PSU Ripple Control/Filtering

[ikrel]

I have  a Transformer with a rectified output of 35V, with my oscilloscope i can measure a 3V@120Hz ripple. I know I can just put a cap across the output but I would actually like to know how I can do this on paper, without having to rely on just watching my scope until the ripple goes away as i swap capacitors. I've read about filters but the concept still escapes me, could anyone enlighten me on this?

[amspire]

Yes, it is pretty simple. First you have to see if you have a half wave rectifier (50/60Hz) or full wave (100/120Hz).

You then start with a very simple assumption - that the capacitor will charge only at the very peak of the waveform and will be discharging for the rest of the cycle.

For a full wave rectifier, this means that the capacitor will be discharging for 1/120 = 8.3333mS.

Next you need to know the load current you need - the voltage does not matter much.  So say the maximum load is 0.5A.

To get a voltage drop Vdrop at a current Imax, the formula will be C = Tperiod *Imax  / Vdrop. ( from I = C * dV/dT)

So for a 3V drop, that means C = 0.0083333 * 0.5/(  3) =  1400uF

Now in practice, the actual drop will be less then this as the resistance of the transformer winding will extend the period of conduction to perhaps 25% of the cycle In this case, you could get away with a capacitor that is 3/4 of the calculated capacitor, but seeing the tolerance of electrolytic capacitors is about 20% at best, it is not a bad ide to just use the calculated value.

The other aspect is that electrolytic capacitors can degrade with age, so it is not a bad idea giving yourself  an extra margin .

Richard.

[ikrel]

Ok, I'm get most of that, why 1/120. how exactly do you get that from "the capacitor will charge at the peak of the waveform"and how would that differ from say the capacitor charging during the top 5% of the waveform? (I would just like a reference to follow your line of thought)

Second I only have up too 1000uf caps, and only 1 or 2 of them, how can i get the ripple under control without such a large capacitor.

[TriodeTiger]

Ok, I'm get most of that, why 1/120

1Hz = 1 cycle per second, the mains cycle swings 60 (or 50, where it is used) times per second from 0 -> (+) -> 0 -> (-) -> 0:


Rectified = 120Hz as there are only positive peaks which are the same.


The reciprocal of 120 (1/120) is the time period, just as the reciprocal of the time period is the frequency.

Second I only have up too 1000uf caps, and only 1 or 2 of them, how can i get the ripple under control without such a large capacitor.

You can put them in parallel (which also lowers ESR), however I will let others explain what you can do better as I am learning PSUs myself

[amspire]

Ok, I'm get most of that, why 1/120.

  120 = 2 * 60Hz. on every mains cycles there are two peaks.

how exactly do you get that from "the capacitor will charge at the peak of the waveform"and how would that differ from say the capacitor charging during the top 5% of the waveform? (I would just like a reference to follow your line of thought)

you could use the top 5%, and it will make the equation slightly more complex and give you a solution lower by 5%. Since the period is almost totally dependant on the transformer specs, I just go for the theoretical worse case and assume that the charging time is negligable. You can never go wrong with a worse case value.

Second I only have up too 1000uf caps, and only 1 or 2 of them, how can i get the ripple under control without such a large capacitor.

You may be able to. It all comes down to the load current. What is the load current? What amount of peak-to-peak ripple is OK?

The other factor that has to be looked at is the RMS ripple current in the capacitors - you cannot exceed the capacitors RMS ripple specification. For 1000uF electrolytic, it could be anywhere between about 500mA RMS ripple current for a cheap 6.3VW cap to 2A or more for quality low ESR 35VW or 50VW cap. As an approximation at 1A output current, the RMS ripple is probably about 0.5 A RMS.

Richard

[amspire]

If you do not know the load current, the quick "Cheating" way is to stick a big capacitor in and measure the peak to peak ripple. The amount of ripple will be approximately inversely proportional to the capacitance value. So if attach 2000uF and get 1.6V p-p ripple, then for 3V ripple, you want 2000uF * 1.6/3 = 1067uF.

For this to work, the initial capacitor has to be big enough to allow the circuit to work properly, so you cannot start off too small.

Richard.

[Mechatrommer]

it depends on your load. with ideal 0ohm load, there's no capacitor can help you (or ideally inf.Farad). so my best bet is the oscilloscope, ampere meter and some load. I = C * dV/dT as amspire gave is the ideal formula, and not to forget the transformer limit as he mentioned. thanks for the reminder.

[electroguy]

Second I only have up too 1000uf caps, and only 1 or 2 of them, how can i get the ripple under control without such a large capacitor.

define "under control" please? 1V? 0.1V? 0.01V?

you could put a regulator after the transformer and caps to filter the output to get a regulated voltage. What voltage and current do you need? Don't forget to check the data sheet of your regulator, it may have a large dropout voltage, so you might need get a 40V transformer input to get regulated 35V output.
If this is for audio amp power supply, then probably forget the regulators and just go buy some more caps and put them in parallel (or get bigger caps). Depends on your design and what you want to achieve?

[SeanB]

Do not forget that as the capacitor value is increased the rectifier has to be a higher current unit. If you are drawing say 1A, and have say 22000uf of capacitor on the output then you will fry a 1A bridge rectifier in short order. This is because the charge current is only flowing from the transformer to charge the capacitor for a very short period, and this needs to be within the repetitive pulse rating of the diodes you use.  For most power supplies a bridge is going to be a lot bigger than it's DC ratings, it is common to have a 8A or a 16A unit on a power supply that delivers 3A, as the bigger units have a lower voltage drop at peak current. Anything over 1A will need a heatsink as well.

[T4P]

Do not forget that as the capacitor value is increased the rectifier has to be a higher current unit. If you are drawing say 1A, and have say 22000uf of capacitor on the output then you will fry a 1A bridge rectifier in short order. This is because the charge current is only flowing from the transformer to charge the capacitor for a very short period, and this needs to be within the repetitive pulse rating of the diodes you use.  For most power supplies a bridge is going to be a lot bigger than it's DC ratings, it is common to have a 8A or a 16A unit on a power supply that delivers 3A, as the bigger units have a lower voltage drop at peak current. Anything over 1A will need a heatsink as well.

Unless it comes in a really big package without any tabs ...

[SeanB]

big enough package is heat sink enough for some, I have some very old selenium bridges that have an integrated heatsink.

[Jad.z]

Here is how I do it.


V_ripple = I_load/(f ×C)

Where:
               V_ripple     Ripple voltage that will be superimposed on the DC voltage. (Volt)
               I_load       Maximum load current. (Amp)
               f           For full-wave rectifier f = 2x the mains frequency. (Hz)
                           For half-wave rectifier f = the mains frequency. (Hz)
              C           Smoothing capacitor value. (Farad)

[ikrel]

Math is not my strong point (I'm working on it) could someone re-arrange that equation V=I/FC to find C I've tried for the past hour and can't figure it out yet. thanks...back to google and Kahn I go:)

[DJKA]

I believe it would be:

C = I_load / (f x V_ripple)

Math is not my strong point (I'm working on it) could someone re-arrange that equation V=I/FC to find C I've tried for the past hour and can't figure it out yet. thanks...back to google and Kahn I go:)