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LM386 Gain...

Hi,

I hooked up an LM386 to a 9v supply (+ve to pin 6, -ve to pin 4).

I left pins 2 and 3 (inputs) open, pin 7 open, and pins 1 and 8 open. I
then measured the voltage at pin 5 and got the expected 4.5v.

I then connected pin 1 to 8 using a wire (to increase the gain from 20
to 200). The voltage at pin 5 dropped by about half a volt.

Can someone explain why this happened???

Thanks

Gareth
 
J

John Popelish

Jan 1, 1970
0
Hi,

I hooked up an LM386 to a 9v supply (+ve to pin 6, -ve to pin 4).

I left pins 2 and 3 (inputs) open, pin 7 open, and pins 1 and 8 open. I
then measured the voltage at pin 5 and got the expected 4.5v.

I then connected pin 1 to 8 using a wire (to increase the gain from 20
to 200). The voltage at pin 5 dropped by about half a volt.

Can someone explain why this happened???

Thanks

Gareth

Even though both inputs are grounded through similar internal 50k
resistors, the input transistors are not perfect duplicates. Any
slight input offset voltage caused by the mismatch is amplifier more
at the higher gain than it is at the lower gain.
 
J

John Popelish

Jan 1, 1970
0
Cool - that makes sense! Thanks.

While I was trying to work out what was going on I was looking at the
schematic (http://www.national.com/ds/LM/LM386.pdf) to see if could
could see what baised the output at Vs/2. I couldn't!! Can someone
explain it?

The two inputs run at zero volts. The differential amplifier stage has
two currents feeding into the common emitters, one from the supply and
one from the output. But there is a resistor between these two
inputs, so any difference between those two currents gets amplified by
the differential darlington input stage. That is, there is a gain
based on the inputs both having zero volts at their bases, but
different voltages on their emitters. Since the current from the
supply into the left emitter has two 15k resistors in series, but the
current from the output to the right emitter has only one 15k
resistor, the only way the two emitters see similar currents is if the
output voltage is 1/2 of the supply voltage.

Of course, the current to the left emitter varies as the supply ripple
bounces around (both from the rectifier pulses and from load current
pulses driving the speaker) so it is a really good idea to bypass pin
3 to ground, so those short term variations are not forced to show up
in the output signal via the bias scheme.

If you want to trim the final output voltage, you can connect pin 3 to
either the supply or to ground through a resistor much higher than 15k
to tweak the left emitter current by trial and error. (or connect
something like a 100k pot from supply to ground and connect the wiper
to pin 3 through a 47k resistor to tweak the current to the left emitter).
 
P

Pooh Bear

Jan 1, 1970
0
Hi,

I hooked up an LM386 to a 9v supply (+ve to pin 6, -ve to pin 4).

I left pins 2 and 3 (inputs) open, pin 7 open, and pins 1 and 8 open. I
then measured the voltage at pin 5 and got the expected 4.5v.

I then connected pin 1 to 8 using a wire (to increase the gain from 20
to 200). The voltage at pin 5 dropped by about half a volt.

Can someone explain why this happened???

It's a crap chip.

Use something more modern. LM386s are notorious for parasitic oscillation
for one thing and modern circuits rarely need the minimum 20x voltage gain.

Graham
 
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