Sensing voltage w/ isolation

I have what I think is a pretty basic electronics question. I need to sense voltage if a switch is tripped. The sensor must be galvanically isolated from the circuit and feeds back to a Due.

The switch in question is 3 terminal SPDT, which controls a constant current source (28V, ~6A). I was thinking of using some kind of optoisolator on the NC side (switch is depressed, need to sense when no longer depressed -> when it goes from NO position to NC). However, I'm not entirely sure how to deal with the CC source. Do I use a current shunt and then feed it into some chonky resistors and send it to ground? Are there optoisolator devices that can withstand that current/voltage directly (I looked and didn't see any, but I am still learning data sheets)?

Advice/suggestions very welcome.

Thank you for your time.

Look at the HCPL3700 isolator, it has a large input range and with a resistor can pick the voltage to trip at.

That is one hell of a constant current source, if you feed that directly into an opt isolator it will fry it. After all and opto is just an LED from the input side.

However the switch that controls the constant current supply is not carrying that sort of current. So do some measurements and find out what voltages are on this switch. Then use a single resistor of the value that will limit current to an acceptable level. Or use some sort of potential divider to get the voltage down.

Maybe a current sensor like ACS712 will be suitable for this purpose - there are variants for +/- 5, 20 and 30 amps. Since the circuit is for 6 amps, it means one of the variants for 20 or 30 amps.

It SOUNDS simple - so we need a schematic

Basically a "constant current" supply can provide a current to a load within its voltage range - so potentially down to zero
so we cant use the voltage to see if its on.

However if the output is O/C it can not provide current
so we cant use the current to see if its on.

If the switch is SPDT you MAY be able to use the "empty" pin to sense its position - depending on what is connected to the common. Alternatively - could you replace it with a DPDT that would provide a completely isolated position?

Hi all, thanks for your responses. Responding from my phone so forgive me for the inevitable typos!

@JohnRob Thanks, I will look into it.

@Grumpy_Mike The switch is carrying the current (it is within an enclosure acting as an interlock. It is a 3 terminal SPDT switch so the NO pin is being used under normal operations). Currently (no pun intended) the NC terminal is unconnected, and I was hoping to use it to know when the interlock is thrown, but I am having trouble coming up with components that can withstand the CC source. I was thinking of using a voltage divider, but most of the resistors seem to be far larger than I could fit within the enclosure.

@flashko Thanks, I will look into it.

@johnerrington I am on holiday until Monday night, but will post a schematic then (I left my laptop at home, maybe good and bad, haha). I was thinking of trying to use the NC terminal so when the switch was tripped it would be sensed. I will also check if going to a DPDT would be an option, that would be easy!

Thanks again all!

You don’t have to have components that handle the whole current. A constant current supply works by increasing the voltage until the set current flows through the load. But that voltage increase can only be up to the maximum voltage that your supply can give.

So if you have a constant current supply that gives 6A at maximum voltage of 28V, and you put a load on it that is 28K, then the voltage ramps up to to 28V and goes no further, the result is that you only have 1mA flowing through that resistor.
So what you need to use is a load resistor that allows the current sufficient to turn in the optical isolator you use. If that is say 10mA then your resistor in series with the LED in the optical isolator need only be 2K8.

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@Grumpy_Mike Doh! You are totally correct, just simple Ohm's law.... for some reason I was thinking that the CC would always drive the set current... Thank you Mike!

@JohnRob Ordered some HCPL3700s, thanks again!

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