Rowland LM3886

Moze

Member
2012-12-02 11:40 pm
There are many LM3886x3 PCBs for sale based on Rowland model 10 they all claim 150w into 8ohms... but in this thread:

http://www.diyaudio.com/forums/group-buys/215211-pcb-3xlm3886-pa150-150w.html

Post#2 he explains that when parallelling these chips, you only get more power output with lower impedance loads. The example he gives is that whe you parallell TWO 3886s, you get 100w into 4ohms but no power increase at 8ohms and if you PA 3 chips you can get 150w into 2.7ohms, and still only 50w into 8ohms..

My question is who is correct? Here is the board on ebay: : http://pages.ebay.com/link/?nav=item.view&id=271564589477&alt=web
 

Mooly

Administrator
Paid Member
2007-09-15 8:14 am
If one chip can deliver 50 watts into 8 ohms then it means it is delivering 20 volts rms at its output. If you parallel the chip with another one, or another ten even, you still have only 20 volts rms available and that is still 50 watts rms into 8 ohms.

What does increase is the ability to deliver current and so the paralled chips can maintain that 20 volts rms into lower impedances.

If one chip can just manage 2.5amps rms (which is 50 watts into 8 ohm) and you connect a 4 ohm load then the output will current limit and distort. Two chips gives a 5 amp capability and so on.
 

Moze

Member
2012-12-02 11:40 pm
If one chip can deliver 50 watts into 8 ohms then it means it is delivering 20 volts rms at its output. If you parallel the chip with another one, or another ten even, you still have only 20 volts rms available and that is still 50 watts rms into 8 ohms.

What does increase is the ability to deliver current and so the paralled chips can maintain that 20 volts rms into lower impedances.

If one chip can just manage 2.5amps rms (which is 50 watts into 8 ohm) and you connect a 4 ohm load then the output will current limit and distort. Two chips gives a 5 amp capability and so on.

Thats what I thought, so how come these boards and kits are being advertised as 150w @ 8ohms?
 

mrWagner

Member
2006-10-27 1:34 pm
Maybe 150W is not possible, but more than 50W should. As with single chip 50W is the max, but if I paralel, I should be able to achive a 70-80W from lets say +-40V.
The doc you linked mentioned that there is no sense in paraleling at 8 ohm, but I just don't really get it why. Shouldn't it increase the continuous capabilities at least a bit?
 
Yes it will increase the capability, but only a very little. The graph which you can use to work out how much is on page 9 of the datasheet. It shows the clipping voltage vs supply rails up to 40V.

When you parallel you effectively halve the output current from each chip, assuming gains are perfectly matched (which of course they never are). I reckon the clip voltage will increase, but perhaps by only 0.5V. Is this really worth adding a parallel chip for?

<edit> On reflection, perhaps I was a little pessimistic, I think if you run at 40V you will gain something on continuous sinewave power. But on music (where the crest factor is higher than for a sinewave) you'll not get much, if any benefit.
 
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billshurv

Member
Paid Member
2014-03-01 11:53 pm
Into 8 Ohms you can get 60W with +/-35V rails.
Into 4 Ohms you can get 68W with +/-28V rails if your heatsink is up to it.

To get significantly louder than either of those you need significantly more power. Say 3dB otherwise you will barely notice it. Parallel allows lower impedance. Bridging allows higher voltage swings, but half the current swing. If you need more than 60W into 8Ohms then I would suggest a different amplifier.
 

tomchr

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Paid Member
2009-02-11 12:58 am
Calgary
www.neurochrome.com
Each LM3886 is guaranteed (by TI) to be able to deliver 7 A of output current. Assuming no other limitations, this means it can deliver a peak output voltage of:

8 Ω: 7 * 8 = 56 V
4 Ω: 7 * 4 = 28 V
2 Ω: 7 * 2 = 14 V

This is why I and many others, including TI, recommend using ±28 V rails with the LM3886. If you are only driving 8 Ω loads, you can increase the supply voltage to ±35 V. Sure, the absolute maximum is ±42 V, but using ±35 V as the design target allows for mains variation, component tolerances, etc.

Most amplifiers will need to be able to drive 4 Ω and 8 Ω loads, so ±28 V is used. This allows for about 38 W into 8 Ω and 65 W into 4 Ω.

For some this is not enough power. The limitation is the output current of the LM3886, so let's use two in parallel. Again, assuming no other limitations the limits on the output swing are as follows:

8 Ω: 2*7 * 8 = 112 V peak
4 Ω: 2*7 * 4 = 56 V peak
2 Ω: 2*7 * 2 = 28 V peak

Now we're cooking. Note that the supply voltage is still limited to ±42 V (use ±35 V as the design target for reasons mentioned above). As shown in the list above, with two LM3886es in parallel, the amp can drive both 8 Ω and 4 Ω loads to the ±35 V rails, but for 2 Ω load, it can only drive to ±28 V rails.

So let's use three LM3886es in parallel:

8 Ω: 3*7 * 8 = 168 V peak
4 Ω: 3*7 * 4 = 84 V peak
2 Ω: 3*7 * 2 = 42 V peak

Now, even the 2 Ω case can be used with ±35 V rails.

That was the first question: Why use multiple chips in parallel. Now, let's look at why you don't get three times the power in the 8 Ω case.

The absolute maximum limit on the supply voltage for the LM3886 is ±42 V when the LM3886 handles a signal. The limit is a bit higher in the MUTE condition, but that's hardly relevant here. As I mentioned before, ±42 V is the absolute maximum and I suggest using ±35 V as the design target to allow for component variation and variation in mains voltage. If you use a regulated supply, you can squeeze closer to the ±42 V limit.

In addition to the limit imposed by the power supply, the LM3886 will have some dropout voltage across it when driving large currents. You can find the dropout voltage specified in the data sheet with a graph showing it for various loads. I'll assume the following numbers in my calculation (based roughly on memory):

8 Ω: Vod = 3 V
4 Ω: Vod = 4 V
2 Ω: Vod = 5 V

This means, for the 8 Ω case, an amp with three LM3886es in parallel will be able to drive to Vsupply-Vod = 35-3 = 32 V peak. So let's do the power math:

P = Vpeak2/(2*R) -- note the voltage is the peak voltage.

8 Ω: P = (35-3)2/(2*8) = 64 W
4 Ω: P = (35-4)2/(2*4) = 120 W
2 Ω: P = (35-5)2/(2*2) = 225 W

As Bill points to above, you do need to make sure that your heat sink is up to the task. I strongly recommend doing the math if you're planning to drive heavy loads with the LM3886. For more information, see my Taming the LM3886 - Thermal Design page.

Now do note that putting LM3886es in parallel is not an easy task. The differences in offset voltage between the LM3886es cause the ICs to "fight" - i.e. set up a standing current in the output ballasting resistors. This reduces the current available for the load and also heavily degrades the THD. It's also why you'll see a bunch of multi turn trimpots in Jeff Rowland's 6xLM3886 amp.
The best way I have found for implementing a parallel LM3886 amp is to use the two channels of an LM4780 in parallel. The two LM3886es inside appear to match better in terms of offset. I don't think the dice are matched by TI, but at least they're at the same temperature which will help. It could also be that they're adjacent dice on the wafer so they could match better than two random LM3886es pulled from a lot. Sadly, TI has decided to discontinue the LM4780 by September 2016 and supplies are becoming limited already.

Tom
 
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Moze

Member
2012-12-02 11:40 pm
What my goal is is to bi amp my center channel and power 2 surround speakers.

Center - BW LCR600S3 8ohm (min 3ohm), need 100w x 2
Surrounds - Plok T15, 8ohms, need 100w ea

So essentially I need my amp to deliver 4 channels of 100w...I understand this can be done by a BR100 config or a BPA config - problem is I cannot find a BPA or BR PCB kit here in the states...
 

billshurv

Member
Paid Member
2014-03-01 11:53 pm
Well I have to be honest, I don't see the point of bi-amping those B&Ws. They are 89dB efficient so a single 100W amp will give you a peak of 109dB, above what you need. another amp for the tweeters will give you 0dB extra.

And I would be suprised if those Polks would handle 100W for long.

All the speakers you link are 89dB/w so need 40W to hit 105dB that THX specifies for movies at 1m. How much power you need above that depends how big your room is.

Now 60W to 100W is only 2.2dB. In general use you are not likely to notice that difference. IMO of couse, YMMV.
 
What my goal is is to bi amp my center channel and power 2 surround speakers.

Center - BW LCR600S3 8ohm (min 3ohm), need 100w x 2
Surrounds - Plok T15, 8ohms, need 100w ea

So essentially I need my amp to deliver 4 channels of 100w...I understand this can be done by a BR100 config or a BPA config - problem is I cannot find a BPA or BR PCB kit here in the states...
If you want amplifiers that can deliver 100W then build/buy amplifiers that CAN deliver 100W into your specified load.

The LME498xx series can easily meet 100W into any load from 4ohms to 16ohms.
 

tomchr

Member
Paid Member
2009-02-11 12:58 am
Calgary
www.neurochrome.com
If you want amplifiers that can deliver 100W then build/buy amplifiers that CAN deliver 100W into your specified load.

Now there's an idea! Another approach, which is equally valid, is to look at your *needs* and design an amp that fits your *needs*. If you do the math on the needed power, I bet you'll find it to be rather less than 100 W.

The LME498xx series can easily meet 100W into any load from 4ohms to 16ohms.

True. However, the LME49810, LME49811, and LME49830 were discontinued by TI along with the LM4780 (and other audio ICs). The official story was a fab closure, but that doesn't make sense at all. I bet it's a combination of elimination of expensive packages, inventory pruning, and maybe moving some products from smaller wafers to larger wafers, which would likely require some form of re-qualification of the process.

In this thread you'll find a link to a document from TI which lists the devices affected by the "fab closure": http://www.diyaudio.com/forums/parts/280990-ti-discontinue-many-ics.html

Tom
 
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