I know that it is possible to generate higher voltage from lower voltage source, I just don't know how ...

Anyway, few days ago I got this:


As you can see, it's volt-meter and that's the mistake I've made. I just needed some kind of analog display that I could control (I want to make something else, not volt-meter). And now I know that I should have bought ampere-meter display, so that I could control it using current, and not voltage.

So, my question is, can I control that 30V volt-meter using 5V components (any ideas are welcome) ?

If there isn't any way (or any practical way), how can I generate 30V from 5V ? Will that be linear ? I mean, will the output voltage linearly depend on input voltage ?

In the end I would want to be able to control that volt-meter display using PWM signal.

Thanks :)

  • \$\begingroup\$ @Fake Name 's answer is good - but be careful testing the meter after removing an internal series resistor. It is liable to be a current meter and very sensitive. So applying eg 3V to it may destroy it. A good idea is to measure any series resistor that you find inside and then add a smaller one in its place or externally and test). eg if you find a 27k internal resistor it is probably a 1 mA movement - try a 10k resistance instead and meter should be about 12V. Then adjust down from there wity lower resistor values. If R is about 270k it is a 100 uA movement - do first test with 100k etc. \$\endgroup\$
    – Russell McMahon
    Feb 19 '12 at 22:36
  • 1
    \$\begingroup\$ BETTER: Take meter as is now & measure a voltage. Higher the better up to 30V. Then (1) ADD a 10k resistor in series and note reading. (2) Then instead add a 100k resistor in series and note reading. If it is a 1 mA movement the 10k will reduce reading by ABOUT 1/3rd and the 100k will make it about zero. If it is a 100 uA movement the 10k will hardly affect it and the 100k will reduce it by ABOUT 1/3rd. Now you know basic sensitivity (approximately) and can go from there. With care and thought you can calculate correct new series resistor using this method. \$\endgroup\$
    – Russell McMahon
    Feb 19 '12 at 22:40
  • \$\begingroup\$ It IS analog meter where 1mA is maximum value.I've opened it as FakeName said and there was one resistor, rated 29.7k. So when I apply the 30V, the current will be 1mA. So I'll just buy 4.7K resistor, and then when I apply 5V there will be 1mA current, and the meter will show maximum value ;) \$\endgroup\$
    – xx77aBs
    Feb 20 '12 at 6:34

It should be possible to convert that meter to a more usable voltage range.

Generally, analog meters use a movement that is somewhere in the range of 100 uA to 1 mA full scale.

In other words, inside the meter is a resistor divider that scales the meter range to 30V, so that when 30V is applied to the terminals, there is 100 uA/1 mA of current flowing through the actual meter coil.

The meter you link to has two screws holding the terminals on the back of the meter on. It is quite likely that you can simply unscrew the back of the meter, and remove/change the scaling resistor divider.

The one exception to the 100 uA/1 mA meter rule is for direct-reading current meters, particularly for AC. Analog current meters in < 10 - 25 A range that do not specify they require an external shunt may not be possible to convert, as they use a different mechanism,

  • \$\begingroup\$ Appl a voltage <= 30V but as high as possible. \$\endgroup\$
    – Russell McMahon
    Feb 20 '12 at 1:48
  • \$\begingroup\$ @Russell McMahon - what? \$\endgroup\$ Feb 20 '12 at 3:05
  • \$\begingroup\$ <=30 but preferably 30 29 28 27 26 25 24 just maybe 23 22 21 20 19 18 17 16 15, getting a bit low if 14 13 12 11 10, and 9V if you must but you can do better than 8 7 6 5 4 3 2 1 or less. Yes? \$\endgroup\$
    – Russell McMahon
    Feb 20 '12 at 11:04


Summarised solution:

  • This method will work in many cases to reduce the maximum voltage value that a panel meter reads. This is mainly for typical modern moving coil meters but will work for some but not all meter types.

  • If you read this and understand it you will be able to use the results obtained for other purposes.

  • The method calculates the internal meter resistance and meter full scale current allowing calculation of the correct series resistance for an other voltage (higher or lower). Meter resistance is allowed for.

A 30 V meter is assumed as the original question related to a 30 V meter.
The text has been altered to be more general as Vfs has been used as meter full scale voltage.

In this specific answer only Vfs = 30V.

(1) Apply a test voltage not more than Vfs (here=30V) but as close as possible to Vfs. eg 30V (= Vfs) = excellent
24 V = Good
12V = acceptable 9V (eg PP3 battery) = OK, higher would be better. 5V = too low really

Read and record voltage shown on meter.
Call this V1.

(2) Connect a resistor in series with the meter and apply V1 again.
Current path is V1+, into resistor, out of resistor, into meter +ve, out of meter -ve to V1-.

Ideally select a resistor such that reading is about half V1 but as long as it is between say 10V and 20V it should be OK.
If reading with resistor is above 20V add more resistance.
If reading with series resistor is below 20 V use a smaller resistor.
When reading is > 10V and < 20V goto next step

Call series resistor R2 (there is NO R1) Call new voltage reading V2

Meter resistance consists of

  • internal series resistor = Rint
  • resistance of meter movement = Rmeter.

2(b) Alternative: You can by pass steps 1 & 2 above by measuring (Rint + Rmeter) directly using an ohm meter.
Ideally use a meter with a 9V battery for ohms testing.
Measure on several ohms ranges.
Best value is probably one not at extreme high or low end of a range.

2(c) Semi-expert alternative to (1) + (2) : If you have an external power supply with an accurate current meter or an external current meter of appropriate range you can measure (Rint + Rmeter) by applying
R=V/I = (V_across_meter/I_through_meter)
Note that if u=you us a series current meter it will drop some voltage. Measuring V_across_meter literally across the meter with a voltmeter may be a good idea. If the series current meter changes value more than very very slightly if at all when you apply or remove the voltmeter across the meter on test EITHER deal with it if you understand how OR uses tests above instead.

(3) Calculate (Rint + Rmeter) = R2 x V2 / (V1 - V2)
This is the effective resistance of the meter as seen by the circuit being measured.

Calculate: I1 = V1 / (Rint + Rmeter)

I1 is the current needed to produce the reading V1.

(4) Calculate: Ifs = Vfs/V1 x I1

Ifs is the full scale meter current.
Ifs MAY be 1 mA or 100 uA or in a few cases 25 uA or some other current.
Regardless of what it is, this is what it takes to deflect meter to full scale.

(5) Dismantle meter.
There will PROBABLY be a single resistor in series with the meter movement.
ie Meter +ve terminal via resistor via meter to meter-ve terminal.
OR meter +ve terminal via meter via resistor to meter -ve terminal

6(a) IF the arrangement is NOT as above with meter + a single resistor, take a photo, record any markings on internal components and report back.

6(b) IF arrangement is as above with meter movement + single resistor then:

  • Remove resistor - this will PROBABLY be best done by cutting the leds near but not at where they are terminated soldered / ... so there is room to solder on a nw resistor. IF the resistors is surface mount either do what seems best to allow a new resistor to be inserted OR report back. DO NO destroy resistor when removing if possible.

  • Measure resistance of removed resistor.
    Call this Rint

(7) To make meter read full scale voltage of your choice = Vnew:

Calculate: R3 = [Vnew / Ifs] - Rint

Install resistor of value R3 in place of old Rint

Note that Vnew may be higher or lower than the original full scale voltage.


Note that the MINIMUM direct voltage that can be set as Vfs is when Inew + 0.

ie Vfs_min = Ifs/Rmeter

Where Rmeter = (Rint + Rm) - Rint

  • \$\begingroup\$ Holy misspelling and formatting issues, batman! \$\endgroup\$ Feb 20 '12 at 3:06
  • \$\begingroup\$ @Fake Name - Exquisite answers, good spelling, proper formatting - choose any one. \$\endgroup\$
    – Russell McMahon
    Feb 20 '12 at 11:06

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