Power Supply Rating / Fuse calculation / Understanding ?

[artquake]

Hi @ all

Initial Situation:
One Channel Amp consisting of two LM4780 in Parallel / Bridged Operation

Power Supply
Vpri = 230V
Vsek = 25VAC
Diode drop = 2 x 0.7V = 1.4 V
Vcc/Vee (+/-) = (25 - 1.4) x sqrt(2) = 33.4V

Operating range DC Output due fluctuation of Vpri:
Vcc/Vee @ -10% Vpri (207 V) = 29.8VDC (<42V)
Vcc/Vee @+10% Vpri (253 V) = 36.9VDC (<42V) à OK

Vpeak at Load = 30V @ 8 Ohm Load
Imax in Load = 3.75A per LM4780 and rail

Ipeak for power supply = 2 x 3.75 = 7.5 A for a pair of LM4780 respectively 2 x 7.5 A per Rail as it has to deliver current for – and +. So the max. required current demand from Transformer is total = 15A.

Transformer:
.. must then be dimensioned with:
P = Vpeak x Ipeak x 2 x 2 = 30V x 3.75A x 2 x 2 = 450 W à Toroid = 500W

Filtering Caps:
f=50Hz which equals to t=0.01 s
If I want VRipple of 1% @ Vcc/Vee the VRipple = 33.4 x 0.01 x 2.828 = 0.945 V.
So I need a filtering cap with a capacity of C = (I x t / Vripple) = (7.5 x 0.01 / 0.945) x 10^6 = 79365 or 80000 uF

Secondary Fuse (DC):
Ifuse = Ipeak = 7.5 A à chosen 8 A

Primary Fuse (AC):
IFuse = (2 x 7.5A) / (230V / 25V) = 1.63 A -> chosen 1.6 A

Questions:
Is this dimensioning approach for transformer and capacitor correct or are there some wrong considerations in there ?
How big would you choose the primary and secondary fuses ?

I appreciate every information ….

Thanks in advance
artQuake

[AndrewT]

it's not how I do it.

Four amps @ 60W into 8r0.
Vpk=31Vpk.
Ipk=3.9Apk.
Secondary fuse ~50% of Ipk ~ F2A. Eight required for the four sets of power supply rails.
Continuous maximum power 60W, allow transformer rating of 1.5times the maximum total output power ~1.5*(4*60)=360VA.
primary fuse~360VA/220Vac*3 ~T5A. If soft start is fitted this can be reduced to T1.6A or maybe lower. Try T1A with 60r in the primary feed.

For good & deep bass from a SS amp I recommend the input filter (DC blocking cap) be set to F-3dB<=2Hz.
This requires the NFB filter to be set to F-3dB<=1.4Hz.
and in turn the PSU F-3dB<=1Hz.
For 8ohm speakers each power amp requires +-20mF to give 1Hz filtering frequency. F-3dB=1/2/Pi/8ohm/0.02F=0.995Hz

Transient and peak currents come from the smoothing caps not from the transformer.

PS,
almost no one agrees with me, make your own decisions.

[artquake]

Thank you andrewT for the fast reply

Originally posted by AndrewT

Four amps @ 60W into 8r0.
Vpk=31Vpk.
Ipk=3.9Apk.
Secondary fuse ~50% of Ipk ~ F2A. Eight required for the four sets of power supply rails.

What's the reason that you choose 50% of Ipk ? Sinus-->RMS ?

Originally posted by AndrewT

If soft start is fitted this can be reduced to T1.6A or maybe lower.

Thats what i forgot to write that i have. A Softstart. So the result is almost the same as mine ... (factor 3.9:3.75)
So when choosing T1.6A is it only "maybe lower" or "necessary lower" then 1.6A ?

Originally posted by AndrewT [/i]

For 8ohm speakers each power amp requires +-20mF to give 1Hz filtering frequency. F-3dB=1/2/Pi/8ohm/0.02F=0.995Hz

Interesting that corresponds nearly to my 1% VRipple ...
But it would be good to know how you calculate the 20mF as it's part of your solution ... So why it requires +-20mF ?

Originally posted by AndrewT [/i]

Transient and peak currents come from the smoothing caps not from the transformer.

Between Bridge and Smoothing Cap or between smoothing cap and Amp ?

You may say that you do not agree, but anyhow your results ar almost the same ... I am delighted to read that ...

... all roads lead to Rome ... any other road out here ?

artQuake

[Juergen Knoop]

you don't need to fuse the LM4780.
Those are selfprotecting.
Choose the primary fuse according to the transformers nominal wattage, if you have an inrush current limiter.
regards

[artquake]

Thanks Juergen Knoop,
I know that the LM4780 has internal protection, as LM3886 does too, but since i blowed a few LM3886 in the past, due of oscillation probs, where the spike protection not really helps that much ... but this was not really the reason i asked at the beginning the following:

Originally posted by artquake
Questions:
Is this dimensioning approach for transformer and capacitor correct or are there some wrong considerations in there ?
How big would you choose the primary and secondary fuses ?
....

I only brought the previous approach as a example with a BPA200 / LM4780.
So let's assume that the approach should be valid also for SS Amp applications ...
My intention of my questions where only to get some echo and thru that to be sure i understood the basics how to proceed in calculating/designing a solid power supply ... and this is only a very small step into it.

artQuake

[AndrewT]

F-3dB=1/2/Pi/8ohm/0.02F=0.995Hz

0.02F is the smoothing capacitance. That is the same as 20mF and = 20,000uF

[megajocke]

"One Channel Amp consisting of two LM4780 in Parallel / Bridged Operation"

Is this with a 8 ohm load bridged? Giving 60W into 8 ohms for each of the 4 chips? Do you want to be able to do sine wave testing at full power?

I'd do it like this:

An Ipk of 3.9A gives a average current draw per rail of 1.24A (3.9A / pi) per rail and chip. Total current draw = 1.24A * 8 = 9.92A DC

The transformer needs to have an AC current rating of about 1.8 times the DC output current of the rectifier (RMS-to-average ratio of the rectifier current). AC current = 1.8 * 9.92 = 17.9A. This gives a VA rating of 25V * 17.9A = 450VA.

If continous full power sine waves are not needed then the transformer rating can be reduced. For 1/3 continous power 450/sqrt(3) = 260VA would be enough. (sqrt(3) because 1/3 output power gives 1/sqrt(3) current draw but at the same voltage) Something like this is probably the smallest that can be used.

Higher VA ratings will give better regulation from the transformer and less droop under load. 500VA would probably be enough for 4 ohm loads and music (don't know if the chips are up to this though at this voltage). 360VA as suggested by AndrewT sounds good for 8 ohm loads to me - a bit better than the minimum required but not some kind of insane overkill.

DC rail fuses of 2A per chip sound correct to me. Output RMS current = peak/sqrt(2). RMS current per rail = total RMS / sqrt(2) -> Rail RMS = Peak/2. But what happens when other chips are still trying to drive one with blown fuses? I'd at least let each bridge halve share fuses. 4A F per rail and side in that case.

If using fuses before the rectifier bridge instead they should be rated the same as the transformer rated secondary current. 10A slow blow for a 500VA 2x25V for example.

Primary fuses need to be bigger than VA/primary voltage suggest due to transformer startup surge if not using softstart. I'm not sure how much is enough. I have often seen 2 times or more used.

[artquake]

Originally posted by AndrewT
0.02F is the smoothing capacitance. That is the same as 20mF and = 20,000uF

and

Originally posted by AndrewT

For 8ohm speakers each power amp requires +-20mF to give 1Hz filtering frequency. F-3dB=1/2/Pi/8ohm/0.02F=0.995Hz

Thanks Andrew, but i still don't understand your statement.
Well, I wanted to know how you estimated the value of 20mF because your 20mF where also a part in your solution/statement ... So why it requires +-20mF ?
Or did you only wanted to show if i'd i take a value of 20mF that i'd get f-3dB @ 1Hz ? I don't think so, if so then every amp with a choosen f-3db of 1Hz @ 8Ohm would require 20mF ...
It looks an interesting way to estimate the value of capacity but you must help me to understand the reason ...

Originally posted by Megajocke
Is this with a 8 ohm load bridged? Giving 60W into 8 ohms for each of the 4 chips? Do you want to be able to do sine wave testing at full power?

@Megajocke first thanks for reply ... i need to check your statement without ruffle or excitement for reply ... but in advance allready some info:

It's for a pair of LM4780 (corresponds also for 4 x LM3886) in bridged/parallel operation @ 8Ohm.
I want to be able to listen "music" with 8 Ohm speakers, but the power supply should be able to deliver sine wave at full power (est. 200W).

Best regards
artQuake

[AndrewT]

Hi,
go back to my original assertion.

Originally posted by AndrewT
For good & deep bass from a SS amp I recommend the input filter (DC blocking cap) be set to F-3dB<=2Hz.
This requires the NFB filter to be set to F-3dB<=1.4Hz.
and in turn the PSU F-3dB<=1Hz.

it necessarily follows if you accept that -3dB @ around 2Hz is one of the requirements for good bass. This is equivalent to many manufacturers bandwidth claims of -1db @ 4Hz.

[artquake]

Andrew, I think i got it now ... initially i did not understood how to link your f-3db statement. But after more then one hour of thinking ... Yes !

Because DC has 0 Hz and the f-3db can also be equated with Vripple (as a sort of loss / reservoir). And we need to make a compromise with that, otherwise C increases to infinity.

Therefore the f-3dB < 1Hz equals to < 2% VRipple

Thanks ... learned again something new with this.

artQuake