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.

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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,

apply KCL,

               ( 0-(-10))/10  + (0-10)/5 =-1A, hence D1 is reverse biased.

At diode D2,

apply KCL,

                (V-(-10))/10 + (V-10)/5  =0

V= 3.3V ,Hence D2 is forward biased

 Also read : Best Books for GATE ECE by Toppers


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.


FREE Daily Quiz on Electronics for
GATE-2017 Preparation
FREE Daily Quiz on Electronics for GATE-2017 Preparation.
Start Quiz  

Questions From GATE :

Question :

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,

  1. 5 and 2.7
  2. 2.7 and 5
  3. -5 and 3.85
  4. 1.3 and 5


Solution :

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

Applying KCL,

= 2.7 + (5-2.7)/2

=2.7 + 1.15 =3.85

Hence C

Also read : Tips for solving questions on 2-Port Netwok

Question :

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



Solution :

 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.

Hence d.


 Best Wishes !!



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