How Do You Design a 12V 2A Unregulated Power Supply?

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Discussion Overview

The discussion focuses on designing a 12V, 2A unregulated power supply, specifically addressing the specifications for the transformer, bridge rectifier, and smoothing capacitor. Participants explore calculations related to voltage, current, and ripple factor, as well as the methodology for determining component ratings.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant calculates the transformer VA rating as exceeding 24VA based on the output voltage and current.
  • The peak secondary voltage is determined to be 17V, derived from the output voltage multiplied by the square root of 2.
  • The capacitor value is calculated using the formula C = I / (2f * Vr), resulting in approximately 24,540 micro-farads.
  • Another participant questions the output voltage, suggesting it may be around 15V DC after accounting for the rectifier voltage drop.
  • Clarification is sought on the capacitor calculation, particularly regarding the ripple voltage conversion.
  • Participants express uncertainty about the methodology used for the calculations, with one participant affirming that it looks correct.

Areas of Agreement / Disagreement

There is no consensus on the correctness of the calculations or methodology, as some participants express uncertainty and seek clarification on specific points.

Contextual Notes

Participants have not fully resolved the assumptions regarding ripple voltage calculations and the impact of the rectifier on output voltage. The discussion includes varying interpretations of the formulas and their application.

BigMan52
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Homework Statement


Specify ratings and values for the transformer, bridge rectifier and smoothing capacitor for a 230 V unregulated power supply with an output voltage of 12 V, output current of 2 A and a ripple factor of <0.024

Homework Equations


V * I
V*sqrt 2
C= I/2f*Vr
Id=(1+sqrt2)*IL

The Attempt at a Solution


The transformer VA rating should exceed 12*2 so 24VA
The peak secondary voltage will be 12 * sqrt 2 = 17V
The value of the capacitor will be: 2/2*50*0.815 = 24,540 micro-farads
Bridge rectifier will need to be able to handle the peak forward current of: (1+sqrt2)*2=4.83amps
Power dissipation is 2*2=4watts (assuming 2 volt drop across the bridge)
Have I approached this is the correct way?
 
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BigMan52 said:

Homework Statement


Specify ratings and values for the transformer, bridge rectifier and smoothing capacitor for a 230 V unregulated power supply with an output voltage of 12 V, output current of 2 A and a ripple factor of <0.024

Homework Equations


V * I
V*sqrt 2
C= I/2f*Vr
Id=(1+sqrt2)*IL

The Attempt at a Solution


The transformer VA rating should exceed 12*2 so 24VA
The peak secondary voltage will be 12 * sqrt 2 = 17V
The value of the capacitor will be: 2/2*50*0.815 = 24,540 micro-farads
Bridge rectifier will need to be able to handle the peak forward current of: (1+sqrt2)*2=4.83amps
Power dissipation is 2*2=4watts (assuming 2 volt drop across the bridge)
Have I approached this is the correct way?
It looks as if your PSU will give 15 volts DC out. 17 volts peak less 2 volts for the rectifier.
Also could you clarify your capacitor calculation - not sure about it.
 
OK thank you I thought that would be the output for the DC.
This is how I calculated the capacitor value:
Using C= I/2f*Vr:
I= 2 amps
f= 2 x 50
Vr = ripple voltage which I calculated by: 0.024 * 12 =0.288 V (rms) converted to peak to peak by 0.288*sqrt2*2 = 0.815 V
 
Is that the correct methodology?
 
BigMan52 said:
Is that the correct methodology?
Looks correct to me.
 

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