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# [SOLVED]Voltage balancing resistors for series capacitors

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#### Robert T

##### Member level 1
I am constructing a loop antenna for low frequency RF. 6 capacitors are in series to divide the 1200V across the antenna.
The current on the antenna is approx 1.5 amps.
How do I determine the wattage and value of resistors to balance the voltage across each capacitor?

With 1200V and 1.5 A, it implies an impedance of 800 ohms operating somewhere. Do the capacitors serve as a matching impedance? Or are they the source of this impedance? I'm not sure.

If it were DC, the balancing resistors would be able to equalize the charge on the caps eventually, regardless of the cap values, and regardless of the variation between them.

In the case of AC, you will select a low enough value for the balancing resistors, so the RC time constant is able to have some effect during a cycle. The ohm value should not be so low that it discharges the capacitors more than a few percent during a cycle.

So if the smallest cap is 80 percent of the largest cap, then it may charge to a voltage 20 percent greater than the largest cap. And then if the resistor value is selected to discharge the smallest cap 20 percent during a cycle, then that may be sufficient for the resistors to do their job of equalizing all charges.

This is only supposition, because obviously the waveform in the antenna makes a lot of difference in the action of the balancing resistors..

Robert T

### Robert T

Points: 2
To use resistors is a waste of aerial Q (unless you want it broadband) and a waste of power. Measure the value of the capacitors then calculate the volt drop across the lowest value. Either pad it out to make it larger in value or increase its voltage rating.
Frank

Robert T

### Robert T

Points: 2
Balancing resistors are needed only if capacitors are under DC voltage. If only AC voltage is applied it will distribute along capacitors inversely proportional to capacitance of each capacitor. Assuming all capacitors in that chain has the same capacitace AC will be divided to equal AC voltages.

Robert T

### Robert T

Points: 2
Balancing resistors are needed only if capacitors are under DC voltage. If only AC voltage is applied it will distribute along capacitors inversely proportional to capacitance of each capacitor. Assuming all capacitors in that chain has the same capacitace AC will be divided to equal AC voltages.

that is what I was thinking too

Thanks all. Will modify antenna to get capacitor values as close as possible, pad out where necessary, and high voltage rating to avoid using resistors.

Brad, from your calculation, the antenna is drawing a lot less power than its 1.5A rating. I assumed that was what was going into the coil. The resistance on the coil is only 8 ohms. Not sure why it wont draw any more. Maybe the inductance on the coil is an issue.

You have given no details of your aerial matching network. But a loop aerial is a inductance in series with a resistance. This means that you need a capacitor across it to null out the inductive bit. This then gives you an effective impedance of L/CR. This has to be matched to your amplifier. The easiest way is to include a large capacitor in series with the earthy end of the resonating capacitor. This acts like a tap on a transformer. So if its 10 times the value of the resonating capacitor, then the L/CR is presented to the amplifier as L/100 X CR. The side effect is that the value of the resonating capacitor is now 10% lower, so its value will have to be readjusted. You will have to go through the loop a couple of times to optimise its value.
Frank

Robert T

### Robert T

Points: 2
I have no problems to imagine 1200V in the series resonant circuit of a loop antenna. I arrived at similar values when calulating a long range 13.56 MHz reader circuit.

You didn't tell the application parameters, so we have to assume that you know how to calculate everything. Borber ant others clarified that you don't need balancing resistors for series connected AC capacitors. I would however try to find high voltage capacitors that handle the voltage without series connection.

Robert T

V
Points: 2

### Robert T

Points: 2
I applied the information provided and the voltage across the antenna increased to 1800V PtP.
Could not readily find any caps for this voltage so used 6 * 10nF, 700VAC capacitors in series to tune the antenna. This was the highest voltage value that was readily available. Using 6 caps I was also able to fine tune the antenna by mixing the values.
The resistance on the coil is now 7 ohms.

Thanks for the information on the impedance matching Frank. I have just started reading up on matching and will try this tomorrow.
Would it be right to understand that if the impedance between the amp and LC antenna are matched, then more current would flow into the antenna?
The caps in series give a value of approx 1.7nF. If I place a 18nF capacitor with one end between the end of the series and the coil, and the other end to ground(if my understanding is correct), will the signal not be lost through the 18nF cap rather than going through the coil?

Thanks, Robert

When the impedances are matched the power transfer is the maximum. No this a valid way of trying to match a high impedance parallel tuned circuit, whose impedance at resonance is high (10 K?) to a low impedance source, such as a 50 ohm signal generator.
I have actually used this technique to generate 400V of RF from using very high Q coils with a 100 PF and a .01MF to resonate it, feeding the signal generator across the .01.
Be careful when its working correctly you may get more then your 1800V!!, start of with a low power.
Frank

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Robert T

### Robert T

Points: 2
Are either of these 2 circuits as per your description Frank. The coil Q is very low (less than 5).

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You forgot the ground reference of the amplifier in both schematics. The circuit doesn't make much sense without it.

I presume it's meaned this way:
- the single ended amplifier is sourcing a voltage against ground
- the "bridge" amplifier outputs a symmetrical voltage referred to ground.

In both cases, the 16 nF capacitor is unwanted because it destroys the series resonant circuit. I have no idea why you put it there.

I assume so far that the operating frequency is around 60 kHz so that L and 1.6 nF act as series resonator. Otherwise please clarify.

Robert T

### Robert T

Points: 2
Your right on both cases FvM. Frequency is at 58kHz.

I am trying to understand how Frank uses a capacitor to lower the impedance of the LC circuit so the current from the amplifier is increased.
At the moment there seems to be a significant impedance that is limiting the voltage across the coil. Did some calculations and the impedance across the series capacitors seems to be significant when compared to the 8 ohm amplifier.
If I reduce the resistance by using a larger diameter wire, the voltage increases but if I increase the turns on the coil, there comes a point where the increased diameter of the wire has not effect and the signal is negligible.

I am not familiar with impedance matching amplifiers to antennas.
(Maybe this should be a different thread?)

Thanks.

You don't need to refer to impedance matching circuits to understand the circuit operation. It's simply so that the negative imaginary impedance of the capacitor cancels the positive imaginary impedance of the coil, so the coil resistance and some capacitor ESR remains as purely real load. Basic AC circuit theory.

A more complex matching circuit (or alternatively a transformer) becomes necessary if the coil R does not match the permitted amplifier load impedance range.

Robert T

### Robert T

Points: 2
If I understand correctly, the maximum current out of the amplifier occurs once the capacitor and coils impedance's are matched.
Currently the amplifier does not even get warm. Tried increasing the amplifier from 15W to 35W but there was little difference in the voltage on the antenna.

Is there any part of the circuit I can modify to increase the voltage across the antenna, apart from the size as that is fixed?

Is there any part of the circuit I can modify to increase the voltage across the antenna, apart from the size as that is fixed?
The number that actually matters for generated field is ampere turns.

Robert T

### Robert T

Points: 2
Maybe I need to change the amplifier then because as I increase the number of turns, the current from the amplifier decreases. Currently I have an audio amplifier TDA7297. It has an internally fixed gain.
Some audio amplifiers have protection against very inductive loads though it is not mentioned on the data sheet for this amplifier.
Is there another type of amplifier that I should be considering.

Maybe I need to change the amplifier then because as I increase the number of turns, the current from the amplifier decreases.
Increasing the ampere turns doesn't not necessarily mean more turns. It can also be achieve by thicker wire (possibly litz wire), resulting in lower ESR and higher current.
Some audio amplifiers have protection against very inductive loads though it is not mentioned on the data sheet for this amplifier.
A tuned series resonant circuit is no inductive load.

The coil and amplifier combination should be designed starting from required ampere turns. There may be other requirements like resonator bandwidth if you intend any kind of modulation. You also didn't mention any inductively coupled load which might affect the resonator behaviour.

V
Points: 2
I have not been able to find a formula to calculate the current in the coil. Ohms law comes up with over 70 amps which cannot be right. (V = 1400 V PtP (495 rms), R = 7 Ohms)
So don't know what the ampere turns is.

The voltage I am aiming for is 1800 PtP.
To increase bandwidth I understand (I think) that I need to increase the series resistance (increase resistance in the coil or include a resistor) or reduce inductance of the tuned antenna. Given the amplifier is only designed for an 8 ohm load, the bandwidth is going to be relatively small. (~150Hz) with the current configuration.

If my understanding is correct, if I want to increase the bandwidth I need an amplifier that will supply into a more resistive load.

Appreciate your assistance with this FvM.
Robert

I feel some misunderstandings with elementary AC network calculation. The coil is surely inductive, but not seen by the amplifier as inductive load, because the inductance is compensated by the series C.

In resonance, the amplifier output voltage is applied to the real load component, which is comprised of the coil DC resistance, skin and proximity effect losses, capacitor ESR and possibly magnetical losses in the vicinity, e.g. eddy currents or an intentional load circuit.

Robert T

Points: 2