# How to indicate Low voltage(2.7kV) and normal voltage(above 2.7kV) in 3.3kV 10uA power supply through LEDs

I have a power supply which gives 3.3kV 10uA output. Now what i need to do is to turn ON a Green LED when the voltage is above 2.7kv and turn red LED ON when it goes below 2.7kV. Could anyone guide me how am i supposed to detect such a huge voltage and indicate it on LEDs. One solution i thought was to use a voltage divider circuit of 2.74Mohm and 10kohm resistor to it to 12v. Is it a good approach. If yes then how can i detect and show it on LEDs that the supply voltage is below 2.7kv or 3.3kv. Thanks in advance.

• Assuming your supply is limited to $10\:\mu\textrm{A}$ and that you do not want to draw off more than 1% of that for the purposes of detection, you are talking about an effective resistance of $\frac{3.3\:\textrm{kV}}{100\:\textrm{nA}}\approx 33\:\textrm{G}\Omega$ to ground. This is difficult to come by. And it suggests that your suggested divider would effectively extinguish your output, as well (trying to pull over $1\:\textrm{mA}$ from your supply rail.)
– jonk
Commented Apr 11, 2017 at 17:33
• Assuming you decided to try a divider, you could consider making it from a long series chain of resistors used to make up such a high resistance divider. But then I'd worry about leakage through any board or mounting materials, which typically have bulk impedance values near the range you are at. You'd need to be careful about designing all your mounting arrangements here. And even with that done successfully, there remains the problem of the final sampling circuitry which should have negligible draw against the divider. There, you might consider a relaxation oscillator.
– jonk
Commented Apr 11, 2017 at 17:42
• Another approach is to consider the Coulomb charges which are required. These charges can be periodically sampled without any significant current draw-down. If interested in seeing the basic idea demonstrated, watch this video: youtube.com/watch?v=8BQM_xw2Rfo So I think this is a good question and I'd enjoy seeing analog ideas presented for you (that include suitable hysteresis, as well.) I have a few thoughts, but I think there are far more experienced folks here who may offer a direction for you. I'll enjoy seeing a good answer.
– jonk
Commented Apr 11, 2017 at 17:51
• Literally just musing out loud but is there something electrostatic...like wrapping several turns of your 3kV3 wire around a neon and seeing if it strikes at a lower voltage with an old-fashioned R-C-neon oscillator and a peak detector... Just thinking of different directions. There's a bit of a jump from there to your 2kV7 detection. Anybody any idea on this sh...stuff? Commented Apr 11, 2017 at 18:06
• @TonyM I think I'd despair of the idea of developing such a specialty transducer. But as someone I know once said, "with enough time and money, anything is possible." I've decided to provide something here, close to where my own mind settled down towards. It might do the job, though I didn't provide the final LED driver (a minor detail.)
– jonk
Commented Apr 11, 2017 at 21:15

Given the low current available and the high voltage, I would think you would want to avoid connecting anything at all directly to the high voltage source.

If I (as a hobbyist) were to try this, I think I would go with an electrostatic detector.

I would place a small metal plate say about 1cm from the wire carrying the 3.3kV. The plate is connected to the gate of a jfet transistor. The drain of the jfet is connected to a low voltage power supply 5V or 9V through a resistor. The source goes to ground.

You should be able to find a resistor value which gives you a drain voltage that varies with changes in the high voltage level. Attach that to a comparator to drive your LEDs.

I'm posting from my phone, so drawing a diagram isn't possible.

The detector is pretty much just the circuit given at this site for an electrometer.

The diagram given at that site:

Remove the ammeter, and measure the voltage at the junction of the 1.5k resistor and the transistor. A comparator there should give you something to work with. You may (most likely will) need to change the 1.5k.

I think the gate will probably charge up and you will end up with fixed voltage. In that case, you use an oscillator to periodically ground the gate, and measure the height of the pulses across the transistor. A simple comparator won't do the job then. You could pass the pulsing signal through a low pass filter and get a DC voltage for the comparator, though.

I'll take a crack at this. Besides being a hobbyist with no electronics training, I also have to admit that it has been decades since I needed to mess around with high voltages and I have zero experience attempting to measure such large voltages and discriminate between values that are as close as you seem to need. Keep in mind my limited perspective when reading the following answer and see if it makes sense to you, as well.

Some preliminary thinking, first.

1. You need to work on a high voltage circuit that is spec'd to supply about $10\:\mu\textrm{A}$ to its own load. You haven't disclosed any details about the load beyond this detail. I have to assume that your current circumstance works: in the sense that you are successfully achieving the required voltage when you have arranged everything correctly and avoided accidental sources of leakage sufficient to drag the supply rail down. I'm guessing here that you merely want some kind of indicator that tells you when the voltage has managed to climb over some threshold sufficient for your purposes and that this voltage is $\ge 2700\:\textrm{V}$.
2. Given that there seems to be time involved in developing the voltage rail and that you need an indicator, I'd imagine that you want to load the power supply as little as possible. I mentioned the figure of 1% as a random suggestion, earlier. But let's make it a firm specification, instead. This means the load cannot be more than $100\:\textrm{nA}$ on the supply. This sets a fairly stringent limitation, suggesting a loading in the neighborhood of about $33\:\textrm{G}\Omega$.

The approach I'll suggest uses actual components you can buy from Digikey as in stock and active parts. My mind immediately went to the Dale (Vishay Dale, today) $1\:\textrm{G}\Omega$ resistors. These are available today in a cheaper version than the older $2\:\textrm{W}$ variety (ROX0501G00FNEL), which is the RNX0381G00FNEE. They still cost about USD 3.00 each. But in the quantity you need, they get closer to USD 2.00 each. These resistors are specified as 1%, which should be good enough for your use. They also drift over temperature, but not terribly: $200\:\frac{\textrm{ppm}}{^\circ\textrm{C}}$.

In laying out these resistors, you need to expect about $100\:\textrm{V}$ across each one and that you will need space for all these resistors, too. Any mechanical tie-downs will need to have very high impedance to whatever casing you use. You could use a plastic box as a housing, I suppose. But you need to be always conscious of leakage paths. In your case, you don't have to go to the crazy extremes I've seen for very high voltage cases (a $250\:\textrm{kV}$ unit I remember seeing comes to mind, now.) But you do need to be conscious about your construction details.

This also usually includes cleanliness. Skin oils are a bane (in my experience) and so you may need to consider a way to keep things clean (and/or to clean them, once they are assembled.) You may want to look for a specialty cleaning fluid for this purpose. (I used to simply immerse the entire circuit in boiling freon... but those days have changed.)

If you can find a precision resistor that can tolerate high voltages across it, to replace some of the above resistors, that's your option. Digikey does offer some resistors with higher values. But for the following circuit I'm not searching further. I have used Dale before and generally trust the parts.

I think you will also need to buffer the final tap before submitting it to a comparator. In this case, I won't use a device actually specified for use as a comparator, though. I don't have good experiences with them for higher voltage cases, and instead prefer to go with opamps for this purpose. It opens up some very good options. The opamp I'll choose here isn't necessarily the best (in any regard) for this use. It merely happens to be the first one I found that seems to be acceptable for this use. I'm suggesting the LT6014, which is a dual package (you need a dual here.) It's not expensive, either. (Especially when you compare its cost to all those resistors I mentioned.)

I've included some hysteresis. The voltage at the node being monitored will be close to $1\:\textrm{V}$ at the maximum input voltage of $3300\:\textrm{V}$, so that's how the rest of the circuit is arranged. I've also chosen to assume a $5\:\textrm{V}$ supply rail can also be made available for the new circuit. These are widely available. Feel free to change it, but also be aware of the need to change other related values, too. (Note the use of a buffer.)

simulate this circuit – Schematic created using CircuitLab

This circuit should have about $200\:\textrm{V}$ of hysteresis ($\pm 100\:\textrm{V}$) around a center-point of about $2600\:\textrm{V}$. (It can be successfully simulated, too.)

Note 1: I've included a link to the $250\:\textrm{kV}$ supply, earlier. They show some details in construction there. I'd also like to include another of their pages: their $15\:\textrm{kV}$ floating output power supply. Review some of their construction details for ideas.

Note 2: I did not attempt a red and green LED display. These are trivial to develop and I assume the OP is capable of achieving that, given the final opamp's output.

Note 3: TonyM suggested the idea of using two $20\:\textrm{M}\Omega$ resistors in parallel, instead of one $10\:\textrm{M}\Omega$ resistor as a protection against the possibility that one of them fails. I think this is reasoned, especially considering the fact that there are very high voltages here and it's not only the resistor's own failure but also possibly the failure of a solder joint that is at issue. So thanks, Tony, for the additional thought.

Note 4: Here's a spice simulation result, showing the hysteresis band for the output of the final opamp versus the input voltage as it transitions up and down through the voltage range of interest:

• Just for the 'reliable design' exercise, I'd make R33 two 20 M resistors in parallel, for Single Point Of Failure (SPOF) protection. Currently, if any component fails short/open-circuit, the circuit is protected except if R33 fails open-circuit. Then the op-amp would be blown apart :-) Commented Apr 11, 2017 at 21:57
• @TonyM Hehe. Got it. Sounds like a worthy consideration!! But the opamp makes a fine fuse, not so? ;) I'm fine adding your suggestion to the text to capture it, if you don't mind. I suppose I could also consider MOV, TVS, zener, etc.
– jonk
Commented Apr 11, 2017 at 22:05
• Should these resistors be of some special type. Could you please tell me the wattage for 20Mohm resistor that i should use. Commented Apr 12, 2017 at 17:37
• @SabheehAli The wattage can be quite low. All the resistors connected to the high voltage line require on the order of $\mu\textrm{W}$. The ones I recommended are capable of six orders of magnitude more. Even if this were all in one resistor, it's still under one milliwatt total. So no worries there. If you decide to replace several of them with higher-valued resistors so that you need fewer of them, then you need to start worrying about their voltage limitations, though. Not a problem with what I wrote. But it could be a problem if you replace several with a single resistor.
– jonk
Commented Apr 12, 2017 at 18:14
• @SabheehAli I provided a direct link to Digikey for the resistors, and the manufacturer's type code. Look at the text and click on the link. You'll see the special type of resistor I mentioned (Dale.) Expensive. But 1% precision (hard to get, otherwise) and I think worth it. You'll enjoy the process of construction, too. Once you build the resistor series, don't build the rest until you've tested it as applied to your HV circuit to make sure that it doesn't load it down. You also still need to work out the LED circuit, in the end. I didn't add that.
– jonk
Commented Apr 12, 2017 at 18:17

I would like to start by pointing out that I'm not a high-voltage person. I have no idea if the following suggestion is safe. Implement at own risk.

I would use a comparator and a resistive devider. Have the output drive your Green LED as well as an inverter. The inverter in turn drives the Red LED.

Use many, many resistors in series for the high-voltage path to make sure you don't put too high a voltage across one single resistor. Use a trimming potentiometer in series with your resistor on the low-voltage side (so the comparator to ground) to trim the values so the system trips at the right point.

The main problem I forsee is that your powersupply can only supply 10uA. This means you will have to use high resistor values in order to not have your detector circuit load down your powersupply.

You might want to use a capacitive divider instead of a resistive one.

simulate this circuit – Schematic created using CircuitLab

The expression for $V_{sense}$ is:

$$V_{sense} = \frac {C_1} {C_1 + C_2}V_1$$

and for $C_2 \gg C_1$ (which is what you want so that $V_{sense}$ is downscaled to just some volts):

$$V_{sense} \approx \frac {C_1}{C_2}V_1$$

$C_2 = 1000 C_1$ could be a good ratio. You might want to choose a ratio that matches your trip point (3.3 kV) to the voltage of a bandgap voltage reference for comparison purposes.

Choose the capacitance values so that the total charge stored in the capacitors $C_1$ and $C_2$ is not too high. Otherwise, they will draw significative current from $V_1$ as it charges, which can be problematic for your application due to the $10 \mu A$ current limit. High-voltage ceramic capacitors below 1 nF might be worth looking for.

Next thing to do would be measuring the SENSE output with a very high input impedance (in the $G\Omega$ range) comparator or op-amp so you don't load the divider.

If you use a comparator with open collector (or open drain) output you might be able to directly pull down an LED serving as an indicator for the 3.3 kV.

DISCLAIMER: These are just some design guidelines. You should carry out your own calculations and simulations to verify the feasibility of the proposed approach.

Note below: Of course, the comparator and the LED will need its own low-voltage power supply.

• The TSX393IDT might work. It's a 2-channel comparator with a 10pA input bias current and 20ma open-collector drain. Supply voltage up to 16v -- \$0.95 qty 1. Commented Apr 11, 2017 at 18:46
• I tried implementing this method in multisim. Used a 15V supply and assigned 2uF and 1uF to C1 and C2 respectively. According to the calculation it should give 10v at Vsense node but the simulation result is in negative fF. Commented Apr 11, 2017 at 19:06
• Try a transient simulation and ramp up the DC source from 0 V up to 15V. Commented Apr 11, 2017 at 20:01
• I'm able to successfully simulate the capacitive divider in LTSpice by using the following directive: .tran 0 1m 0 1u startup. Commented Apr 11, 2017 at 21:16