Calculating the Voltage Drop of the LM386

"Archimedes"

** Nonsense question.

** That would tell you the voltage supplied to the LM386.

Now, why don't you say what you are REALLY after ?

..... Phil

What exactly do you mean by "the voltage drop of the LM386"?

The LM386 is an audio power amplifier, and I can't think how the term "voltage drop" would apply to it.

Measuring between pins 4 and 6 will give you the power supply voltage, which is determined by whatever is supplying power to the circuit, not by the LM386.

--
Peter Bennett, VE7CEI  
peterbb4 (at) interchange.ubc.ca  
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

Hi

What I want to determine is how much voltage is being used by this IC

- thats all.

Cheers Shelton.

I think you need to know Ohms law and the output impedence and current? Can you not measure the output voltage swing across your load at full power?

CHEERS

In , P Bennett said

I suspect that Shelton was inquiring about how far short of the supply rails can the load input voltage be.

The datasheet is at:

One little bit of relevance here that I see in the datasheet:

The graph in the right column, middle row, Page 4, "Distortion Vs. Output Power"

Conditions: 6 volt supply, 8 ohm load

There is a "base/root end of takeoff" of distortion upticking with output power at close enough to .2 watt, arguably as low as .18 watt.

The worse distortion one of these figures has THD roughly .3%, and the better of these is for THD upticking just a little at about .2% or so.

These wattages for sinewave applied to an 8 ohm load are 3.58 and 3.39 volts peak-to-peak respectively. Subtract these from 6 volts and divide by 2, and respectively for these different argued degrees of distortion threshold 1.21 volts and 1.305 volts respectively is average of the 2 directions; negative/positive; for LM386 to need to have a sinewave to have peaks "short of the supply rails" in order to have distortion "not significantly upticking" or "barely upticking".

In usual "music duty" and "voice duty", I seem to think that peaks getting distorted by the amplifier to 2% are very acceptable. In the above graph, the curve is sharply taking off through the 2% point at about

1/4 watt. With an 8 ohm load, a 1/4 watt sinewave has peak-to-peak voltage 4 volts - with 6 volt supply this has average of the 2 shortfalls from the rails being exactly 1 volt. I would like to comment that the distortion will probably be in the form of flattening the peaks - so I suspect that for good lower distortion the output signal needs to have peak-to-peak level more than 2 volts less than the voltage across the supply rails - I would say about 2.5 volts.

- Don Klipstein ( snipped-for-privacy@misty.com)

"Archimedes"

** Another nonsense question.

YOU supply the DC voltage to the IC.

It will then USE a varying mount of current.

.... Phil

I wouldn't say "the voltage used by this IC" - the voltage between pins 4 and 6 is the voltage supplied to the LM386 by the surrounding circuit.

What is it that you really want to know? How much _power_ is used/disippated by the LM386, perhaps?

--
Peter Bennett, VE7CEI  
peterbb4 (at) interchange.ubc.ca  
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

According to the datasheet, available from

there are several versions of this part. The -1 and -3 versions are recommended for operation from 4 - 12 volts, while the -4 is recommended for 5 - 18 volts.

--
Peter Bennett, VE7CEI  
peterbb4 (at) interchange.ubc.ca  
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca