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I have a simple component that plays a song when I pull the input pin high. It operates at 3V and everything works fine when I provide power from my bench power supply(I can also see that it draws 0.15A current when operating). I want to build it into a product that has a 5V 2.1A adapter as a power supply so I added an LM317T voltage regulator. I configured the regulator with R1=390Ω and R2=560Ω according to Figure 6. of the datasheet and when I test the output without load it does provide 3V but when I use it to drive the audio component the output voltage drops to ~2V and the audio is played in a slow and distorted way. At first I just had these two resistors but since then I also added all the recommended capacitors(but not the recommended protection diodes, Figure 7.) but it still has the same behavior. According to the datasheet the LM317T should be able to provide 2.2A of current. My questions are: what am I doing wrong and how to fix it?

Edit: the schematic is the same as Figure 6. Tried to draw it in kicad if it helps(this is the simpler version, I've tried with the caps as well from Figure 7 without the protection diodes): enter image description here

@Justme: Normal speaker not a piezo(8Ω 0.5w)

@brhans: I took a picture(it's the left regulator): enter image description here

@audioguru: I'll test that and get back to you but I doubt it, it should be able to provide 2.1A

Took a picture of the component as well, it's from an old alarm clock. The cables are: red=3v, gray: GND, green: input signal(3V), and the two yellow ones are going into the speaker(+/-). enter image description here

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    \$\begingroup\$ Show us the schematic. \$\endgroup\$
    – Hearth
    Jun 27, 2021 at 14:43
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    \$\begingroup\$ @erdeszt Indeed you got the answer all by yourself. Page 5 of the datasheet shows TYPICAL minimum drop-out voltages. It's not guaranteed. And they hover around 2V. The LM317 is NOT an "LDO" (not a 'low voltage dropout" regulator). You are asking it to drop less voltage than it's capable of. It's a VERY old design (40 years?) Modern LDO's would have NO PROBLEM handling this. You can easily find one that's "fixed" at 3.0V. Further, note the LM317 has a MINIMUM output current ... If you're not pulling enough from it, the output floats around. Modern LDO don't do that either. \$\endgroup\$
    – Kyle B
    Jun 27, 2021 at 15:48
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    \$\begingroup\$ I knew that was your problem just reading your first paragraph. I was quite happy to see you'd found the solution yourself. Well done :) BTW, 3.3V is a standard "digital" voltage, and PROBABLY would be OK here. But if you don't KNOW that to be true, you do take some small risk by using it. \$\endgroup\$
    – Kyle B
    Jun 27, 2021 at 15:51
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    \$\begingroup\$ Another option, much simplier, is use 3 pieces 1n4148 diode in series. Each would drop about 0.7V, so you'd get very close to your desired 3V (without needing capacitors). You'd want to add maybe 1k from 3V to GND, just to make sure there is SOME current flowing in the diodes all the time (keep them at 0.7V drop). With zero current flowing, they'll not drop quite so much voltage. SInce you're on a power supply, not batteries, this trickle current shouldn't even be noticed. \$\endgroup\$
    – Kyle B
    Jun 27, 2021 at 15:56
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    \$\begingroup\$ HI Brother. THx for that offer. I'm not here for the points... just to help my fellow man out. Good luck!!!!! \$\endgroup\$
    – Kyle B
    Jun 27, 2021 at 16:08

1 Answer 1

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@erdeszt Indeed you got the answer all by yourself. Page 5 of the datasheet shows TYPICAL minimum drop-out voltages. It's not guaranteed. And they hover around 2V. The LM317 is NOT an "LDO" (not a 'low voltage dropout" regulator). You are asking it to drop less voltage than it's capable of. It's a VERY old design (40 years?) Modern LDO's would have NO PROBLEM handling this. You can easily find one that's "fixed" at 3.0V. Further, note the LM317 has a MINIMUM output current ... If you're not pulling enough from it, the output floats around. Modern LDO don't do that either.

I knew that was your problem just reading your first paragraph. I was quite happy to see you'd found the solution yourself. Well done :) BTW, 3.3V is a standard "digital" voltage, and PROBABLY would be OK here. But if you don't KNOW that to be true, you do take some small risk by using it.

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