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Can I use resistor instead of pot with LM386?

Discussion in 'Audio' started by rahulb, Apr 14, 2018.

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  1. rahulb

    rahulb

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    Mar 14, 2018
    Hi,

    I am using a LM386 amplifier to amplify um66 output. Speaker is 8 ohm 1 watt. Supply voltage in 9v

    I don't want to vary the sound level so, the pot is not needed here. Just I want maximum of sound.

    So, my question is , can I use 10k resistor instead of a 10k pot in the circuit given below:

    [​IMG]

    thanks
     
  2. davenn

    davenn Moderator

    13,597
    1,875
    Sep 5, 2009
    a 10k resistor will give you minimum audio out

    Personally, I wouldn't aim for max out as that is likely to overdrive the LM386 and it will go into distortion
    try a 1k resistor for a start and see how it goes. You may eventually, after testing, drop to a 470 Ohm resistor

    Dave
     
  3. kellys_eye

    kellys_eye

    4,276
    1,146
    Jun 25, 2010
    Use a PRESET potentiometer and set the volume how you need it to be - at least you can easily change the volume of there are problems in the future. Some presets are hardly larger than the resistors you'd otherwise be using.
     
    davenn likes this.
  4. WHONOES

    WHONOES

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    May 20, 2017
    Using a 10K resistor instead of a pot will work fine. A fixed resistor of ant value(within reason) will not affect the gain of you LM386. The only effect of using different values would be the roll off frequency from its combination with C1.
    As an aside, I don't like the way the wiper of VR1 is connected directly to the input of the LM386. When the pot is adjusted it will cause disturbances to its output which will inevitably appear at the speaker.
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    I would replace the pot with a trimpot. That way you can set the appropriate gain and leave it alone.
     
  6. WHONOES

    WHONOES

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    May 20, 2017
    Putting a trimpot as per VR1 does not adjust the gain off the amplifier. It's gain is set internally to 20. VR1 just adjusts the input level.
     
  7. Audioguru

    Audioguru

    2,868
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    Sep 24, 2016
    The datasheet for the UM66 shows an output of 0.8V peak at 600uA. The maximum peak output of an LM386 driving an 8 ohm speaker when the supply is exactly 9V is 3V peak. Then the maximum gain must be 3V/0.8V= 3.75 times but less when the battery voltage drops.
    But the LM386 in the circuit with C2 has a gain of 200 times (for a low level microphone).

    A logarithmic volume control set to one-third cuts the signal 50 times to 16mV which when multiplied 200 times gives an output of 3.2V peak which is more than the maximum of 3V peak so the loudest sounds will be distorted. A preset variable resistor is linear, not logarithmic so it must be turned down to almost zero.
    Why not remove C2 so that the gain is only 20 times? Then two series resistors as a voltage divider can cut the 0.8V
    (3V/0.8V)= 3.75 times.

    The output of an UM66 might be positive pulses only then only half of the total output swing of the LM386 will be used. But demos on You Tube of the "melody" sound like a squeaky little buzzer, very distorted anyway.
     
  8. davenn

    davenn Moderator

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    Sep 5, 2009
    and that is what is wanted ;)
     
  9. Audioguru

    Audioguru

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    Sep 24, 2016
    Guess what I noticed:
     

    Attached Files:

    davenn likes this.
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Ok, it attenuates the signal appropriately before applying a fixed gain so that the overall gain of the circuit is as desired.
     
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Hahahaha well spotted.
     
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