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According to this answer the LM7805 has no minimum load. The data sheet also indicates that the output voltage should remain between 4.8 and 5.2 V for output currents of between 5 mA and 1 A.

I have a LM7805CA from an unknown manufacturer (I don't recognize the logo, which is an upward-pointing triangle to the left of a downward facing triangle enclosed by lines on all but the bottom side), but the additional markings appear to be "GOOKJ V6" and "MAR 918." (Unfortunately the printing on it is such that it's difficult to read and near-impossible to photograph.)

I've got the output hooked up to a constant-current variable load and the input is around 9 V (I see similar behaviour throughout a 7 V to 12 V input range) from a bench power supply, like so:

LM7805-test

The LM7805 is underneath the red alligator clip (you can just see the edge peeking out the top) and test clips you see there lead to my multimeter; I'm taking all voltage readings from that since it's more accurate than the load's display. (The actual output voltage in the piacture is 4.96 V.)

As you can see in the image, with about 170 mA of load the output voltage is within spec. As I increase the load to 200 mA the output voltage increases to around 5.02 V and stays there even up past 500 mA.

However, when I reduce the load to around 120 mA the voltage drops to 4.84 V (still within spec, but barely) and at 80 mA the output voltage drops to 4.72 V. The load won't reliably sink much less than this (I have seen 4.66 V at around 60 mA), but earlier, playing with just an LED and 330 Ω resistor in series between the output and ground (on a breadboard), I saw the output voltage drop even further, to around 4 V.

So what's going on here? Is this device really going out of spec, or is it something to do with the way I'm measuring things?

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    \$\begingroup\$ Have you followed the designs in the datasheet for the LM7805? \$\endgroup\$
    – HandyHowie
    Commented Mar 6, 2020 at 9:30
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    \$\begingroup\$ "If it is going out of spec, is that just expected for random no-name parts from AliExpress?" You can have absolutely no expectations whatsoever from Ali. This is a standard component, highly available around the world. Just buy one from a well-known silicon vendor. But of course it's "cheaper" to buy unknown crap and trouble-shoot for hours than to pay $1 for a proper component, right? \$\endgroup\$
    – Lundin
    Commented Mar 6, 2020 at 10:10
  • \$\begingroup\$ @Lundin I'd love to pay $1 for a proper component. The problem is, by the time it gets to my door from Mouser or similar vendors that $1 component has turned into a $20 component. \$\endgroup\$
    – cjs
    Commented Mar 6, 2020 at 11:53
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    \$\begingroup\$ If you are going to buy components that are, as you said "random, no-name parts", then you just have to live with what you get. It isn't reasonable to come here and expect a good answer when your parts could be counterfeit or damaged or factory rejects. We get weary reading questions like this, sorry. \$\endgroup\$
    – Elliot Alderson
    Commented Mar 6, 2020 at 13:14
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    \$\begingroup\$ And to explain Elliot's remark; we get this a lot. Someone buys parts from AliExpress / DealExtreme / Ebay / whathaveyou, tries something with them, and something doesn't work as expected. They ask a question here, people put in effort trying to debug the problem. Turns out some cheaply obtained part is junk. Mildly amusing the first time, less so the next thousand times. Because this happens frequently, people are less inclined to help when there's lottery parts involved, and instead they just tell you to buy proper parts first. I can't blame them. \$\endgroup\$
    – marcelm
    Commented Mar 6, 2020 at 17:12

1 Answer 1

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Measuring with an LED and 330 Ω resistor is okay. So if you did that measurement correctly and it only provides 4V then it doesn't meet the spec in the data sheet you linked. (Whether that's the spec for that actual part is unknown since the manufacturer is unknown.)

When measuring with that variable load device, you may need capacitors depending on what exactly the load is doing.

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