IndiaBIX

Electronics - Diodes and Applications - Discussion

Discussion Forum : Diodes and Applications - General Questions (Q.No. 33)
Diodes and Applications
33.
What is the peak output voltage for this half-wave rectifier?

1 V
7.8 V
10.9 V
15.6 V
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
17 comments Page 1 of 2.
Narendra Singh said:   1 decade ago
what is the formula?
Prashil & Vijay said:   1 decade ago
V2/V1 = N2/N1

V2 - voltage at secondary
V1 - voltage at primary
N2 - no of turns of secondary coil
N1 - no of turns of primary coil
Nabhdeep chauhan said:   1 decade ago
v2= n2/n1*v1
=1/20*169.706
so v2=8.4853 volts

so output voltage will be 8.4853-0.7
Vo=7.7853
so option 'B' is correct 7.8 volts
Prveen said:   1 decade ago
Is diode forward bias. ?
Alok kumar said:   1 decade ago
Diode is foreword biased practically no voltage drop across the diode but due to silicon or germanium diode 0.7v or 0.3v drop across the diode secondary voltage of the transformer is approximately 8.48 volts. But subtract the drop across the diode 0.7v =7.78 volts.
Ismail said:   1 decade ago
Vs/Vp=Ns/Np.

Vs=169.706*1/20.

Vs=8.485v.

Peak output voltage=Vs-0.7=7.78v.
Ralph said:   1 decade ago
Yes, but voltage is always given as RMS unless otherwise stated, so the correct answer is not listed.

Vpk = 169.706Vrms / .707 = 240Vpk / 20 = 12Vpk - .7V = 11.3 Vpk.
Mohamed said:   1 decade ago
What is the output characteristics ?
Adco said:   1 decade ago
I agree with @Ralph. It is not indicated that it is vdc. You will always assume the source to be in RMS.
Shah said:   1 decade ago
From wahab, comment from previous question.
Vs/Vp = Ns/Np,

Vs = 169.706*1/20.
Vs = 8.485v.

Peak output voltage = Vs-0.7 = 7.78v.
Post your comments here:

Your comments will be displayed after verification.

Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers