filter capacitor sizing calcs

As a rule of thumb, I use 1000 to 2000 uF per amp. But that was for a nominal 12 VDC supply.

Another way to look at it is using the RC time constant, which allows the voltage to drop to 63% of peak. Also, you must account for the entire time the sine wave is below peak, so use 1/120 Sec. So for 4 ohms,

C = (1/120)/4 = ~2000 uF. Three time constants are needed for 5% drop, so that would be ~6000 uF.

For a 12 ohm load, you get 690 uF. So three times that is 2000 uF, for 1 amp at 12 volts.

For an amplifier power supply, the load is not quite the same as the output load. It will depend on the efficiency and the frequency you are amplifying.

You should do an LTSpice simulation and check the voltage waveform as well as the RMS current from the power supply (transformer) and peak current through the diodes. You might be surprised at what you find. Also check the RMS current in the capacitors (with real values for source impedance and capacitor ESR), and make sure they will handle the ripple current. And check for overall power dissipation to see what sort of efficiency you get. Switching supplies start looking pretty good.

Paul

You will want to know what the peak current draw is=20 for the amplifier. The speaker impedance (4 ohms)=20 is not really relevant to the calculation. The load on the power supply is the amplifier.

You lost a set of parentheses. You should have (1/240) / (4* ln(Vo / 0.95 Vo)) = 0.02 F, or 20 mF, or 20,000 uF.

But the time should be 1/120, or 8.3 ms, since worst case, and desiring only a small drop, you have to wait almost from one peak to the next, so that would be 40,000 uF. Also, note that the 4 ohm rating for a speaker is an impedance and may not be that high for the lower frequencies. Another way to look at it is to figure out your peak current and maximum allowable voltage drop. Since you don't want much percentage voltage drop, you can use a linear approximation instead of an exponential. Then, Q = C * V --> C = Amps * seconds / volts. If you have a supply of 30 V, and you want no more than 5% drop, or 1.5 V, and you have 7.5 A (about 4 ohm load), full-wave, you have C = 7.5 A * 8.3 ms / 1.5 V = .

-- John

Absolutely right! My goodness, I'm losing the ability to do basic algebra too... I messed up right after ln(Vcap/Vo) = -t/RC.

Thanks.

Michael

You merely calculate the drop in voltage in the period when the cap is NOT being charged by the transformer.

So, setting " dt " at 6 mS and "dv" at 2 volt and I at 4 amps average.

C = 4 x 0.006 / 2

= 0.012 F

= 12,000 uF

...... Phil

The formula is C=3DIT/E where I is the current, T is the time, and E is drop in voltage, and C is the capacitor. If it's 60Hz, full wave rectified, the time is 8.33 mS. If the voltage is 1 volt and the load is 4 ohms, the current will be 250mA. and a 5% voltage drop will be

-Bill

I was a bit wrong on this. According to an LTSpice simulation, my 6000 uF capacitor would have a voltage dropoff of about 20%. Just about 30,000 uF gives a 5% drop in voltage, using a 12 VAC source and a 4 ohm load. The peak voltage is 15.57, and RMS output is 15.2 volts. The load current is

With Schottky diodes, there is 61.5 watts input and 57.8 watts output, for

Paul

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The formula is C=IT/E where I is the current, T is the time, and E is drop in voltage, and C is the capacitor. If it's 60Hz, full wave rectified, the time is 8.33 mS. If the voltage is 1 volt and the load is 4 ohms, the current will be 250mA. and a 5% voltage drop will be

-Bill

==========================

I did an LTSpice simulation and found that 30,000 uF was just about right. The time is actually a little less than 0.0083 seconds because the discharge time starts when the peak of the AC waveform drops below the actual peak output. There is about 1.5 mSec charging time and 6.8 mSec discharge time, resulting in a calculated capacitance of 34,000. It is quite high, but really you don't normally need to hold within 5% of the peak. My capacitor value of 6000 uF, based on 3 TCs with the 4 ohm resistor, gives about 20% P-P ripple, which is usually good enough. It's probably better to add an inductor if you want to get better ripple without huge capacitors and high peak diode currents.

But I still say a switcher is the way to go for anything over about 10 watts, or if you are concerned about efficiency. Don't allow your power supply to contribute to Global Warmongering!

Paul

Several.

The time isn't 8.33 ms or even 10ms where 50Hz current is the norm. It'll be around 80-75% of that figure, since for a good 20-25% of the half cycle, the cap is actually being charged.

Also, your calculation assumes that the 4 ohms is driven straight from the supply. In audio, the polarity is constantly reversing, so power is drawn from BOTH supplies, so you can halve the current for starters. Also, with a sinewave signal, the current drawn will be reduced also.

Graham

"Paul E. Schoen Monumental F****it "

SMPS are far less reliable and generally LESS efficient than transformer/ rectifier PSUs.

...... Phil

Actually, the speaker is never driven from both supplies at once. (If both positive and negative output devices were on at the same time, they'd be shorting both supplies together!) So you don't have half the current, you have the same (peak) current but it is drawn from alternate supplies on alternate half-cycles of the signal.

Best regards,

Bob Masta DAQARTA v3.50 Data AcQuisition And Real-Time Analysis

be

he

Yes, it's 72 degrees where the voltage falls 5% so the discharge time would be 80 percent (72/90=3D 0.8) of the cycle, or 6.66 mS since the frequency is 120 Hz. So the new figure would be C=3D(.25 * .00666) / .05 =3D 33,000uF. But the fact the amplifier uses both positive and negative supplies doesn't change that much since a low 100Hz bass beat will be

-Bill

Well ... if you insist on doing a worst case analysis ... ! ;~)

You also forgot that a sinewave won't demand as much current as a square wave though.

I also reckon the 5% ripple is unrealistic. 10% is a more sensible number to target. Put all that into the equation and 10,000 uF is more than adequate.

Graham

Yes, and with realistic values of transformer resistance, capacitor ESR, and diode characteristics, you really can't get much better than that. You need almost ideal components to get enough charge during that sine wave peak. Even 15-20% is OK, and you can use reasonable size capacitors.

If load is fairly constant, an inductor can really help, but the output voltage will be closer to the RMS value of the AC input under load, and can become unstable (with oscillations) and poorly regulated with light loads.

Paul