LM1876 from a pc souncard - Can it be moded? - Page 2 - diyAudio
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Old 21st October 2011, 03:06 AM   #11
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It means the power supply is capable of 800mA on the -12 V supply and when you exceed that (VERY easily) something will go into 'protect' or may just fail. Asymmetrical power supply current will cause some bizarre distortions as the positive supply stays steady and the negative 'comes and goes'.

The chip will only draw whatever current is required for the signal and load combination. Higher signal = more current. Lower ohms load = more current = more problems.

G
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Old 21st October 2011, 07:44 AM   #12
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If the negative rail goes very low the amp may draw more from the positive and get DC offset at the output. However i think the power supply will not allow it to go very low, when it shuts down you'll know why.

If the current would be increased the chip will of course draw more power (and supply more power to the speaker) if it's called for.
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Old 21st October 2011, 10:39 AM   #13
AndrewT is offline AndrewT  Scotland
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if the -ve supply is limited to 400mApk and there is little decoupling on the sound card PCB then the maximum output current will be ~-800mApk.
The maximum power into a 4ohms speaker will be limited to ~ 0.8V^2 * 4ohms / 2 = 1.28W.
If you connect an 8ohm speaker the maximum power output rises to 2.56W.
This assumes the sound card uses no current and the whole 8-00mApk is available to meet the current demand of the speaker.
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Last edited by AndrewT; 21st October 2011 at 10:41 AM.
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Old 2nd November 2011, 12:27 AM   #14
dmx2020 is offline dmx2020  New Zealand
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AndrewT

Not sure what you mean when you say 'if the -ve supply is limited to 400mApk...then the maximum output currrent will be ~-800mApk' ???

The -12V rail is rated at 0.8A max (and the +12v rail is rated at 4.2A)

So do you mean that the total output current will be (0.8Ax2) 1.6A ???

From this the max power into 8 Ohms per channel will be P= I^2 * R /2

which is (1.6A^2 * 8)/2 = 10.24 W (max per channel roughly)

Is this correct ???
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Old 2nd November 2011, 06:34 AM   #15
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Hi DMX.

I would first check to see if the LM1876 is configured to run on a single ended power supply. IMO this is the most likley configuration due to the voltage rails available from common PC power supply. And also realise that the LM1876 will not swing rail to rail so the max output voltage will be lower by at least 3 volts .

Regards Ian

Last edited by madtecchy; 2nd November 2011 at 06:43 AM. Reason: Typo
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Old 2nd November 2011, 10:32 AM   #16
AndrewT is offline AndrewT  Scotland
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Quote:
Originally Posted by dmx2020 View Post

Is this correct ???
no. Both my answer and your interpretation are wrong.
Thinking about it again, if the -Ve supply is limited to 400mA then the maximum demand of the load must also be limited to the same 400mApk, not the 800mApk quoted earlier.

For a symmetrical AC sinewaveform, that limits the output power and output voltage to minuscule values.
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Old 2nd November 2011, 10:37 PM   #17
dmx2020 is offline dmx2020  New Zealand
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where does 400mApk come from??? It is rated at 0.8A at -12V so it should be 800mApk on the -ve rail and similarly in the +ve rail ...isnt it?

So the max current from both rails is 1.6A??

Last edited by dmx2020; 2nd November 2011 at 10:39 PM.
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Old 3rd November 2011, 12:32 AM   #18
AndrewT is offline AndrewT  Scotland
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I don't know. Maybe there was a typing error that no-one noticed.
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Old 4th November 2011, 03:06 PM   #19
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Worst case (very low frequencies, buffer cap doesn't do much):
-12V @ -800mApk (~=-283 mArms), hence 1600mApp ~= 566mArms max (~= 2.5W @ 8 ohms or 1.25 W @ 4 ohms)

Best case (high frequencies, peaks entirely absorbed by buffer cap):
-12V @-800mArms, so you get 1600 mArms total max (~=20 W @ 8 ohms, except you won't get more than 6..7 W on +/-12V, so 10 W @ 4 ohms is more likely to be it)

Now considering where most power is commonly needed - in the bass - the importance of adequately sized buffer caps becomes obvious.

Basically I can see two options here:
1. Give the '1876 a new PCB, heatsink and power supply, which should yield a nice 2x 20 W power amp.
2. Reduce gain to maybe 10..14 dB (with a small cap across the inputs to keep it stable, usually double digit pF range) and use it as a competent headphone driver. Maybe the gain already is set no higher than 20 dB, I don't know.
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Old 6th November 2011, 10:49 PM   #20
dmx2020 is offline dmx2020  New Zealand
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Thank you for the answers guys!
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