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I have a CC/CV DC to DC converter I bought on aliexpress.

I does not come with a screen so I attached to it an LED voltmeter ammeter.

Anyway everything works fine.

One thing though, I don't know at what value the CC potentiometer is at. To adjust before attaching a load, I short the output and the ammeter gives me the Amps the CC is at. So I can adjust it.

My question is. I want to add a push button which would short the output so I can see the CC value. A pushbutton normally open.

Would that cause problem if there's a load currently attached to the output as well. I'm thinking about sparks and bounces and voltage spikes.

I am a software engineer just tinkering with electronics.

Thanks!

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  • \$\begingroup\$ @jsotola because the load may not be pulling enough current to put the thing in CC mode. \$\endgroup\$
    – TimWescott
    Commented Feb 27, 2019 at 19:40
  • \$\begingroup\$ @jsotola exactly... maybe the load is pulling 250mA but I want to set the CC at 1A \$\endgroup\$
    – Mike Gleason jr Couturier
    Commented Feb 27, 2019 at 19:41
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    \$\begingroup\$ @jsotola: This is a problem with some of the lower cost bench power supplies. We got some in work by mistake. You can't read the current limit before switching on so the only way to do it is to short the terminals, switch on, set the current limit and then remove the short. OP seems to have a similar problem. \$\endgroup\$
    – Transistor
    Commented Feb 27, 2019 at 19:46
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    \$\begingroup\$ install a spring loaded SPDT toggle switch ..... COM connects to V+ .... NC connects to load ..... NO connects to GND \$\endgroup\$
    – jsotola
    Commented Feb 27, 2019 at 20:57
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    \$\begingroup\$ something like this switch .... amazon.com/Safety-Toggle-Switch-Center-Momentary/dp/B006WQR3R4 ......... google switch spdt on-on one side momentary \$\endgroup\$
    – jsotola
    Commented Feb 27, 2019 at 21:11

2 Answers 2

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Yes, it would, but probably not too severe -- and "too severe" depends on your situation, so you'd need to test.

You could make it a bit better by putting a low-value resistor in series with the button.

You could make it better yet by using a momentary-contact, break-before-make SPDT switch that disconnects the load and then shorts the terminal. I'd still use the above-mentioned resistor, though.

To actually find the switch you'll need to go shopping -- they're out there, I can't recommend a part number, and it just feels like this is the sort of thing that'll be more readily available as a toggle switch than a button, but that's for you to discover.

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  • \$\begingroup\$ good suggestions, thanks for your answer. you make me think also there are 3 way switches, so I could use one where the bottom position is used for the load, the center does nothing and the upper position would be the CC adjust. The switching would be more "isolated". I liked the idea of the push button because you can't "forget it", it re-opens the short by itself. \$\endgroup\$
    – Mike Gleason jr Couturier
    Commented Feb 27, 2019 at 19:44
  • \$\begingroup\$ I would advise against using a break before make and recommend using a push to make button. If you use a break switch the PSU output voltage will rise to the open-circuit value and charge up any output capacitors to max voltage. This will then get dumped into the load when the load contact makes. \$\endgroup\$
    – Transistor
    Commented Feb 27, 2019 at 19:48
  • \$\begingroup\$ @Transistor: I'm assuming he'd use the switch when the thing is in CV mode, so breaking the circuit isn't going to result in a great increase in output voltage (unless the regulation is crappy). \$\endgroup\$
    – TimWescott
    Commented Feb 27, 2019 at 19:49
  • \$\begingroup\$ @Transistor I guess the same thing would happen even without a switch. If I manually connect the load, disconnect the load, then reconnect it. No? \$\endgroup\$
    – Mike Gleason jr Couturier
    Commented Feb 27, 2019 at 22:17
  • \$\begingroup\$ The added benefit of the switch @jsotola is talking about in another comment is that I would be able to turn off the load without disconnecting the wires. Useful. \$\endgroup\$
    – Mike Gleason jr Couturier
    Commented Feb 27, 2019 at 22:20
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All that should happen to the load if you place a short circuit in parallel with it is it will see zero volts. Whether or not the load can tolerate this depends on what the load is. If you're just powering some LEDs, then nothing bad should happen, however if you're charging a battery, then as soon as you short the output, the battery will output whatever it's short circuit current is, directly into the switch, likely destroying either the switch, the battery, or both.

There may be some inductive spikes when the short circuit is "broken" when you let go of the button, but I wouldn't worry about that unless you're dealing with large currents (above a few amps), or long lengths of wire (dozens of feet). If you find that inductive spikes are a problem, you may add a diode next to the switch like so:

schematic

simulate this circuit – Schematic created using CircuitLab

Be sure to choose a push-button that can take tolerate both the output voltage of the converter, as well as the maximum CC current output that the converter can output, otherwise you may fry your switch.

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    \$\begingroup\$ Thank you very much for your answer! Didn't thought about the battery scenario. Good catch. \$\endgroup\$
    – Mike Gleason jr Couturier
    Commented Feb 27, 2019 at 19:48

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