfrac{V_{CC} - V_{CE(sat)}}{R_C}$$

Reasoning and calculation

Saturation means $\beta I_B$ is more than $I_{C(sat)}$.

The base current asks for more than the circuit can give.

A bigger $I_B$ only raises $\beta I_B$.

The real current is the smaller of the two.

$$I_C = I_{C(sat)}$$

So the collector current does not rise and does not fall.

It stays at $I_{C(sat)}$.

The switch is already ON, so its state does not change either.

Reading the options

Compare the options:

(1) and (2) say the switch changes.

(3) and (4) say $I_C$ changes.

Only (5) says $I_C$ does not change.

Why the other options fail

In saturation the switch is already ON.

The OFF state needs $I_B$ to be about zero.

It uses $I_C = \beta I_B$ in saturation too.

It thinks a bigger $I_B$ lowers $I_C$.

Checks

$9.8$

$I_{C(sat)} = \dfrac{10 - 0.2}{1000}$

$15$

$\beta I_B = 100 \times 150\ \mu\text{A}$

$I_C = I_{C(sat)}$

$I_C = \min(30,\ 9.8)$

Conclusion

$I_C = I_{C(sat)} = \frac{V_{CC} - V_{CE(sat)}}{R_C}$

In saturation $I_C$ is limited by $V_{CC}$ and $R_C$.

More base current does not change the collector current.

So the answer is option (5).

$I_C = \beta I_B$ holds only in the active region.

In saturation questions first find the biggest collector current, $\dfrac{V_{CC} - 0.2}{R_C}$.

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