Neutral potential ground difference

[grasscut]

Im studying a application where multiple 12V input voltage are coming into a circuit board, i was told that optocouplers are use to neutral the potential ground difference of the input 12V from various appliances. can someone explain to me how does the potential ground of each appliances varies and how the optocoupler helps to neutral the potential difference.
thank you!

 

[.Scott]

Most appliance that use or generate a 12VDC power or signal will already have that signal isolated. In other words, if you measured the impedance from the 12V signal or its local ground to Earth ground, you would discover that it was very high - well over 1MOhms.

However, there are conditions where these 12VDC signals may not share the same ground and may be indirectly tied together though shared circuits or through grounding issues.

So, for example, if you look at the signal from source "A" by comparing the source "A" signal (Sa) with the source "A" ground (Ga) you will get the signal you expect. And the same for signals "B" (Sb,Gb) and "C" (Sc,Gc). But is you try to tie those grounds together, you could get some very high current conditions. For example, you might discover that there is a 120VAC difference between Ag and Bg.

So, by using opto-isolators, you can encode each signal into an IR level and read that IR level using circuitry with your local ground. Since there are no electrical connection between your signal monitoring circuit and circuits A, B, and C, there is no problem with possible currents among those circuits.

Let's take a specific example:
Let's say you have 16 12V batteries connected in series. What you might normally do is 16:1 MUX those sixteen signals using a common ground and then pass the result to an ADC (Analog to Digital Converter). But tying all the battery negatives together would create dead shorts across 15 of the batteries - and associated fireworks. So instead, you opto-isolate each one, and then use the MUX and ADC.

 

[grasscut]

Most appliance that use or generate a 12VDC power or signal will already have that signal isolated. In other words, if you measured the impedance from the 12V signal or its local ground to Earth ground, you would discover that it was very high - well over 1MOhms.

However, there are conditions where these 12VDC signals may not share the same ground and may be indirectly tied together though shared circuits or through grounding issues.

So, for example, if you look at the signal from source "A" by comparing the source "A" signal (Sa) with the source "A" ground (Ga) you will get the signal you expect. And the same for signals "B" (Sb,Gb) and "C" (Sc,Gc). But is you try to tie those grounds together, you could get some very high current conditions. For example, you might discover that there is a 120VAC difference between Ag and Bg.

So, by using opto-isolators, you can encode each signal into an IR level and read that IR level using circuitry with your local ground. Since there are no electrical connection between your signal monitoring circuit and circuits A, B, and C, there is no problem with possible currents among those circuits.

Let's take a specific example:
Let's say you have 16 12V batteries connected in series. What you might normally do is 16:1 MUX those sixteen signals using a common ground and then pass the result to an ADC (Analog to Digital Converter). But tying all the battery negatives together would create dead shorts across 15 of the batteries - and associated fireworks. So instead, you opto-isolate each one, and then use the MUX and ADC.

Thank you for your reply sir! Am i right to say that, in order not to build up the current when all the sources are tied up, a opto-isolator is better as it reads the IR level rather than the amount of voltage coming from each sources? I was told that because each source might have a different ground potential, so maybe one source which supposed to input 12VDC, didnt actually input 12 VDC, say maybe input around 10VDC instead. thus, the opto-couplers were use to neutralise the potential ground difference in difference source.
Example of my application is source A, B, C will all input a 12VDC and gnd to a PCB where each has its input tied to a opto-coupler before sending the output to a IC. so I am wondering what's the effect of the opto-coupler that is able to neutralise the potential ground difference to read all the input as 12VDC.
would be great to learn more from you :)

 

[Rive]

I was told that because each source might have a different ground potential, so maybe one source which supposed to input 12VDC, didnt actually input 12 VDC, say maybe input around 10VDC instead.

If those are same kind of sources then they will all send you 12V, but only relative to their own ground.
But the grounds of the different sources might be on different level than the ground of your instrument.
If you tie together grounds at different level then you might get currents high enough to damage your equipment or you can create dangerous situations.

What an optocoupler does is to create ground independent (light) signal from the input, and let you sense it on the output, relative to an another ground level, without actually connecting the grounds. So you can keep the grounds separated, while you can receive the signal.

You have to be careful, since most optocouplers works as a LED on the input and as a transistor on the output. You have to design the circuits carefully.
Also, there is a bandwidth limit for optocouplers, you can't use them for high frequencies.

 

[.Scott]

Am i right to say that, in order not to build up the current when all the sources are tied up, a opto-isolator is better as it reads the IR level rather than the amount of voltage coming from each sources?

You might have the right idea - but, no, you're not saying it right.

A voltage measurement is actually a difference of the charge available at two points. So if you hold a battery in your hand and I put a volt meter across the battery terminals, I may read 1.5 volts. But if I put the red probe on the plus battery terminal and the black probe on my hand, on a dry winter day I might measure thousands of volts - because of a static charge between myself and the person holding the battery.

You are measuring voltages from circuits A, B, C, D, ... with you measuring circuit M. A voltage is measured across two points. So without the opto-isolators, you would connect all the grounds together - allowing the circuits to share a common ground. Then any voltage measurements would be relative to that common ground.

But that could be a problem. One of your 12 volt sources could be a 12-volt battery hanging from a 6000 volt power transmission line. Connecting that to the ground of you measuring circuit could be a big problem. So instead, you connect one end of the battery to a resistor, then the other end of the resistor to a infrared LED, then the other end of the IR LED to the other end of the battery. This way, the greater the voltage, the brighter the IR light. Now you can put a photo-detector on the circuit board of your measuring circuit and measure that brightness without getting your measuring circuit anywhere near the 6000 volts.

The opto-isolator (or opto-coupler) is that IR LED / IR photo detector combination. Those two devices are put into a single package with a transparent tunnel between them. As long as the voltage difference (6000 volts in our example) is within the specification of the opto-isolator, there is no way for an electric current to pass across that IR tunnel.

 

[grasscut]

Thank you Rive and .Scott for the explanation! truly appreciate it!