-5v to 5v into 0v to 3.3v ...

I have a Minelab Sovereign GT metal detector that has a Target ID meter.

The meter is an LCD display that is at times hard to read and quite basic.

I'm looking at creating a new one from scratch with an OLED display and a bit of functionality.

There is very little info on the wiring but I figured since the LCD is powered from the detector and gets the target ID value it must be possible to get an Arduino powered from it and read the value.

Normally 4 wires go from the detector to the coil. With the digital meter 5 come from the detector and 4 leave as the digital meter has a cable that is used to connect to the meter with an extra wire. The extra one is the TID or target ID which straight from the detector has (after using a multi-meter) -5v to +5v which would correspond directly to the values on the meter.

I pulled the ID meter apart and did some testing/investigation.

It seems the 4 wires that usually leave are 2 pairs that make up the 2 coils. The voltage in these fluctuated quite a bit and was small.

There was a very small circuit board between the 5 incoming wires and the LCD screen and I understand what it does... but not how.

Basically somehow it creates a stable 8.75v that the LCD runs off and a signal that seems to be 0 to 0.5v (instead of the original -5 to +5v) that feeds into the din pin on it.

As an experiment I was able to power my Arduino Uno from the detector using the 8.75V on the barrel connector. As expected... reading the value from it's input gave 0 as the full range now is so small.

The circuit has a POT that allows the input signal to be tuned before coming in.

It has 4 diodes which I'd guess turn the +ve and -ve signals into all +ve.

Has 3 capacitors which I'd be guessing allow a small device to be powered. (100nF @25v).

Assuming I can copy the circuit (as it's 1 sided and handful of components) I should be able to power the Arduino.

I however are not sure how to scale the -5v to +5v into either 0v to 5v or 0v to 3.3v.

For 5v Arduino I need to divide the input voltage by 2 and shift it +2.5v and something similar if I go for a 3.3v MCU.

I'd attach a copy of the circuit in the meter... except my brother has borrowed the detector for the week.

Link below shows how an external multimeter can be connected and used. However I want to be able to use an OLED, not have external batteries and have a few buttons to allow for some options.


I understand how a voltage divider works and are fine with the calculations... but it's the shifting that's throwing me. If it was 0v to 10v I could scale that 0v to 5v or 0 to 3.3v easily.

To convert -5 to +5 into 0 to +5 is very easy

just 2 equal resistors in series. - say 10k. One end of R1 +5, junction R1,R2 to output, low end R2 to input signal.

to convert to 0.3.3v, 3 resistors :R1 to 3.3, junction to out, R2 lowerend to input as before

but upper resistor R1 (to +3.3) is 10k, lower R2 15k. Add R3 30k from junction to ground

good 'ole kirchoff!



I append a schematic to make things clear…


kirchoff1.pdf (14 KB)

:o Ahhhhh thanks Allan. Very clear. I suppose in general I can use the vref pin which will give 0-5v on 5v machines and 0-3.3v on 3.3v machines.

I'm going to knock up a schematic for when the detector gets back. I'd almost bet the output from the 2 coils is an AC voltage that averages ~0 but in essence it's prob more like a sinewave and can hence be fed into capacitors after being made into +ve only voltage.

I'm from a programming background more than electronics but getting a better understanding of how they hang together.

I suppose in general I can use the vref pin which will give 0-5v on 5v machines and 0-3.3v on 3.3v machines.

I think it's not wise to use the Vref pin for that.
Just use the 5volt pin (or the 3.3volt pin).

"I however are not sure how to scale the -5v to +5v into either 0v to 5v or 0v to 3.3v."
A full wave bridge rectifier is an excellent means of taking an input of any polarity and resulting an output of one polarity. They aren't fancy, but they are effective. They are available in "IC" (DIL) form.

1/ Leo... correct
2/ Runaway Pancake- yes a bridge will rectify but you lose 2 diode drops - approx 1.3-1.4 volts. Not much good for accurate measrements.

regards. Allan.

I thought he wanted 3 volts out of the deal.

Yes, but a linear mapping. The bridge will give a large dead spot.

and you can't get much cheaper than 2 or 3 resistors.....



I append a schematic to make things clear...


Great!! I downloaded this schematic, printed it out and hung it over my workbench.