Electronics - Diodes and Applications - Discussion

Discussion Forum : Diodes and Applications - General Questions (Q.No. 4)
What is wrong with this diode?

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Answer: Option
No answer description is available. Let's discuss.
16 comments Page 1 of 2.

Jyoti Kushwaha said:   1 decade ago
Diode is in revers biased and hence reflecting the resistance in Megohm. It is normal behavior of a diode, nothing is wrong with Diode. So answer should be [C] nothing.

Arvind kumar said:   1 decade ago
Diode is in reverse biased so it provide higher resistance,so it is normal behavior of diode,so nothing wrong with diode so answer is [c]

Pawan said:   1 decade ago
Since the negative terminal connected to p side the diode must be open circuited.

Puneet Thakral said:   1 decade ago
Answer [c] is correct. But actual reason behind this we are using multimeter not a voltage source or excitation source so neither it is in the state of forward bias nor reverse bias it acting as a normal component in which we don't apply any input.

So answer[c] is correct.

Pavani said:   1 decade ago
Puneet thakral explaination is correct.

Vshwas choudhary said:   1 decade ago
DIODE neither reverse bias nor forward bias. The reason behind is no source (voltage or current) connected. So diode act as a normal component.

So answer "C" is correct.

SJCEian said:   1 decade ago
Diode is reverse biased. So, there might be no fault with the diode.

Akanksha rao said:   1 decade ago
Option c is correct because here we are not using the voltage or current devices so diode is not reverse bias as well as not in forward biased so there is nothing wrong with diode.

Trishul said:   1 decade ago
In a multimeter we can check resistance of any component. To find resistance it applies some excitation (voltage) across component, then some current flows.

Therefore it calculate R=V/I. In here diode n terminal will have more +ve than p terminal. Therefore reverse biased. Some reverse current(Io) will flows (not open). Resistance =v/Io.

Ravina jain said:   10 years ago
Answer C is correct because diode without any input acts as a normal component and it is in reverse bias according to connection as shown above so reverse bias gives high resistance.

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