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 LM386 input level
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 24th March 2015, 09:01 PM #1 diyAudio Member   Join Date: Dec 2014 Location: Liverpool LM386 input level I have recently built a power amp using the LM386. I have built several of the circuits from the datasheet including the one below which has a voltage gain of 20. As I want to feed line level to the input, this would exceed the amplifiers maximum input voltage of 0.8V peak to peak. Would the potentiometer at the amplifiers input be able to reduce the voltage level that reaches the amplifiers input so that the signal does not distort?
 24th March 2015, 09:12 PM #2 diyAudio Member   Join Date: Feb 2015 Yes, it will, but you can also add a resistive divider ahead of the potentiometer to further attenuate the signal.
 24th March 2015, 10:00 PM #3 diyAudio Member   Join Date: Dec 2014 Location: Liverpool Thanks! If a signal has a voltage of 0.8V peak to peak, does this mean that the amplifier must be able to handle an input power of 0.8V, or 0.4V?
 24th March 2015, 10:31 PM #4 diyAudio Member   Join Date: Jul 2014 . Here ya go. Before you read anything else scroll down and look at the first image. Or come to think of it don't read anything at all, the picture tells the story. Measuring the Sine Wave Your meter reads "calculated RMS" voltage, which for practical purposes is the same as the RMS value referenced. Sudden thought. "Line level" is nominally one volt. Do you have good reason to believe that the 0.8 volt figure you mention is, in fact, peak-to-peak? If it's the voltage your meter showed you then it's not, unless you have a peak-to-peak meter. In audio we're not often concerned with peak-to-peak voltage, except in power supplies. A transformer's AC output voltage rectifies into DC at the AC peak value. That is: AC output voltage (meter reading) x 1.414 = AC peak output voltage = Rectified DC Level I can't offhand think of any other need to know peak voltages. Note that 1.414 is the inverse of 0.707. . Last edited by bentsnake; 24th March 2015 at 10:46 PM.
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Join Date: Jul 2014
Quote:
 Originally Posted by kevnaff If a signal has a voltage of 0.8V peak to peak, does this mean that the amplifier must be able to handle an input power of 0.8V, or 0.4V?
The entire audio world orbits around the magical-mystery figure of line level, which is nominally (i.e. approximately) one volt. Nearly everything either/and/or expects a line level input, or has a line level output. That's a one-volt meter reading, peak-to-peak doesn't enter in.

Rule of thumb: Unless otherwise stated amplifiers develop their full output when the input is line level.

If you already have a line level signal then you wouldn't preamp it. You'd either send it to a mixer or other device, or directly to an amplifier.

<< If a signal has a voltage of 0.8V peak to peak, does this mean that the amplifier must be able to handle an input power of 0.8V, or 0.4V? >>

It would mean 0.8 volts, except...

...it means neither, because it doesn't work that way. You don't deal with peak-to-peak voltages. You use your meter reading, or the published specifications for the equipment.

By the way, power and voltage are two different things. Best not to mix terms.
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Last edited by bentsnake; 24th March 2015 at 11:34 PM.

diyAudio Member

Join Date: Jul 2014
Quote:
 Originally Posted by kevnaff Would the potentiometer at the amplifiers input be able to reduce the voltage level that reaches the amplifiers input so that the signal does not distort?
Getting back to your original question, as patlaw pointed out, sure it will.

But if you find yourself running with the volume control turned waaay down toward zero, so adjustment is limited and difficult, here are a couple of ways to implement the voltage divider that patlaw also mentioned. The shown resistance values are arbitrary.
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Attached Images
 volume-level-control-dropping-resistor-v3b.jpg (19.7 KB, 108 views) volume-level-control-dropping-resistor-box-v1c.jpg (41.5 KB, 106 views)

Last edited by bentsnake; 24th March 2015 at 11:24 PM.

 25th March 2015, 03:51 PM #7 diyAudio Member   Join Date: Mar 2009 Unless you can control the level of the audio going in, you really need a potentiometer on the input. Also use a input coupling capacitor to keep any possible DC out of the amplifier. If your speakers are capable of any bass, increase the output cap to 1,000uf. Do not use the LM386 beyond 9 volts with 8 Ohm loads. It has weak current capability and power collapses under higher loads. This chip is very sensitive to poor ground layout so separate power ground (supply voltage return, output return) and input & SVR returns into separate traces right at pin 4!
 25th March 2015, 07:51 PM #8 diyAudio Member   Join Date: Dec 2014 Location: Liverpool Thanks for the responses! If I were to use a laptops headphone output as the input of the amplifier, would this give me around 0.775V RMS at the laptops maximum output? I should have checked with my multimeter at the time but don't have it anymore!
 25th March 2015, 08:46 PM #9 diyAudio Member   Join Date: Mar 2009 Headphone outputs trend to be a little low for driving the LM386 at default gain. You might want the gain at around 40db. If strictly used for headphone devices, you don't have to have a volume pot since most devices have their own. You should also add a 1K resistor across the input so the amp doesn't buzz or oscillate when unplugged from the headphone device.
 25th March 2015, 09:01 PM #10 diyAudio Member   Join Date: Dec 2014 Location: Liverpool I also built the LM386 with a gain of 46dB and this worked fine. What I am really trying to figure out is this.... I have attempted to build a studio monitor with a line input for a school project using the LM386. The output from the monitor both sounds and tests reasonably well, however I have to justify why I chose the LM386. With the LM386 having a maximum input voltage of 0.8V peak to peak, it would only really be suitable for amplifying domestic equipment and not something that would be found in a studio If I was to attenuate the input level using a potentiometer to below 0.8V peak to peak so that there is no clipping of the signal at the output, how would this affect the output? For example would attenuating the signal at the input and then amplifying up to the desired level, be the same as having an amplifier than can handle say 3V peak to peak? Last edited by kevnaff; 25th March 2015 at 09:01 PM. Reason: Wrong info

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