How to make a voltage limiter?

strantor

Joined Oct 3, 2010
5,025
I want to have an LED illuminate if I apply anywhere from 2vdc to 1000vdc to it, without burning up the LED. Is there a way or a component to add that will allow me split the voltage, 2V fixed always to my LED and the remaining 998V (or however many volts) back to ground or elsewhere? I tried using a volatge divider, but If I set it up to not burn out my LED at 1kv, then it won't illuminate below 480v.
Thanks

blueroomelectronics

Joined Jul 22, 2007
1,758
Too large a range to be practical.

Audioguru

Joined Dec 20, 2007
11,251
You need to make a high voltage current regulator circuit, but its minimum input voltage will be about 5V.
If the LED current is 20mA and the supply voltage is 1000V then the current regulator must dissipate 998V x 20mA= 20W!

Pich

Joined Mar 11, 2008
119
I wonder what the application for the high voltage would be? Probably the only way to do this would be a current source. You will need to use high voltage transistors.

tyblu

Joined Nov 29, 2010
199
Are you sure you will be working with 1000 Vdc? Is there isolation? It is possible to make an auto-ranging circuit, like a DMM, but safety would have to be paramount in the design. I suggest a manual range device, with 100mA fuses on each leg, followed by an isolating DC-DC converter. Too complicated? Lower the maximum DC voltage.

strantor

Joined Oct 3, 2010
5,025
Too large a range to be practical, yes. I'm not sure how to get around it though. I am redesigning a system (made in 1968) in a production facility that shuts down a cable-making machine if continuity between all the conductors is not present, or if leakage through insulation to ground is detected (while the cable is being made). The high voltage is needed to detect indirect shorts. The led is for an opto-isolator for my controller.

blueroomelectronics

Joined Jul 22, 2007
1,758
How did they do it in 1968?

Kermit2

Joined Feb 5, 2010
4,158
You are talking about a high voltage, very low current source( like a DC megohmeter)

a solution with opamps will probably be best to protect the LED end of the circuit from wide range voltage swings. A fet input one would be needed, since you want to detect the voltage change when you have a short, but don't want any DC current to be drawn by the circuit. A multimeg ohm range divider network feeding a comparator with a fixed reference voltage would be my first attempt at making a working solution. Some experimenting with values might be needed to get the operational range you want.

whatsthatsmell

Joined Oct 9, 2009
102

Audioguru

Joined Dec 20, 2007
11,251
I worked for Philips in the '60s. They showed a "diode" that made red light.
I first thought that its junction was red hot but it felt cool.

eblc1388

Joined Nov 28, 2008
1,543
I want to have an LED illuminate if I apply anywhere from 2vdc to 1000vdc to it, without burning up the LED.
I come up with the following idea. It has not been tested but simulation shows good results. I don't know if you can obtain the MOSFET in where you live.

The circuit is basically a constant current source therefore will drive the LED with constant current for any input voltage(V2) over 2V. You can't light an LED using 2V or less because the LED requires at least that much voltage to light. One can alter the LED current by varying the resistor.

The 9V battery can be connected permanently as there is no current drain so it will last for a period same as its shelf life.

Take care of MOSFET mounting insulation as high voltage will be present on Drain terminal of MOSFET.

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strantor

Joined Oct 3, 2010
5,025
How did they do it in 1968?
Well, I'm about 2/3 sure about that because 2/3 is all the drawings I have. The rest have cigarette burns and coffee stains too thick to make out. It looks like a bunch of RTL/TTL/relay logic with a healthy dose of nonsense. The part showing the "leakage fault detector" is most crucial, and it's gone. It would take (me) days examining each physical component and
make a new drawing to get a full understanding of what's going on, but I don't see the point. It has been plagued with problems since it was made. The last tech who was babysitting this thing for the past 35yrs is gone and he didn't leave any notes. So, my solution is to make make a new system, as simple as possible, with commonly available parts.

strantor

Joined Oct 3, 2010
5,025
I come up with the following idea. It has not been tested but simulation shows good results. I don't know if you can obtain the MOSFET in where you live.

The circuit is basically a constant current source therefore will drive the LED with constant current for any input voltage(V2) over 2V. You can't light an LED using 2V or less because the LED requires at least that much voltage to light. One can alter the LED current by varying the resistor.

The 9V battery can be connected permanently as there is no current drain so it will last for a period same as its shelf life.

Take care of MOSFET mounting insulation as high voltage will be present on Drain terminal of MOSFET.

This looks good but I don't fully understand it...
1. Will this work if the load "D1" is "down stream" of the MOSFET? In researching, I have always seen the load drawn "Upstream" of the MOSFET.
2. Assuming the answer to #1 is yes, will the voltage applied to "D1" be the 9V from the battery ("V1") or will it be the 1000v from "V2"?

Thanks!

Kermit2

Joined Feb 5, 2010
4,158
So the testing voltage is continuously variable(a pot.), or does it have a switch that you set to use a specific voltage; say 500V?

Could you not just use a meg-ohm meter and hard wire it into your test points for the machine?

edit: ebcl1388 has a good LED circuit, if that's what you are after.

strantor

Joined Oct 3, 2010
5,025
So the testing voltage is continuously variable(a pot.), or does it have a switch that you set to use a specific voltage; say 500V?

Could you not just use a meg-ohm meter and hard wire it into your test points for the machine?

edit: ebcl1388 has a good LED circuit, if that's what you are after.
It will be a fixed 1000v. I could use a megomhmeter for the voltage source, as I doing currently for testing, but there needs to be some logic performed, which the megger can't do. I will have 2 LEDs (LEDs are for opto-isolator inputs to my programmable microcontroller BTW) that need to be lit continuously (indicating continuity OK) and if one turns off (indicating an open in the conductor) then the microcontroller will activate a relay to shut down the machine. the 3rd and final LED will need to remain off, indicating good insulation/no shorts (direct or indirect). If this LED comes on then it will result in a shutdown also. For some reason I was mistakenly thinking (when I posted this question) that if I were to detect an indirect short (very thin insulation or an air bridge) that I would detect <1000v, which is why I wanted the LED to respond to any voltage above 2V. I think now that I should be able just to use 1.5MΩ 2W potentiometer as a voltage divider in the same way that I am using for the 2 continuity inputs. The only thing I concerned about with that idea, is that if the 1KV faces such a high resistance on the other side of an air bridge, maybe it won't make the jump. What do you think?

eblc1388

Joined Nov 28, 2008
1,543
Please provide a sketch of what you are suppose to test and the setup.

Sometimes words alone can't convey sufficient information for others who are not familiar with what you are actually doing.

Are you detecting voltage ranging from 2 to 1000V and try to light a LED when the voltage is above/below certain level? That would be completely different to a problem of lighting an LED with a supply which can be anywhere from 2V to 1000V.

strantor

Joined Oct 3, 2010
5,025
Please provide a sketch of what you are suppose to test and the setup.

Sometimes words alone can't convey sufficient information for others who are not familiar with what you are actually doing.

Are you detecting voltage ranging from 2 to 1000V and try to light a LED when the voltage is above/below certain level? That would be completely different to a problem of lighting an LED with a supply which can be anywhere from 2V to 1000V.
Right, Sorry I had been neglecting making a drawing, but it's got to be done sometime! Here you go (ignore the 4 symbols at bottom):

The top-most thing might need more explaining: that is a 7 conductor length of cable (usually around 30Kft). Not drawn, but to the right end, is a reel paying off this cable (7 insulated conductors, wrapped in a thin layer of tape); Going left...In the center of the length are a series of bobbins which pay off a thick steel armor wire; At the left end (slightly before the slip ring) is an area where all these thick steel armor wires get twisted onto the cable, and potential exists for one to cut completely or slightly through the conductor insulation. These armor wires are grounded to the machine, which is grounded to earth. I want to have the 3rd LED light if this happens. After the twisting convergence place, there is another reel which wraps up the newly-armored cable, which is where the slip rings are.

The area in question is boxed in Red. I have tested the "1kv-to pot(voltage divider)- to LED- to photoresistor- to controller" thing and it works as long as voltage stays at 1000v. I am just worried that having too much resitance before ground will prevent the 1kv from arcing and me from being able to register that arc as a short/insulation fault.

Last edited:

strantor

Joined Oct 3, 2010
5,025

Revision on my capacitor arrangement

eblc1388

Joined Nov 28, 2008
1,543
Let see if I have gotten your configuration correct.

1. 7-core cable with one core(supply core) energized to about 1000V via a slip ring

2. another two slip rings monitor separately the remaining 3+3 cores for continuity or insulation damage. If the core in (1 above) breaks, no high voltage come back to the other two slip rings. If either one of the (3+3) cores breaks, only one slip ring return high voltage.

3. if there is leakage or short to earth caused by the armoring wires on the core insulation of any cores, high voltage from source and two slip rings will be much lower than 1000V or will at zero volt.

4. If all the above are correct then you cannot distinguish if the core in (1) above has broken or there is an earthed condition on any of the cores. Therefore you must also monitor the level of the source 1000V voltage, which you have not done so in the drawing you have provided.
Rich (BB code):
Fault Condition                           Result
==================           ===============================
supply core broken               source hv normal, no hv at both slip rings
other core broken                source hv normal, one slip ring has no hv
any core insulation damaged     source hv low or lost, same volt on sliprings
5. So you would need a total of three input conditioning circuits to feed the proper level into the PLC, with electrical isolation. The inputs to the PLC should be digital 1 or 0.

6. for abnormal continuity condition, the hv will drop to zero and this is easy to detect using the same detector in (7) below.

7. for insulation damage, you will have to decide now on how low the 1000V will be if the insulation has been nicked or damaged by the armor wire. You can find out by putting a resistor across this 1000V supply and see how many volt it will drop to. That value of resistor is the lowest resistance on the cable that you can accept without triggering an alarm. Many mega-ohmmeters will drop their output voltage if the insulation is not good as there is no real need to put 1000V across a 20KΩ insulation.

In other words, what is the resistance to ground for the cable core before an alarm is raised.

8. once you have an idea of how low the 1000V falls, then three voltage detector can be constructed to provide proper signaling to the PLC input.

strantor

Joined Oct 3, 2010
5,025
Let see if I have gotten your configuration correct.

1. 7-core cable with one core(supply core) energized to about 1000V via a slip ring - Correct

2. another two slip rings monitor separately the remaining 3+3 cores for continuity or insulation damage. If the core in (1 above) breaks, no high voltage come back to the other two slip rings. If either one of the (3+3) cores breaks, only one slip ring return high voltage. - Correct

3. if there is leakage or short to earth caused by the armoring wires on the core insulation of any cores, high voltage from source and two slip rings will be much lower than 1000V or will at zero volt. I had carefully selected the value of 1.5MΩ for the "short/leakage detector" voltage divider as well as the other 2 voltage dividers, so that if all 3 of my conditions are true (Continuity check 1, continuity check 2, and short) then I will not exceed the maximum power of the supply (1.5W, = 1.5mA), therefore voltage will not drop if a short is detected, AND I will still have enough current to power all 3 of my LEDs (3mm, red, .48mA)

4. If all the above are correct then you cannot distinguish if the core in (1) above has broken or there is an earthed condition on any of the cores. Therefore you must also monitor the level of the source 1000V voltage, which you have not done so in the drawing you have provided. I agree, this is a much better solution. This way will provide the lowest potential possible, a direct path back to (-) without any 1.5MΩ resistance in the way.
Rich (BB code):
Fault Condition                           Result
==================           ===============================
supply core broken               source hv normal, no hv at both slip rings
other core broken                source hv normal, one slip ring has no hv
any core insulation damaged     source hv low or lost, same volt on sliprings
5. So you would need a total of three input conditioning circuits to feed the proper level into the PLC, with electrical isolation. The inputs to the PLC should be digital 1 or 0.

6. for abnormal continuity condition, the hv will drop to zero and this is easy to detect using the same detector in (7) below.

7. for insulation damage, you will have to decide now on how low the 1000V will be if the insulation has been nicked or damaged by the armor wire. You can find out by putting a resistor across this 1000V supply and see how many volt it will drop to. That value of resistor is the lowest resistance on the cable that you can accept without triggering an alarm. Many mega-ohmmeters will drop their output voltage if the insulation is not good as there is no real need to put 1000V across a 20KΩ insulation.

In other words, what is the resistance to ground for the cable core before an alarm is raised. Right, I get what you are saying. I will have to resume with this tomorrow and find out.

8. once you have an idea of how low the 1000V falls, then three voltage detector can be constructed to provide proper signaling to the PLC input.
You have been very helpful, and I am very grateful!