Confusion on voltage divider and high voltage measurement - Page 1
Forum section: electronics / projects

author: toyonline
author_rank: Contributor
timestamp: April 07, 2014, 02:49:04 am
content: Hi, I am measuring and monitoring a AC signal from LC circuit. The real output is could be tuned from several volts to several kV. A serial capacitors are used to form a voltage divider. C1=4.7pF (7200Volts), C2=4.7nF. According to voltage divider formula, the voltage of monitoring output and real output should have a ratio of C1/(C1+C2), which is ~1:1000.That's what I understand. But when I connect those two port to a HP oscilliscope, the observed Vpp voltage ratio of monitor and real output is not 1:1000, instead of 1:700. Anyone kindly help me to find out reasons for the reduced voltage ratio? Did I miss some important knowledge on voltage divider?And another finding is, only when the monitoring and real output ports are connected to oscilliscope at the same time, the measured frequencies are the same. But if those two ports are measured one by one, a frequency shift will be read from the scope. That brings a problem of which frequency is real?Thanks very much.Felix

author: pa2ees
author_rank: Contributor
timestamp: April 07, 2014, 03:14:16 am
content: I would check the capacitance of your scope probe. I bet you're introducing more than 10pF capacitance with your probe, thus you might be changing the divider characteristics.Sorry I don't have any insight on the frequency shift. It could be due to the scope capacitance.

author: m12lrpv
author_rank: Regular Contributor
timestamp: April 07, 2014, 05:54:04 am
content: As I understand it caps don't voltage divide. Resistors do.Sure I'm no expert but I know Dave covered this in a power supply teardown where there were resistors in parallel in order to balance the caps (they were equal value) because they couldn't do it properly on their own.

author: Niklas
author_rank: Frequent Contributor
timestamp: April 07, 2014, 06:06:21 am
content: Capacitors can be used to divide voltage but the result is not as predictable as for resistors. Capacitance tolerance is usually more likely to be in the range 10-20% compared to 1-5% for resistors. That could also change depending on dielectric material, temperature, applied voltage and aging.The phase shift could be related to non-ideal properties, such as ESR. See if you can download a SPICE-model from the capacitor manufacturer and check how many different subcircuits the complete capacitor is based on.

author: digsys
author_rank: Unknown
timestamp: April 07, 2014, 06:43:53 am
content: As others have said, with that capacitive divider setup, ANYTHING can happen !! You'll get differences by just about anything,ie swapping leads, moving the leads, coiling them up etc etc. You really need a HV resistive divider followed by a BUFFER amp.Assuming 10KV > 10Mohm divider = 1mA = 10W, so 100Mohm for 1W is more reasonable, but 100Mohm into a 10MohmCRO will give you a very non linear derated input as well, but at least you can calculate that.IF you have to go with caps only, then you'll need to up them at LEAST 10X, to get past all the parasitics and other nastiesthat you're seeing, stuff you have VERY LITTLE control over.

author: toyonline
author_rank: Contributor
timestamp: April 07, 2014, 06:49:40 am
content: Quote from: pa2ees on April 07, 2014, 03:14:16 amI used one probe of 12pF to connect output port with the scope. And for monitoring port, it was connected through 75 ohms coax to the scope.Do you mean that once I connected those two ports with probes (or coax) will actually add new impedance. And thus will change the overall impedance of each port. So, actually the scope will see some different impedance from the output and monitor port. Therefore the voltage ratio read from the scope was according to (Zprobe+C1)/(Zprobe+C1+Zcoax+C2)? Am I correct?This means when measuring the voltage ratio (with scope), I will inevitably change the ratio, because of additional impedance from probes (or even the scope). So if I want to know the real voltage at the output port, I only need to measure the monitor port, and multiplied by divider ratio (C1+C2)/C1, which is ~1000.For example, if I measured a 100mVpp from monitor port, the real value at the output port should be ~100Vpp (although not directly measured). But if I used my probes and coax to connect those two ports together to the scope, maybe I will see only 70Vpp for the real output port.

author: pa2ees
author_rank: Contributor
timestamp: April 07, 2014, 02:32:49 pm
content: In the Schematic you just provided, your probe would introduce a 12pF capacitor from AC to GND. You would also introduce that 75Ohm impedance from your monitor port to ground. Yeah, I think your measurement tools are messing up the readings.I would do as the other posters have suggested, and up your capacitance, and add in a very low input capacitance buffer amp (something like a FET input with 0.5pF to 2pF of input capacitance). Caution, those come with caveats of their own, including layout issues, and they're horrendous on breadboards.

author: toyonline
author_rank: Contributor
timestamp: April 07, 2014, 02:52:55 pm
content: Quote from: digsys on April 07, 2014, 06:43:53 amThanks!I know I should learn more knowledge on this. The reason for up capacitance is to make those two as main loading and thus less affected by probes or coax?Change the capacitors will possibly change the characteristics of the upstream AC circuit, since it is a LC tuned circuit.Let me put it another way. If I could rule out imperfection of the capacitors, such as insulator resistance, ESR etc., and measure those capacitance to obtain the capacitance ratio. And then read a voltage from monitoring port. Could I know the voltage at output by multiply those numbers? instead of measuring that.

author: toyonline
author_rank: Contributor
timestamp: April 07, 2014, 02:57:55 pm
content: The capacitor C1 I used have a resistance >10^5 Mohms, and C2 a insulator resistance 1000 ohmfarad.

author: ejeffrey
author_rank: Super Contributor
timestamp: April 07, 2014, 05:01:20 pm
content: Hi.Capacitive dividers work fine for AC signals, but you have to design it right. First off, you need capacitors that are designed for signal use. Low-k ceramic (C0G / NP0) are probably the best bet here. High-k ceramics (X7R or similar) capacitors will change capacitance with voltage. Film capacitors are also OK, but whatever you choose has to have high enough voltage rating.Second, you need to make sure the impedance at the signal frequency is reasonable -- what is the signal frequency? You want the impedance of the lower capacitor to be small compared to 1 megaohm (the scope impedance), and the impedance of the upper capacitor to be large compared to the source so as to not load the resonator unduly.Assuming you are operating at relatively low frequency and have the scope set to 1 megaohm input impedance, you need to keep the coax as short as possible and realize that the coax capacitance will add in parallel to the capacitance to ground, as will the scopes input capacitance. Those effects will tend to increase the division ratio, not decrease it. If you have the scope terminated at 75 or 50 ohms, it will be loading the voltage divider more heavily. A better bet is to build the capacitive divider and then use a 10x scope probe to measure at the divided port. That causes another factor of 10 division, but you avoid standing waves in your coax. You still have to account for the probe capacitance loading the divider, but that should be relatively small.Finally, you need to measure the actual capacitor values. Tolerances on capacitors are bad. Probably the error in your division ratio is simply that the capacitors are not their rated values.The reason the frequency shifts is that when you connect the scope directly to the output, the scope input capacitance loads the LC circuit you are measuring, changing its resonant frequency. The (more) correct value is when you measure the monitor only, although the voltage divider will also pull the frequency.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: April 07, 2014, 07:39:25 pm
content: IMO, you should study the Boonton Q-meter, or others. I think HP made one too. They basically create a tank circuit from the DUT and measure the output. The problem is exactly the same as yours. The Boonton will give you a new appreciation for vacuum tubes.

author: toyonline
author_rank: Contributor
timestamp: April 08, 2014, 08:46:15 am
content: Quote from: ejeffrey on April 07, 2014, 05:01:20 pmThanks a lot.The frequency I am using is around 1MHz.I have measured capacitance of C1 and C2. C1=4.7pF, C2=4.1nF (almost 10 percent off). So now the estimated ratio is ~870:1, much closer to my observation.I think that frequency shift may possibly due to your explanation. The loading impedance of probe or scope changes the resonant.

author: tszaboo
author_rank: Super Contributor
timestamp: April 08, 2014, 11:21:59 am
content: Capacitors have voltage coefficient. Also, they have temperature coefficient. if you want to make a 1:10 divider, I suggest to start with capacitor from the same manufacturer, same series, same size. And order a lot of them, as probably 90% of them are not going to work as intended. Or ultimately, go with somethink like this:http://www.vishay.com/capacitors/ceramic/voltage-multipliers/And a safety suggestion: add bleeding resistors to the system!

author: toyonline
author_rank: Contributor
timestamp: April 08, 2014, 12:14:11 pm
content: Quote from: NANDBlog on April 08, 2014, 11:21:59 amThanks for the suggestions!