# Calculating the Voltage Drop of the LM386

Discussion in 'Electronic Basics' started by Archimedes, Mar 30, 2008.

1. ### ArchimedesGuest

Hi All

How do I calculate the voltage drop of the LM386 IC?

Do i simply make a measurement across the Vss pin and the Ground Pin?
In this chip this is Pin 6 and Pin 4.

Regards
Shelton.

2. ### Phil AllisonGuest

"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

3. ### Peter BennettGuest

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.

4. ### ArchimedesGuest

Hi

What I want to determine is how much voltage is being used by this IC
- thats all.

Cheers
Shelton.

5. ### VaractorGuest

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

6. ### Don KlipsteinGuest

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

The datasheet is at:

http://www.national.com/ds/LM/LM386.pdf

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 ()

7. ### Phil AllisonGuest

"Archimedes"

** Another nonsense question.

YOU supply the DC voltage to the IC.

It will then USE a varying mount of current.

..... Phil

8. ### Peter BennettGuest

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?

9. ### ArchimedesGuest

Ok Phil
We are not all as smart as you - maybe you should go fuk your dog -
low life scum.

10. ### Phil AllisonGuest

"Archimedes" <

** Shelton -

I usually call people like you stupid, arrogant pricks.

Bu I reckon you are a genuine mental defective who cannot help himself.

So, be a good fellow and piss off.

Or I will give you hell.

....... Phil