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

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Discussion Overview

The discussion revolves around the design and analysis of a full-wave rectifier circuit using two P-N junction diodes and a transformer with two secondary windings. Participants are tasked with explaining the circuit operation and calculating various current values based on a specified output voltage and resistive load.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Participants are asked to draw a circuit diagram and explain the operation of a full-wave rectifier using two P-N junction diodes.
  • One participant inquires about how to solve for peak current, mean current, and RMS current given a peak output voltage of 25V and a 1000W load.
  • Another participant suggests that a sketch of the schematic would be beneficial for clarity, emphasizing the importance of showing work before seeking help.
  • Calculations presented include peak current as Vpk/1000, RMS voltage as Vpk/sqrt(2), and mean voltage as Vrms multiplied by a factor of 0.9.
  • One participant corrects another's misunderstanding regarding the representation of peak voltage and current, clarifying that Vpk represents voltage, not current.
  • There is a suggestion to use Ohm's law (V/R=I) for calculating the peak current and other current values based on the known voltages.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the correctness of their calculations and interpretations of voltage and current relationships. There is no consensus on the accuracy of the results presented, and some participants challenge earlier claims without reaching a definitive resolution.

Contextual Notes

Participants have not provided sketches of their circuits, which may limit the clarity of their explanations. There are also unresolved issues regarding the correct interpretation of power and resistance values in the context of the calculations.

Who May Find This Useful

This discussion may be useful for students or individuals interested in electrical engineering concepts, particularly those studying rectifier circuits and related calculations.

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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