# Driving a 1930s multimeter / voltmeter dial from a software controlled source

#### cisor

Mar 17, 2022
7
Summary: The problem I think I have - Multiplying DC 0 - 3.3v volt signal to 0 - 8v range
Is there a means to produce the alternating current required for a doubler circuit either with a voltage under 0.5v or with separate power / signal channels?
The PIFCO will also measure 0 - 240v AC, this might give more options?!

Detail: What I'm trying to achieve and why I'm probably asking the wrong questions!...

Hi,
I have this beautiful, Bakelite, c. 1930s, all-in-one multimeter built by PIFCO.
I'd like to display it, but I'd also like it do something!

The specific something barely matters maybe a CPU usage indicator, maybe internet bandwidth, maybe soil humidity for my greenhouse.
Any of this data I can grab from my various smart home devices and can be pushed to, or pulled from, a controller through various software means.
The software side is something I can comfortably do.

The electronics is where I need advice!

I have several Raspberry Pi (of various vintages) lying around so I'd like to use this for the control.
I have proven the concept of software controlled voltage as shown in the video below.
(Edit: I'm being told by the forum that I cannot post an embedded video or video link. I'm sure you can picture the needle moving around points between 0 & 3.3v! https://i.imgur.com/xUpYkLe.mp4. Edit to the edit, apparently I can paste it plain, but not use the link function of the editor!)

BUT I'd like to use the whole range of the dial.

Googling suggests I need a voltage multiplier circuit. These seem to need either an AC source or a chopper device.

A 555 timer controlled voltage doubler / tripler / multiplier appears to be the answer at first look.
The circuits I have seen seem to suggest the powering of this IC needs to be minimum 5v. IC operating power and the signal to be multiplied appear to be one and the same.

Is there a means to produce the alternating current required for a doubler circuit either with a voltage under 0.5v or with separate power / signal channels?

The PIFCO will also measure 0 - 240v AC, this might give more options?!

Limitations:
• I cannot see any way to open the device without damaging it, so I'm not going to do this. Needs to work "as-is"
• Guage needs to cover the range 0 - 8v DC or 0 - 240v AC
• Raspberry Pi can produce only 0 - 3.3v
• Single, concealable, slender power wire to supply both controller (Rasp Pi) and any additional circuitry
• Raspbery Pi can also simultaneously supply a constant 5v (no programmatic control)
• £50 GBP (approx \$60 USD) budget for additional components
• I have the knowledge of 2 weeks electronics at school 25 years ago! Basically an idiot!

Last edited:

#### kellys_eye

Jun 25, 2010
5,142
Send the Raspi output to a simple op-amp configured non-inverting multiplier (x2.4-ish) to get 0-8V. Don't bother with the 240V option as it requires dangerous voltage levels at aren't actually required for your application.

#### cisor

Mar 17, 2022
7
Is this what you mean? (image attached, couldn't embed)
Taken from here

Can this be used to just triple the voltage? I can control the exact range software side by just considering 2.66 as the 100% value

#### Attachments

• op-amp-non-inverting-amplifier-01.png
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Moderator
Nov 17, 2011
12,774

Mar 17, 2022
7

#### cisor

Mar 17, 2022
7
Given that apparently Vin = Vout x R1 / (R1 + R2)
I'm guessing I need to build the following. Does this look right?

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,774
You'll need a rail-to-rail amplifier. Here's a selection.

#### cisor

Mar 17, 2022
7
Thanks!
Is the circuit otherwise sensible?

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,774
The meter shown in your image shows "R = 200 Ω" for the 8 V range. This means any amplifier you use needs to be able to drive this load which translates to >= 40 mA. The ones from the filtered list in my link should be good.

#### Alec_t

Jul 7, 2015
3,250
Is the circuit otherwise sensible?
The op-amp can't output a voltage higher than its supply voltage. It will generally be somewhat less (though not much less for a so-called 'rail-to-rail' op-amp).

#### cisor

Mar 17, 2022
7

The op-amp can't output a voltage higher than its supply voltage. It will generally be somewhat less (though not much less for a so-called 'rail-to-rail' op-amp).

Any ideas how to achieve this then?

Do I need to include something like this on the supply side then?
https://www.ebay.co.uk/itm/223564731153

#### crutschow

May 7, 2021
520
Is the circuit otherwise sensible?
The input signal and the power supply must share ground (common).

#### Alec_t

Jul 7, 2015
3,250
The DC 5V to 12V boost converter in the link looks ok for your purpose.

#### cisor

Mar 17, 2022
7
Just dropping back in to say thanks for your help everyone

I don't have a working solution yet, but I'm getting there, I hope!

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
12,774
Let me propose this idea:
1. Use an adjustable boost converter (example) to create 8 V from the 5 V you have available.
2. Create a 3.3 V PWM signal by the Raspberry Pi (I think this is what the Rasberry Pi does anyway when creating an "analog" output).
3. Use a logic level MOSFET (examples) to convert the 3.3 V PWM signal to a 8 V PWM signal to drive the gauge
Like so:

At a duty cycle of the PWM of 0 % the MOSFET will be permanently off, output to the Gauge will be 0 V.
At a duty cycle of 100 % the MOSFET will be permanently on, output to the Gauge will be 8 V.
For any duty cycle in between the output of the MOSFET will switch rapidly between 0 V and 8 V with a mean voltage of 8 V × duty_cycle

With the PWM frequency above 1 kHz (for example) the gauge will average the switched waveform and indicate an analog value between 0 and 8.

No opamp required.

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