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.
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.
Bhiksham said:
1 decade ago
I think if something is not mentioned we have to take it as Si as default.
Ram said:
1 decade ago
Yes if nothing is mentioned we need to assume it is a silicon diode.
GUNA SEKAR said:
1 decade ago
Ya if nothing is mentioned we need to assume it is a silicon diode ram and bhiksham is correct answer.
Sharvi said:
1 decade ago
P=V*I
= 0.7 * 40mA
= 28 mA.
= 0.7 * 40mA
= 28 mA.
KIRAN said:
1 decade ago
P=V*I
= 0.7V*40mA
= 28mW.
Where 0.7V is the cut in voltage for the silicon diode.
No need to mention the name of the diode. we can take it for granted as Si is the best material for making electronic devices.
= 0.7V*40mA
= 28mW.
Where 0.7V is the cut in voltage for the silicon diode.
No need to mention the name of the diode. we can take it for granted as Si is the best material for making electronic devices.
Jey said:
1 decade ago
0.7 voltage is default.
Prabhu said:
1 decade ago
But here not specific silicon diode.
SURENDRA SONI said:
1 decade ago
If we calculate power dissipation for ge type diode then it will be 0.3*40 = 12 mW.
If we calculate power dissipation for si type diode then it will be 0.7*40 = 28 mW.
So answer is A.
If we calculate power dissipation for si type diode then it will be 0.7*40 = 28 mW.
So answer is A.
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