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Old 8th September 2011, 08:18 PM   #1
epilot is offline epilot  United States
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Default LM 3886 INPUT volatge

Hello guys,

Just want to know what is the maximum P-P input volatge which we can give to the LM3886 amplifier while the gain is 21 and the load is 4 ohms and the ouput power is 68W?

How to calculate the maximum p-p input voltage yet be sure that the amplifier does not go to cliping or saturation?

Thanks
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Old 8th September 2011, 08:33 PM   #2
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If you have +- 35 V DC you can get +-33 V out I should think. Divide this by 21. Around 1 Vrms is max.
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Old 8th September 2011, 08:47 PM   #3
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Originally Posted by peranders View Post
If you have +- 35 V DC you can get +-33 V out I should think. Divide this by 21. Around 1 Vrms is max.
Thanks sir,

The power supply volatge is +-28V.
I do not know why you used the volatge of power supply to calculte what I wanted?!!
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Old 8th September 2011, 09:01 PM   #4
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The supply voltage will determine the max output voltage so if you wil have 28 V then the peak voltage will be a couple of volts less.

This is stated in the datasheet under the parameter Vod and note 3 on page 3.
http://www.national.com/ds/LM/LM3886.pdf
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Old 8th September 2011, 10:35 PM   #5
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Originally Posted by epilot View Post
Hello guys,

Just want to know what is the maximum P-P input volatge which we can give to the LM3886 amplifier while the gain is 21 and the load is 4 ohms and the ouput power is 68W?

How to calculate the maximum p-p input voltage yet be sure that the amplifier does not go to cliping or saturation?

Thanks
You are needing to know minimum p-p input voltage. Having a certain gain and ohm load does not in itself cause a higher voltage to clip sooner. It would cause a lower voltage to clip sooner. The max voltage is that spec'd as max in the datasheet, but once you calculate minimum voltage for your gain and load, it would just be a waste of the transformer VA capacity to pick one with higher voltage than needed, rather than higher current capability, and a higher voltage than needed will cause the chip to run hotter than it otherwise would.
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Old 8th September 2011, 11:13 PM   #6
epilot is offline epilot  United States
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Originally Posted by ! View Post
You are needing to know minimum p-p input voltage. Having a certain gain and ohm load does not in itself cause a higher voltage to clip sooner. It would cause a lower voltage to clip sooner. The max voltage is that spec'd as max in the datasheet, but once you calculate minimum voltage for your gain and load, it would just be a waste of the transformer VA capacity to pick one with higher voltage than needed, rather than higher current capability, and a higher voltage than needed will cause the chip to run hotter than it otherwise would.
To be honest, I did not understand your first ans second above sentences, Plz explain it to me.


Ok Let me explain what I mean.
I told that the load is 4 ohms, the supply voltage is +-28V and of course the output power as the datasheet says for such a load and supply voltage is 68W.
Now I calculate the output RMS voltage which is:
Vrms=sqrt P x R= sqrt 68 x 4=16.49V, so the P-P output voltage will be:
Vp-p =2(16.49 x 1.414)=46.63V
OK now if the gain of the amplifier happens to be known (here as I told is 21)
we can have the p-p input voltage as:
Vin p-p =Vout p-p / gain so Vin p-p = 46.63/21 = 2.22V
So I calculated the input p-p voltage for the the output p-p voltage of 46.63V and the gain of 21. I call that input p-p voltage the maximum input voltage which I am able to give to the chip white the load is 4 ohms, gain is 21 and the output power is 68W, right???? It reflects to me that any input p-p voltage bigger than 2.22V will cause the output to go to saturation and clip...

What is your idea?
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Old 9th September 2011, 12:09 AM   #7
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No, not right. You seem to have the false impression that higher voltage than absolutely necessary causes clipping. Higher does not cause clipping. Clipping is caused by too LOW an input voltage, when input signal x gain > available output voltage after forward drop across the amp chip (loss after PSU input voltage) or current limiting protection.
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Old 9th September 2011, 05:16 AM   #8
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To simplify for you: A normal signal source is what you can have together with a LM3886 with the gain set to 21.

28 V supply voltage => 26 V peak out => 1.24 V peak in => 875 mVrms is clipping voltage.
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Old 9th September 2011, 06:00 AM   #9
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Maybe I need more sleep, we seem to be talking about two different things so disregard what I wrote above.
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Old 9th September 2011, 08:38 AM   #10
epilot is offline epilot  United States
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Originally Posted by ! View Post
No, not right. You seem to have the false impression that higher voltage than absolutely necessary causes clipping. Higher does not cause clipping. Clipping is caused by too LOW an input voltage, when input signal x gain > available output voltage after forward drop across the amp chip (loss after PSU input voltage) or current limiting protection.
Ops!
exactly opposite!

Suppose you have an ideal op-amp which its power supply is 10V and its gain is 5. Any input voltage LARGER than 2V causes the output of the op-amp to go to saturation (2 x 5= 10V). In this cause if the input signal happens to be sine, It would be changed to Square wave after clipping.
Actually I do not know what you are talking about? An op-amp must be able to amplify ANY LOW voltage before it goes to saturation...
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