Hi everybody,

I'm working on a high voltage LED indicator installed on a main circuit which charges a capacitors bank.

The indicator is simply a combination of a zener diode (which sets reference voltage), a pair of voltage dividers (one for voltage sensing, the other for voltage reference) and a comparator at the output of which is connected an LED with its balancing resistor (see schematic...VDCin is about 850 volts).

The purpose of all this is to turn on the LED when the charging voltage reaches (and stays above) a certain threshold (around 650 volts).

Now, while the voltage between the LED pins (1.2 volts) is well under the spec limit and the behaviour of the comparator is as expected (a surge without hysteresis when the voltage reaches the target threshold), the current flowing out of the comparator is absolutely not enough to continually and sensibly light on the LED, in fact it lays in the range of the MICRO amperes (see graphs).

I tried to change the configuration feeding the comparator with a 9 volts battery but the same problem arised: very tiny amperage at the output still. However I'd like to keep feeding the comparator without recurring to an external supply.

Are the voltage dividers not so well configured? (not enough current through them...)

Is the type of comparator which is not suited for this application?
(though I think that a comparator is still better than a transistor given the voltages involved...I was thinking to an LM339)

Maybe a neon lamp replacing the LED would be better? (what type of lamp...)?

Or...what else?

Thank you for taking the time to answer me.

 

This may be a forbidden topic here due to the mains voltage involved.

Not sure about that particular comparator, but many have an open collector output, meaning they go "open" when the comparator output goes high, and complete a path to ground when the output goes low. So the LED will never light in the configuration you've drawn. You need to power the LED from somewhere and sink current - not source - from the comparator. Or use a transistor whose base is pulled high when the output is high (open) and gets grounded when low.

 

This may be a forbidden topic here due to the mains voltage involved.

I'm sorry. I'll move this post to the general electronic chat but I cannot delete this one.

You need to power the LED from somewhere and sink current - not source - from the comparator. Or use a transistor whose base is pulled high when the output is high (open) and gets grounded when low.

I don't completely get the point here: Should I power up the LED through an external power source? Given the extremely low saturation voltage of a transistor's base, should I rethink the voltage dividers configuration accordingly?

 

LTC1716 does not seem to be an open-collector output device, (see datasheet), but if it is powered by 1Mohm from the input, it will obviously collapse its supply if it tries to drive an LED.
http://cds.linear.com/docs/Datasheet/1716fa.pdf

Also, there is no visible link between the amplifier ground and Vin-, so it is not clear what potential will be applied to the LED

 

After some discussion the moderators have decided to close this thread. Trying to power a 10-15V circuit from 800VDC is not safe, and I do not believe it can be made so.

You will need specialized resistors if you connect to such a line, conventional resistors are generally rated for 250V.

If I were to try to make a monitor circuit I would put 10 or so resistors in series, then insulate them several ways with heat shrink and possibly liquid tape (a specialized insulation paint available from hardware stores).

Be very careful though, 800V is enough voltage to reach out and bite you, you don't have to directly touch it. Combination of these issues make it a safety issue.