Ac to DC

[jdahl]

Hey,

When converting form AC to DC through a rectifier-bridge, is't nessasary to use a cap to reduse the ripple, right, I'm wonder how big does it have to be? are there a formula? its up to the load right? And do I also need some filters to remove unwanted noise?

And the dc voltage is the AC peak?! so 24VAC is about 31VDC?

johannes

 

[k7elp60]

Hey,

When converting form AC to DC through a rectifier-bridge, is't nessasary to use a cap to reduse the ripple, right, I'm wonder how big does it have to be? are there a formula? its up to the load right? And do I also need some filters to remove unwanted noise?

And the dc voltage is the AC peak?! so 24VAC is about 31VDC?

johannes

Yes what you say about the output voltage of the bridge is pretty close when considering the voltage drop of the bridge rectifier. To calculate the value of the filter capacitor need, you must have load current,and the maximum ripple voltage. This formula works pretty well:
Vripple= Iload/(2FC).
Where Iload= load current
F= mains frequency in hz
C= capacitance in Farads.

 

[Roff]

Yes what you say about the output voltage of the bridge is pretty close when considering the voltage drop of the bridge rectifier. To calculate the value of the filter capacitor need, you must have load current,and the maximum ripple voltage. This formula works pretty well:
Vripple= Iload/(2FC).
Where Iload= load current
F= ripple frequency in hz(with a full wave rectifier=twice mains frequency)
C= capacitance in Farads.

F is the mains frequency, not the ripple frequency. You have already accounted for the frequency doubling with the '2' in the equation. This gives you a very conservative answer. The ripple will always be less than the equation predicts.

 

[k7elp60]

F is the mains frequency, not the ripple frequency. You have already accounted for the frequency doubling with the '2' in the equation. This gives you a very conservative answer. The ripple will always be less than the equation predicts.

Roff is correct, I have edited my post. I have used a rule of thumb of 3000uF/A of load, and recently found this formula and found it interesting.
I would assume that this formula is for a full wave rectifier.

 

[Speakerguy]

Simulators are better than a formula. Essentially because they are *lots* of formulas.

https://www.duncanamps.com/psud2/index.html

Best power supply designer I've found.

 

[Roff]

Roff is correct, I have edited my post. I have used a rule of thumb of 3000uF/A of load, and recently found this formula and found it interesting.
I would assume that this formula is for a full wave rectifier.

I'll try to explain the equation, and its shortcomings:
The capacitor charges through the rectifier until the sine wave peak passes. Then the diode becomes reverse-biased, and ceases to conduct. The capacitor then starts to discharge (assuming a constant current load) at the rate of v=I*t/C. At some point during the discharge, the voltage on the anode of the other diode in the rectifier becomes more positive than the load voltage, causing the cap to begin to charge again. If the discharge had proceeded for a complete half cycle (that's the "1/2F" term in the equation), the equation would be exact. However, as I said, the discharge ramp is terminated when the rectifier gets forward biased. As the ripple gets larger, this happens earlier in the half-cycle, causing the calculated error to be greater as the ripple increases.
Did that make sense?

 

[k7elp60]

I'll try to explain the equation, and its shortcomings:
The capacitor charges through the rectifier until the sine wave peak passes. Then the diode becomes reverse-biased, and ceases to conduct. The capacitor then starts to discharge (assuming a constant current load) at the rate of v=I*t/C. At some point during the discharge, the voltage on the anode of the other diode in the rectifier becomes more positive than the load voltage, causing the cap to begin to charge again. If the discharge had proceeded for a complete half cycle (that's the "1/2F" term in the equation), the equation would be exact. However, as I said, the discharge ramp is terminated when the rectifier gets forward biased. As the ripple gets larger, this happens earlier in the half-cycle, causing the calculated error to be greater as the ripple increases.
Did that make sense?

Thanks Roff, I understand what you are saying.

 

[Hero999]

Hey,

When converting form AC to DC through a rectifier-bridge, is't nessasary to use a cap to reduse the ripple, right, I'm wonder how big does it have to be? are there a formula? its up to the load right? And do I also need some filters to remove unwanted noise?

And the dc voltage is the AC peak?! so 24VAC is about 31VDC?

johannes

It depends on what you're powering.

A DC motor doesn't need a filtering capacitor.

A regulator does and it needs to be large enough to ensure the voltage does not drop below the lowest input voltage required for it to regulate properly.

 

[quixotron]

I thought you were referring to the rock band.