How to isolate from high voltages

rogerm123

New Member
Each hour I am checking a fuse to see if it is blown or not. I send a short duration voltage out and during this same duration look for the in pin being "high". All works well with the strategy until I actually turn on power to the fuse side. Yes (and I laughed as well) the chip smoked.
1. My first misunderstanding was that I could use an isolated power system for the chip(batteries) and it would not be effected by the voltage across the fuse. I would appreciate some one explaining why that isolation did not work?

2. Would opto-isolators solve this problem on the out and in pins?
Thanks.



25947
 

AllyCat

Senior Member
Hi,

Firstly, how are you intending for the PICaxe to "signal" if the fuse is blown or not (or what is it to do) ? Presumably not Smoke Signals? ;)

Secondly, what is the source and nature of this "5kV pulse". How long does it last, ns, us, ms or seconds?

Thirdly, is the "0v" connected to the "Earth" symbol at lower left? If not, it probably should be. If it can't be, then that may be where an Opto-coupler is appropriate - Perhaps rated to 5 kV !

Of course the 5 kV pulse can only appear as shown when the fuse is blown, but then you appear to be applying 5 kV directly across two PICaxe pins, so it's not surprising if the "Magic Smoke" rises!

Next, you (should) have the two sides of the fuse connected to Inputs, NOT Outputs, and in most cases (certainly this one) it is recomended to put a resistor in series with each pin, that should limit the input current to the rating of the "(Electrostatic) Protection" diodes inside the PICaxe, which is 20 mA. So at least for the "Pulse" input, no less than 5 kV / 20 mA = 250 kohms. And even that amounts to a transient power dissipation of 100 Watts in the resistor!

Basically, it seems you just need to test if there is a "short circuit" across the fuse terminals, but to protect the PICaxe from any high voltages. If this 5 kV pulse is always positive, then a 5 kV diode (with cathode to the fuse and a pull-down on the Anode) might be sufficient. Or if a 5 kV diode is a problem then start with a High Impedance potential divider across the fuse. Then the PICaxe input could "look" for the pulse, or try to pull the fuse voltage up to 5v and see if it actually rises.

Cheers, Alan.
 

hippy

Technical Support
Staff member
Applying any 5kV signal directly to a PICAXE is like to end in tears. The recommendation would be to never do that. In fact., for anything related to more than low voltages our recommendation would be -

Don't do that

How to make a PICAXE immune to 5kV isn't something I have any experience in.

To be able to measure whether the fuse was intact I would probably switch the fuse between PICAXE and 5kV side using a DPDT relay, but you would have to ensure the relay is up to the task and that the 5kV was isolated from the PICAXE in all circumstances. I would be thinking of something like -
Code:
                   .---------.  .-------.
                   |         O  O       |
                   |           /        |
                   |          O         |
           2 x R   |       ___|___      |
            ___    |      |_______|     |
       O---|___|---'        |   |       `---O
PICAXE      ___        Fuse |   |             5kV
       O---|___|---.       _|___|_      .---O
                   |      |_______|     |
                   |          |         |
                   |          O         |
                   |           \        |
                   |        O   O       |
                   `--------'   `-------'
"Fuse blown indication" for 250V lighting circuits is often done by putting a neon in parallel with the fuse. If the fuse blows the neon will light but reduce the current in the circuit to a minimal level. I don't know if something similar could be done here. An opto-sensor monitoring the neon should provide good electrical isolation., can be separated by a transparent material to ensure that.

I guess the question is; what is this 5kV pulse, and why you are running it straight to some ground via a fuse ?

Is it intended as some kind of simplistic lightning detector ?
 

rogerm123

New Member
Some background.
Plug in livestock electric fence chargers have to be protected from voltage surges(usually lightening related) coming in on the 115V side (1/2-1 amp, 240V fuse) and back through the fence side (1amp, 240V). When a storm comes through and if it blows a fuse; you could be unprotected for longer than you would want. Fence Charger pulses are regulated to be no longer than (I think) one second and then be off for one second. What I notice is a pulse much less than a sec . The output voltage can be 5-10kv depending on the miles of fence it will charge. The amps I am sure are regulated to some standard to prevent bodily harm. Since an animal makes the circuit between fence and earth; the charger is highly grounded with several earth rods.
I monitor both fuses for continuity on a frequency, and if detected blown, use a stepper motor to rotate in another good fuse. My thought I would be isolated as I was not tied to ground or in anyway to the other circuit.
I had toyed with using a non contact current sensor but was unsure if the fence side would really have any current flow if not grounded somewhere. Where I have fences that are where I can see them easily I use a device that hangs on the fence (grounds to nothing) and if the pulse is not seen it turns on and blinks red. I am assuming they are using something like a Jfet or MosFet to detect.
 

hippy

Technical Support
Staff member
Plug in livestock electric fence chargers
Many thanks and I think I've grasped the big picture ...
Code:
                                                        _
                                                      _|_|_
                                .-----------------     (O)   Ouch!
         .-----.   .-------.    }----------------- -O--|V|\
115V >---| PSU |===| PULSE |----^-----------------     | | |
 AC  >---|     |   |  GEN  |----.    Acme Fencing      / \
         `-----'   `-------'  __|__  _ _ _ _ _ _ _  __|___|__
                              /////                 /////////
The only thing I am not clear on is where the fuse you are monitoring is within the above, what it's connected between.
 
Last edited:

AllyCat

Senior Member
Hi,

Beware that a neon "strikes" at a few hundred volts and then falls back to perhaps 100 volts at low impedance, so a series resistor is absolutely essential. Normally it might be inside the neon's assembly (sleeved wire) or the lamp holder, but if you're going to apply 5 - 10 kV then the resistor must be suitably rated for the voltage and power as in #2 above.

However, I'm not convinced that the fuse will see as much as 5 kV because it's probably before the step-up / pulse generator stage. But the PICaxe will still need quite generous protection/isolation, unless you're using a pure optical coupling such as a Neon + LDR. Since both need a series resistor, I don't know if a Neon gives any higher output brightness than an efficient LED at the same current. Note that LDRs are generally quite "slow" devices, so may not detect narrow pulses very well.

If you're monitoring the voltage across the fuse, then you' won't see anything (a.c. or d.c.) if the fuse is unbroken. But if the fuse is blown, then there probably won't be a pulse either, so you probably will detect the presence of the supply voltage (which is still far higher than an unprotected PICaxe pin can tolerate).

Alternatively, you might detect the 5 kV pulse itself via a very high ratio "Capacitive Potential Divider". The High Voltage (top) capacitor probably needs to be only tens of pF, which might be reasonably available, or you could fabricate one from a length of Coaxial (e.g. Antenna) cable. This would tell you specifically whether the fence is "working" rather than whether the fuse is blown.

Cheers, Alan.
 

rogerm123

New Member
A neon will detect HV, then detect the neon optically and go from there.
Just a thought.
Good luck......
I had not thought of that. Actually hand manual fence checkers do use a neon , or several neon bulbs with different resistors and a ground wire to stick into the ground to show the strength of the voltage pulse.
 

rogerm123

New Member
Many thanks and I think I've grasped the big picture ...
Code:
                                                        _
                                                      _|_|_
                                .-----------------     (O)   Ouch!
         .-----.   .-------.    }----------------- -O--|V|\
115V >---| PSU |===| PULSE |----^-----------------     | | |
AC  >---|     |   |  GEN  |----.    Acme Fencing      / \
         `-----'   `-------'  __|__  _ _ _ _ _ _ _  __|___|__
                              /////                 /////////
The only thing I am not clear on is where the fuse you are monitoring is within the above, what it's connected between.
one fuse is in series with the 115V supply as it first enters the charger. the other is is just before the output to the fence wire
 

rogerm123

New Member
Hi,

Beware that a neon "strikes" at a few hundred volts and then falls back to perhaps 100 volts at low impedance, so a series resistor is absolutely essential. Normally it might be inside the neon's assembly (sleeved wire) or the lamp holder, but if you're going to apply 5 - 10 kV then the resistor must be suitably rated for the voltage and power as in #2 above.

However, I'm not convinced that the fuse will see as much as 5 kV because it's probably before the step-up / pulse generator stage. But the PICaxe will still need quite generous protection/isolation, unless you're using a pure optical coupling such as a Neon + LDR. Since both need a series resistor, I don't know if a Neon gives any higher output brightness than an efficient LED at the same current. Note that LDRs are generally quite "slow" devices, so may not detect narrow pulses very well.

If you're monitoring the voltage across the fuse, then you' won't see anything (a.c. or d.c.) if the fuse is unbroken. But if the fuse is blown, then there probably won't be a pulse either, so you probably will detect the presence of the supply voltage (which is still far higher than an unprotected PICaxe pin can tolerate).

Alternatively, you might detect the 5 kV pulse itself via a very high ratio "Capacitive Potential Divider". The High Voltage (top) capacitor probably needs to be only tens of pF, which might be reasonably available, or you could fabricate one from a length of Coaxial (e.g. Antenna) cable. This would tell you specifically whether the fence is "working" rather than whether the fuse is blown.

Cheers, Alan.
Not familiar with Capacitive Potential Divider but I will research. Can you comment on proposed opto coupler use below?
 

Attachments

papaof2

Senior Member
I think you may have a hard (or very expensive) search for a solid state opto-coupler at 5kV. That's a place for a DIY opto-coupler unless you find a commercial model which can handle that voltage level. DIY being a neon bulb with proper current limiting resistors (a series of resistors to spread the dissipation and increase the overall voltage rating) and whatever spacing provides proper voltage isolation for the sensor - think neon bulb at one end of a 3 inch (75mm) length of black heatshrink tubing just big enough to secure the neon bulb and the light sensor the correct distance from each other.

Do NOT be misled by descriptions which include things such as 4000 volt standoff or 4000 volt isolation. That's telling you the potential that can exist between the light source and the light sensor, not the voltage that can be applied to either one.

You might find something commercial designed for mains voltage (100-240 volts AC) but anything rated for higher input volts is likely to be custom made and very expensive - your electric power provider probably has devices for monitoring distribution lines at the 80 to 100kV levels but I would expect the price to be in the hundreds of dollars or pounds - and probably UP.
 

AllyCat

Senior Member
Hi,
.... Can you comment on proposed opto coupler use below?
Firstly, the OptoCouplers are not providing any useful isolation because both sides are wired to the 0 / 5v rails.

Secondly the "Input" pin will probably always indicate "High" because there is no pull-down resistor.

Exactly what happens when the fuse is "good" is difficult to predict (the two Opto LEDs appear to be in series across 5 volts), but if the fuse blows then the "High voltage" will be applied either across the Opto LED (destroying it) or across the resistor shown just below the "0" (probably over-Watting it, depending on its value).

Cheers, Alan.
 

tmfkam

Senior Member
Having repaired far too many electric fence energisers, assume NOTHING. They are 'potentially' (sorry) LETHAL. Much of the safety is the assumption that animal hair/fur will limit the current delivered. Be very careful.

I might try using a small coil placed 'close' to the output wire to see if it will detect the EMP from the fence wire. Keeping a safe distance at all times...
 

hippy

Technical Support
Staff member
one fuse is in series with the 115V supply as it first enters the charger. the other is is just before the output to the fence wire
So like this, which makes sense to me ...
Code:
                                             .---------.---.
          F1    .-----.   .-------.    F2    }---------   .|.
115V >---O==O---| PSU |===| PULSE |---O==O---^---------   | | Very
 AC  >----------|     |   |  GEN  |----------.            |_| High
                `-----'   `-------'        __|__  _ _ _  __|__
                                           /////         /////
That 'very high' resistance is air, reduced by rain, livestock leaning on the fence to try and turn themselves into burgers, and jokers who may throw metal chains onto the fence. That, plus lightning strikes, may blow the F2 fuse, and that's what we need to check is blown or not..

There are two ways; checking across the fuse or by checking if the voltage pulse is reaching the fence.

I can understand why it was thought having a separate battery supply for the PICAXE and measuring across the fuse looked to be safe, seemed to be isolated. But it is not, not when one pin of the PICAXE also connects directly to ground, nor when connected to the other side of the fuse, is floating in air is connected to ground by a very high resistance.

Some sort of 'neon parallel to the fuse' could work, but emulating one of those pulse detectors which simply hangs on the fence is probably easier, better, and safer. The software is simple; if no pulse is detected for a couple of seconds turn the LED on. That's just a few lines of PICAXE code.

The challenge then is detecting the pulse. For 'hang-on' monitors I would have guessed there's some sort of inductor which produces a pulse when the fence is pulsed, maybe some sort of in-line capacitor circuit. The magic solution being a single signal pulse of low enough voltage that the PICAXE can detect without damage and large enough that it can tell that signal is there, relative to its own isolated supply - no direct connection to ground, except through air.

In practice, most circuits I found used a direct connection to the fence, via some large value mega-ohm resistor voltage dividers, direct or via an op-amp comparator to a digital input or ADC. Avoiding false pulse detection seems to be challenging in some cases, from mains or radio, so there may need to be some experimentation and the code might not be quite so simple as first imagined.

I would still favour some sort of opto-isolator in there as that guarantees the PICAXE never being exposed to excess voltage.
 

rogerm123

New Member
Appreciate all for the helpful input. I know now why I became a mechanical engineer vs electrical. Based on above I am going to change my strategy to try to detect the electric field or use neons to supply the logic to change the fuses.
Thanks
 

Buzby

Senior Member
If your 'hang on' detectors do what you want, buy a few more and hack them !.

Just add a circuit that uses the existing LED as a trigger for whatever you want it to do. Anything from strobe on a tall pole, or a radio linked alarm.

A lot simpler and safer than trying to build your own HV detector.
 

hippy

Technical Support
Staff member
If your 'hang on' detectors do what you want, buy a few more and hack them !.
That would be my preferred approach. Units you can trust will likely have had far more effort put into making and proving them safe and performant than almost anything designed by those without such extensive experience of kV voltages.

And there's never anything wrong in standing on the shoulders of others to achieve your ultimate goal. Get it working then improve it if needed is always a legitimate approach.
 

piclt

Member
Most "hang on" testers have an earth lead and spike. Detect the current pulse in the earth lead or by using a 1 or 2 ohm resistor in series in the lead measure the voltage pulse across it. All this will be "very low voltage". Then use micro controller to trigger mobile phone and warn if no pulses.
 

Buzby

Senior Member
The 'hang on' probably has a very high resistance, it maybe even be a capacitive device, so the LV side will have a very tiny current. This will be difficult to detect reliably.

Worse still, if the earth spike fails, the whole of your LV detector system might be raised to 8kV !.

A much safer way is to use a phototransistor to detect the LED, with no connection to any of the HV side of things. The phototransistor will need to be shielded from ambient light, so mounting will need some thought.

A more reliable way would be to replace the LED with an optocoupler, thus eliminating any problems with external light. However, the optocoupler must be rated for 8kV, or whatever, and fitted in such a way that the integrity of the existing enclosure is not compromised.
 

piclt

Member
Yes, There are earthed and non-earthed type "non-contact" hangers-on, The non-contact one just detect the "electric field" surrounding the high voltage wire. Google ......there are plenty of examples of home made non-contact with neons and sounders, fets with gate circuit open etc etc.
 
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