I am having problems correctly setting up my volt/ammeters. I am attempting to use them to monitor a boost converter I recently purchased. I would like to have one set up on the input of the boost converter and the second on the output. I have tried many ways of wiring but for some reason the V/A meter on the input will not read the amperage. Also when input V/A meter is connected the output V/A meter will show an extremely high amp reading (I'm assuming it's taking the reading from both V/A meters' shunts instead of just it's own). I have switched the meters to rule out a faulty meter. They both read voltage fine and the output will read amperage fine (when input meter is disconnected. I have to assume I am wiring something wrong, so I have included a wiring diagram (forgive me if it's bad, I've never made one before). Also I have included info on the parts I used. How I wired everything

Following Information added after original post Boost Converter was advertised as 1200w, 20amp, input 8v-60v with output of 12v to 83v Boost Converter

V/A Meters Used

-My 12v power supply was a modified computer switch mode power supply.

-The actual load was electrolyte solution with copper as both anode and cathode. (I use electrolysis for de-rusting items).

Just to summarize, V/A Meter 1 (Input Meter) won't read amperage, but does read voltage. V/A Meter 2 (Output Meter) reads both voltage and amperage, but gives incorrect amperage reading if Meter 1 is connected. Why is this? How do I fix this and/or wire this correctly?

  • 2
    \$\begingroup\$ Those meters are probably supposed to be run on an isolated supply. You're running both from the one supply so you have a sneak path from your output to your input. Try running with only the input meter and then only the output meter. If that works try running both with isolated supplies. \$\endgroup\$
    – Transistor
    Commented Aug 22, 2019 at 7:47
  • 1
    \$\begingroup\$ I will definitely try that, and now that you mention that, when i first got the meters, there was an ohms reading between the black shunt wire and the black wire for the meters power source. So that would mean, the electrical path could (or does) go through the meters power source ground wire, therefore bypassing the shunt and resulting in "0" amps reading on the input meter... I have been trying to figure this out for days and although i haven't yet tested your idea, im feel like your spot on. Thank you! And i will post my results once i am able to test \$\endgroup\$
    – ICEMAN
    Commented Aug 22, 2019 at 7:59
  • \$\begingroup\$ If the bipolar junction is correct power the input off of 12V and output 5V. \$\endgroup\$ Commented Aug 23, 2019 at 1:58
  • \$\begingroup\$ @ICEMAN - After reviewing this question, and re-reading it, I seemed to have previously missed the part where you hypothesized the reason for your extremely high current readings on the output meter. You were completely correct, and you deserve props for that! With the ground path of the output meter being shorted to the power supply side of the input shunt, your output meter was, in fact, measuring the voltage drop(which it uses to calculate the current) through both shunts and the wiring between them, which would actually show more than double the current draw. Good catch!!!! \$\endgroup\$
    – Hitek
    Commented Aug 24, 2019 at 5:36
  • \$\begingroup\$ @hitek Thank you for the compliment! I'm am still very new at everything electrical. I'm just hobbyist tinkering in my freetime and it's nice to know I might actually be learning something. And the corrected diagram you posted in your answer below will serve my purpose perfect. I will test that configuration and post if it works. Thank you! \$\endgroup\$
    – ICEMAN
    Commented Aug 24, 2019 at 7:28

1 Answer 1


You have shorted the input shunt:

Shorted Shunt

Both the power input and current measurement connectors(as indicated by the "dangling" ground wire of the power connector in the left meter wiring diagram) share a common ground. At very least, you should completely disconnect the five volt power source(including ground) from the input meter(and 12 Volt power source), and connect the meter's power input to the 12 Volt power source along side the input voltage measurement wire, also indicated in the left-most wiring diagram for the meters.

Ideally, if the output of the boost converter is less than 30 Volts, you would connect the output meter's power and ground to the output of the boost converter just like you did with the input meter, eliminating the 5 Volt source completely. If the output of the boost converter is too high, then you could use some other regulator connected only to the output of the boost converter to get the supply voltage down to the ~28 Volts allowed by the meter.

You might be able to just bypass the regulator completely and connect the output meter's power input to the 12 Volt input of the boost converter. This would need to be tested for accuracy{*}, however, as there will be a slight difference in potential between the grounds of the two meters, being separated by the shunt.

No Short

If that doesn't work, and your 5 Volt source is actually it's own regulator of some sort and it's input can handle the input or output voltage of your boost converter, then you could repurpose that by connecting it's ground to the boost converter's side of the ground, making sure it is in no way directly connected to the 12 Volt source's ground.

If you need all of this to be connected "in the same box with common grounds", then you should be able to move the ground of the input meter to the boost converter side of the shunt, and the current sense wire to the 12 Volt power supply side of the input shunt, but then the input meter will show negative current rather than positive.

In any case:

TLDR; The 12 Volt power source's ground and the boost converter's ground MUST be completely isolated from each other for this to work.

{*}Although I have three of these meters myself, I have not tested them in this configuration specifically. I did, however, test the forward and reverse current aspects through the shunts.

  • \$\begingroup\$ Thank you for this answer. I was finally able to test your configuration and it worked flawlessly. To test I used 2 vehicle headlights in series as the load. Both meters read accurately and with all measurements. With input at 11.6v @ 3.64a the output was 17.2v @ 2.31a. So 42.2 watts in equals 39.7 watts out. That would make it roughly 94% efficient? Did I do that right? Also, what exactly is meant by forward and reverse currents? Like switching which way the current flows through the shunt? \$\endgroup\$
    – ICEMAN
    Commented Aug 25, 2019 at 9:16
  • \$\begingroup\$ @ICEMAN - That sounds perfect! Although, I should say, when I mentioned verification, I meant comparing the outputs of the dedicated meters to the values measured using a more accurate hand-held or bench DMM. The readings should be very close, even though I admit that these all-in-one meters didn't track well in measurement tests compared to a 6 digit meter I used to compare them. They did track within a fifth of a volt(0.2 Volt difference Max) in my tests, so if you are getting similar results, then you should be there! \$\endgroup\$
    – Hitek
    Commented Aug 25, 2019 at 9:25
  • \$\begingroup\$ @ICEMAN - Sorry I missed the question about the current. The meters that I have(which are several years old but appear identical to yours) measure the voltage drop across the shunt, and will operate in the positive and negative voltage range. If, for example, you connected the shunt inline with the battery on your vehicle, they will show negative current flow from your battery when it's not running but there are still active loads(like the key being on, and/or chargers in the lighter ports). Whenever the vehicle is running and charging, they will show positive current flow to the battery... \$\endgroup\$
    – Hitek
    Commented Aug 25, 2019 at 9:45
  • \$\begingroup\$ After disconnecting the black ground input wire it started showing correct current!! I was powering the meter from same input source as the buck converter. \$\endgroup\$
    – the Hutt
    Commented Jun 22, 2023 at 6:18

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