Any PCB based capacitive divider guidelines?

Discussion in 'General Electronics Chat' started by toozie21, Feb 8, 2016.

Oct 4, 2012
43
1
Somehow I missed learning about capacitor dividers and saw one when reading a tear-down for an agricultural electric fence charger.

It got me thinking about them and wondering what kind of rules there are for them (I can find very little info on them for the questions I have). For instance, if I expanded the article I read to be for a 15kV pulse, could I create a 30pF on the FR4 (so let's call it 2 in^2), and then add in a /0.1uF through-hole capacitor to divide down to and output of 4.5V?

The math seems to make sense, but I wasn't sure if I was defying the laws of physics (or maybe over-stating what FR4 is capable of dropping voltage wise). Any insight or articles to this kind of stuff would be dine as well.

2. #12 Expert

Nov 30, 2010
16,705
7,358
The math works for voltage, but the current is going to be so small that I can't imagine what you would use it for.

3. kubeek AAC Fanatic!

Sep 20, 2005
4,691
806
That probably would work, but how thick will the pcb be between the two plates?

Oct 4, 2012
43
1
What I would be interested in is measuring the voltage at the middle junction. So I would need much current (and there isn't much to begin with on a fence charger), but it might not be enough still (sadly).

Well, I guess that would be something that would need to be worked out (and something I am not fluent on). I guess the standard 2-layer size was what I was thinking

5. kubeek AAC Fanatic!

Sep 20, 2005
4,691
806
Ok, that solves the issue of isoaltion voltage, but then a 30pf capacitor at 1.6mm pcb will be roughly 25*50mm.

Oct 4, 2012
43
1
7. kubeek AAC Fanatic!

Sep 20, 2005
4,691
806
Yes that is correct, I just wanted to point out that 2 square inch is huge board space, and you will need more because you cannot have the two plates right on the edges of the board, you need to leave enough material so the voltage doesnt jump across.

Oct 4, 2012
43
1
OK, maybe that is where my ignorance comes in. I was thinking that there was only one plate, and it would be 2 in^2. Then I would have a trace from it running to the second cap (which is not a PCB cap). Is that trace what you are referring to needing to have decent space? I can imagine it would need a fat trace and a nice gap between the two caps.

9. kubeek AAC Fanatic!

Sep 20, 2005
4,691
806
To make a capacitor you need a top and a bottom plate, and what counts for the capacitance is only the area where those are overlapping. So you need two 2 in^2 foils, each on one side of the board, then you can run a trace far enough to the standard capacitor.

Oct 4, 2012
43
1
Bingo, makes perfect sense now. Two 2in x 1in plates on both sides of the board perfectly aligned with each other. Then a trace from one of those to the next capacitor.

Now how do I figure out how little current I will have? I guess I could try to model the total resistance of the two capacitors and look at the voltage drop voltage?

11. kubeek AAC Fanatic!

Sep 20, 2005
4,691
806
That depends on how fast are the edges of the signal you are measuring.

Oct 4, 2012
43
1

Oct 4, 2012
43
1
SO I tried to work the math a bit more today, but I am really not sure how to work the frequency. With a C1 of 30pF and a C2 of 0.1uF, I have a Ct of 29.991pF.

I also have a VC1 of 14995.5V and a VC2 of 4.5V.

I know that I=V/Xc, which is I=15kV/Xc/

Xc =1/(2*pi*f*Ct) => so combing that with the above and filling in I get: I = 15kV*(2*3.14159*f*29.991pF) => I = 94247.7*f*2.9991e-11 =>
I = 8.995e-11 * f.

So how to I come up with the frequency of an impulse to derive the current?

14. kubeek AAC Fanatic!

Sep 20, 2005
4,691
806
You cannot work out the frequency, but you can work out how fast the edge is. I=C*dV/dt. If the rise time is 1us and you neglect the 100nF, then I=30pF * 15kV / 1us = 450mA flowing through the two caps in series.
I think that any reasonable loading on that divider will have little effect on the measured voltage. That 741 is a crappy choice instead of a proper comparator.
Also I noticed that the circuit is powered directly from mains, which is a major safety hazard.

Oct 4, 2012
43
1
Thanks, I think that that makes sense. I will only need a mA or two (I assume), so that is good to know that I will have more than enough. I was considering trying to implement something with an LF398 to be a sample and hold triggered off of the pulse itself (I am more of a digital guy, so op amps are a little bit of a curve for me).

I would be powering off a batter, so no worries there .

16. kubeek AAC Fanatic!

Sep 20, 2005
4,691
806
With batteries you will need to find some other way to control the transistor, as the counter relies on the 50Hz mains frequency. Or use some low voltage AC transformer.

Oct 4, 2012
43
1
Well I am only using his circuit as an example for how to read the voltage. I don't want to create a fence energizer, I was just interested in reading the voltage of a current fence with a microcontroller. I've tried with a resistor voltage divider in the past and had minimal success, so I thought that this might be a good way to go as a next step. I am not sure that it will work any better though.

Oct 4, 2012
43
1
Is there any particular reason to go with an odd-shaped PCB-capacitor like in the instructable as opposed to a rectangle?

19. kubeek AAC Fanatic!

Sep 20, 2005
4,691
806
All the cutouts look like a radius from the soldered pins to keep isolation distance. But that one trace looks to be completely ignoring all that.