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

AI Thread Summary
To design a 12V 2A unregulated power supply, the transformer should have a VA rating of at least 24VA, with a peak secondary voltage calculated at 17V. The smoothing capacitor value is determined to be approximately 24,540 micro-farads using the formula C = I/(2f*Vr), where the ripple voltage is calculated as 0.288V peak-to-peak. The bridge rectifier must handle a peak forward current of about 4.83A, considering a 2V drop across it, leading to a power dissipation of 4 watts. The calculations presented appear to be correct, confirming the methodology used for determining the component ratings.
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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