Converting 2 wires cheap voltmeter to 3 wires


You can find these very cheap voltmeters all over ebay.



They have a single wire for both power supply and volt measurement.
No surprise, they need enough voltage to power up the circuit and display, that is ~ 3.3 v

As a result they are  useless to display lower voltage.

If we flip them, the power supply part is simple enough :

On the Top left,there is a diode marked as D1 (reverse polarity protection), followed by a 7133H  voltage regulator on the bottom left to get 3.3 v.

Just remove the diode and connect 5v power supply there

Done !
Now you can measure much lower voltage. It might be a good idea to put back a reverse polarity protection diode somewhere ....

Comments

  1. DanieleNovember 29, 2019 at 3:57 AM

    Hi Meanx, I was searching something about issues of this device, due often it burns voltage regulator. I use these display only to read the charge of a battery (12v 20amp here the data sheet model EA20
    http://www.go-aliant.com/lithium/images/datasheet/Datasheet_Aliant_EA_Series_EN_201612_E.pdf
    Any suggestion on about how I can avoid this issue? Many Thanks in advance.

    ReplyDelete
  2. meanXNovember 29, 2019 at 4:31 AM

    Normally they should be ok up to 24v
    If not, you could try using a big 7805 to lower the voltage on the VCC pin of the voltmeter
    That will lower the heat the 7133H has to dissipate by a lot

    12V => 7805 => Power supply of the voltmeter
    ==================> measure pin of the voltmeter

    You still need to use a 3 pins voltmeter or convert a 2 pin to a 3 pins as explained above

    ReplyDelete
    Replies
    1. DanieleNovember 29, 2019 at 4:39 AM

      Many thanks Meanx. I buught these display and now Im having issues to my customers. I just discovered (im not an electronics) that the diode I add on the circuit (blue 12vdc), make rise up the current consumption from 12mA to more than 30mA...it could be the cause too?

      Delete
  3. meanXNovember 29, 2019 at 6:59 AM

    It depends how you connected it.

    If you draw 30mA from the regulator it's too much
    The regulator is MAX 30mA and since it is linear i'd be dissipating
    (12-3.3)*30 mA>250 mW which is a bit too much, depending on the package it"s either 200 mW max or 500 mW max

    If you connected the LED directly to the battery, no problem

    ReplyDelete

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