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I just bought this power supply for a project:

https://uk.farnell.com/mean-well/gst120a20-p1m/adaptor-ac-dc-20v-6a/dp/2815845

It is supposed to power two somewhat larger stepper motors (and an Arduino), so I needed something with a beefier amp rating.

After attaching it to a panel mount DC connector, I just went in with a multimeter to test the voltage and PINs, which gives me a 20V reading as expected.

However, after I remove the power supply from the wall, I still see 20V initially, and then I see the voltage dropping quite slowly, over a fairly long period; I'd say around half a volt per second, or even slower.

  1. I'm a beginner, and don't know what to make of this? Is this a normal phenomenon and does it have a name?
  2. Is my power supply broken or low-quality (it didn't come cheap!) Did I measure it wrong?
  3. Might this not be an issue with devices that expect a certain minimum voltage?
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  • \$\begingroup\$ This is probably just leftover energy in the energy-storage components that are needed to make the power supply operate reliably. However, a bit more info could help us: How are you taking the measurement? Is the load (Arduino plus steppers) attached when you take the measurement? \$\endgroup\$
    – nanofarad
    Dec 10, 2019 at 15:10
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    \$\begingroup\$ You might notice that the voltage drops to zero much faster if it's connected to your system (or any other load). \$\endgroup\$
    – Hearth
    Dec 10, 2019 at 15:27

2 Answers 2

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It is a normal phenomenon.

The reason it happens is because there is a capacitor which can hold a charge. Even when the power is off, it can keep it, or release it (which it does).

Sometimes you see it at a laptop connection that has in the power cable a small (ring type) LED, that glows when attached to mains, and when disconnected it slowly decreases in brightness until off; this can take also tens of seconds.

I don't know if there is an official term for this, but the releasing is called 'capacitor discharge'.

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    \$\begingroup\$ Even a phone charger can demonstrate this in some cases (like keeping the "fully charged but plugged in" LED lit on one of my bike lights for a couple of seconds). \$\endgroup\$
    – Chris H
    Dec 10, 2019 at 15:51
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    \$\begingroup\$ Of course. So if the device is actually charging the caps will empty in nothing flat, but if the charge circuit has turned off they'll keep an indicator on for a bit \$\endgroup\$
    – Chris H
    Dec 10, 2019 at 16:44
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    \$\begingroup\$ @ChrisH Than my understanding was equal … I always like watching until the LED fully stops shining, but cannot recall a device that took more than about 10 to 20 seconds. \$\endgroup\$
    – Michel Keijzers
    Dec 10, 2019 at 17:13
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    \$\begingroup\$ It can be even dangerous in the case of transformless supplied devices. \$\endgroup\$
    – Circuit fantasist
    Dec 10, 2019 at 18:37
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    \$\begingroup\$ @Circuitfantasist Also in a (small) charger? I know for televisions it can be hazardous because of the high voltage. \$\endgroup\$
    – Michel Keijzers
    Dec 10, 2019 at 18:41
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Perfectly normal on a power supply with no loads on it.

It's a capacitor, which is like a buffer. It job is to smooth out the DC power by resisting changes in voltage. The capacitor is trying to keep the voltage at 20V even though you turned it off.

If there were an actual load on this power supply, the load would instantly consume this buffer of energy. However, since there is no load (or the loads are switched off), the capacitor's charge just sits there, waiting, oblivious that you have turned off the power.

In fact, an unsuspecting technician can get nailed by this stored energy! So they add a feature called a "bleed-down resistor" that drains the capacitor once the power is off. That's why you see the voltage decline. It's slow because the resistor is actually bleeding even when it's on, so it's kept small/slow to save energy.

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