Electronic Devices - Semiconductor Diodes - Discussion
Discussion Forum : Semiconductor Diodes - General Questions (Q.No. 20)
20.
Calculate the power dissipation of a diode having ID = 40 mA.
Discussion:
20 comments Page 1 of 2.
Dipenti Kumari said:
5 years ago
We know that;
Pd = vd.Id.
= 0.7v * 40mA.
= 28mW.
Pd = vd.Id.
= 0.7v * 40mA.
= 28mW.
(2)
Niranjan Kumar said:
7 years ago
When not mention which material is used for diode, then Si should be assume.
So, power dissipation across the diode is Vd * Id =(0.7*0.040)w=0.028w = 28mw.
Where 0.7 is silicon cut in voltage.
So, power dissipation across the diode is Vd * Id =(0.7*0.040)w=0.028w = 28mw.
Where 0.7 is silicon cut in voltage.
(1)
M krishna naik said:
9 years ago
Silicon is the most commonly used diode hence we assume silicon if nothing is mentioned regarding diode name.
Vd * Id = (0.7 * 0.040)W = 0.028 W = 28 mW.
Vd * Id = (0.7 * 0.040)W = 0.028 W = 28 mW.
Tracy said:
9 years ago
The answer is A. Yeah, if it isn't stated, you have to use Silicon for able to get the 28 mW.
Rao said:
9 years ago
Since no specification of Diode, option D also could be the answer. If it's assumed Si diode then option A is right.
Bhanu said:
10 years ago
How we assume voltage 0.7 because there is no mentioned si (or) germanium.
Hemalatha k said:
1 decade ago
Voltage drop of diode is 0.7. So it is correct.
Rahul said:
1 decade ago
By default assume it si diode.
Tejas said:
1 decade ago
Si is the most commonly used diode hence we assume Si if nothing is mentioned regarding diode name.
Jasmine said:
1 decade ago
Voltage drop of a diode is 0.7V.
Power dissipated = V x I = 0.7V x 40mA = 28mW.
Power dissipated = V x I = 0.7V x 40mA = 28mW.
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