Torroid Calculations Question. (And Class)

[Macaba]

I have seen this thread:
What toroid transformer to use

But unfortunately it doesn't make any reference to the actual maths.


Say I look at a torroid thats rated as such:

300VA - 25V + 25V@6.0A

How can I work out how many LM3886's this will sustain. To make my question more explanatory, here is a calculation that I could go through:

25Vrms * 1.41 = 35.25Vdc
35.25Vdc * 6A = 211 Watt capability.

Now the real problem is that do the LM3886 draw equal amounts of current on both rails, and the theoretical 68W RMS maximum mean approx 68W draw on BOTH rails? Or just one?
Even worse, am I wrongly assuming that a 68Wrms output would mean 68W load on the power supply?

I've had a good look through many threads, I couldn't find any solid reference to transformer selection.

Next question:

What class is the LM3886 constituted as? And any comments on 'biasing' it into Class A.
Oh- if the Lm3886 is an 'op-amp', surely this means it would naturally run as Class A, as an ideal op-amp can go into the -ve rails just as easily as the +ve rails?

Thanks for any help.

[sangram]

One is the actual math, which you've worked out correctly.

I use a rough (unscientific) rule-of-thumb that you get roughly half the power from a transformer into your speaker cable, so a 300 VA transformer would be good for upto 150 watt power output (given the voltage and current calculations are satisfied), and so on.

Therefore I would expect this toroid to be sufficient for upto two LM 3886 chips. But in theory you should be able to run about 6 chips off it, if the PS and chips were 100% efficient, which they are not.

AFAIK 68 watts is 1KHz audio output power at 1% THD, but I'm not too sure about that. Assuming amp efficiency of 80%, it would need 85 watts from the supply. If PS efficiency is 80%, it would need 106 watts from the transformer. If transformer efficiency is 80%, it would need 132 watts from the mains... In reality things may be a little more effcient, but it shows the basics of the 2:1 calculation which I use, may be oversizing it quite a bit.

Mass-market commercial stuff is usually rated for 70% of sinewave power, so a 100 watt amp may be powered by 70VA transformer... I was reading up some specs of a 2x200 watt RMS amp that proudly claimed to use a '300watt toroidal transformer'. An amp that generates its own power, how about that!

I would think a 300VA would run 3 amps but I would not go below 100VA/channel after accounting for voltage...

Edit: Sorry, I didn't answer the whole question...

Power output will not be half the supply voltage x current, but a bit less. That's because the amp does not swing all the way up and down to rail voltage. You need to start with desired output power (like in the datasheet), find out the required voltage swing, then size the power supply accordingly for voltage and current. Then size the transformer depending on how much of a margin you need. Usually 100 VA is adequate, and 200 is overkill. Somewhere between the two is where you need to be, depending on how comfortable and loaded you are.

As for Class A, there used to be a mod which consisted of connecting a 1.5K 10 watt resistor from the negative rail to the output of the amp. I don't know what that did, but a search might help. I don't even know if it was a Class A mod, but was touted as such, I remember a debate going on at the time.

I believe National calls it a Class AB amp, so yes at low outputs it would operate in Class A AFAIK.

[Nordic]

Firstly, I will assume you want to drive 8 ohm loads, as that voltage is too extreme for 4 ohms...

Using this assumption and your voltage, refer to the AN-1192 pdf
"2. VCC = ±35V, RL = 8Ù
Pdmax = VCCtot2/2ð2RL = (±35V)2/2ð2(8Ù) = 31.0W"

So if optimaly configured you will only produce 31W per chip....

Now go do your sums again... and report on what conclusion you came to.

[mhennessy]

Originally posted by Macaba
I have seen this thread:
[url]25Vrms * 1.41 = 35.25Vdc
35.25Vdc * 6A = 211 Watt capability.

Not quite - you're right about the voltage going up root-2 (x1.414), but the current has to go down - a bridge rectifier and smoothing capacitor can't create energy!

At the very least current goes down by root-2 (x0.707) but for other reasons it's nearer 0.6. Which means you get between 150 and 120W approx from the 150VA winding.

Luckily with audio, you can take liberties (as has already been said). But the advantage of DIY is you're not worried about commercial pressures to save every penny, so why not over-engineer the PSU? Personally, I like to have at least double transformer VA per watt - in other words a 50 watt amp should have at least 100VA of transformer behind it. As an example, I recently built a 4-channel LM4780 amp that gives 4 times 50W. The transformer is a 400VA device. When I operate the amp in bridged mode, it can give around 2 times 150W. The transformer is still big enough, but a little smaller than I'd like, based on my "rule of thumb". However, music has such a high peak-to-mean ratio that this is unlikely to cause any problems unless I start running it for disco use!

If it helps, I have a "power supplies 101" article on my site. Although I suspect it might be too basic for you as it doesn't cover the maths in any detail... http://www.mhennessy2.f9.co.uk/artic...r_supplies.htm

To work out the consumption on each rail, you would need to know the exact shape of the musical waveform. However, real signals tend to be highly symmetrical when averaged, so it's safe to assume they're equal. And a "perfect" class B amplifier is assumed to be around 78% efficient, meaning that if you deliver a 68W sinewave to a load, the power supply would have to deliver around 87W to the amplifier. This is the total consumption, so around 44W per rail, if you want to think of it in those terms...

As an example, take a 50WPC amp, using a 100VA TX. Each winding gives 25V at 2A. After the rectifier and smoothing caps, the voltage becomes around 35V and the available current becomes 1.2A. Total power available to each rail: 42W. Total power available to amp: 84W. As 78% of 84 is 65W, you've got a little "headroom" in hand. You might be alarmed by the 1.2A but remember this is continuous average current - the peaks can be much higher as these are supplied by the PSU caps. By rights, 50W into 8 ohms requires 2.5A RMS, which would actually require a 200VA transformer! FWIW, my original gainclone prototypes had a 160VA TX per IC - they're about 14UKP each so no big deal really....

These are all just rough "rules of thumb" - you can get a hell of a lot more theoretical about this stuff, but there's no need for DIY IMO. Remember, torroidal transformers are "conservatively" rated, and can withstand occasional abuse, plus music generally has a high PMR (peak-mean ratio), so if you ran a 50W amp at the verge of clipping, the average power delivered to the load is liable to be more like 5W (assuming a PMR of 10dB).

Hope this helps,

Mark

[mhennessy]

Originally posted by Macaba
I have seen this thread:
[url]What class is the LM3886 constituted as? And any comments on 'biasing' it into Class A.
Oh- if the Lm3886 is an 'op-amp', surely this means it would naturally run as Class A, as an ideal op-amp can go into the -ve rails just as easily as the +ve rails?

LM3886's are essentially Class B.

Op-amps don't "naturally" run in class A. The class definition isn't based on how near to the rails the output swings, rather how long do the positive and negative output devices conduct for. For class A, both devices conduct 100% of the cycle, 50% for class B, between 50 and 100% means class AB. Less than 50% is class C - not desirable in an audio amp, and not pleasant to listen to!

Most amps tend to conduct for slightly more than 50%, making them class AB strictly speaking, but a lot of manufacturers set the bias on the high side to ensure that temperature variations don't cause unwanted shifting towards class C operation with the resulting crossover distortion. Stabilising the operating point is one of the trickiest practical problems when making a power amplifier, but actually this is really easy in a chip because all the transistors are thermally linked - one of the benefits of chip amps...

Adding a 1.5K resistor between the rail and o/p does not make for a class A amp! Suppose there are 35V rails, this resistor can pass 23mA when the o/p is at 0V. Ignoring the fact this current will change with signal, this results in class A operation up to 8mW (into 8 ohms)! Beyond this, it's class AB. However, this will work for a preamp, where the opamp is driving a (say) 22K load. Just make sure the PSU rails are clean!

To make an LM3866 class A up to nearly 50W, you'll need something like a 1.7A current source between the output and one of the rails. Think of the heat this would produce - around 120 watts! And should the load impedance be less than 8 ohms, the amount of "class A power" falls accordingly...

Cheers,

Mark

[sangram]

Originally posted by Nordic


So if optimaly configured you will only produce 31W per chip....

That's power dissipation, not power output... Not really the same thing. Not familiar with the 3886 datasheet, but the others all a have a voltage-power output graph somewhere in the datasheet, and usually separate graphs for each load impedance. I would imagine an LM3886 with 35 volt rails should be able to deliver its entire power output, or pretty close (my 4780 datasheet shows it as 62 watts for 8 ohms and 35 volt rails)

[sangram]

Originally posted by mhennessy


Adding a 1.5K resistor between the rail and o/p does not make for a class A amp! This results in class A operation up to 8mW (into 8 ohms)! Beyond this, it's class AB.

Great info!

And how would I calculate the resistor value if I wanted Class A upto, say, 1 watt (My normal listening level)?

[peranders]

Macaba, you don't need to do any particular deep calculations if you are thinking music in your amp. 25-50 VA per channel is enough. 50-150 VA , normal and over that overkill.

.. so 80-120 VA is good, more than 300 VA is overkill.

If the amp is at 100% output power it consumes 130-150% of the output power.

The theorethical limit of effiency is 79% and a more real value is 60-65% I should guess.

Notice that the transformer consists of iron and this makes the transformer able to "averaging" to power. The only that counts is the temperature inside the transformer. The size of the transformer is therefore determined how the output power over time looks like. If you have no temperature protection and you want to play sinus at 100% more than 10-15 minutes you must have a bigger transformer but if it's about "normal" music you can settle for less.

Your class A ideas is not so applicable. Mostly because of the cooling properties of the IC. If you really want class A, it's probely better to build a dedicated class A amp.

I really recommend that you read the AN-1192 page 3.
http://www.national.com/appinfo/audi...gn_Guide13.xls
http://www.national.com/appinfo/audi...sign_Guide.pdf
http://www.national.com/an/AN/AN-1192.pdf
http://cache.national.com/ds/LM/LM3886.pdf

[Narcisse91]

Originally posted by Nordic
Firstly, I will assume you want to drive 8 ohm loads, as that voltage is too extreme for 4 ohms...

Using this assumption and your voltage, refer to the AN-1192 pdf
"2. VCC = ±35V, RL = 8Ù
Pdmax = VCCtot2/2ð2RL = (±35V)2/2ð2(8Ù) = 31.0W"

So if optimaly configured you will only produce 31W per chip....

Now go do your sums again... and report on what conclusion you came to.

Pdmax is power dissipation... That equation is really meant for calculating cooling requirements. For those voltage and load conditions, output power will still be whatever National claims (50W typical, I think).

[mhennessy]

Originally posted by sangram


Great info!

And how would I calculate the resistor value if I wanted Class A upto, say, 1 watt (My normal listening level)?

The idea is to take the standing current and double it (because this is how much current can go to the load before the other o/p device switches off) - this gives you the peak current available within class A operation. To find the RMS, divide by root-2. Then, to find the power, it's I-square times R

Re-arranging, Iq=0.5 ( sqrt(2) x sqrt( P / Rload ) )

For 1 watt into 8 ohms, this is a standing current of 0.25A

I don't know what your supply rails are, so I'll assume the 35V I used previously. Ohms law (R=V/I) is 35/0.25 = 140 ohms. How hot will this resistor get? Vsquared over R - nearly 9 watts! That's a lot of heat to loose.

I don't recommend a simple resistor here - as the output swings beyond the 2.83VRMS the current in this resistor changes. It might also be low enough to couple nasties on the PSU rails onto the output. Far better would be a current source - just a couple of transistors would do.

Hope this helps,

Mark