Electronic Devices - Diode Applications - Discussion
Discussion Forum : Diode Applications - General Questions (Q.No. 9)
9.
Determine the peak value of the output waveform.
Discussion:
13 comments Page 1 of 2.
Silambu BE said:
1 decade ago
It is an clipper circuit with
-5v i/p = 20-5 = 15
-5v i/p = 20-5 = 15
Sameer said:
1 decade ago
Since +ve signal applied to the anode of the diode and the magnitude is greater than the Vt,the diode is forward biased and conducts thus Vo=Vi-Vt=20-5=15V.
Padma said:
1 decade ago
It is clipper circuit so output voltage is Vo=Vi-Vt=20-5=15.
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.
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.
20-5 - 0.7 = 14.3v.
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).
Singh said:
10 years ago
Diode is in reverse bias. So diode voltage won't be taken into account.
(1)
Blk said:
7 years ago
@Singh.
How is it in reverse bias?
How is it in reverse bias?
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.
Rajan said:
7 years ago
@Jamar.
But the input voltage is greater in magnitude than the battery. The diode will still be forward-biased.
But the input voltage is greater in magnitude than the battery. The diode will still be forward-biased.
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