Parallel LM3886 output resistor(s)

[markiemrboo]

Hi all,

I have decided to make the basic circuit LM3886 in a parallel configuration, with three chips in parallel. Possibly pointless, but I have chips to spare and want to use them

I saw something very similar to what I want to do on shine7.com. In the two chip parallel version he used 3x0.5R 1% 1W metal film (I think) resistors as they seem to be easier to find than a 3W 0.1 ohm 1% resistor.

I have some 0.47R 5% 2W carbon resistors, which I could hand match, but I have found some 0.33R 1W 1% "metal oxide" resistors for sale. Would three of these matched in parallel be worth it over three of the 0.47R carbons?

Three 2W 0.47R's would be 0.157R and 6W
Three 1W 0.33R's would be 0.11R but 3W

The three 0.33R's are closer to the appnote circuit, which shows 0.1R and would lose slightly less output power I guess?




Also, I wouldn't mind knowing what kind of power output I am looking at with three chips in parallel per channel with a +/-35v supply. It's confusing me quite a lot!

Supposing I can reach +/-30v on the output then in to 8 ohms (assuming pure resistive non realistic speaker) this would be 112W (peak?). But the appnote says that two chips, or even a single chip, can "only" give 50W in to 8 ohms at this supply voltage.

Is this because at 35V the SOA graph indicates that the collector current is ~2.5A. I guess this is *2, 2.5A for the negative and 2.5A for the positive side?

Ohms law says 20v in 8 ohms = 50W = 2.5A, so this seems to work out right and explain the limit.

But then I thought with parallel it increased the current handling, so it would be capable, in an ideal situation, 5A per side? Which should mean it would give:

30v in 8 ohms = 112W = 3.75A

Then I would be voltage limited rather than current limited as before? No?

[AndrewT]

Hi,
your output resistors could be matched even closer than 1%.
The small values you plan to use are difficult to measure directly.
String them together in series and pass 200 to 400mA through them and measure the voltage drop across each. I use a pair of 12 way terminal blocks to temporarily mount these low value resistors. Keep the first one as a reference to measure the next eleven, until you find sufficient for matched sets.

The parallel sets don't all need to be identical.

Suppose you were using 10off 1r0 in parallel to give 0r1. Then choosing three that are a close match and selecting for for each amp. Find the next three that match and split them between the three amps. etc.

BTW,
buying 100off 1r0 600mV metal film will cost about £1.50, convert that to local currency. It's peanuts and even ten only come to £0.15 for an equivalent 6W resistor that can be matched to much better than 1%.

You are almost there with your power calculation because you have already realised it's voltage and resistance that matters and the limit is the resulting current.

Let's take your +-35Vdc supply and assume you can get 28V into your load. P=V*V/2/R = 28*28/2/8= 49W for an 8ohm load
Current into that 8ohm load is 28/8=3.5Apk.
Three amplifiers in parallel each need to deliver just under 1.2Apk to meet this demand.
Let's look at real speakers.
the minimum impedance of a reactive speaker is about 6ohms (we are not using a 4 to 8ohm speaker for this example).
the peak current into the real speaker is 28/6=4.7Apk and still draws less than 1.6Apk per amplifier.
But that 6ohm (about 60 to 70% of nominal) only applies to continuous signals. Non continuous signal and particularly fast stopping and starting signals can make the speaker appear as 35% of nominal impedamce for very fast transients.
The fast transient current into an 8ohm speaker is 28/8/0,35=10Apk. or about 3.3Apk per amplifier. The 3886 has a specified peak current in the range of 7A to 11A. so your three parallel amps can easily drive the worst case 8ohm speaker.
Do the same calculations for a 4ohm speaker and the peak current into the worst case 4ohm speaker will be about 6.7Apk per amplifier, again this is below the minimum value of peak current from the poorest performing amp in the spec sheet.

Most builders use a parallel pair for 4ohm loading and this arrangement can limit if the amps can only manage 7Apk each. Your three parallel amps will manage these worst case loads no problem.
If your PSU has a slightly higher voltage, the 3// should still manage OK at about +-38Vdc into 4ohm speakers, and obviously any 6ohm or 4 to 8ohm can be driven if you choose.

[markiemrboo]

Originally posted by AndrewT
Hi,

Hi Andrew,

Thanks for your reply You are always really helpful!

Originally posted by AndrewT
your output resistors could be matched even closer than 1%.
The small values you plan to use are difficult to measure directly.
String them together in series and pass 200 to 400mA through them and measure the voltage drop across each. I use a pair of 12 way terminal blocks to temporarily mount these low value resistors. Keep the first one as a reference to measure the next eleven, until you find sufficient for matched sets.

I made an "LMS Impedance Bridge" ages a go which I was hoping to use to try and match the resistors.

http://www.arrl.org/qex/2005/Steber.pdf

The PDF describes it quite well. The program seems to have two decimal places after the resistance, but I am not sure how accurate it is etc.

The parallel sets don't all need to be identical.

Suppose you were using 10off 1r0 in parallel to give 0r1. Then choosing three that are a close match and selecting for for each amp. Find the next three that match and split them between the three amps. etc.

BTW,
buying 100off 1r0 600mV metal film will cost about £1.50, convert that to local currency. It's peanuts and even ten only come to £0.15 for an equivalent 6W resistor that can be matched to much better than 1%.

This would be quite a good idea actually. I think I might well do that, thanks.

You are almost there with your power calculation because you have already realised it's voltage and resistance that matters and the limit is the resulting current.

Let's take your +-35Vdc supply and assume you can get 28V into your load. P=V*V/2/R = 28*28/2/8= 49W for an 8ohm load

Ah hold on. Would I be right in saying that the calculation I did was 'peak' power, and that is RMS average power?

I can understand if it's the proper way to state in RMS continuous average power, but (please bear with me) would I be right in saying that three LM3886's in parallel could actually maintain the peak ~100W in to 8 ohms continuously, and it's not just a misleading 1ms figure?

I hope that's right, because as simple as the concept is the RMS vs peak wattage thing has managed to confuse me a little bit.

Current into that 8ohm load is 28/8=3.5Apk.
Three amplifiers in parallel each need to deliver just under 1.2Apk to meet this demand.
Let's look at real speakers.
the minimum impedance of a reactive speaker is about 6ohms (we are not using a 4 to 8ohm speaker for this example).
the peak current into the real speaker is 28/6=4.7Apk and still draws less than 1.6Apk per amplifier.
But that 6ohm (about 60 to 70% of nominal) only applies to continuous signals.

Non continuous signal and particularly fast stopping and starting signals can make the speaker appear as 35% of nominal impedamce for very fast transients.
The fast transient current into an 8ohm speaker is 28/8/0,35=10Apk. or about 3.3Apk per amplifier.

Yikes! I didn't know this could happen! So paralleling might have a small benefit even for 8 ohm 'nominal' impedance speakers then? Nice.

The 3886 has a specified peak current in the range of 7A to 11A. so your three parallel amps can easily drive the worst case 8ohm speaker.

Good good! My current speakers are 8 ohm nominal so it shouldn't have any problem at all with those.

Do the same calculations for a 4ohm speaker and the peak current into the worst case 4ohm speaker will be about 6.7Apk per amplifier, again this is below the minimum value of peak current from the poorest performing amp in the spec sheet.

Most builders use a parallel pair for 4ohm loading and this arrangement can limit if the amps can only manage 7Apk each. Your three parallel amps will manage these worst case loads no problem.

This is also good to know.

If your PSU has a slightly higher voltage, the 3// should still manage OK at about +-38Vdc into 4ohm speakers, and obviously any 6ohm or 4 to 8ohm can be driven if you choose.

From what I remember the mains here seems a bit high, and a 25vac transformer ends up at about 37-38vdc after rectifying and filtering.

I imagine that three in parallel should even manage a reasonable job on a 2 ohm speaker, not that I actually intend to buy a 2 ohm speaker



One last question if I may. I'm not going with a regulated supply this time. I have a dual secondary transformer, and was just wondering if this is the correct way to 'convert' it in to a center tapped secondary:

vac -----| | ---- +ve 1
            | |
            | | ---- gnd 1
            | |
            | | ---- +ve 2
            | |
vac -----| | ---- gnd 2

I would join "gnd 1" and "+ve 2" together to form the common ground point (center tap), right? then gnd2 would end up being the negative rail when referenced from this common ground point?

I think that's right, but I just wanted to make sure!

[AndrewT]

Hi,
You probably all know my opinion on chipamps and low value load impedance.
I'll restate it for those who don't know me nor my opinions.

Do Not even Think about 2ohm (reactive) load and chipamps. They do not live together at all well.

If it helps, try to forget about peak power ratings.

You can get sinewaves and these can be measured accurately.
These sinewaves are very predictable and much of electronic design is based on that very predictability.
You can measure current in Amperes (A) and when it's sinewave current it can be represented as A, or Aac, or Arms. The context of the usage usually helps identify whether it's rms or not. I rarely use anything other than A when defining (non peak) current.

The same applies to voltage. The slight difference being that I prefer to always define exactly what voltage I am referring to; Vdc, Vac, Vpk. You can use and often see Vrms.

Drive a resitive load with a sinewave voltage and measure the average power over a whole (or multiples of) cycle and the result is Watts (W).
Do not apply rms to power measurements, the term does not exist.

If you do wish to refer to peak ratings, I find it less ambiguous to use Ipk and Vpk and forget about trying to convert to equivalent power during the instanataneous moment of power AT THE PEAK.
It is very usefull to know peak voltage and peak current during operation of a circuit. But the peak in power is much less useful, mostly because it is brief and thus has little (zero) heating effect. To develop any heat the power MUST be dissipated over a period of time (however brief that period is).

I do have an exception to this peak power, I use it for SOAR but over defined time periods eg. 100mS or 1uS etc. i expect ther are quite a few other exceptions, but in general you can avoid peak power and it's association with exaggerated power claims from less than reputable retailers/manufacturers.

Transformer connections.
Yes, your conversion diagram from dual to centre tapped is correct.
It is very easy to check that the correct wires are connected.
Measure the voltage across each winding. join two tappings together. Measure the voltage at the two open ends, it should be exactly double the single winding voltage. If the voltage measures zero or near zero then one of the joined wires is incorrect. However as long as the other two ends are open no damage will ensue. Not even a blown fuse.

BUT,
please use a light bulb tester every time you plug any new project into the mains.
An example:- If the dual primary is wired in error then the peak current in the line can be many hundreds of amps. If your lucky only a fuse or circuit breaker will open. I don't want to even think about being unlucky.
If you look back a couple of years, you can find me enquiring about a safe method of checking that the dual primary was correctly wired. I learned later that I was advised a lot of nonsense and that the test bulb is the easy and safe way to do it.

[markiemrboo]

Originally posted by AndrewT
Hi,
You probably all know my opinion on chipamps and low value load impedance.
I'll restate it for those who don't know me nor my opinions.

Do Not even Think about 2ohm (reactive) load and chipamps. They do not live together at all well.

Oh It just seemed logical that if three in parallel could handle the worst 8 and 4 ohm speakers, it would at least cope fairly well with a 2 ohm speaker. I wasn't intending on ever using it with a 2 ohm speaker anyway though. It was pure thought!

If it helps, try to forget about peak power ratings.

Ok.

You can get sinewaves and these can be measured accurately.
These sinewaves are very predictable and much of electronic design is based on that very predictability.
You can measure current in Amperes (A) and when it's sinewave current it can be represented as A, or Aac, or Arms. The context of the usage usually helps identify whether it's rms or not. I rarely use anything other than A when defining (non peak) current.

The same applies to voltage. The slight difference being that I prefer to always define exactly what voltage I am referring to; Vdc, Vac, Vpk. You can use and often see Vrms.

Drive a resitive load with a sinewave voltage and measure the average power over a whole (or multiples of) cycle and the result is Watts (W).
Do not apply rms to power measurements, the term does not exist.

I just read up on this and I think I now understand what's going on and where I was getting confused. To get the power you use the RMS voltage and current, as you said, so if I were to see a 30v "peak" - that is from the crossover 0v point of the wave to the very tip of the wave - pure sinewave on the output, then this would be 30 * 0.707 = 21.21v RMS, and in to an 8 ohm load this would then be 56W of power output. This is correct I hope?

If you do wish to refer to peak ratings, I find it less ambiguous to use Ipk and Vpk and forget about trying to convert to equivalent power during the instanataneous moment of power AT THE PEAK.
It is very usefull to know peak voltage and peak current during operation of a circuit. But the peak in power is much less useful, mostly because it is brief and thus has little (zero) heating effect. To develop any heat the power MUST be dissipated over a period of time (however brief that period is).

Point taken, thanks.

I do have an exception to this peak power, I use it for SOAR but over defined time periods eg. 100mS or 1uS etc. i expect ther are quite a few other exceptions, but in general you can avoid peak power and it's association with exaggerated power claims from less than reputable retailers/manufacturers.

You've lost me there, I will just forget about peak power as you advised!

Transformer connections.
Yes, your conversion diagram from dual to centre tapped is correct.
It is very easy to check that the correct wires are connected.
Measure the voltage across each winding. join two tappings together. Measure the voltage at the two open ends, it should be exactly double the single winding voltage. If the voltage measures zero or near zero then one of the joined wires is incorrect. However as long as the other two ends are open no damage will ensue. Not even a blown fuse.

Excellent. Thanks for the confirmation! I will test it with the joined secondaries not connected to a load (the power supply) first with the multimeter to make sure.

BUT,
please use a light bulb tester every time you plug any new project into the mains.
An example:- If the dual primary is wired in error then the peak current in the line can be many hundreds of amps. If your lucky only a fuse or circuit breaker will open. I don't want to even think about being unlucky.
If you look back a couple of years, you can find me enquiring about a safe method of checking that the dual primary was correctly wired. I learned later that I was advised a lot of nonsense and that the test bulb is the easy and safe way to do it.

I will do

Thanks once again!