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Continuity tester protection

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anpe11

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Hello!

I want to build my own continuity tester, a rather simple one actually.. My idea is
to send a 5V-signal from a PIC through the "probes" and back to a
optocoupler (4N33) that tells the PIC that we have connection (continuity)! :rolleyes:

Now the problem is.. what if someone was to put mains voltage on these probes?
Here in my country (Sweden) we have 230V/50Hz, and I don't think the
5V-circuit would like this so much.. so does anyone have a good idea how to
1) protect the circuit from being damaged and 2) tell the user that a
continuity test cannot be performed due to the voltage applied on the probes?
And just to keep it simple, by "telling the user" I mean light up a LED or something like that.

Thank you very much! :)
 
Use rail clamp diodes and a high value resistor to limit the current to protect the diodes.

I could give specific component values if you would define what current you were limiting to and what voltage drop you would consider continuity. Some things like power filaments (heaters/light bulbs) have to be under some moderately decent load to measure properly. So defining what exactly you're trying to test would be helpful.
 
Thanks for your quick answer. To be honest I don't really now the exact values, but the main purpose with the tester is to
test wires in a regular house installation, such as continuity between the neutral and ground wires. My plan is to make this
tester some way automated or operated via RF, that's the reason I don't want use to my regular one (Elma 2000) in this particular case :)

Edit: spelling....
 
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Any ideas? I did a test on my Elma 2000-instrument.. I measured the current and voltage
that this unit puts out to test continuity, and the results are 1.2V/1.2 uA. So maybe these
values are ideal to test continuity without harming any components that may be
in contact with the tester's voltage and current? There is a small tag on the unit
that says "Cont.: 0..400 kOhm", if that tells you anything..

Thanks!
 
As Sceadwian says, but use X rated capacitors to induce a low frequency AC coded signal onto the wires and pick it off elsewhere. This will give you basic connectivity and isolate the HV signals. Do also consider Opto isolators for the interface.
 
1.2ua will fail to test loads that are based on higher currents, that was my main point.

WHAT you are testing is more important than how you are testing, because what you need to do is determine the method used based on the application. There is no one single answer. What you want to do, will determine how you do it.

If the answer to the load question, under what specific circumstances 'continuity' is declared is not able to be specifically defined, then there can be no answer!

It's like asking the meaning of life! From what perspective is the first response I would give.
 
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You have an optocoupler in your circuit but it's not clear how it's connected. :confused: Are you applying voltage directly from the PIC and then using the coupler for the detection voltage? Can you post a diagram of your proposed circuit.
 
Okay, this is what I've come up with so far.

View attachment 66154

When I run the simulation everything looks nice but I am not convinced that this is the way to go...

The two 100K resistors in the circuit are there to limit the current through the zener to a maximum of 2 mA when 400 Vrms is applied on the probes. As I said, in my country we have 230V/50Hz, and 400V between the phases. This should protect the circuit from blowing
up if someone was to put the probes on either LIVE and NEUTRAL, or L1-L2, L1-L3 or L2-L3.

I may need to replace the resistors with several others (wich add up to 200K) since
the power dissipation in this case will be 0.4W per resistor.

So, how about the simulation? Well, when I put 5V on the input named "From PIC" in the circuit, and short the probes, I got 500 mV on the output "To PIC input", telling the PIC that we have continuity.

If I put an AC source (230 Vrms/50Hz) between the probes, and a "zero" from the PIC
on "From PIC", I got 0V on the output.

If I in this case put a "one" (5V) from the PIC on "From PIC", the output ("To PIC input") starts
to pulse out 4.95V at 50 Hz. See picture.


View attachment 66155

I am very sorry for the confusing way to explain all this, but I do not
master your language that well..! Now to the questions! ;)

1) Can someone explain to me how the output can pulse 4.95V/50Hz when
5V is applied on PROBE 1 when also an AC source is present at the same time?

2) Is there any way to do this better? I'm sure there is, but will this work in
real life? I don't want to build this circuit and hear a BOOOM! when I plug the
AC in :D

Thanks to everyone that has helped me so far...!

Edit: I may also mention that Vdd = 5V.
 
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