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Voltage Indicator

Hello, I want o make a Voltage Indicator circuit which can turn on LED when voltage is 1800V to 50V. One option is resistor but the power loss will be more.
Circuit should be simple/robust and should take power supply from this Dc voltage itself i.e. 50v to 1800v. Any help will be appreciated.
Thanks
 

alec_t

Well-Known Member
Most Helpful Member
Even if the LED were bright enough with 1mA current, the volltage dropper from 1800V would have to dissipate 1.8W. What is the source of this 1800V and can it power 1.8W without the 1800V sagging significantly?
 
Even if the LED were bright enough with 1mA current, the volltage dropper from 1800V would have to dissipate 1.8W. What is the source of this 1800V and can it power 1.8W without the 1800V sagging significantly?
Yes this dc voltage is very stiff. It is dc link of ac-dc- ac converter.
 

ronsimpson

Well-Known Member
Most Helpful Member
How much maximum voltage it can withstand?
Never put voltage on a Neon. You must have a resistor. It pull zero current until the voltage hits the firing voltage. 50 or 60 or 80 depending on which type you have. After firing the voltage drops to the hold voltage. There must be a current limit resistor.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
How much maximum voltage it can withstand?
It's like an LED - it requires a certain voltage to light (in a neons case around 70V) and after that a current limiting resistor to stop it going BOOOOMMMM!!!!!.

The advantages over an LED are that the higher voltage means you don't have to drop 'quite so much', but more than that it's current requirements are less so the resistor chain can be much higher and waste less energy.
 
Never put voltage on a Neon. You must have a resistor. It pull zero current until the voltage hits the firing voltage. 50 or 60 or 80 depending on which type you have. After firing the voltage drops to the hold voltage. There must be a current limit resistor.
ok but will it work for the full range i.e. 1800V to 50V ?
 
It's like an LED - it requires a certain voltage to light (in a neons case around 70V) and after that a current limiting resistor to stop it going BOOOOMMMM!!!!!.

The advantages over an LED are that the higher voltage means you don't have to drop 'quite so much', but more than that it's current requirements are less so the resistor chain can be much higher and waste less energy.
I think they are not very robust. Is it true? Also I will be connecting DC supply which will be degrading the life span (as mentioned in datasheet)
 

ronsimpson

Well-Known Member
Most Helpful Member
One more try.
This is a constant current source. In spice I used a 3N45 which is only good for 450 volts. I see there is/was a IXTA3N250 which is a 2500V transistor.
R1 sets the current limit.
M1 will get warm/hot. 1mA 1000V = 1 watt.
I have a collection of 2000V transistors but they are not in production now. With a little looking I think there are some available.
120694
 

alec_t

Well-Known Member
Most Helpful Member
How about using a good old-fashioned analogue voltmeter? Analogue multimeters typically have a 50uA moving-coil meter movement. Such a multimeter meter in series with a ~40 megohm shunt made from a chain of resistors would give a visible indication over the whole voltage range.
 

ronsimpson

Well-Known Member
Most Helpful Member
Using lower voltage MOSFETs.
Years ago I built amplifiers that had 2000 P_P voltage outputs.
Here M1, M2 share the voltage. So two 1000 volt transistors can 2000 volts.
Same as the old circuit M1 is a constant current source. M2 is set to take 1/2 the voltage.
120702
Blue trace is input voltage 0-1kV. Red is at the 1/2 point. Green is the current that holds within 10%.
120703
 

Nigel Goodwin

Super Moderator
Most Helpful Member
How about using a good old-fashioned analogue voltmeter? Analogue multimeters typically have a 50uA moving-coil meter movement. Such a multimeter meter in series with a ~40 megohm shunt made from a chain of resistors would give a visible indication over the whole voltage range.
Problem is the resolution, at 50V it's only moving 1/36th of what it does at 1800V - so barely noticeable.

Must admit I like Ron's constant current idea :D
 

Diver300

Well-Known Member
I know that 1800 V will need some careful handling, and maybe this whole thread should be in the high-voltage section.

I would be very cautious about relying on potential dividers that use 100 MΩ resistances. It is really easy to get leakage paths with a fraction of that resistance, which would mean that the divider's ratio is more dependent on the humidity in the room than it is on the values printed on the resistors.
 

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