So currently I am trying to read the current draw from a battery that is being run through a transformer to generate an arc to light a gas burner. I have a roughly 1 ohm piece of wire that I am measure the voltage drop across by having it at the end of the circuit between the ignition system and the battery where I have an Arduino uno analog input reading the voltage drop form one end while the other is connected to ground.

The issue I am having is that the transformer pops up the voltage to a several thousand and then the arc occurs like a normal igniter and the burner has a ground wire screwed into it that connects back to the ignition system/ battery but the voltage across the small wire I have is still showing up as a extremely high voltage and consistently screws with my micro-controller causing me to have to restart it.

I was looking at using optocouplers to measure voltage but it still doesn’t seem like it would help me finding the actual draw on the battery (when I took a current measurement-connected from the battery positive to the ignition system I read upwards to 20 A when the igniter is sparking but that doesn’t seem right for 4 AA batteries.

On the hand of optocouplers I was also looking at if they would actual give me an accurate voltage reading of the system or if the voltage would be too low when it wasn’t sparking to be read from(below 1 V I suspect).

If I can just get the voltage to a reasonable to expect value (considering right before it enters the battery’s negative node It should be roughly Zero volts if I connect the negative end also to the ground Pin if my microcontrollers. Then I have a 16 bit ADC that shouldn’t be able to read accuracy of up to 7.8125 uV or defiant at roughly 61 uV. This also makes me concerned though if the amount of amperage that “apparently” is in this circuit would fry the ADC. The multimeter was saying my analog input pins were hitting almost 1 A but again that seems way out of order considering my board isn’t a smoldering heap yet.

So really my questions are: Any idea what’s causing this high voltage/current that my multimeter is reading coming in and out of the battery?

Is this the actual voltage/current or is something off?

Would an optocouplers help reduce effect on the circuit from the transformer and if reading extremely low voltage values will that actually care across as a proportionate value to the other side of the optocouplers?

Thanks for the help!

Edit: I am measuring the current by taking the measured voltage across the 1 ohm wire with Ohms law on the microcontrollers. However the 20 A I measured was from a multimeter. Along with the 1 A I was getting through the analog pins.

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    \$\begingroup\$ A circuit diagram would be helpful. \$\endgroup\$ Jun 11, 2019 at 22:25
  • \$\begingroup\$ So it’s a commercial gas fireplace burner so I’m not sure the exact schematic. But the wiring is Battery positive———Burner/igniter——analog input pin/1 ohm wire——Battery negative \$\endgroup\$
    – Brandon
    Jun 11, 2019 at 23:24

1 Answer 1


You probably won’t be able to measure this without extraordinary skills at EMI proofing everything. The arc current rise time will be in excess of 1GHz BW and this means you need RF experience to make a current shunt with 50 Ohm coax and 50 Ohm terminators with shielding and filtering on all I/O lines and semi-rigid or microwave coax.

Basically the air coil used in igniters has an inductance which ramps current up at a rate of dI=V/L*dt so your pulse width controls the peak current. Then when released the voltage rises to V=LI/dt with dt being very small in xx nsec due to stray capacitance. When the air ionizes and arcs that same current now collapses from 1kV/mm (sharp point) with a negative resistance in sub-nanosecond times with ringing. When your wire has 10nH/cm this magnetic impulse induces a high voltage magnetic field that resets your uC.

My furnace igniter can draw a 2cm arc easily which puts that around 20kV. The current could be reduced with 20kOhm car spark plug wire and still ignite but the flame would burn the carbon wire and insulation.

If you are having ignition problems it isn’t the current that ignites it but the voltage necessary to draw an arc in the path of the natural gas flow with air mix. So I would want to know a lot more about your igniter design. Cars need about 30kV so that a 22 thou gap at 8 atmospheres of piston pressure to create an arc with (0.55mm gap ) and breakdown now increased to about > 30kV/mm when cold.


If you use a low side current shunt for 100mV max current and use an INA diff amp with gain next to battery current shunt and all impulse generation shielded from this circuit, you may see more accurate results. But you need to isolate this arc current from inductive coupling into your board with balanced signals and shielding.

  • \$\begingroup\$ Your mileage may vary. \$\endgroup\$ Jun 12, 2019 at 2:06
  • \$\begingroup\$ My question would then by why does the voltage across the battery and current get that high? Shouldn’t the voltage difference be just the 6 V from the AA batteries? I’m not even wanting to measure the arc voltage, just the battery power consumption when the burner is in use. \$\endgroup\$
    – Brandon
    Jun 12, 2019 at 3:25
  • \$\begingroup\$ It’s basically a false reading due to inductance! V=LdI/dt so unless you have no inductance or microwave controlled impedance with shielding for everything, you are going to see errors. \$\endgroup\$ Jun 12, 2019 at 3:32
  • \$\begingroup\$ Would it be possible then to set up2 normal 5 V relays to just break the connection between the analog inputs and the measurement resistor then after I know the pilot is ignited I can close the connection again without reading a spike since it would act as a electromechanical switch? \$\endgroup\$
    – Brandon
    Jun 12, 2019 at 12:01
  • \$\begingroup\$ Also would you say that an optocoupler would be necessary then if I use a relay to wait for the pilot to finish igniting since I wouldn’t have to look at spikes? \$\endgroup\$
    – Brandon
    Jun 12, 2019 at 12:08

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