Electronics - Diodes and Applications - Discussion
Discussion Forum : Diodes and Applications - General Questions (Q.No. 33)
33.
What is the peak output voltage for this half-wave rectifier?
Discussion:
17 comments Page 1 of 2.
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.
Claude said:
4 years ago
@All.
If you draw the output waveform of the HW- RECTIFIER given an input of 8.485 V(since you don't really use the peak voltage value as an input but rather the rms value) the output WAVEFORM would have a peak of (8.485-0.7) = 7.8 V,
So, B would be correct.
If you draw the output waveform of the HW- RECTIFIER given an input of 8.485 V(since you don't really use the peak voltage value as an input but rather the rms value) the output WAVEFORM would have a peak of (8.485-0.7) = 7.8 V,
So, B would be correct.
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.
Vpk = 169.706Vrms / .707 = 240Vpk / 20 = 12Vpk - .7V = 11.3 Vpk.
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
=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
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
V2 - voltage at secondary
V1 - voltage at primary
N2 - no of turns of secondary coil
N1 - no of turns of primary coil
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.
Vs/Vp = Ns/Np,
Vs = 169.706*1/20.
Vs = 8.485v.
Peak output voltage = Vs-0.7 = 7.78v.
ABC said:
4 years ago
Vpeak across the resistor is 11.3V.
Vrms is 7.78V, since it says Vpeak, therefore the answer is not listed.
Vrms is 7.78V, since it says Vpeak, therefore the answer is not listed.
(2)
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.
Junior said:
7 years ago
V(sec)=n*V(pri).
V(sec)=1/20*169.706 ,
V (sec)=8.4853.
V(out)= V(sec) - 0.7.
=7.8v.
V(sec)=1/20*169.706 ,
V (sec)=8.4853.
V(out)= V(sec) - 0.7.
=7.8v.
MARKANDEPPA said:
1 decade ago
But they will not mention it's diode silicon or germanium, we should calculate as a ideal.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers