# Microelectronics Questions and Answers – Modes of a BJT

This set of Microelectronics Multiple Choice Questions & Answers (MCQs) focuses on “Modes of a BJT”.

1. A D.C. voltage of 800mV is given to the base of a npn transistor, the reverse saturation current 5*10-17A while the emitter is grounded. The collector is attached to a 1KΩ resistor which connects to a 10V power supply. If VCE(sat) is 200mV, choose the correct option.
a) The transistor is in inverse-active mode
b) The transistor is in saturation
c) The transistor is in active mode
d) The transistor is in cut-off

Explanation: The collector current is 1.15mA. VCE=10-1.15=8.85. The transistor is in the active mode since VBC is less than 0.

2. Check the mode of operation. Assume the collector current is 1mA.

a) Active
b) Saturation
c) Cut-off
d) Destroyed

Explanation: The transistor is deep into the active region. It has a collector voltage of 9.9V.

3. A D.C. voltage of 850mV is given to the base of a npn transistor, the reverse saturation current 5*10-17A while the emitter is grounded. The collector is attached to a 1KΩ resistor which connects to a 10V power supply. If VCE(sat) is 200mV, choose the maximum VBE to drive the device into saturation at room temperature.
a) 631mV
b) 636mV
c) 635mV
d) Data not given

Explanation: The maximum VBE is 634.71mV. The collector current at saturation is derived as VCE=IC * RC and hence, it results to 2μA. Thereafter, we use the relation VBE=VT*ln($$\frac {I_c}{I_s}$$) and we know VT is 26mV at room temperature.

4. The large signal model is mostly used when the signal strength is very large. How large does it need to be?
a) Sincerely larger than the thermal voltage
b) Larger than 700mV
c) Larger than 800mV
d) Larger than the threshold voltage

Explanation: The small signal model of the B.J.T. is applicable when the input voltage is much greater than the thermal voltage.

5. The threshold voltage of the diode is 0.5V while that of the transistor is 0.7V. For what value of V2 does Q1 reach the active region in the following case?

a) At 0.5V
b) At 0.7V
c) At 1.2V
d) Data not given

Explanation: Q1 turns on at 1.2V since that would be the minimum voltage to turn the transistor ON. T=he minimum voltage drop across D1 & Q1 is found out to be 1.2 V, for both the elements to conduct current, by using KVL.

6. The threshold voltage of the diode is 0.8V while that of the transistor is 0.7V. Look at the following circuit and a graph of the node voltages vs V2. Choose the perfect event justifying the graph.

a) The base emitter junction creates a bias drop
b) The diode creates a bias drop
c) The diode and the base emitter junction creates a bias drop
d) Cannot be determined

Explanation: The diode creates a voltage drop of about 0.47mV while the transistor creates a drop of about 0.8V. Thereafter, it results in a shift to the voltage source which is connected to the node n003 in the above circuit.

7. The threshold voltage of the diode is 0.7V while that of the transistor is 0.7V. Look at the following circuit and a graph of the diode current vs input voltage. Choose the perfect event justifying the graph.

a) The output resistance is 0
b) The transconductance is increased in the active region
c) The gain is 0
d) The diode is damaged

Explanation: The transconductance of the Q1 increases significantly in the active mode. The diode would get damaged but that isn’t justified by the graph.

8. Each transistor results in a current Ic=1mA. Calculate the collector resistance so that each transistor remains in saturation. Vcc is 5V while VCE(Sat)=0.2V.

a) 8K
b) 6K
c) 4.8K
d) 3K

Explanation: If the resistor is higher than 5k, the transistor goes into cut-off. If it’s greater than or equal to 4.8K, it’ll remain in saturation. If it’s less than 4.8K, the transistor gets driven into active mode since VCE becomes greater than VCE(Sat). This can be readily observed from the fact that VCC – I * R1 = VCE and this is constant for both transistors while I is the sum of the currents coming out form each transistor.

9. The pnp transistor is kept in saturation mode. Identify the conditions required for such a situation.
a) VBE < 0
b) VBE < 0 & VCB > 0
c) VBE < 0 & VCB < 0
d) VBE > 0

Explanation: For a pnp transistor, VCB is the voltage across an n-p region. To make it drive into saturation mode, we need to make VCB > 0 so that the region becomes forward biased.

10. The pnp transistor is kept in inverse-active mode. Identify the conditions required for such a situation.
a) VBE < 0 & VCB < 0
b) VBE < 0 & VCB > 0
c) VBE > 0 & VCB < 0
d) VBE > 0 & VCB > 0

Explanation: In the inverse active mode, the BC junction is forward biased while the BE junction is reverse biased. Only VBE > 0 & VCB > 0 helps to satisfy the condition.

11. In the inverse active mode, for an npn transistor, which carrier flows dominantly from the collector to the base?
a) Majority carrier electron
b) Majority carrier holes
c) Minority carrier electron
d) Minority carrier holes

Explanation: In the inverse active mode, the B-C junction is forward biased. Electrons are the minority carriers in the base while holes are the minority carrier in the collector for an npn transistor. More electrons travel from the collector to the bases which area majority charge carrier. The collector current is thus reversed than that in active mode.

12. The minority carrier concentration gradient, in the base region, remains in the same direction in saturation and active mode.
a) True
b) False

Explanation: The minority carrier concentration gradient in the base is same in active and saturation mode. This is because the transistor does allow a bit of current under saturation.

Sanfoundry Global Education & Learning Series – Microelectronics.

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