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P-N junction

  • Thread starter Solidsam
  • Start date
  • #1
23
0
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.
 

Answers and Replies

  • #2
1,097
2
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!
 
  • #3
23
0
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 dont have a scanner so cant post sketches
 
  • #4
1,097
2
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 dont have a scanner so cant post sketches
You have Vpk = 0.25A. Vpk represents a voltage not a current.

[tex]P = I_{pk}V_{pk}[/tex]

[tex]I_{pk} = \frac{P}{V_{pk}}[/tex]
 
  • #5
23
0
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?
 
  • #6
1,097
2
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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