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#1 |
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diyAudio Member
Join Date: Jan 2004
Location: Indiana
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I bought one of these modules off Ebay that is a prebuilt stereo amp with (2) lm3886 on board.
Looks like I can run these in parallel pretty easily with the addition of the output resistors. 1/10th ohm at 3w on each output and then into a 4 ohm speaker. I've seen in some schematics as well, the components regarding the noninverting input being shown. Looks like a 20k feedback resistor and then a 1K and a 100uf electrolytic to ground. Will I need to adjust those components for parallel operation? I don't have the schematic on this amp and like I say, it's setup for isolated stereo operation. (I'll trace the PCB when I have time -- but thought I'd ask). Also, it's calling for a dual 28vac transformer for stereo -- what would be the right voltage for parallel mode? Thanks. |
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#2 |
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diyAudio Member
Join Date: May 2006
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A 2x28V AC tansformer is too much voltage , that will give you +/-42v DC which is the absolute maximum the chip can take but if the Voltage in your area happens to fluctuate 10% it will bring your voltage over the limit , you will also dissapate more heat so you will need a Bigger heatsink ....
I would suggest a 2x24v AC transformer for 8 ohm loads and a 2X18v AC for 4ohm loads (each chip) .... For going paralell , I would try to match the Gain set resistors on each chip to .01% , I would also match the Output resistors to 1% and maybe use a 5w instead of 3w as a hotter resistor will tend to drift in value .... Good luck |
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#3 | |||
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diyAudio Member
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The isolated LM3886TF would indeed mean.. Quote:
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__________________
If you've always done it like that, then it's probably wrong. (Henry Ford) |
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#4 |
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diyAudio Member
Join Date: Jan 2004
Location: Indiana
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Ok --
So really running the LM3886 in parallel is really a no brainer. 4 or 8 ohm speakers are fine with a 24vac transformer and really I don't see any need to change the feedback resistors that they have on the board for normal stereo operation -- I see they are 1% -- I imagine they're matched well. Just put the .1 ohm 5 watts in series with each output and that's it. And parallel the inputs of course. |
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#5 |
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diyAudio Member
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The datasheet specifies 0,1 %. If you think, the National engineers did not specify that correctly, explain why. Don't recommend to ignore that specification based on your imagination only.
__________________
If you've always done it like that, then it's probably wrong. (Henry Ford) |
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#6 |
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diyAudio Member
Join Date: Sep 2007
Location: Hangzhou - Marco Polo's 'most beautiful city'. 700yrs is a long time though...
Blog Entries: 62
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To get reasonable current sharing between two paralleled devices, the gain setting resistors matter a whole lot more than the current sharing ones. Those on the outputs can be 5% without problems.
As I see it you're likely to run into poor sharing at low frequencies with those component values (100uF and 1k). This is because of the relatively poor tolerances of aluminium electrolytics (20% typically). A 20% change in the 100uF cap gives of the order of a 0.5% change in the gain at 20Hz, so your relatively expensive 0.1% resistors are rather wasted. The solution is simple - up the 100uF to 1000uF. Edit - I've just had a look at the AN-1192 and it seems they didn't get this one proof-read by Bob Pease. 0.1% resistors are most certainly not required on the + inputs to the amp chips. 1% output resistors are also overkill. Last edited by abraxalito; 19th October 2009 at 03:35 PM. |
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#7 | ||
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diyAudio Member
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The DC offset should also be similar on both ICs. That is why the blocking resistors are also 0,1 % types. Looks like overkill, but maybe Bob Pease did proof-read and he knows something about the ICs we don't. 1000 µF in the gain loop makes little sense, if the DC blocking cap on the non-inverting input is not adjusted accordingly. And even then it is not really a practical value. The 1 % output resistors must be judged with the power dissipation in mind. Paralleling chipamps only makes sense, if you want to take advantage of the amplifier's power limits. The nearer you get to the limits, the smaller an imbalance can be tolerated. If the rails stay well clear off the LM3886's rating, you can use bigger tolerance. With a 2x28 V transformer 1 % seems more recommendable, while 5 % is certainly sufficient for a 2x18 V transformer. But then again, with such a low voltage there is no need for paralleling.
__________________
If you've always done it like that, then it's probably wrong. (Henry Ford) |
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#8 | ||||||||||
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diyAudio Member
Join Date: Sep 2007
Location: Hangzhou - Marco Polo's 'most beautiful city'. 700yrs is a long time though...
Blog Entries: 62
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Do show your calculations. I'm guessing they went something like this :
A 100uF cap has an impedance at 20Hz of about -j80 ohms. At the tolerance extreme, this goes up to -j100 ohms. This difference of j20 ohms is indeed j2% of the 1k resistor. So is that how you reached your answer? Quote:
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#9 | |||||||||
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diyAudio Member
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Wow, what a lecture!
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80 µF at 20 Hz = j99,472 Ohm 120 µF at 20 Hz = j66,314 Ohm Worst case for 0,1 % resistors is Ci1 = j66,314 Ohm, Ri1 = 999 Ohm, -> Xi1 = 1001,199 Ohm, Rf1 = 20020 Ohm -> a = 20,996 Ci2 = j99,472 Ohm, Ri2 = 1001 Ohm, -> Xi2 = 1005,930 Ohm, Rf2 = 19980 Ohm -> a = 20,862 a average = 20,929, a error = ±0,32 % Worst case for 1 % resistors is Ci1 = j66,314 Ohm, Ri1 = 990 Ohm, -> Xi1 = 992,219 Ohm, Rf1 = 20020 Ohm -> a = 21,358 Ci2 = j99,472 Ohm, Ri2 = 1010 Ohm -> Xi2 = 1014,887 Ohm, Rf2 = 19800 Ohm -> a = 20,510 a average = 20,934, a error = ±2,03 % Quote:
Your previous answer shows that you already followed my reasoning. I won't answer to nit-picking. Quote:
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What do you base yours upon that he didn't? Infallibility, because his name is well-known? Quote:
It makes little sense to filter the non-inverting input at 3,38 Hz and the inverting input at 0,16 Hz. Those two filters should have similar corner frequencies and add up to a reasonable quality factor. More than half an octave of separation between them should be an exception. It has no advantages that justify its size and cost. You will find few amplifiers with a 1 mF capacitor in the inverting leg, if any. Are you prepared to accept the accumulated experience of the majority of amplifier designer's as proof? Quote:
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__________________
If you've always done it like that, then it's probably wrong. (Henry Ford) |
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#10 | ||||
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diyAudio Member
Join Date: Sep 2007
Location: Hangzhou - Marco Polo's 'most beautiful city'. 700yrs is a long time though...
Blog Entries: 62
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Not really, since a lecture is generally unidirectional. Here I'm interacting with your remarks so its more of a tutorial. But I agree its rather long - that's because you've made so many mistakes. So here's what I'll do - I'll just deal with one group of your mistakes at a time to keep the post to a manageable length. So for now, I'll just deal with the calculations for the gain matching of the two amps.
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C1a = j6.6 ohm, R1 = 999 ohm; X1 = 999.02 ohm, Rf1 = 20020 ohm; a = 21.04 C1a = j9.9 ohm, R1 = 1001 ohm; X1 = 1001.05 ohm, Rf1 = 19980 ohm; a = 20.96 average a = 21, error in a ±0.04 giving ±0.2% This is the same (to the number of significant figures in the calculation) as having no capacitor at all, so the value of the relatively expensive 0.1% resistors hasn't been thrown away. Quote:
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