Guru Help!

[Space Varmint]

I know you don't like them, sorry. But if I'm putting in about an eighth to a quarter watt into a '386, what sort of output should I expect? Watts and speaker audio.

or anyone, thanks

 

[Boncuk]

Hi Frank,

as far as I know you have put in a signal at medium to high impedance (4.7 to 10K), meaning low power input. The LM386 is programmable for different amplification factors of 20, 50 and 200.

What you get out depends on the maximum power the chip can dissipate and of course the impedance of your speaker.

Regards

Hans

 

[Mikebits]

At those input levels, 1/8, 1/4W you will severely overdrive the 386 and have a completely distorted output even at the lowest gain setting. Perhaps you should rethink your amp input level.

Maybe restating input levels at a volt PP value into a given impedance as 1/4W is vague with no reference.

 

[Space Varmint]

At those input levels, 1/8, 1/4W you will severely overdrive the 386 and have a completely distorted output even at the lowest gain setting. Perhaps you should rethink your amp input level.

Maybe restating input levels at a volt PP value into a given impedance as 1/4W is vague with no reference.

Well that tells me I have plenty of drive. I really don't think I have that much. Looking for output power.

Hey Hans

 

[Mikebits]

Output power is a function of amp gain and input voltage. It would seem to me that your first requirement would be an output requirement into a known impedance and then determine input level from there.

 

[Hero999]

It depends on the gain and input impedance of the LM386.

If the gain is 20 then an output power of 0.25W RMS into an 8R load will be enough to cause clipping.

The output voltage will be 2.83V peak.

The input voltage will be 2.83/20 = 141mV peak.

If the input impedance is 50k, the input power will be 0.141²/50000 = 400nW peak, 200nW RMS.

 

[Space Varmint]

It depends on the gain and input impedance of the LM386.

If the gain is 20 then an output power of 0.25W RMS into an 8R load will be enough to cause clipping.

The output voltage will be 2.83V peak.

The input voltage will be 2.83/20 = 141mV peak.

If the input impedance is 50k, the input power will be 0.141²/50000 = 400nW peak, 200nW RMS.

Right on the money Hero. Thanks. That's what I was looking for.