Electronic Devices - Diode Applications - Discussion
Discussion Forum : Diode Applications - General Questions (Q.No. 8)
8.
Determine the peak value of the current through the load resistor.
Discussion:
13 comments Page 1 of 2.
Sumanth said:
9 years ago
Answer should be 2.5 mA rather than 2.325 mA.
Gurpreet said:
9 years ago
When d1 is off then it acts as open. It means load resistance and other r1 resistance are in series. Thats why we take load resistance +2k.
Waheed ktmlk said:
10 years ago
Can anyone explain?
Divya sree said:
1 decade ago
How can get vt is 0.7?
Narsingh said:
1 decade ago
What is 0.7?
Anurima das said:
1 decade ago
There are 3 resistors. Why the effective resistance is 4k?
Junayd said:
1 decade ago
@Rajashekhar.
Answer is correct as 9.3v after the diode consider two branches one is having load resistor and the series resistor of 2k while the other branch has 2k only. So load resistor branch has 4k total. So 9.3/4k = 2.325mA.
Answer is correct as 9.3v after the diode consider two branches one is having load resistor and the series resistor of 2k while the other branch has 2k only. So load resistor branch has 4k total. So 9.3/4k = 2.325mA.
Rajashekhar said:
1 decade ago
@Every1 at +ve half cycle d1 is rev. biased.
10-.7 = 9.3.
9.3/4k = 2.325mv.
But we need only voltage across 2k hence.
2.325m*2 = 4.65.
Since they didn't mention the cut off voltage we have to take the diode as ideal if we repeat the above same process we get the answer 5v hence the ans must b.
10-.7 = 9.3.
9.3/4k = 2.325mv.
But we need only voltage across 2k hence.
2.325m*2 = 4.65.
Since they didn't mention the cut off voltage we have to take the diode as ideal if we repeat the above same process we get the answer 5v hence the ans must b.
(2)
Putta said:
1 decade ago
@Sumanth.
You are correct but here some voltage drop passing through diode ie)0.7v
10-0.7=9.3v.
9.3/4k=2.325 mA.
You are correct but here some voltage drop passing through diode ie)0.7v
10-0.7=9.3v.
9.3/4k=2.325 mA.
Sameer said:
1 decade ago
On +ve signal,diode1 is open and diode2 is forward biased and hence the op Volt vo=vi-vt
=10-0.7=9.3V
Reff=load resistance+2k
2k+2k=4k
Hence I=vo/reff=9.3/(4*10^3)=2.325mA.
=10-0.7=9.3V
Reff=load resistance+2k
2k+2k=4k
Hence I=vo/reff=9.3/(4*10^3)=2.325mA.
(1)
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