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.
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)
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.
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)
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.
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.
Shrikrishna said:
1 decade ago
Actually resistance is RL+2K=4k
Now voltage across load is 10-.7v=9.3v
so i=9.3/4
2.325
Now voltage across load is 10-.7v=9.3v
so i=9.3/4
2.325
Digvijay said:
1 decade ago
Answer is correct, 10V-0.7V(drop across diode)/4k(2k series with 2k)
Anurima das said:
1 decade ago
There are 3 resistors. Why the effective resistance is 4k?
Sumanth said:
1 decade ago
Answer should be 2.5 mA rather than 2.325 mA.
Sumanth said:
9 years ago
Answer should be 2.5 mA rather than 2.325 mA.
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