Confusion on voltage divider and high voltage measurement

[toyonline]

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

 

[crutschow]

What is the frequency of the signal?

If you are not using a 10:1 scope probe and the frequency is near line frequency, then the typical scope input impedance of 1 megohm is loading down the capacitive voltage divider. You should use a 10:1 probe for the measurement. The scope input capacitance is not a problem since it is in parallel with the C2 capacitor and is much smaller than 4.7nF, however the connecting cable capacitance could be a factor if you have a long coax cable.

It also appears that the loading of the oscilloscope is loading the circuit under test which causes a frequency change. Again you should be using a 10:1 probe, if you are not.

 

[toyonline]

In scope probes, R1C1=R2C2 but since the R ratio is unknown unless you have done a leakage test to verify it is much more than 1 GOhm . Is it possible your insulator leakage is 700MHOhm?
Ideally the R Ratio would match the Cap ratio for broad spectrum and low frequency , otherwise R ratio will over-ride the cap impedance ratio at this low frequency.

The second problem is your 4.7pF Output is being loaded by the measurement capacitance of the probe and amp.

From the specification, the 4.7pF should have a resistance 10^5 Mohms at 25C, 500VDC. For the 4.7nF, the insulation resistance is 1000 ohmfarad. If I using this data to do a Spice simulation, I didn't see any apparent distortion on the voltage ratio of real output and monitoring port.

Is it possible due to the connection cable and probes? Since I used a 12pF scope probe to connect the real output to the scope, and a 75 ohm coax to monitor port. If considering impedance from probes and coax, the ratio seen by the scope should be (Zprobe+C1)/(Zprobe+C1+Zcoax+C2).

This means my measurement device add some distortion to the voltage ratio. So, if I only care about the voltage on the output port, I should read voltage of the monitor port, and multiplied by (C1+C2)/C1, i.e. ~1000 to give me the output voltage (instead of measuring it?).

Am I correct?

 

[FvM]

Without a frequency information, all calculations are void.

 

[toyonline]

What is the frequency of the signal?

If you are not using a 10:1 scope probe and the frequency is near line frequency, then the typical scope input impedance of 1 megohm is loading down the capacitive voltage divider. You should use a 10:1 probe for the measurement. The scope input capacitance is not a problem since it is in parallel with the C2 capacitor and is much smaller than 4.7nF, however the connecting cable capacitance could be a factor if you have a long coax cable.

It also appears that the loading of the oscilloscope is loading the circuit under test which causes a frequency change. Again you should be using a 10:1 probe, if you are not.

Yes, it is a HP10073A 10:1 probe, with 12pF, 1Mohms.

There are two frequencies from source, one 750kHz, the other is 1.5MHz.

I wonder if I could simply measure monitoring port voltage, and simply multiplied (C1+C2)/C1 to obtain the real output voltage? Since it seems to me the measurement device and cable will introduce some distortion.

 

[FvM]

At 1 MHz, capacitor initial tolerance (typicaly 5 or 10 %) is the dominant error source, second is cable capacitance of about 65 pF/m depending on the length.

Obviously the measured circuit is affected by the probe impedance. The effect could be better understood if we knew it's equivalent circuit. The question "which measurement is real" has to be answered by yourself based on application requirements.

 

[chuckey]

With a circuit like this, do not hang anything across the whole tuned circuit, because you will damp the Q and detune it with the added capacitance. Just measure the 4.7 PF and 4.7 NF carefully to get their ratios. Scopes are not precision voltage measuring devices use a valve voltmeter (or its solid state equivalent ).
Years ago I used this system to calibrate real valve voltmeters against a lab standard, by using a very high Q coil (Q > 470), a 100 PF and a 10 NF capacitors and a standard lab RF signal generator. I could get > 500V @ 1MHZ across the coil.
Frank

 

[toyonline]

With a circuit like this, do not hang anything across the whole tuned circuit, because you will damp the Q and detune it with the added capacitance. Just measure the 4.7 PF and 4.7 NF carefully to get their ratios. Scopes are not precision voltage measuring devices use a valve voltmeter (or its solid state equivalent ).
Years ago I used this system to calibrate real valve voltmeters against a lab standard, by using a very high Q coil (Q > 470), a 100 PF and a 10 NF capacitors and a standard lab RF signal generator. I could get > 500V @ 1MHZ across the coil.
Frank

Thanks Frank!

That means once I rule out imperfections on capacitors, such as a high insulation resistance, low ESR, etc., the real output (without any load) voltage should be obtained from that capacitance ratio.

Possibilities could change that output voltage include: 1) loading impedance, which will detuned the circuit; 2) measurement device connected to the output, through probe or coax. Actually this could also be viewed as kind of loading impedance.

Since I don't have a valve voltmeter at hand, the best way for me to estimate that output voltage is to measure the capacitor ratio, and measure monitoring voltage (through Spice simulation, I noticed monitoring port is less affected by measuring device). Then calculate the real output (without loading).


I would like to put this argument one more step. Even I connect some simple loading device (for example a pair of small sized parallel electrode plates) with the output, the voltage across C1 capacitor could also be calculated from that ratio and monitoring port. Since if I focused on that two series capacitors, the voltage across them would always be proportional to C1/(C1+C2) (if there is no significant leakage duet to insulation resistance or ESR). Am I correct?

- - - Updated - - -

With a circuit like this, do not hang anything across the whole tuned circuit, because you will damp the Q and detune it with the added capacitance. Just measure the 4.7 PF and 4.7 NF carefully to get their ratios. Scopes are not precision voltage measuring devices use a valve voltmeter (or its solid state equivalent ).
Years ago I used this system to calibrate real valve voltmeters against a lab standard, by using a very high Q coil (Q > 470), a 100 PF and a 10 NF capacitors and a standard lab RF signal generator. I could get > 500V @ 1MHZ across the coil.
Frank

Hi Frank, may I know how do you obtained such a high Q coil? Do you make it by yourself? Thank you.

Felix

 

[D.A.(Tony)Stewart]

We don't know anything about your objective, nor input or output impedances.

All we know is you are tuning 4.7pF with 0.5 & 1MHz
, somehow, but we don't know how or why.

Obviously anything in the pF range is easily loaded but at 1MHz, but you are looking for a 30% measurement error.

for reference @1MHz
1pF = 160 kΩ (= 10e6/6.28)
and 1000pF = 160 Ω

Thus the cap ratio dominates any leakage ratio or ESR which can be neglected at 1MHz ( but not 60Hz, which I assumed at first)

However there may be other capacitance on your layout not shown in your schematic which can introduce errors.

In order to design a fixed cap divider ratio, all leakage capacitance must be known, or you use a known signal level and determine where the stray leakage is coming from then determine your equivalent capacitance and whatever ratio... e.g. 700:1, then only use the monitor.

 

[chuckey]

Just a bit of history Felix. I was test engineer at a manufacturer of airfield radio beacons. These are LF (190 -400 KHZ) MCW transmitters BUT because they were used actually on landing strips, the aerials were 5m high whips (else the planes would crash into them). These very short aerials were simulated by a 10 ohm resistor and a 100 PF capacitor. So the aerial matching units were basically a series inductor.
These inductors were made from 50mm long sections of 125mm tubing with many "pins (6mm diam) sticking out of them, so in the height there would have been , say 5 rows, and around the circumference say, 20. Err, that's 100 pins per section. The winding were 119 X 42 SWG Litz wire, basket wound, using the pins as guides. This was to keep the interlayer capacitance as low as possible, I suppose there would have been about 20 turns in all.
When we tested the complete transmitter and aerial tuning unit, the transmitters would have delivered 80 watts into a 50 ohm load, but only delivered 10 watts into the dummy aerial, the balance being lost in the ATU. The whole construction of the ATU was non metallic (no Q damping here), I could draw 50mm sparks from the coils8-O .
Frank

 

[toyonline]

We don't know anything about your objective, nor input or output impedances.

All we know is you are tuning 4.7pF with 0.5 & 1MHz
, somehow, but we don't know how or why.

Obviously anything in the pF range is easily loaded but at 1MHz, but you are looking for a 30% measurement error.

for reference @1MHz
1pF = 160 kΩ (= 10e6/6.28)
and 1000pF = 160 Ω

Thus the cap ratio dominates any leakage ratio or ESR which can be neglected at 1MHz ( but not 60Hz, which I assumed at first)

However there may be other capacitance on your layout not shown in your schematic which can introduce errors.

In order to design a fixed cap divider ratio, all leakage capacitance must be known, or you use a known signal level and determine where the stray leakage is coming from then determine your equivalent capacitance and whatever ratio... e.g. 700:1, then only use the monitor.

Here I attached the circuit I am using. The circuit was used to magnify signal from a linear amp (source). I have checked the insulation resistance of the caps. To me, it seems caps leakage is not the reason. Yet I think the problem comes from my measurement, e.g. probe impedance.

I don't quite sure whether upstream caps will affect that ratio much. In my understanding, since all signals are finally applied across the caps C1 and C2 divider, the ratio error will possibly come from 1) caps leakage of C1 or C2, and 2) measurement impedance, which means the scope doesn't see real numbers at exactly C1 and C2 position.

Since I am working at ~1MHz, it seems caps leakage is not that significant. So does this means I need to forget ratio read from the scope, but instead of measuring a caps ratio of C1 and C2, plus readings at monitor port?

Thank you.

Felix

 

[chuckey]

I do not know what your "matching" network is mean't to do. First of all short out C4 and remove R1, they only waste power and provide no useful purpose. Given the values from your circuit, the input section is 42 PF + 530µH and the output section, 70 PF +268µH. These sections both resonate at about 1 MHZ, so the input section looks like a short circuit !! So the output impedance of the generator is across the main circuit. I think to get the best bang for your buck would be to put a 70 X 50 = 3500 PF capacitor in series with the earthy end of the 70 PF and feed your RF in here. You should get loads of RF across the main coil - Like I did see above.
Frank

 

[toyonline]

I do not know what your "matching" network is mean't to do. First of all short out C4 and remove R1, they only waste power and provide no useful purpose. Given the values from your circuit, the input section is 42 PF + 530µH and the output section, 70 PF +268µH. These sections both resonate at about 1 MHZ, so the input section looks like a short circuit !! So the output impedance of the generator is across the main circuit. I think to get the best bang for your buck would be to put a 70 X 50 = 3500 PF capacitor in series with the earthy end of the 70 PF and feed your RF in here. You should get loads of RF across the main coil - Like I did see above.
Frank

C4 could be removed in this situation. But the output voltage was across R1, so R1 was not designed to be removed here. The purpose of whole circuit was to magnify voltage through source (linear amp). The output was connected to a parallel electrode plates to create a high electrical field. So the capacitance of the loading electrode plates also change the resonant and output voltage.

So the purpose of the divider was actually to monitor that output voltage, through a scope. The reason for using a scope is besides the voltage value, I also need to monitor waveform at output port.

The divider was designed to have a ratio ~1000:1. But then I want to see what voltage is really produced at the output port, as well as the waveform. So I connect it with a probe to the scope, and read that 700:1.

 

[chuckey]

I have been giving your problem some more thought. Your output voltage is dependent on the Q of the output circuit,which is primarily determined by the loss of the inductor and any resistive loading. I think the reactance of the 268µH at 1 MHZ is 1800 ohms, so if its got a Q of 100, there is a loss resistance of 1800 X 100 across the tuned circuit = 180 KΩ, in addition you have a 1M5 across it. The load is just capacitive, lets say the impedance of the circuit is 180KΩ -7% ~ 165K. So you have to match your 50 ohm PA to a load of 165K, or ~3000 :1 so the tapping point should be 1/√3000 or 1/55, as your tuning cap is 70 the bottom cap should be 55 X 70 = 3850 PF. My guess at 3500 PF seems pretty accurate. Of course if your Q of the inductor was 200, the loss resistance would be twice as much, say 360 K, less the 1M5 = 280 K, the Z match, 280K/50 = 5600, tapping = √5600 ~75, so the capacitor would now be 70 X 75 ~ 5000PF . . You can see where this is going
Frank

 

[D.A.(Tony)Stewart]

There are some very critical values to resonate exactly at 750kHz and 1500 kHz. The cap divider ratio is not critical as your middle cap C3 = 70pF dwarfs the 4.7pF. Note I increased the value to tune it.

If YOU MUST verify the cap ratio, DO it with standard 50 Ohm generator and 2 caps only, otherwise just use predicted ratio.

Then tune the other parts to optimize the center freq for each. Notice that I put a grid resistor value of 100 Ohm for the source. the lower the higher the Q> anything higher than 500R combines the two resonances into one in between. What do you use?

Note L ratio should be 2 and note the other critical C values changed.

Total peak gain with 100R source is >25dB balanced for each. How much gain do you need?


If you want to play with this... go here and enable Java Applet
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[toyonline]

There are some very critical values to resonate exactly at 750kHz and 1500 kHz. The cap divider ratio is not critical as your middle cap C3 = 70pF dwarfs the 4.7pF. Note I increased the value to tune it.

If YOU MUST verify the cap ratio, DO it with standard 50 Ohm generator and 2 caps only, otherwise just use predicted ratio.

Then tune the other parts to optimize the center freq for each. Notice that I put a grid resistor value of 100 Ohm for the source. the lower the higher the Q> anything higher than 500R combines the two resonances into one in between. What do you use?

Note L ratio should be 2 and note the other critical C values changed.

Total peak gain with 100R source is >25dB balanced for each. How much gain do you need?

Thanks a lot.

The variable cap is C3. Since in my applications (not measure voltage here), the output was connected to a parallel electrode plates, which will add new capacitance and C3 need to be retuned for maxmium gain.

I didn't use any grid resistor R. Since the impedance of the LC part is small compared with that R, an inclusion of resistor R would decrease my output voltage. For example, in your simulation, a 100ohm R would give 25dB gain, rather than 45dB gain without a R.

I would like to generate a output voltage of ~7kVpp by using this circuit. I am using a signal generator of output 300mVpp amplified by a linear amp in between the generator and the LC circuit. I am not quite sure of how much gain by such a voltage. But up to now, I could only obtain ~6kVpp (read from the monitor port through a scope, if the ratio was ideally 1000:1). Above 6kVpp will heating the circuit much, and further decrease the voltage.

 

[toyonline]

I have been giving your problem some more thought. Your output voltage is dependent on the Q of the output circuit,which is primarily determined by the loss of the inductor and any resistive loading. I think the reactance of the 268µH at 1 MHZ is 1800 ohms, so if its got a Q of 100, there is a loss resistance of 1800 X 100 across the tuned circuit = 180 KΩ, in addition you have a 1M5 across it. The load is just capacitive, lets say the impedance of the circuit is 180KΩ -7% ~ 165K. So you have to match your 50 ohm PA to a load of 165K, or ~3000 :1 so the tapping point should be 1/√3000 or 1/55, as your tuning cap is 70 the bottom cap should be 55 X 70 = 3850 PF. My guess at 3500 PF seems pretty accurate. Of course if your Q of the inductor was 200, the loss resistance would be twice as much, say 360 K, less the 1M5 = 280 K, the Z match, 280K/50 = 5600, tapping = √5600 ~75, so the capacitor would now be 70 X 75 ~ 5000PF . . You can see where this is going
Frank

Thanks Frank!

Exactly my output is dependent on Q of the circuit. And the Q of the circuit is mainly depended on inductors.

I do not understand well in your calculation of loss resistance, by multiplying Q with reactance of inductors. I know that Q could be represented by inductor reactance divided by its resistance, Q=X(L)/R(L), which means R(L)=X(L)/Q. In my situation, R(L)=18 ohms, if Q=100. (I am not sure about Q of my inductors, maybe <<100 I guess)

But I think your calculation on the capacitance of the lower cap explains why a 4.7nF cap was chosen here to monitor the circuit. I do not know theory behind that, i.e., matching and tapping point. I need to learn more.

I am looking for methods to improve voltage gain from my output. That's why I am interested in your high Q coils mentioned before. Thanks for sharing!

Felix

 

[D.A.(Tony)Stewart]

6kV into 1.5M load is presumably non-inductive carbon or ceramic 25W load?

How did you make a zero Ohm source impedance?

If you slide the cap values you can see the spectrum shift almost instantly. on my simulator.

 

[toyonline]

6kV into 1.5M load is presumably non-inductive carbon or ceramic 25W load?

How did you make a zero Ohm source impedance?

If you slide the cap values you can see the spectrum shift almost instantly. on my simulator.

It's a non-inductive resistor. Since I have no clear idea of what to use for that loading, simply choosing a most common one in our lab.

The power comsumption of loading 1M5, for 6kV, is ~24W. I dont know does this will mean something in my circuit? or voltage gain at output?

Ah, yes. Source output impedance. That could be either from the signal generator or my linear amp. It could possibly be 50 ohms. That means I would expect a reduced voltage gain from the LC circuit due to source impedance. Thanks for pointing out.

By which caps do you mean in that frequency shift test? I mean 27p,15p, 82p will all affect the frequency.

 

[D.A.(Tony)Stewart]

It sounds like an interesting problem but too poorly defined. I for one, need overall purpose and more accurate Specs of the design to assist further... input and output, specs, ... physical, environmental, electrical, mechanical..... each will disturb your results. Is it a variable f with a tolerance? Exactly how the Linear amp is designed affects the overall gain , (Q) and thus output amplitude and also to some extent the resonant frequency. The coils and the ESR and self resonant frequency (SRF) are critical. Do you have a part number, or is it custom ferrite with Litz wire . YES all the caps 27,15,82 are critical shift the two resonant frequencies together and slightly together or apart.

Is this a model , a prototype or a national security secret?

What is the desired result? To heat up a 50W carbon 1M5 resistor ( which is not common in any lab)