One or two questions are expected from diodes in GATE.The article provides tips to solve questions on application of Diodes.Learn how to solve these questions.
A diode is forward when anode is more +ve than cathode, otherwise it is reverse biased.
If I > 0 => V> 0 [short circuit , forward biased ]
V < 0 => I =0 [open circuit ,Reverse biased]
Tips to Remember :
Check whether the diode is forward biased or not.
I= (5-3 )/2 =1 A (forward biased)
I = (-2)- (-5) =3/5 A (forward biased)
Also read : Tips for solving problems on zener diode
Reduce the circuit using thevenin’s rule where possible.The basic procedure fo Thevenin’s Theorem is remove the load resistor RL or component concerned . Find Resultant resistance by shorting all voltage sources or by open circuiting all the current sources. Find equalent source voltage by usual circuit analysis methods. Find the current flowing through the load Resistor.
Replace all diodes with short ckt. Take a current I from anode to cathode at any test diode. Find I .If current is +ve , then the diode is forward biased.
At diode D1,
( 0-(-10))/10 + (0-10)/5 =-1A, hence D1 is reverse biased.
At diode D2,
(V-(-10))/10 + (V-10)/5 =0
V= 3.3V ,Hence D2 is forward biased
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Step 1 : To solve clamper circuit , find the capacitor voltage Vc in its steady state.
Step 2 : Replace the diode with open circuit and draw the output form.
Capacitor always charges to its maximum open circuited voltage in its steady state.
Questions From GATE :
The diode in the circuit shown, if Von = 0.7 volts but is ideal otherwise. If Vi = 5 sin(ωt) volts, the minimum and maximum values of Vo (in volts) are, respectively,
- 5 and 2.7
- 2.7 and 5
- -5 and 3.85
- 1.3 and 5
Sinwt value ranges from -1 to 1,hence Vmin = -5V ,Vmax = +5V
Vo minimum will be there when Vi is minimum, means we are applying -5V so it is reverse biased.
As it is reverse biased ,no current is flowing through resister,so no voltage drop at resister,Hence Vo miniumum is -5V.
For V0 maximum
V0 maximum = 2+0.7 + I*1K
= 2.7 + I * 1K
= 2.7 + (5-2.7)/2
=2.7 + 1.15 =3.85
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In the figure, assume that the forward voltage drops to the PN diode D1 and Schottky diode D2 are 0.7 volts and 0.3 volts respectively. If ON denotes conducting state of the diode and OFF denotes the non conducting state of the diode, then in the circuit,
a Both are ON
b. D1 is ON and D2 is OFF
c. Both are OFF
d. D1 is OFF and D2 is ON
We assume D2 is on and D1 is off, as D2 need less cut-in voltage to get forward biased.If D1 is at off all voltage is given to D2 so it is in On State.
Voltage at(D1) = I ( 20 ohm) +0.3V
= ((10- 0.3V)/1020 )*20 + 0.3V = 0.5V
It is less than cut-in voltage ,so it is in off state.
Best Wishes !!