The following simple question appeared in Kerala Medical Entrance 2006 Test paper:A PN junction diode is connected to a battery of emf 5.5 V and external resistance 5.1 kΩ. The barrier potential in the diode is 0.4 V. The current in the circuit is
(a) 1.08 mA (b) 0.08 mA (c) 1 mA (d) 1 A (e) 2 mA
(a) 1.08 mA (b) 0.08 mA (c) 1 mA (d) 1 A (e) 2 mA
Out of 5.5 volts, 0.4 volt (barrier potential) is dropped across the forward biased diode and the remaining 5.1 volts appears across the 5.1 kΩ resistance. So, the current through the resistance is (5.1V/5.1kΩ), which is equal to 1mA [Option (c)].
The following MCQ on half wave rectifier also is simple:
A sinusoidal voltage of peak to peak value 310 V is connected in series with a diode and a load resistance R so that half wave rectification occurs. If the diode has negligible forward resistance and very high reverse resistance, the r.m.s. voltage across the load resistance is
(a) 310 V (b) 155 V (c) 219 V (d)109.5 V (e) 77.5 V
If Vm is the peak value (maximum value) of an alternating voltage, its r.m.s. value is Vm/√2. The r.m.s. value of full wave rectifier output also is Vm/√2. Therefore, the mean square value of the full wave rectifier output is Vm2/2. The mean square value of the half wave rectifier output is half of this, which is equal to Vm2/4. Therefore, r.m.s. value of half wave rectifier output voltage is Vm/2.
Note that the peak to peak value (310 V) is given in the question. The peak value Vm is 155 volts and hence the r.m.s. value is 155/2 = 77.5 V.
Here is a question which may confuse you and you may be tempted to pick out the wrong answer:
A common emitter low frequency amplifier has a collector supply voltage (VCC) of 9 V. The amplifier has input resistance 1kΩ and collector load resistance 5 kΩ. The transistor used has a common emitter current gain (β) of 100. What will be the peak to peak output signal voltage if an input peak to peak signal voltage of 50 mV from a low impedence source is applied to this amplifier?
(a) 250 V (b) 25 V (c) 18 V (d) 12.5 V (e) 9 V
The voltage gain (amplification) of the amplifier is, Av = β (RL/Ri) where β is the common emitter current gain, RL is the load resistance and Ri is the input resistance.Therefore, Av = 100×5/1 = 500.
( Note that questions on calculation of voltage gain are often seen in Medical and Engineering Entrance test papers. They are simple to solve at Higher Secondary and Plus Two levels, as we hve done above).
The important thing to remember (especially in the context of the present question) is that you will get the full gain only if the input voltage is small. If the input voltage were 5 mV peak to peak for instance, you would have obtained a peak to peak output voltage of 500 times the input voltage, which is 500×5 mV = 2500 mV = 2.5 V.
But since the input is 50 mV, you cannot obtain 500×50 mV (= 25000 mV = 25 V) for the simple reason that the collector supply voltage is 9 volts only. The maximum possible peak to peak signal voltage will be 9 volts only given in option (e).
[The transistor will swing between saturation and cut off as the signal voltage swings between the positive and negative peaks and you will get a clipped output signal which has a peak to peak value of approximately 9 volts in the present case].
The following MCQ on half wave rectifier also is simple:
A sinusoidal voltage of peak to peak value 310 V is connected in series with a diode and a load resistance R so that half wave rectification occurs. If the diode has negligible forward resistance and very high reverse resistance, the r.m.s. voltage across the load resistance is
(a) 310 V (b) 155 V (c) 219 V (d)109.5 V (e) 77.5 V
If Vm is the peak value (maximum value) of an alternating voltage, its r.m.s. value is Vm/√2. The r.m.s. value of full wave rectifier output also is Vm/√2. Therefore, the mean square value of the full wave rectifier output is Vm2/2. The mean square value of the half wave rectifier output is half of this, which is equal to Vm2/4. Therefore, r.m.s. value of half wave rectifier output voltage is Vm/2.
Note that the peak to peak value (310 V) is given in the question. The peak value Vm is 155 volts and hence the r.m.s. value is 155/2 = 77.5 V.
Here is a question which may confuse you and you may be tempted to pick out the wrong answer:
A common emitter low frequency amplifier has a collector supply voltage (VCC) of 9 V. The amplifier has input resistance 1kΩ and collector load resistance 5 kΩ. The transistor used has a common emitter current gain (β) of 100. What will be the peak to peak output signal voltage if an input peak to peak signal voltage of 50 mV from a low impedence source is applied to this amplifier?
(a) 250 V (b) 25 V (c) 18 V (d) 12.5 V (e) 9 V
The voltage gain (amplification) of the amplifier is, Av = β (RL/Ri) where β is the common emitter current gain, RL is the load resistance and Ri is the input resistance.Therefore, Av = 100×5/1 = 500.
( Note that questions on calculation of voltage gain are often seen in Medical and Engineering Entrance test papers. They are simple to solve at Higher Secondary and Plus Two levels, as we hve done above).
The important thing to remember (especially in the context of the present question) is that you will get the full gain only if the input voltage is small. If the input voltage were 5 mV peak to peak for instance, you would have obtained a peak to peak output voltage of 500 times the input voltage, which is 500×5 mV = 2500 mV = 2.5 V.
But since the input is 50 mV, you cannot obtain 500×50 mV (= 25000 mV = 25 V) for the simple reason that the collector supply voltage is 9 volts only. The maximum possible peak to peak signal voltage will be 9 volts only given in option (e).
[The transistor will swing between saturation and cut off as the signal voltage swings between the positive and negative peaks and you will get a clipped output signal which has a peak to peak value of approximately 9 volts in the present case].
You will find more multiple choice questions in this section at physicsplus: Multiple Choice Questions from Electronics
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