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 2 of 2.
Bhanu said:
10 years ago
How we assume voltage 0.7 because there is no mentioned si (or) germanium.
Ram said:
1 decade ago
Yes if nothing is mentioned we need to assume it is a silicon diode.
Dipenti Kumari said:
5 years ago
We know that;
Pd = vd.Id.
= 0.7v * 40mA.
= 28mW.
Pd = vd.Id.
= 0.7v * 40mA.
= 28mW.
(2)
Hemalatha k said:
1 decade ago
Voltage drop of diode is 0.7. So it is correct.
Arindam said:
1 decade ago
Vd * Id = (0.7 * .040)W = .028 W = 28 mW
PRASANNA said:
1 decade ago
But they have not stated its si diode.
Prabhu said:
1 decade ago
But here not specific silicon diode.
Sharvi said:
1 decade ago
P=V*I
= 0.7 * 40mA
= 28 mA.
= 0.7 * 40mA
= 28 mA.
Rahul said:
1 decade ago
By default assume it si diode.
Jey said:
1 decade ago
0.7 voltage is default.
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