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
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?
A very old idea. Neon lamp/indicator.
NE-38s or 2ML turns on at 65V and is rated for 300uA. Also "A1B" or A1D.
look at 5AB These are 50V parts. 300uA.
The idea is to reduce the current compared to a LED.
A very old idea. Neon lamp/indicator.
NE-38s or 2ML turns on at 65V and is rated for 300uA. Also "A1B" or A1D.
look at 5AB These are 50V parts. 300uA.
The idea is to reduce the current compared to a LED.
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.
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.
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.
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.
That's not going to happen - Ron's solution is quite impressive, but again obviously includes active components, and relies on fairly esoteric devices.
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.
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.
Blue trace is input voltage 0-1kV. Red is at the 1/2 point. Green is the current that holds within 10%.
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.
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.