How to isolate and to pass voltage signal between two isolated circuits.

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DX400

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Hi,

What is the best (and quite simple and not very expensive) way to isolate circuit1 from circuit3 by some buffer2?
The circuit1 is some voltage source with some load. The voltage on load might vary from 0.1 to 1 V, and the maximal current on load - up to 30 mA.
The circuit3 is some ADC input of the PC controlled board to measure the load voltage.



I have to “connect” the load to the ADC input to measure the load voltage, but I am unable to physically connect it. So I need some buffer circuit “2” based, may be, on optocoupler or galvanic (coil) to pass the voltage signal (current no matter) from the load to the ADC input. So the passed through “2”signal have to be equal or proportional to the voltage on load.

I have looked for the optical isolation by some optocoupler, but I haven’t found the suitable example of such circuit.

Give me, please some example(or reference) of the circuit to resolve my problem.
 
Probably one of the easiest ways is to move the ADC to the primary side, using a small 8 pin PIC, then use that to read the analogue value, and send serial data through an opto-isolator.

You can get analogue opto-isolators, but you would probably need to amplify the signal first as it's so small, and I don't think they are terribly linear?.
 
As suggested by Nigel, an 8 pin pic could work well. What resolution do you need for the ADC?

Another possible solution would be a voltage to frequency converter feeding the opto. Again, what resolution is required?

Mike.
 
Thank you very much for the answers.

The ADC resolution should be of 12 bit. What is “8 pin PIC”? It’s some kind of microcontroller?

Please explain how the PIC microcontroller can help here. Or give me some reference if such solution exists.
 
An 8 pin pic is a microcontroller with a 10 bit ADC. Requiring 12 bits is going to make it more difficult. On a 1V signal 1 bit will equal approximately 0.25mV. Do you really need such prescision?

Mike.
 
Last edited:
Pommie said:
An 8 pin pic is a microcontroller with a 10 bit ADC. Requiring 12 bits is going to make it more difficult. On a 1V signal 1 bit will equal -025mV. Do you really need such prescision?
Click to expand...
No I don't, the 10 bit resolution is enough. Please explain how to use 10 bit ADC PIC microcontroller here.
 
DX400 said:
No I don't, the 10 bit resolution is enough. Please explain how to use 10 bit ADC PIC microcontroller here.
Click to expand...

As with any micro-controller, you have to program it for it to do anything, but it's pretty easy to read the ADC and send the data out as an RS232 type serial data stream
 
You can get ic's made for the job, but not neccessarily cheap.
You could use the voltage to control the amplitude of a certain frequency, then pass that through a transfomer and detect it again afterwards.
 
There are quite a few integrated current sense plus isolation ICs.

This is a good example:

HCPL-7520 Broadcom / Avago | Mouser

HCPL-7520 Broadcom / Avago Optically Isolated Amplifiers 4.5 - 5.5 SV +/-5% datasheet, inventory & pricing.
www.mouser.co.uk


All you need is a 5V reg from the 24V supply at the load sense side and you get the sense voltage from a totally isolated set of pins at the other side of the IC - with anything up to 500V or more difference between the sides.
 
LEM do some isolated voltage transducers, such as https://www.lem.com/en/lv-25200

Of course, if the OP is actually trying to measure current, all of the current transducers that measure the magnetic field are isolated anyhow and that would be the way to go.
 
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