Minimizing Current Consumption in High Voltage Divider Circuit

[Zaim01]

Hello everyone,

I'm working on a circuit for monitoring the bias voltage of a GM tube, which requires a high voltage of 100V. I am currently using a OA1NP22C op-amp configured as a unity-gain buffer to read the divided voltage. To minimize current consumption, I'm considering using a voltage divider with 1GΩ and 12MΩ resistors. However, this configuration still consumes approximately 9.8µA, which is higher than I would like for my low-power application.

Here is a simplified version of my circuit:

100V Bias Voltage (GM Tube)
|
R1 (1GΩ)
|
+----> To OA1NP22C Op-Amp (unity gain buffer) > MCU
|
R2 (12MΩ)
|
GND

My questions are:

1. How can I further reduce the current consumption of this voltage divider to the nanoampere range?
2. Would switching out the voltage divider using a transistor be a viable solution? If so, what configuration and components would you recommend for achieving this?
3. Are there alternative methods or components that could help in significantly reducing the power consumption while maintaining accurate voltage monitoring?

Any insights, suggestions, or examples of similar implementations would be greatly appreciated!

Thank you!

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Additional Context:

• Control Signal: 2.9V from a microcontroller.
• Power Efficiency: Critical to keep the overall system power consumption as low as possible.

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Feel free to ask for further details if needed.
Zaim

[Zaim01]

Hi Barry,

Thank you for your concern and valuable insights.

I understand the importance of safety when dealing with high-voltage circuits. Yes, I do have experience designing HV circuits and I did before, and I am also consulting with experts in this field to ensure the design's safety and reliability.

This boost converter is very low power, so the current is minimal. The technique I'm using is called biasing, which is commonly used to bias LCDs, APDs, photodiodes, GM tubes, and more. The circuit is designed with safety in mind, and the PCB will be assembled and insulated with conformal coating.

Using GΩ resistors is still common in medical devices. In the event of a failure, our software algorithms are designed to detect anomalies and immediately shut down the oscillator, effectively closing the high-voltage system.

I appreciate your advice on seeking additional guidance from forums dedicated to HV applications. It's always good to get a wide range of inputs to ensure the design is both effective and safe.

Thank you again for your concern and advice.

Regards,  

Zaim

[AScha.3]

Hi,

i think, if you really need to measure the voltage, the 1GΩ / 12MΩ resistor is best, simple solution.

All photocouplers (AQVxx series) or mosfet at 100V have 100nA or more leakage current - so useless here.

Just - afaik - most ("professional" ) circuits anyway producing a regulated HV , so there is a point with some divider there, to check the voltage ; if you can access this, it would be the best solution, consuming no extra power.

btw  GM working at only 100V ? typical is 400...700V .

[Zaim01]

Hi,

You raise some good points. Since my voltage is 100V, I can't use a MOSFET with a Vds of 100V; it should have some margin, so at least 120V is necessary. I searched for a suitable MOSFET on Digi-Key but couldn't find one that meets these specifications with low leakage current.

Regarding the GM tubes, you are correct; their working voltage plateau is typically from 350V to 2000V. In my design, the boost circuit initially steps up 3.3V to 100V, and from there, I use a 4-stage Cockcroft-Walton voltage multiplier to reach the required biasing voltage of 400V.

To minimize current consumption, instead of using a voltage divider at 400V, I take feedback from the input of the first stage of the multiplier, which is 100V. This approach reduces the power consumed by the divider resistors significantly.

Given that all photocouplers and MOSFETs at 100V have leakage currents of 100nA or more, the 1GΩ / 12MΩ resistor solution appears to be the best and simplest approach for this application.

Thank you for your insights and suggestions.

Regards,
Zaim

[BarryWhit]

Would a N.O. Solid-State Relay (SSR) work in your application instead of a FET? (I've never used one, but it looks like a maybe)

[AScha.3]

Hi,

>To minimize current consumption, instead of using a voltage divider at 400V, I take feedback from the input of the first stage of the multiplier, which is 100V.

Now i know. 🙂

+ Yes, here the most simple solution is the 1GΩ / 12MΩ resistors .

[Peter BENSCH]

@MM..1 Right, I had indeed read G instead of M for the second resistor, sorry.

@Zaim01 Once again on the subject of switching off the voltage divider: this only makes sense on the high-voltage side, as switching it off on the GND side shifts the measurement potential to 100V. However, in order to be able to switch off the voltage divider on the high-voltage side without having to use a charge pump with its current consumption, a p-FET would have to be used, which is controlled by an n-FET against GND. To ensure that the p-FET can also fully switch off, a gate-source resistor is required there, the current of which is added to the current through the voltage divider during the switch-on time. While the two FETs are switched off, the leakage current of the n-FET would be added to the leakage current of the p-FET.

It should also be noted that with such small currents, the circuit design has a significant influence on the results. For example, due to the possible board leakage currents, a conventional circuit board without insulation milling is poorly suited for this purpose.

[Zaim01]

Hi Barry,

Thank you for your suggestion! Using a Solid-State Relay (SSR) could indeed be a viable alternative to using a FET for enabling and disabling the voltage divider in my application. I will get some samples and check their behaviors.

[AScha.3]

No..i didnt find any, thats even close to the needed off-state current.

Same for the AQVxx opto-mos (solid-state...) , i have some of them here, but also much more off-state current than is needed here.

What you improve, if the off-state current is 0...1uA (depending on temp.), if you have only 100nA to switch off ?

-> useless here.

[Zaim01]

Yes, I searched extensively for SSRs and found that the minimum off-state leakage current is around 100nA, which is comparable to the leakage current of the voltage divider itself. Therefore, using the divider without additional components may be the optimal solution.

[Zaim01 · Accepted solution]

You mean a circuit like the one below:

The issue with this circuit is the leakage of the MOSFET and the current consumption through R1. When increasing the value of R1, the P-MOSFET T1 may not turn off completely, and if R1 is too small, it could draw unnecessary current. Could you please recommend some values based on the schematic above?

Or, if I have misunderstood your point, please clarify.

Thanks in advance,
Zaim