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Need help with LM386 amp

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vcuauhtemoc

New Member
I'm making a circuit that takes in an audio signal, amplifies it, and outputs it to an LED which flashes in response to the audio signal. The current setup I have with the LM386 amp is working fine, but the problem is I plan to implement this into a pair of headphones, and the IC is getting power from a 9V battery which I would never be able to fit in the earphones. However, I do have a 3v 90mAh button battery which would work, but I can't figure out what the resistance should be between the IC and the battery. Currently with the 9V I have a 1K ohm resistor between the IC and the power.
 

MrDEB

Well-Known Member
Read your data sheet!!

the LM 386 has a min voltage requirement of 4v
9v w/ a 1 k resistor??
need schematic
 

c36041254

Member
Well, Id don't think 3V battery can run LM 386 as the minimum voltage required is 4V. I presume you'll be using three of the button cells in series. All you need to do is to make the current to be of 4mA (See the data sheet of LM 386) or higher so just apply the simple ohms law and you'll have the value of resistance you need (depending upon how many voltages you are giving to power the IC)
 

c36041254

Member
I don't think you need the 1K resistor at all , the load of the IC will take only that much current as much is required. There is a diode in the IC which protects the IC from getting "over current", you can put another diode in forward bias with pin 6 if you want to make it "double protected ". That's it !

Have fun !
 

Hero999

Banned

vcuauhtemoc

New Member
I don't think you need the 1K resistor at all , the load of the IC will take only that much current as much is required. There is a diode in the IC which protects the IC from getting "over current", you can put another diode in forward bias with pin 6 if you want to make it "double protected ". That's it !

Have fun !
I've tried removing the resistor, but what happens then is the LED is lit continuously, regardless of whether or not the amp is getting any input from the audio. A diode doesn't keep the LED off either.
 

Hero999

Banned
Here's the simplest circuit on the datasheet.


You don't have to be a genius to spot the differences between it, and your circuit.
 

audioguru

Well-Known Member
Most Helpful Member
The DC output of the LM386 is half the supply voltage. So of course an LED at its output to ground will light.
You need to bias the input so that the output voltage is low without a signal and the signal will cause the output voltage to rise and light the LED.

The LED will look like a dim blur unless you add a peak detector diode and capacitor.

Do it like this:
 

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vcuauhtemoc

New Member
Well, Id don't think 3V battery can run LM 386 as the minimum voltage required is 4V. I presume you'll be using three of the button cells in series. All you need to do is to make the current to be of 4mA (See the data sheet of LM 386) or higher so just apply the simple ohms law and you'll have the value of resistance you need (depending upon how many voltages you are giving to power the IC)
Thanks, it came out to 150 ohms, which works perfectly. Problem solved.
 

Mr RB

Well-Known Member
Nice for minimum parts count, but you can improve energy efficiency;

Arrange the 1M resistor from Q1 base to gnd, keeping it just off. Replace Q2 with a PNP (obviously move the 47R resistor) so if Q1 turns on it turns Q2 on. Same parts count but ZERO power use unless sound input makes the LED light up.
 
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colin55

Well-Known Member
The low-current LED indicator is a good design. I will add it to the list of 200 Transistor Circuits on the websote.
 
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