Source: http://www.physicsplus.in/2011/02/
Timestamp: 2019-04-24 23:56:33+00:00

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(1) In the circuit shown, D1 is a silicon diode which has a voltage drop of 0.7 V while in full conduction under forward bias. The zener diode D2 has a breakdown voltage of 6.8 V. What is the current through the 450 Ω resistor?
You can use the voltage drop across a forward biased ordinary silicon diode as a reference voltage in voltage regulator circuits. In the circuit shown, a reverse biased zener diode (of breakdown voltage 6.8 V) and a forward biased silicon diode (of voltage drop 0.7 V) in series make a reference voltage of 7.5 V. The output voltage of the circuit is thus 7.5 V.
Since the input voltage to the regulator circuit is 12 V, the voltage drop across the 450 Ω resistor is 12 V – 7.5 V = 4.5 V.
The current through the 450 Ω resistor is (4.5 V)/(450 Ω) = 0.01 A = 10 mA.
(2) The circuit shown in the adjoining figure is the simplest shunt voltage regulator using a zener diode of breakdown voltage 6 V. What is the power dissipated in the zener diode?
Since the input voltage to the regulator circuit is 10 V and the regulated output voltage is 6 V, the voltage drop across the 40 Ω resistor is 10 V – 6 V = 4 V.
Therefore, the current through the 40 Ω resistor (current limiting resistor) is (4 V)/ 40 Ω = 0.1 A = 100 mA. This is the total current flowing into the parallel combination of the zener diode and the100 Ω load resistor.
The current through the load resistor of 100 Ω is (6 V)/(100 Ω) = 0.06 A = 60 mA.
Therefore, the current flowing through the zener diode is 100 mA – 60 mA = 40 mA.
Power dissipated in the zener diode is 6 V×40 mA = 240 mW.
The channel width is twice the highest modulating signal frequency and is therefore equal to 2×5000 Hz = 10000 Hz = 10 kHz.
[Remember that in the standard AM sound broadcast systems the channel band width allotted to a station is 10 kHz].
fc = 9 N1/2 where N is the maximum electron number density.

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