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Old 5th February 2010, 10:07 PM   #1
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Default Max voltage input lm3886

hi, one mouth ago i finished to built my power amp based on LM3886s in parallel configuration (100W on 4Ohm load),LM3886 power supply is +30V and -30V.... now , i've just started to built a new preamp but i can't understad what is the maximus voltage input of each lm3886... preamp power supply is +30 - 0 V, not dual. how can i determinate preamp gain, my signal input is from my guitar (about 5mV rms, i think)?
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Old 5th February 2010, 11:26 PM   #2
tomchr is offline tomchr  United States
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Gain is defined as the output voltage (versus frequency) divided by the input voltage (versus frequency). Measure those two and calculate the gain.

...or post a schematic of your circuits then it's possible for others to calculate the gains for you.

~Tom
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Old 9th February 2010, 03:51 AM   #3
ryan750 is offline ryan750  Philippines
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download
national.com/appinfo/audio/files/Overture_Design_Guide15.xls

plug it in your components value then you'll see how sensitive your LM3886, from there you can calculate the needed gain by your preamp,

by the way, how's your parallel LM3886? planning to build one soon,
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Old 9th February 2010, 07:08 AM   #4
ratza is offline ratza  Romania
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Taking the LM3886 datasheet, the maximum voltage swing at the output is in the limits (V+ - 3V) to (V- + 3V). In your case the maximum voltage swing is +/-27V or 54Vpp. I will assume that you use a gain of 23 for your amp (22k and 1k feedback resistors). So, Aamp = 23.

Knowing your gain of the amplifier so the maximum peak to peak input voltage for maximum output voltage will be Uout / Aamp. The gain and has the following expression:
Aamp = 1 + (Rf1/Ri) = 1 +22/1 = 23 (27dB)

Using the first equation will result that the input voltage is:
Uin = Uout / Aamp = 54Vpp / 23 = 2.348Vpp

Now, for the preamplifier is just the same, except you need to calculate the gain instead of voltage, and afterwards choose the values for the feedback resistors. You know that the input voltage is 5mVrms, which means that you have:
Uin1 = 5mVrms * 2 *sqrt(2) = 14.14mVpp

Gain of the preamp will be:
Apre = Uin / Uin1 = 2.348Vpp / 14.14mVpp = 166
________________________________
The above is the hard way The easy way is like this -> know input voltage, know output voltage, calculate total gain:
Aglobal = Uout / Uin = 54Vpp / 14.14mVpp = 3818.9

But Aglobal = Apre * Aamp, so the gain of the preamplifier will therefore be:
Apre = Aglobal / Aamp = 3818.9 / 23 = 166
________________________________


The minimum preamplifier voltage supply must be chosen bigger than Uin (voltage at preamplifier output) with the values shown in the opamp datasheet. Your preamp is powered at 30V single rail, which is more than enough, even 6V might do the job. I hope I cleared up the things a little bit.
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Old 10th February 2010, 01:04 PM   #5
AndrewT is offline AndrewT  Scotland
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Hi,
the maximum input to your power amp, assuming the same as Ratza showed, is 2.35Vpp.
I like to see 10 to 20dB of signal overhead capability in the preceding stages.
10dB requires 3.2 times as much voltage and 20dB requires 10 times as much voltage. Since we will never need to use this for a valid audio frequency signal, it does not matter if the signal is becoming grossly distorted at this artificially high value.
So, let's settle for 10dB of overhead giving the maximum output requirement for you pre-amp of about 3*2.35 = 7.4Vpp. This is just about achievable from a 9Vdc supply.
Now, we'll look at 20dB of overhead. 10 * 2.35 = 23.5Vpp.
This should be achievable from a 26Vdc supply.
A supply of 30Vdc would be ideal for a 20dB overhead on a maximum output audio signal of 2.35Vpp
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