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Old 21st January 2004, 08:11 AM   #1
Rarkov is offline Rarkov  England
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Default Formula for CRC Filter

Hi,
Does anyone have a formula to calculate the R @ ?Watts in a CRC filter?

I am using a +-55V transformer, so I expect about 80V.

I will be using around 18000uF in total per channel.

I'd also like to know the final voltage.

Any ideas? Thanks.

Gaz
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Old 21st January 2004, 09:14 AM   #2
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Hi, is this in a class a, or class AB amp, as this will make a bit of difference to things, as what actualy matters here is the current flowing throught the resistor, not the voltage of the supply.

If it is a class a (or AB if you use the maximum currnet to find the maximum voltage drop and power dissipated) amp, you would simply use the following equation to work out the power dissiapted:

P = I^2 x R

So, for a 0.33ohm resistor, and a 4A bias current, it would give you:

P = 4*4*0.33 = 5.28W

To work out the voltage drop you can just youse good ol' V = IR, so in this case it would be 1.32V.
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Old 21st January 2004, 01:46 PM   #3
Rarkov is offline Rarkov  England
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Hi,
Sorry - should have given more info. It's for a Class T Tripath monobloc amplifier using a TA0104 module. The transformer is a 1kVA 2x55V.

P = I^2 * R = (9.09 * 9.09) *.33 = 27.26W

Vloss = 3V

Is this possible?

Thanks,
Gaz
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Old 21st January 2004, 02:17 PM   #4
UrSv is offline UrSv  Sweden
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For this I'd use only C and not CRC.
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Old 21st January 2004, 02:37 PM   #5
Rarkov is offline Rarkov  England
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OK then...

Looking at this PSU - I have a 55V secondary.

This will goto about 77V unloaded. Assume 78.5V with rectifier losses and maybe back to 77.5V with losses from the capacitors.

The maximum rated voltage for the EB-TA0104 module is +-75V when using it in bridged mode (as I want to mainly - but with the option of biamping if I want to).

Do you think that loading will pull this under 75V? Should do shouldn't it?

I've ordered the 100 x 1500uF @ 68V Caps to make up the power supply. I will be using 96 of them in four banks (two 24 banks per monoblock), using two in series.

That means (1500uF / 2) * 12 = 9000uF per rail and 18000uF per channel.

Thanks,
Gaz
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Old 21st January 2004, 02:52 PM   #6
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55 VAC would give about 78 VDC without taking into account the no load increase or line variations. Say that you have about 4 % load variation on that transformers (a BIG transformer) and 5 % line variation then that will give you 85 VDC without diode losses at no load. Your 68 V caps (odd voltage BTW) will perhaps feel a bit short of the task. If you then crank it and play something with lots of bass then sure the PSU will sag to maybe 70 VDC but that's just for the odd peaks. Mostly your PSU will be at 75-80 V which is too high IMHO.
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Old 21st January 2004, 03:45 PM   #7
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I think I would have to agree here, your line voltages will probably be too high (especialy for your caps, I wold not like to be around when you pluged it in as I'd probably get a fright when they started to go pop), and putting resistors into the supply won't reduce the voltages by much unless you intend to run at full power (and I doubt you would want to do that too often, especialy bridged - although I think you might run out of current before too long, as 9 amps is only good for just over 600W into 8 ohms, and about 300W into four which you would hardly need things to be bridged for). So, when it is running at low powers, you will still have your lines being a fair bit over the module ratting and way too far over the caps.
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Old 21st January 2004, 08:22 PM   #8
Rarkov is offline Rarkov  England
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Sorry - I did try and explain (not very well though!)

The 1500uF 68V caps will have two in series, actually making a 750uF 136V capacitor. There will be 12 pairs of caps per rail, making 9000uF @ 136V.

Even so, if I forget the bridged bit, they'll allow upto 90V on the EB-TA0104 boards.

Maybe I should just forget the bridged bit.

Thanks,
Gaz
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