Full-Wave Rectifier w/ 2 P-N Junction Diodes & Transformer

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SUMMARY

This discussion focuses on the implementation of a full-wave rectifier using two P-N junction diodes and a transformer with two secondary windings. The output specifications include a peak voltage of 25V at 50 Hz applied to a 1000W resistive load. Key calculations discussed include determining the peak current (0.25A), mean current (0.016A), and RMS current (0.0177A) based on the rectifier's output voltage and load resistance.

PREREQUISITES
  • Understanding of P-N junction diodes and their function in rectification
  • Knowledge of transformer operation and secondary winding configurations
  • Familiarity with electrical current calculations (peak, mean, RMS)
  • Basic circuit diagram drawing skills
NEXT STEPS
  • Study the operation of full-wave rectifiers using various diode configurations
  • Learn about transformer specifications and their impact on rectifier performance
  • Explore detailed calculations for electrical parameters in AC circuits
  • Research waveform analysis techniques for rectified outputs
USEFUL FOR

Electrical engineering students, hobbyists designing power supplies, and professionals working with rectification circuits will benefit from this discussion.

Solidsam
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Show how two P-N junction diodes can be used in conjunction with a
transformer that has two secondary windings, to form a full-wave rectifier.
(Draw a circuit diagram including connections to the mains transformer).
Explain the operation of your circuit using sketches of waveforms. (3 marks)

(ii) If the output from your rectifier is 25V(PEAK) at 50 Hz and is applied to a 1000W
resistive load, find:

a) The peak current in the load

b) The mean current (3 marks)

c) The RMS current



how do i solve a,b and c? i know how to sketch the p-n junction.
 
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Solidsam said:
Show how two P-N junction diodes can be used in conjunction with a
transformer that has two secondary windings, to form a full-wave rectifier.
(Draw a circuit diagram including connections to the mains transformer).
Explain the operation of your circuit using sketches of waveforms. (3 marks)

(ii) If the output from your rectifier is 25V(PEAK) at 50 Hz and is applied to a 1000W
resistive load, find:

a) The peak current in the load

b) The mean current (3 marks)

c) The RMS current


how do i solve a,b and c? i know how to sketch the p-n junction.

Why don't you post us a sketch of your schematic? I realize you've said you understand that part, but a sketch of the circuit is always a good place to start.

You need to show us your work and a reasonable attempt at the solution before we can help!
 
1000w should be 1000ohms


Vpk=25/1000=0.25A

Vrms=25/sqroot2=17.68v Irms=17.68/1000=0.0177Arms

Vmean=Vrms*0.9=15.9V Imean=15.9/1000=0.016Amean


Are these result and equations correct? sry don't have a scanner so can't post sketches
 
Solidsam said:
1000w should be 1000ohms


Vpk=25/1000=0.25A

Vrms=25/sqroot2=17.68v Irms=17.68/1000=0.0177Arms

Vmean=Vrms*0.9=15.9V Imean=15.9/1000=0.016Amean


Are these result and equations correct? sry don't have a scanner so can't post sketches

You have Vpk = 0.25A. Vpk represents a voltage not a current.

P = I_{pk}V_{pk}

I_{pk} = \frac{P}{V_{pk}}
 
ok well since i have the mean voltage, rms voltage and peak voltage can i no just use V/R=I


so Vpk/1000ohms=Ipk


then do the same for the rest of them?
 
Solidsam said:
ok well since i have the mean voltage, rms voltage and peak voltage can i no just use V/R=I


so Vpk/1000ohms=Ipk


then do the same for the rest of them?


Oh you where pointing out a typo, not your actual reasoning? (This with regards to the 1000W or 1000 ohms)
 

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