This is the complete wiring done inside the PSU. Picture shows the DC DC Buck converter module that I am using to make the PSU

I am new to this forum. I am a Diploma in Electronics from India and a passionate DIY hobbyist too.

I have a question regarding cheap 10 amp ammeter and a voltmeter module I bought off aliexpress.

I am building a variable DC PSU using a 24 volt 10 ampere SMPS as the main power source and feeding its output to a cheap XL4016E1 based buck converter which has CC CV Control. I have done all connections right and the PSU works fine on 230 V AC. Now I want to run the buck converter on external DC input (I have installed protection diodes to prevent clashing of the SMPS output with the external DC input) and was wondering whether the voltmeter ammeter module will need an isolated DC DC converter module to power it up to prevent damage to it while I run the entire PSU on external DC input, or should I simply think of installing a 9 volt battery to power the voltmeter ammeter separately and activate battery power only when external DC supply is detected at the DC jack by using a relay.

PSU Photo 1 PSU Photo 2

  • \$\begingroup\$ Welcome to EE.SE, but how can anyone answer this without a schematic of your circuit and a link to the datasheet for the module you bought? No datasheet? Then see what to check for when buying an electronic component or module. You should upload your photos to the site instead of having your readers follow links to them. \$\endgroup\$
    – Transistor
    Jul 4, 2020 at 16:34
  • \$\begingroup\$ I am new to blogs and stuff and still figuring out how these things work. I will upload a schematic of my PSU soon. \$\endgroup\$
    – Alex
    Jul 4, 2020 at 16:48

1 Answer 1


The voltmeter/ammeter needs a voltage in a certain range to operate properly. That's the thin black and red wires.

Usually claimed to be something like 4-30V but you might want to be a bit more conservative.

Your two 'or' diodes should be upstream of the regulator module, and should be applied to the POSITIVE input, not the negative. So from that junction you have 24V - 0.7 or so if powered by the 24V supply, and whatever your external input is minus a diode drop (whichever is higher at the moment). If that fills the specification of your meter module you are fine.

Note that the current meter measures via the return current lead. The two black leads (thick and thin) are directly internally connected.


simulate this circuit – Schematic created using CircuitLab

Now, if you REQUIRE a common between input DC ground and output ground then you have a problem (that an external supply or battery will not solve) because the modules measure current on the return side. High-side current measurement would cost considerably more money than the $0.01 op-amp they use.


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