# Frequency response of "purely ohmic" voltage divider?

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1. Nov 12, 2015

### Andrew B

I am trying to understand why/when a purely ohmic voltage divider (impedences Z1 and Z2 are both resistors) would have a delayed response for high-speed, transient changes in an input voltage? Whether the input is high-freq AC, or simply a square wave with a sharp rise or sharp drop, I've been reading that a voltage divider with two resistors hits a limit in response speed and therefore resolution into the input signal, and yet there is no mention of frequency response in the idealized, basic equations -- e.g. V_out = (R2 / (R1 + R2)) * V_in.

Is there a way to quantify when a resistive divider would start losing resolution into a transient voltage source I am trying to measure?

2. Nov 12, 2015

### Staff: Mentor

Real life devices are never pure. There are always parasitic capacitances and inductances that become significant at high frequencies.

What frequencies or time ranges are you looking at?

3. Nov 12, 2015

### CWatters

An ideal voltage divider wouldn't "delay" or otherwise modify the waveform. However what follows the divider may well have capacitance and the combination of the two has to be considered. Try replacing R2 in your equation with R2//C.

4. Nov 12, 2015

### meBigGuy

I hate the "I've been reading" stuff. What have you been reading? It's totally useless to say "I read that ..." following by something that makes no sense without context.

Either what you read is totally wrong, or it is qualified in the sections you did not tell us about.

The only way a voltage divider with two resistors can hit a speed limit is if you are driving a capacitive load. (or the resistors are not ideal)

5. Nov 12, 2015

### Andrew B

Thanks all -- OK this makes sense so far, that non-idealities lead to unexpected capacitances. And please bear with me, I'm from a mech eng. background not EE, so trying to get some new concepts straight without a good vocabulary for it.

@meBigGuy -- My comment was from Wikipedia's intro to Voltage Dividers (but also some qualitative feedback I'd gotten from talking to other people). Quoting the Wiki:
"For direct current and relatively low frequencies, a voltage divider may be sufficiently accurate if made only of resistors; where frequency response over a wide range is required [...] a voltage divider may have capacitive elements added to compensate load capacitance".

@anorluna -- I am charging a capacitive load until it hits dielectric breakdown. It charges within 10's of milliseconds before breakdown initiates. I am interested to get decent accuracy on the peak voltage that initiated the breakdown.

Also the input voltage is very high (>100kV), where probes get very big, so for space considerations I am trying to find a simple circuit I might construct as a voltage divider. I started with the simplest thing (the two resistor divider), but was surprised to find it is more than meets the eye

6. Nov 13, 2015

### meBigGuy

I've never done a 100kV voltage divider. No idea where to get a 100KV resistor, even.

It would help to understand the nature of the capacitive load (amount of capacitance, etc), aprox. over what voltage range breakdown will occur, and how you detect that breakdown occurs. You can use RC equations to determine the voltage with respect to time, but that also requires knowing the capacitance.

Why a voltage divider rather than just a series R? And anything else you can tell us.

I can't quite parse this. Are you saying that once you apply 100KV, it is 10's of milliseconds until breakdown? Is that a requirement? Does it need to be that fast?

7. Nov 13, 2015

### CWatters

I've never done one either. Google suggests this book covers high voltage dividers including the effects of stray capacitance on resistive dividers.

High-Voltage Test and Measuring Techniques By Wolfgang Hauschild, Eberhard Lemke

8. Nov 13, 2015

### rbelli1

I agree with meBigGuy. You will likely be having problems with input capacitance on your measurement device. The R1 of your divider is likely tens or hundreds of meg-ohms. The RC on that with even a few pico-farads is a sizable part of your waveform.

BoB

9. Nov 13, 2015

### donpacino

How large are the resistors?

if you have a significant amount of cable length with a large current you could get some large inductance.

Also how are you generating that power. My point is how do you know your measurement is even accurate? for all you know it is accurate and the source is just different than what you thought.

give us the ENTIRE picture. what you are using to measure the voltage. how you are generating the voltage. values. part numbers. block diagram.

10. Nov 13, 2015

### Andrew B

I've previously given some ballpark numbers, but hopefully the attached image gives the necessary clarity. The voltage supply will be custom made by a HV supply producer. I know some info initially. Due to its design as a DC supply -- not an impulse generator -- its internal voltage measurement is not high speed. Also it of course is measures voltage at the source / tool, not at the capacitive load being tested.

Hence the ultimate goal is to probe the voltage vs. time at the test specimen.

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11. Nov 13, 2015

### meBigGuy

The voltage divider would be a nice solution since the waveform tells all. But it might be difficult to implement.

You need to consider the load that the divider presents to the supply at 200KV. Even a 200M resistor would draw 1ma and dissipate 200 watts

I might try to go about this differently. The HV supply could be controlled by an external voltage. Since you know the relationship between the control voltage and the output voltage you can measure the control voltage at breakdown. (but you would have to sense breakdown somehow)

The resistance. capacitance and inductance of the cable would probably not be an issue at those speeds.

The other possibility is some sort of non contact sensing (but it would have to not break down at 200KV).

Tough problem.

12. Nov 13, 2015

### Averagesupernova

Back when tube type TVs were actually repaired they had high voltage probes that could measure the voltage on the second anode of the CRT. A typical large color TV would have about 30 KV on this second anode. Several schemes existed or measuring this voltage. One such probe had a small meter movement inside it and a number of standard multiplier resistors in series. Another method was just a probe with a bunch of series resistors that was made to go in series with a regular voltmeter.
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See this to see how they physically look: https://www.google.com/search?q=hig...ved=0CAcQ_AUoAmoVChMIp6bQ1YqOyQIVFFFjCh0PFgge
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I can see the reason for the voltage divider. How do you capture a fast event like this at such high impedances when it is just a series resistor? If we load the hi-pot tester to just before its max current we cannot have a load less than about 33 meg. To scale 100 KV down to 10 volts is 10000:1 so a voltage divider with a 33 meg in series with a 3300 ohm resistor with a 10 meg scope probe across the 3300 ohm resistor won't load the divider too much in terms of capacitance. However, how do you get a 33 meg resistor that is physically short enough to have negligible inductance? You can't. You will have to have some compensating capacitors in parallel with the 33 meg. This assumes the resistors can handle that kind of power. If you increase resistance to reduce wattage the problems only get worse as far as capacitive loading goes. The link I provided shows a few images of this. In DC work where we are unconcerned with fast edges compensating capacitors are not really needed. Good luck.

13. Nov 13, 2015

### Averagesupernova

One last thing. Automotive analyzers will have a way to see the high voltage on the spark plug. It is a non-contact probe that wraps around the coil or plug wire. The waveform is displayed on a CRT or LCD. You can actually see the voltage rise and the air ionize (voltage falls) and when the air/fuel charge actually ignites. All this from just a capacitive sensor on the plug wire. Look into how those work.

14. Nov 13, 2015

### nsaspook

We routinely make measurements in this voltage range but It's always with special equipment over 40KeV from a company like Ross if we need precision measurements.
http://www.rossengineeringcorp.com/...viders/hv-voltage-dividers-detailed-info.html

See if you can find a copy of this book, "High Voltage Engineering Fundamentals" E. Kuffel, W.S. Zaengl.
Ch 3.3 is HV measurement with voltage dividers

Last edited: Nov 13, 2015
15. Nov 13, 2015

### meBigGuy

I forgot about the insulation on the wire. You can use an inductive pickup on or near the feedwire as Averagesupernova suggested.

Calibration might be tricky. You could measure the breakdown with a voltage the DC supply can track, and use that to calibrate the pickup when you use the faster waveform.
The fast rising waveform might have a different breakdown than a slow DC that you can measure with the supply.

The collapsing waveform when the breakdown occurs can be picked up easily, although it will have ringing and overshoot.

16. Nov 13, 2015

### donpacino

I'm going to pose a different question... why does the rise time for the voltage supply have to be so fast?

17. Nov 13, 2015

### donpacino

at 5 nF (the smallest capacitance load)
300 ms (the largest time step you're allowed)
and max voltage

your current draw (with no resistor load) will be 3.3 mA. Is that 3.3 mA rating only for steady state or transient?

18. Nov 13, 2015

### meBigGuy

Good catch
I missed the 5-15pf load capacitance. The current draw for 15pf would be worst-case at 9.9ma.

9.9ma @ 200KV is 1980 watts of power. It seemed to me that the supply was ramping up in 300ms, meaning that the supply needs to be capable of dynamically supplying that amount of current for that period of time.

19. Nov 14, 2015

### donpacino

is that 50-300 ms the test spec or the supply characteristic. I was assuming it was a test spec.

When we run HiPOT we typically do timed steps at various intervals. If you are running a hipot it should be consistent each time. You should NOT rely on the characteristics of the supply. If you cannot control the ramp speed of the supply electronically, i suggest you make a system that can do so.

also if this is for an industrial application and you plan on using this for production purposes, you might want to just buy a hipot machine to do all of this for you.
They cost \$, but remember that time is money.

20. Nov 14, 2015

### meBigGuy

I like the idea of timed steps controlled by the supply. Depends on how accurate you really need to be.