2022 A/L Physics Past Paper · Question 44 (English Medium)

Answer: (3) $5\ \text{k}\Omega$, $7.5\ \text{k}\Omega$

Given information

What we know:

A silicon transistor has its emitter joined to the ground, $0\ \text{V}$. The supply is $V_{cc} = +10\ \text{V}$.

A $100\ \text{k}\Omega$ resistor joins the supply to the base. A variable resistor $R$ joins the base to the ground.

The value of $R_c$ is not given. We do not need it, because the base voltage decides the region.

We keep the letter $R$ and try the values in the options.

What the question asks

We want: the value of $R$ for cut-off and the value of $R$ for the active region.

What each letter means:

$V_{cc}$ ($\text{V}$): the supply voltage, $+10\ \text{V}$.

$R$ ($\text{k}\Omega$): the variable resistor from the base to the ground.

$V_B$ ($\text{V}$): the voltage at the base.

$V_{BE}$ ($\text{V}$): the voltage between the base and the emitter.

Relevant theory

What is happening:

The $100\ \text{k}\Omega$ resistor and $R$ are in a row between $+10\ \text{V}$ and $0\ \text{V}$. They share the $10\ \text{V}$. This is a potential divider.

The base is joined between them. So the base voltage depends on $R$.

The emitter is at $0\ \text{V}$, so $V_{BE}$ is the same as $V_B$.

The region of the transistor depends on $V_{BE}$. We find $V_B$ for each value of $R$ and compare it with the values in the book.

When VBE becomes less than 0.5 V and greater than 0.8 V the transistor is in the cut-off and saturation states respectively.

for a silicon transistor, and for a germanium transistor the relevant potentials are lower than the stated values.

So for a silicon transistor, the transistor is cut off if $V_{BE}$ is below $0.5\ \text{V}$, and it is saturated if $V_{BE}$ is above $0.8\ \text{V}$.

Between $0.5\ \text{V}$ and $0.8\ \text{V}$ the transistor is neither cut off nor saturated. That is the active region.

For example, a tap on a pipe. A small turn of the handle gives no water at first. That is like cut-off. Then more turning gives more water. That is like the active region. At a wide open tap the water cannot grow any more, like saturation.

Required equations

$V_B = V_{cc} \times \dfrac{R}{100 + R}$

$V_B < 0.5\ \text{V}$

$0.5\ \text{V} < V_B < 0.8\ \text{V}$

Applying it to this question

In $V_B = V_{cc} \times \dfrac{R}{100 + R}$ the two resistances are both in $\text{k}\Omega$, so the fraction has no unit and $V_B$ is in $\text{V}$.

For the small values of $R$ in the options, the current into the base is small, so the divider formula gives a good estimate.

$0.48$

$V_B = 10 \times \dfrac{5}{100 + 5}$

$0.48\ \text{V}$ is below $0.5\ \text{V}$. So $R = 5\ \text{k}\Omega$ puts the transistor in cut-off.

$0.10$

$V_B = 10 \times \dfrac{1.0}{100 + 1.0}$

$0.24$

$V_B = 10 \times \dfrac{2.5}{100 + 2.5}$

$0.70$

$V_B = 10 \times \dfrac{7.5}{100 + 7.5}$

Reasoning and calculation

$0.10\ \text{V}$ and $0.24\ \text{V}$ are below $0.5\ \text{V}$: cut-off.

$0.70\ \text{V}$ is between $0.5\ \text{V}$ and $0.8\ \text{V}$: the active region.

So the values are $5\ \text{k}\Omega$ for cut-off and $7.5\ \text{k}\Omega$ for the active region. That is option (3).

Checks

$10 * r / (100 + r)$

Does it make sense?

A bigger $R$ gets a bigger share of the $10\ \text{V}$, so $V_B$ is bigger. $5\ \text{k}\Omega$ gives $0.48\ \text{V}$ and $7.5\ \text{k}\Omega$ gives $0.70\ \text{V}$, which go in the order cut-off, then active.

The book gives these voltages for a silicon transistor. A germanium transistor changes at lower voltages.

Conclusion

$5\ \text{k}\Omega,\ 7.5\ \text{k}\Omega$

$R = 5\ \text{k}\Omega$ gives cut-off and $R = 7.5\ \text{k}\Omega$ gives the active region.

The answer is (3).

Find the base voltage from the potential divider. The emitter is at $0\ \text{V}$, so $V_{BE} = V_B$.

For a silicon transistor the book gives cut-off below $0.5\ \text{V}$ and saturation above $0.8\ \text{V}$. Between them is the active region.

Exam tip