2
\$\begingroup\$

I would like to measure a High-Voltage pulse (20kV) with a duration of a few tens of nanoseconds using an oscilloscope and a simple 1000:1 Resistor Voltage divider.

Is it possible to do this without capacitors, considering that it is a simple pulse and not high-frequency?

\$\endgroup\$
  • 2
    \$\begingroup\$ Make sure that the resistors can handle the power (20 kV * … A). \$\endgroup\$ – Michel Keijzers Oct 30 '19 at 16:10
  • 4
    \$\begingroup\$ 'A few 10s of nS' is high frequency. You might be lucky and find that the self capacitance of your two resistors is in the same ratio as their resistances, but I wouldn't put money on it. That's why you add capacitors, preferably one of them adjustable, to keep the 1000:1 division ratio that's easy to get at DC up to higher in frequency. 10s of nS === 10s of MHz for the fundamental, 100s of MHz for the edges. \$\endgroup\$ – Neil_UK Oct 30 '19 at 16:23
  • \$\begingroup\$ Thank you both for your replies. @Neil_UK That was my concern as well, thank you for clarifying this issue to me. \$\endgroup\$ – Cez Chi Oct 30 '19 at 16:26
  • \$\begingroup\$ No, not even close. You will end up with super low ESL and very tight tolerance capacitors to achieve that tens of ns-range. \$\endgroup\$ – winny Oct 30 '19 at 17:26
  • \$\begingroup\$ Not a hope in h*ll. You need the capacitors and and that means more loading of your source. As an estimate, required bandwidth in GHz is 0.35/rise time in ns. So if your rise time is 5ns (for a pulse width of a few tens of ns), the bandwidth required would be 70MHz. \$\endgroup\$ – Spehro Pefhany Oct 30 '19 at 18:12
-2
\$\begingroup\$

A high resistor bridge would be fine, no capacitors needed. Make sure the 1000:1 attenuator is built from high voltage resistors (such as these) to ensure that arcing doesn't occur and destroy the scope. (most 'run of time mill' probes are only good to values like 600 VDC or 1000 VDC, check yours). The< problem that you might run into is the rise times might not be in the ns range because the parasitic inductance of the larger resistors is higher then something like an smt resistor. Using smaller attenuators would be better (like these 20 kV 1000:1 EBG resistors which tout a non-inductive design)

Another more expensive option would be to buy an HV probe for your scope instead of constructing something.

Edit:

My intuition was way off.

I simulated resistors with leads, in my mind the inductance (I used 10 nH which would be in the range of a resistor with large leads) If there is inductance then it does create a pole and blocks high frequencies (vout1 has rolloff in the 1 kHz range). However with capacitance (just about everything has a few pF between it, the inductance does not matter and you should not have a problem with a resistor attenuation.

enter image description here

Apologize for the confusion

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ I beg to differ. \$\endgroup\$ – winny Oct 30 '19 at 17:27
  • 3
    \$\begingroup\$ In a high-voltage high resistance application such as this the parasitic inductance is of no consequence and will not have any effect on operation. The parasitic capacitance however will have a major effect. With reasonable value resistors (many megohms) the bigger issue will be the capacitance load of the scope and cable causing a rise time of many microseconds (could be hundreds). There is no way that nanosecond pulses could be faithfully attenuated. \$\endgroup\$ – Kevin White Oct 30 '19 at 17:30
  • \$\begingroup\$ Inductance is usually the first enemy of high speed, parallel capacitance over the resistor will only help the pulses travel through, there is little parallel capacitance and a lot of inductance (probably in the 10nH range) In an oscilloscope passive probe, they have a variable capacitor over the resistors to negate the inductance in the probe. \$\endgroup\$ – Voltage Spike Oct 30 '19 at 17:49
  • \$\begingroup\$ @VoltageSpike While you are correct that topically inductance is the biggest enemy of high speed, in this case the stray capacitance has much more of an effect on the divider ratio as even 10fF can shunt R1 enough to lower the ratio by a factor of 5-10. \$\endgroup\$ – Ben Watson Oct 30 '19 at 18:21
  • 1
    \$\begingroup\$ I don't think I've ever seen multi-downvoted answer voted as best before. \$\endgroup\$ – DKNguyen Oct 30 '19 at 18:23
2
\$\begingroup\$

As others have said, 10's of nano seconds is high frequency. If you wanted a purely resistive divider and chose values of lets say \$R1 = 10 M\Omega\$ and \$R2 = 10k\Omega\$ which would create a ~1000:1 ratio. Unfortunately, even just a 1pF parallel capacitance which is well within the possibility of stray capacitance, would cause your Vin and Vout to look like this:

enter image description here

schematic

simulate this circuit – Schematic created using CircuitLab

(Note that C2 out to be 1pF NOT 1uF. The simulation was done with C2=1pF)

A better option, if you only need to measure a short pulse, would be a purely capacitive divider. Of course if you need to measure DC this would not be an option. Going this route you would be concerned with series inductance which would be hard to mitigate using discrete components. Measuring high voltage pulses this fast is far from trivial. That is why high voltage scope probes with 50-100Mhz bandwidth are so expensive. You could go the route of making a capacitive divider where the physical construction of your probe gives you the capacitance needed (see this paper). This makes minimizing the inductance much more feasible.

enter image description here

enter image description here

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.