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

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 ?

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:

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]

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

and

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 ...

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.

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

 

[megajocke]

If you want to be able to get full sinewave out continous then you will probably need a transformer of about 400-500VA. Hard to say exactly how big because of unknowns like supply impedance, transformer winding resistance, leakage inductance, size of caps and so on.

All these will affect the peakiness and the factor I assumed to be 1.8 in my calculations. Sometimes the transformer manufacturer gives hints on how big transformer is needed for the capacitor-input rectifier arrangement.

It also depends on what ambient temperature the transformer is rated at and what the real ambient temperature is, airflow etc. The transformer rating is purely thermal so if the transformer has better cooling than it was specified for more power can be drawn from it and vice versa. It also takes a long time for transformer temperature to stabilize - gross overload for a couple of minutes is no problem at all.

 

[megajocke]

If I'm thinking correctly now 40mF will be needed on each rail to get the power supply filter at the frequency suggested by AndrewT. The situation where this matters most is when the amp is clipping - now the speaker will be connected between + and - rail. Thus it is the seriesed capacitors that should equal 20mF in total.

 

[AndrewT]

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

Thanks ... learned again something new with this.

thank you, I have learned as well because I had not tried to link ripple to smoothing.

If I'm thinking correctly now 40mF will be needed on each rail to get the power supply filter at the frequency suggested by AndrewT. The situation where this matters most is when the amp is clipping - now the speaker will be connected between + and - rail. Thus it is the seriesed capacitors that should equal 20mF in total.

It's not the way it was explained to me by a few posters.
+-20mF for F-3dB=2Hz into 8ohm and +-40mF for F-3dB=2Hz into 4ohms.

 

[megajocke]

This could be seen as 2 channels driving 4 ohms so at first glance it looks like +-80mF might be needed. But as this is a bridge the ground isn't even used and all caps are used all the time instead of just half the time. This would suggest +-40mF.

 

[artquake]

Yes when using f-3dB <= 1Hz or VRipple <= 2% i also get the same result.

But:

This could be seen as 2 channels driving 4 ohms

What does this statement say us when comparing it with AndrewT's f-3dB formula ? It follows when using 4 Ohm that would increase C 80 mF per Rail ...

Is that correct ? Does not every LM4780 see 8 Ohm as load ?

BTW: Do you guys have some good Book suggestions about this power supply topic ? (Power Supply Cookbook 2nd Ed i have but it's almost for Swiching PSU's) ....

regards
artQuake

 

[megajocke]

"Is that correct ? Does not every LM4780 see 8 Ohm as load ?"

Each opamp sees an 8 ohm load, thus each LM4780 sees 4 ohms. This is due to the middle of the load being at virtual ground. BUT there is a difference:

If two channels drive the same signal into two 4 ohm loads from a +- supply then the loads (total 2 ohms) will be connected to either the positive or negative rail, only one rail at a time. -> 80mF per rail needed.

But when bridged one 4 ohm load will be connected to positive and the other to the negative rail during a half cycle and vice versa -> 40mF per rail.

The power supply capacitor utilization is better in a bridged amp. Another possibility is to have one of the channels inverted. I've seen this on a power amp I have. The inverted channel has the red binding post grounded and black is output. This way the power supply is better utilized when channels are playing in phase - which they mostly will be at least for large low frequency signals where it matters most. Easy to bridge too - just feed the same signal to both amps.

 

[artquake]

BUT there is a difference:
If two channels drive the same signal into two 4 ohm loads from a +- supply then the loads (total 2 ohms) will be connected to either the positive or negative rail, only one rail at a time. -> 80mF per rail needed.

Do you mean that when i use two LM4780 in Bridged parallel Operation and feeding the inputs of each LM4780 with balanced signal, then 8 Ohms will be seen as 2 Ohms ?
Therefore AndrewT's -3db would correspond with 1% Ripple ... That balanced combination/operation i exactly have here, then the 80mF of my first calc would be correct ...
And what about the secondary fuses now ? 2A per LM4780 and Rail still correct ? I think so ...

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).

megajocke could you please explain me more up on this ? I'd really appreciate it. In any case, thanks for informations up to now.

Best regards
artQuake

 

[AndrewT]

Hi,
two chipamps in parallel mode each supply half the load current (if well matched/balanced).
They see an effective load of double the actual load.

Two chipamps in bridge mode supply double the voltage to the load and thus supply double the current.
They see an effective load of half the actual load.

Now combine the four chipamps into a bridged parallel configuration and each chipamp sees the actual load as the effective load.
Design each chipamp for the actual load you will use.

the 4780 has a multitude of V+ and V- pins. These should all be fed from a common supply voltage, since Nationl do not specify which parts of the circuitry are supplied from each pin.
In bridged mode one of the internal amps will draw current from one rail to load, while the other amp will sink the same current from the load to the other rail. This means that the fuse supplying each rail does not get a rest and so needs to be double the rating that a single chipamp supplying 60W needs.
The smoothing capacitance is working supplying both bridged channels, but each polarity only sees the current loading of one chipamp at a time.

If you bridge a chip then the fuses need to be doubled but the capacitance remains the same.

If you parallel a chip then the fuses need to be doubled AND the capacitance needs to be doubled.

 

[megajocke]

"Design each chipamp for the actual load you will use."

And this would then suggest +-20mF per chipamp for the calculated power supply frequency. +-80mF total. But the bridge connection uses all caps all the time instead of just half the caps at a time - so +-40mF for the suggested frequency.

Designing for some arbitrary ripple % doesn't make much sense to me. What about frequencies below the rectifier output frequency? Ripple % doesn't make much sense then. And why 1%? Why not 10%? That would be more in line with what is usually used.

"megajocke could you please explain me more up on this ? I'd really appreciate it. In any case, thanks for informations up to now."

Here are some recommendations from a transformer manufacturer (don't know which though):;
http://documents.rs-components.com/EITC/UK/infozone/rectifier_relationships.pdf

This link states 1.8 for this factor for the standard rectifier connection. The value will in reality vary between different transformers and according to the used capacitance. Little capacitance -> smaller factor and vice versa. This doesn't matter too much though, 1.8 is a usable approximation.

 

[artquake]

And this would then suggest +-20mF per chipamp for the calculated power supply frequency. +-80mF total. But the bridge connection uses all caps all the time instead of just half the caps at a time - so +-40mF for the suggested frequency.

In my case each Rail (+ and -) have their own fullwave rectifier bridges. And the pair of LM4780 are driven with a balanced input signal into this bridged parallel configuration. So every Amp inside the LM4780 sees a 8 Ohm load, and therfore every amp needs +-20mF. That's 4 x 20 = 80mF per Rail. I tought that i understood it right up to now, but why it is now +-40mF ?

Is this because of that, when the input signal is positive at a point t, then one LM4780 needs (+) while the other in the same moment needs (-) current ? Then this would also need +-80mF.

Did I likely somewhere make a logical flaw ?

Maybee i'd better go to swimm instead of thinking in this heat ... anyhow good brainfood up to now and thanks for the .pdf ...