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First filter cap - Big or Small?
First filter cap - Big or Small?
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Old 19th November 2010, 09:57 PM   #71
megajocke is offline megajocke  Sweden
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I guess it's most often not a problem because the high capacitance types you need to get sane ripple voltage usually have pretty high ripple current specs. But it doesn't hurt to check...
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Old 22nd August 2013, 04:00 AM   #72
Ranchu32 is offline Ranchu32  Australia
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Originally Posted by Sch3mat1c View Post
I don't know of any general purpose aluminum electrolytic capacitors that aren't rated for line frequency rectifier duty.

The only possible reason you'd need high-ripple caps would be if you used a ridiculous amount of input capacitance, which as noted, results in absurd RMS currents, and is an all-around bad idea anyway.

The "2000uF/amp" rule of thumb must be qualified by ripple frequency and supply voltage. After all, you don't want to use 2000uF at 1kV, 1A, that's a seriously dangerous amount of energy! The correct value is:
C = 2.88 * Amps / (Volts * Frequency)
(Interesting, the 2.88 is dimensionless. If you include the units, it's farad-volt-hertz per ampere.)
So at 1A, 12V, 120Hz, you get:
C = 2.88 * 1A / (12V * 120Hz) = 0.002F = 2000uF
So the number is right.

A 6L6 PP amp might want around 400V and 200mA, or
C = 2.88 * 0.2A / (400V * 120Hz) = 12uF
Which is on the small side. Tube rectifiers can use somewhat more, because internal resistance keeps power factor high, at the expense of terrible regulation..

A really big switching power supply might need 160VDC at 10A, or
C = 2.88 * 10A / (160V * 120Hz) = 1500uF

A big audio amp might need +/-60V at 8A, or
C = 2.88 * 8A / (60V * 120Hz) = 3200uF
That's per side.

A TV flyback supply might need 30kV at 1mA, 15.7kHz, or
C = 2.88 * 0.001A / (30kV * 15.7kHz) = 6pF
Of course, this is limited by diode capacitance, which is comparable, and by allowable ripple, which is much smaller than 10%. Also, power factor isn't an issue for flyback supplies, because the current is constant. Large capacitances can be used at will. Picture tubes are on the order of 500pF, IIRC.

This forumula produces values that to me seem completely contrary to the accepted rules of thumb. By my estimates, a 250W amp driving a 8R load would require ~70V rails, deliver 2.5Arms into the load and dissipate about ~1.7A (if biased into class A/B).

C = 2.88 * Amps / (Volts * Frequency)
C = 2.88 * (2.5A + 1.7A) / (70V * 50hz) // Australian 50hz mains
C = ~3500uF

How can 3500uF total filtering capacitance per rail possibly suffice in this case?

Last edited by Ranchu32; 22nd August 2013 at 04:02 AM.
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Old 22nd August 2013, 09:29 AM   #73
DF96 is offline DF96  England
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Yes. I don't know where the 2.88 comes from (it looks suspiciously exact!) but I think 10 would be a better figure - for 10% ripple. 20 would give 5% ripple. I would use peak signal current of 8A, and remember that a full-wave rectifier gives twice the ripple frequency.

Then C = 10 * 8A /( 70V * 100Hz ) = 11400uF
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Old 24th August 2013, 10:49 AM   #74
stormsonic is offline stormsonic  Slovenia
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Recently I have worked on similar problem, what should I put between rectifier and regulator. Spent a lot of time on scope, measuring ripple, currents, waveforms,....
Current draw 1A, only positive voltage needed, not too much voltage should be dropped before regulator. Arrived to DRLCRC solution.
Schottky diode - resistor - inductor - capacitor - resistor - capacitor
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Old 24th August 2013, 08:41 PM   #75
Jsixis is offline Jsixis  United States
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well this is proving to be extremely great reading.
Great debate, I have always wondered why I see small capacitors in commercial amps and huge (as in big plus many) caps in the DIY world.
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Old 25th August 2013, 01:01 AM   #76
Pano is offline Pano  United States
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First filter cap - Big or Small?
Originally Posted by DF96 View Post
Then C = 10 * 8A /( 70V * 100Hz ) = 11400uF
Hmmmm...... looks like I've been sizing my first cap about right, according to this formula.
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Old 25th August 2013, 11:16 AM   #77
stormsonic is offline stormsonic  Slovenia
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Originally Posted by DF96
Then C = 10 * 8A /( 70V * 100Hz ) = 11400uF
My load is 1A at 5V, then

C = 10 * 1A / (5V * 100Hz) = 20000uF

According to this formula, 11400uF will suffice for 8A @ 70V.
And 20000uF is needed for 1A @ 5V.
And only 1000uF for 1A @ 100V.

Somehow I don't trust this formula.

Last edited by stormsonic; 25th August 2013 at 11:31 AM.
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Old 25th August 2013, 03:26 PM   #78
AndrewT is offline AndrewT  Scotland
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as you reduce the current drawn the required capacitance also reduces.

As you reduce the acceptable voltage ripple the required capacitance increases.

You must understand the software to enable you to assess if it is predicting the correct answers.

It comes back to my previous advice.
You must be able to check your results to ensure they are plausible.
regards Andrew T.
Sent from my desktop computer using a keyboard
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