Electronic Devices - Diode Applications - Discussion

Discussion Forum : Diode Applications - General Questions (Q.No. 9)
9.
Determine the peak value of the output waveform.

25 V
15 V
–25 V
–15 V
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

Millize said:   4 years ago
@All.

Here assume that the diode is ideal hence no voltage drop of 0.7 V and the answer would be 15 V.

Also it is not in reverse bias since there is the voltage appearing across the anode is 20V-5V = 15 V it would be forward biased.
(1)

Nirali said:   5 years ago
In the ideal condition of the diode, we do need to 0.7 volt of the diode. that's why 15v is the answer.

Emengrrr said:   7 years ago
The diode is forward bias in the positive cycle but reverse in the negative cycle.

I think it must be 14.3V or if making the diode ideal, it will result to 15V.

Rajan said:   7 years ago
@Jamar.

But the input voltage is greater in magnitude than the battery. The diode will still be forward-biased.

Jamar said:   7 years ago
The negative side of the battery is on the positive side of the diode. That's why it is in reverse bias.

Blk said:   7 years ago
@Singh.

How is it in reverse bias?

Singh said:   10 years ago
Diode is in reverse bias. So diode voltage won't be taken into account.
(1)

Sihle said:   1 decade ago
But then how is that possible, cause we having Vbias in series with the diode, are not suppose to have an expression like 20-(Vbias-0.7V).

Dyab dabbeek said:   1 decade ago
I agree with Ashu because the diode has voltage 0.7v so that:

20-5 - 0.7 = 14.3v.

Ashu said:   1 decade ago
Sir, we need to consider diode vtg as well which is 0.7V constant. So the answer is 20-5-0.7 = 14.3V.


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