How to choose a diode for my audio amplifier?

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Petes21

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Hello I'm making an LM386 audio amplifier, the amp has the option to use DC or a battery (not at once of course), the problem is that the battery is getting drained really fast, I believe that's because the votage from the battery is getting back into the voltage regulator so I figured out I should use a blocking diode.

I have a 12v voltage regulator. What diode could I use for blocking 9v from getting into the voltage regulator?

What other info do I need to choose the right diode? Maybe you can recommend me some standard diode that will work for this?
 

May also use Low drop-out types like 1N5819. Will result in less power losses when operating from battery alone.
 
or just use a switch; spdt

the middle goes to your 386, one side to the battery, and the other goes to the DC power supply

of course the grounds must be connected
 

Yes, I thought I could use a switch but there are three already in the circuit, it would be too much. The DC IN has a coaxial power input so it already comes with a switch that disconnects the battery automatically, I wanted to take advantage of that.
 

An LM386 works well driving an 8 ohm speaker with a 6V to 9V supply. When the supply is as high as 12V then a lot of the power is wasted by heating the LM386 amplifier.
 
It depends upon the current. If less than 1 A use IN4007.If 3 A use 5402.If 4 A use 5408
 

An LM386 amplifier with a 12V supply has a maximum peak output current of only 400mA into an 8 ohm speaker so a 1A diode will be fine even for a stereo amplifier with two LM386 ICs.

You said the DC jack has a switch that disconnects the battery when DC is used then why do you need a diode?
 


I thought someone might ask that :grin:.

It's kind of difficult describing it to me but the circuit has a main on/off switch that controls both inputs (battery and coaxial input) through their grounds.

What the coaxial input switch does is disconnect the battery from the coaxial input when connected so the battery won't get charged (and potentially leak or explode).

So both battery and the dc input are wired together (with a coaxial input switch and a main switch). That's why the circuit needs the diode.

About the diodes, I got the 5408 because I couldn't find the 5819 but after buying it I figured out those leads are really thick and the diode is really big, so I checked the datasheet and it actually is a 3amp 1000volt max diode...

So to make the story short I'll keep looking for the 5819, or the 4007. I found a bunch of 4004s that I salvaged from a power supply, they are 400volt max and 1amp, but I don't know if they'll work?
 
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you should be able to test the diodes with a multi-meter [VOM or DVMM]. Connect the leads to the diode, get an open or short reading. Reverse the leads, and get an opposite reading.
 

The coaxial power supply input should always be connected to the circuit or to the on-off power switch. Then the battery is connected to the switch in the coax jack.
When an external power supply is not connected to the jack then the battery is switched to power the circuit.
The switch in the jack disconnects the battery when an external power supply is used and a diode is not needed.

The on-off power switch should switch the positive supply, not the grounds.
 

Audioguru said:
The coaxial power supply input should always be connected to the circuit or to the on-off power switch. Then the battery is connected to the switch in the coax jack.
Click to expand...

That describes what my circuit does. The coaxial power supply is connected to the on-off power switch. The battery is connected to the switch in the coax jack.

Audioguru said:
When an external power supply is not connected to the jack then the battery is switched to power the circuit.
Click to expand...

Correct.

Audioguru said:
The switch in the jack disconnects the battery when an external power supply is used and a diode is not needed.
Click to expand...

There's exactly where the problem lies because even though the battery is connected to the circuit and the jack is disconnected, internally the battery is still connected to the voltage regulator (to the regulator's output, so the battery voltage goes backwards into it).

Audioguru said:
The on-off power switch should switch the positive supply, not the grounds.
Click to expand...

Will do.

kam1787 The 4004s do work, what I didn't knew was if they would work for the circuit.
 
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I do not know why the external power supply is higher than 12V and you have a 12V regulator (the minimum input to an ordinary 12V regulator is 14V).
An LM386 amplifier works perfectly from an unregulated 9V supply then a regulator and a diode are not needed.
 

OK, Thanks pradeep!

Audioguru, that is because of all the wall adapters I have, ranging from cordless phones to laptop ones. I wanted to experiment with some of the features the IC includes (gain with a potentiometer, bass switch, 4 to 12 volts, etc...).

I will change the voltage regulator to a 9v after experimenting a bit with it so the IC doesn't fries (I noticed it already gets hot at 9v).

Thanks for all the replies guys!

**broken link removed**
 
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The datasheet for the LM386 amplifier IC shows that it needs 10 ohms in series with 0.05uF (47nF today) from its output to ground to prevent oscillating at a high frequency that makes it very hot.

With a 9V supply, an 8 ohm load and if the output is at full blast (0.7W but badly distorted) then the heating is 0.5W so its chip is about 100 degrees C which is hot but below its maximum allowed temperature of 150 degrees C.
NOBODY plays music (acid rock is not music) at full blast all the time so the chip will be much cooler.

With a 9V supply and an 8 ohm speaker the output of an LM386 is only 0.45W with low distortion. When bass boost is added then the higher frequencies have their levels reduced to only 0.225W which is not very loud.
 
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