2019 A/L Physics Past Paper · Question 37 (English Medium)

A transistor driving an LED: R_B and R_C

For optimum operation of a light emitting diode (LED), forward voltage and current should be 2 V2\ \text{V} and 10 mA10\ \text{mA}, respectively. Transistor is having VBE=0.7 VV_{BE} = 0.7\ \text{V}, current gain β=100\beta = 100, and VCE(sat)=0.1 VV_{CE(sat)} = 0.1\ \text{V}. For the circuit shown in the figure, what are the values of RBR_B and RCR_C for the optimum operation of the LED?

  1. (1) RB=100R_B = 100 Ω, RC=1R_C = 1 kΩ
  2. (2) RB=1R_B = 1 kΩ, RC=1R_C = 1 kΩ
  3. (3) RB=1R_B = 1 kΩ, RC=290R_C = 290 Ω
  4. (4) RB=10R_B = 10 kΩ, RC=1R_C = 1 kΩ
  5. (5) RB=10R_B = 10 kΩ, RC=290R_C = 290 Ω
Answer and explanation

Answer: (5) RB=10R_B = 10 kΩ, RC=290R_C = 290 Ω

Given information

Given:

VCC=5V_{CC} = 5 V.

The LED: 2 V and 10 mA.

VBE=0.7V_{BE} = 0.7 V, β=100\beta = 100 and VCE(sat)=0.1V_{CE(sat)} = 0.1 V.

The divider at the base: 3.3 kΩ3.3\ \text{k}\Omega and 1.7 kΩ1.7\ \text{k}\Omega.

What the question asks

We want:

RBR_B and RCR_C so that the LED gets its best 2 V and 10 mA.

Relevant theory

When the transistor works as a switch it is in saturation, so VCE=VCE(sat)V_{CE} = V_{CE(sat)}.

The collector current is set by RCR_C.

At the edge of the active region IC=βIBI_C = \beta I_B. That gives the least base current we need.

The divider gives V=VCCR2R1+R2V = V_{CC}\dfrac{R_2}{R_1 + R_2}.

When the biasing potential is greater than 0.8 V, IB comes to a maximum bringing the transistor to its saturation.

Required equations

VCC=ICRC+VLED+VCE(sat)V_{CC} = I_CR_C + V_{LED} + V_{CE(sat)}

IB=ICβI_B = \frac{I_C}{\beta}

RB=Vdiv−VBEIBR_B = \frac{V_{div} - V_{BE}}{I_B}

Applying it to this question

The plan:

First the voltage of the divider.

Then RCR_C from the LED current.

Then the base current we need.

Last RBR_B.

Reasoning and calculation

1.71.7

Vdiv=5×1.73.3+1.7V_{div} = 5 \times \frac{1.7}{3.3 + 1.7}

290290

RC=5−2−0.10.010R_C = \frac{5 - 2 - 0.1}{0.010}

0.10.1

IB=10 mA100I_B = \frac{10\ \text{mA}}{100}

1010

RB=1.7−0.70.1×10−3R_B = \frac{1.7 - 0.7}{0.1 \times 10^{-3}}

Checks

Unit check:

VA=Ω\dfrac{\text{V}}{\text{A}} = \Omega

In RCR_C we have 2.9 V0.01 A\dfrac{2.9\ \text{V}}{0.01\ \text{A}}.

In RBR_B we have 1 V10−4 A\dfrac{1\ \text{V}}{10^{-4}\ \text{A}}.

Both give ohms.

A note:

The answer (5) treats the divider as a fixed 1.7 V source.

A real divider has a resistance too, 3.3×1.75=1.12 kΩ\dfrac{3.3 \times 1.7}{5} = 1.12\ \text{k}\Omega.

With it, RB=10 kΩR_B = 10\ \text{k}\Omega gives IB≈0.09I_B \approx 0.09 mA. That is a little below the saturation edge.

A smaller RBR_B such as 1 kΩ1\ \text{k}\Omega, option (3), also saturates the transistor firmly.

But the question wants the base current from β\beta. So the answer is (5).

Teacher's note

Does it make sense?

With RC=290 ΩR_C = 290\ \Omega the LED gets 10 mA.

With RB=10 kΩR_B = 10\ \text{k}\Omega we get βIB=10\beta I_B = 10 mA.

Both agree.

(5−2−0.1)/0.01(5 - 2 - 0.1) / 0.01

Conclusion

RB=10 kΩ, RC=290 ΩR_B = 10\ \text{k}\Omega,\ R_C = 290\ \Omega

RC=2.9 V10 mA=290 ΩR_C = \dfrac{2.9\ \text{V}}{10\ \text{mA}} = 290\ \Omega.

IB=ICβ=0.1I_B = \dfrac{I_C}{\beta} = 0.1 mA.

RB=1.7−0.70.1 mA=10 kΩR_B = \dfrac{1.7 - 0.7}{0.1\ \text{mA}} = 10\ \text{k}\Omega.

So the answer is option (5).

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