Unregulated power supply

In summary, 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, the transformer VA rating should exceed 24VA, the peak secondary voltage will be 17V, the value of the capacitor should be 24,540 micro-farads, and the bridge rectifier should be able to handle a peak forward current of 4.83 amps with a power dissipation of 4 watts. The output voltage for the DC will be 15 volts, and the methodology for calculating the capacitor value is using C= I/2f*Vr.
  • #1
BigMan52
15
0

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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  • #2
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.
 
  • #3
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
 
  • #4
Is that the correct methodology?
 
  • #5
BigMan52 said:
Is that the correct methodology?
Looks correct to me.
 

1. What is an unregulated power supply?

An unregulated power supply is a type of power supply that does not have any internal circuitry to maintain a constant output voltage. It simply converts the incoming AC voltage into DC voltage, but the output voltage may vary depending on the load and input voltage.

2. How does an unregulated power supply work?

An unregulated power supply uses a transformer to convert the AC voltage into a lower AC voltage. This is then rectified and filtered to produce a DC voltage. However, since there is no regulating circuitry, the output voltage may fluctuate depending on the load and input voltage.

3. What are the advantages of using an unregulated power supply?

An unregulated power supply is relatively simple and inexpensive to build compared to a regulated power supply. It is also more efficient as there is no regulating circuitry that consumes power. Additionally, it can handle a wide range of input voltages.

4. What are the disadvantages of using an unregulated power supply?

The main disadvantage of an unregulated power supply is that the output voltage can vary significantly depending on the load and input voltage. This can cause problems for sensitive electronic devices that require a stable voltage. It also does not provide any protection against overloading or short circuits.

5. How can I improve the performance of an unregulated power supply?

To improve the performance of an unregulated power supply, you can add a voltage regulator circuit to the output. This will help maintain a constant output voltage even with varying loads and input voltages. You can also use a larger transformer to reduce voltage fluctuations and add protective circuitry to prevent damage from overloading or short circuits.

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