# The ionization chamber as a capacitor

1. Apr 5, 2012

### DragonPetter

An ionization chamber is like a capacitor in many ways. The gas inside is the dielectric, and if ionization occurs, the ionized dielectric atom or molecule will no longer be able to store the same energy at a voltage potential. The chamber will have dielectric breakdown during ionization, where the mainstream DC method for measuring this is to collect the charge and essentially measure the leakage current of the "capacitor". Because the dielectric is breaking down and the electric field is altered from the displaced ionization charges, the capacitance value should also be altered, at least while radiation is present.

Many capacitive measurement techniques are not accurate and rely on charging the capacitor over a time period to measure the time constant; leakage would change the accuracy of this technique, and the technique might not even measure a change in capacitance. A more accurate capacitive measurement is to use a tuned wien-bridge with an AC signal applied.

I am curious if the idea to place an ionization chamber as the measured capacitor in a wien bridge would be a feasible alternative to the DC+electrometer method. If the proper frequency and amplitude of an AC signal is applied to the bridge, the ionized particles will oscillate to this frequency to some degree, where both a lower voltage or a higher frequency will cause less displacement (these factors I think would play a part in recombination). The bridge should measure a change in capacitance from this ionization, where it falls out of balance and produces a voltage.

If a high enough frequency were selected for the wien bridge to be very sensitive, I wonder if this would be a feasible way to detect ionization in the chamber.

I attached a block diagram of the idea I'm considering. The frequency difference measurement would be in the case of closed loop, while the voltage measurement would be used if the loop were open. Is this a round about way of measuring the same thing that the DC method will give, and would there be no benefit from this alternative? I'm expecting it to be easily shot down, but I cant really stop thinking about it until I know why it won't work.

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• ###### AC chamber idea block diagram.pdf
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Last edited: Apr 6, 2012
2. Apr 6, 2012

### DragonPetter

Hmm . . I think this should be moved to nuclear engineering.

Can a nice mod do that? :-)

3. Apr 6, 2012

### DragonPetter

Also, I should mention that the block diagram is for conceptual purposes. I know that i would have to work that out with a little more thought for it to do anything.

4. Apr 6, 2012

### Bob S

The IPDIC (integrating pulse discharge ion chamber) ion chamber was used as an integrating capacitor at Brookhaven National Lab in the late 1960s. See http://www.orau.org/ptp/collection/surveymeters/ipdic.htm
The unit required extremely low leakage NE-2E pigtail neon bulbs. The electronics was entirely digital.

5. Apr 7, 2012

### DragonPetter

Thanks for that link. That instrument is really interesting. I never knew a neon bulb could be used as a signal source like that; old technology is designed so cleverly. I think this capacitance measurement is similar to the first method I mentioned in my first post about measuring capacitance, where a charge or discharge time constant is measured. I was worried that this is much more influenced by the leakage component in the chamber, rather than purely the dielectric/capacitance change.

I think the importance of your link to me is that this is showing that it is feasible to use a capacitance measurement to indirectly measure ionization rather than trying to collect every charge at the electrodes. The reason I began thinking of using an alternative to the HVDC/electrometer method is because I want to get away from all of the practical issues of leakage, parasitics, and extremely low current that seems to make the method difficult to get right. I hope a capacitance measurement would move away from those strict limitations. I also think I might be able to get away with using a much lower voltage level than the HVDC bias that is usually used.

At first, I was considering measuring the capacitance as an integration, but I figured this would be really hard to do because of leakage. Actually, I put a small "rodent" chamber in a radiation field with just a fluke multimeter set to measure capacitance to see if I could measure it, but I didn't observe any significant capacitance change, even transiently. I am thinking the multimeter uses the charge/discharge time constant method to measure capacitance, so I moved to look at the wien-bridge method.

Do you think that an AC bias/wien bridge circuit would have any merit? I am trying to resolve the question if the ionized dielectric will result in a large enough change in energy storage that can be detected. Assume one ion pair is formed . . I doubt any capacitance measurement method would ever be sensitive enough for measuring that incremental change in dielectric, and I'm not sure how much ionization would actually need to occur before the capacitance starts to make a measurable change.

Thanks again so much for your help.

Last edited: Apr 7, 2012
6. Apr 7, 2012

### Bob S

Radiation cannot change the capacitance of an ion chamber, but it can discharge it, just like the pocket ion chamber dosimeters.

The Brookhaven IPDIC chamber works on the principle of a negative resistance relaxation
oscillator (see ).

Here is a simple calculation of the expected pulse rate for a 1 liter argon-filled 1-atm ion chamber with 50 pF capacitance in a 1 mRad/hr radiation field. Read attachment. From Eq (5) on page 4 we get for 1 liter argon
$$Current = \frac{700 pC}{cm^3 Rad}\frac{1000cm^3}{liter}\frac{.001 Rads}{hr}=700pC/hr = 0.194 \space picoamps$$
So the current is about 0.194 picoamps, and with a 50 pF chamber, the pulse rate with a NE-2 oscillator would be about 1 pulse per hour.

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• ###### BeamLossMonitoring5A.pdf
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7. Apr 7, 2012

### DragonPetter

But if the dielectric (filling gas) is altered, which it is by ionization, isn't its effective capacitance altered?

8. Apr 7, 2012

### Bob S

The presence of a gas in an ionization chamber has a negligble effect on its capacitance. The capacitance of a cylindrical is chamber of inner radius a, outer radius b, and length $\ell$ is
$$Farads = \frac{2\pi\epsilon_o \ell}{Ln(b/a)}$$
When dE/dx ionization occurs in an ion chamber with voltage on it, the electrons and ions drift to the electrodes and discharge it, assuming there is enough voltage on it to minimize recombination.

9. Apr 7, 2012

### DragonPetter

But the relative permittivity in the cylindrical capacitance formula is dependent on the gas present. I am believing you though that it is negligible, and I think this is what I needed to be able to get the idea out of my head.

10. Apr 7, 2012

### Bob S

Look up relative dielectric constant ε of air and nitrogen in http://www.rafoeg.de/20,Dokumentenarchiv/20,Daten/dielectric_chart.pdf [Broken], and put it in formula.
$$Farads = \frac{2\pi \epsilon \epsilon_o \ell}{Ln(b/a)}$$
Regardless of how much ionization is in the gas, no atom of gas ever leaves the chamber. So how would the capacitance change?

Last edited by a moderator: May 5, 2017
11. Apr 7, 2012

### DragonPetter

Well, I have been thinking of it in terms of the dielectric model. I assumed the energy stored in the molecule or atom from the electric field is dependent on the electrons bound to the atom/molecule, and when the electrons are stripped, the atom can no longer store energy, or at least the same amount, as the electric field changes.

Last edited by a moderator: May 5, 2017
12. Apr 9, 2012

### DragonPetter

I guess another way to explain what I mean by this post is if we consider nitrogen. When it is ionized, it will no longer be a neutral nitrogen atom, it will now be an $N^{+}$, and I believe the way the dielectric model is explained, a capcitor full of $N^{+}$ atoms will not have the same dielectric permittivity as a capacitor with $N$ atoms, especially if the N atoms were bound together in some way such as $N_{2}$. Its capacitance will have a change in value according to the formula.

I realize that a capacitor full of $N^{+}$ would not make a very good capacitor since it would conduct between the platess, but, for example, if half of the atoms were ionized and the other half were not, you would have half as many neutral N atoms to store energy from the electric field, which means its dielectric capability has reduced.

Last edited: Apr 9, 2012
13. Apr 9, 2012

### DragonPetter

Another example I could think of refers to the chart of dielectrics you provided. It is an impractical example, but I think it makes the point of how I think the dielectric properties can change from ionization.

Consider a capacitor with 2 plates, insulating layers for the plates, and between the insulated plates is salt. Salt has a relative permittivity constant of 6.1, which means the capacitor will have a specific capacitance based on this number.

Now, salt can be ionized into sodium and chlorine ions (although in radiation detection, the ionization potential might not be practical). Chlorine by itself only has a relative permittivity constant of 1.7. So, the capacitance must change when a different relative permittivity is now present, regardless of if all of the atoms never leave.

edit: when I talk about the "dielectric model" I'm referring to the idea that when an electric field is placed across a dielectric, the atom or molecule's internal structure will be displaced, and create its own mini electric field, and this electric field is where the potential energy related to the capacitance is stored.

Last edited: Apr 9, 2012
14. Apr 9, 2012

### Bob S

Two questions:

1) If a solid like salt in a charged capacitor is ionized, the electron and ion will not migrate to the electrodes (because salt is an insulator), so there is no induced signal in the plates. So where does the signal come from?

2) Suppose radiation ionized half the gas in a 1 liter nitrogen ion chamber (recall that there are 6 x 1023 molecules in a standard volume, 22.41 liters).. How many Coulombs would be ionized? How many rads would be needed to ionize this?

15. Apr 9, 2012

### DragonPetter

To answer your first question, the signal comes from the modulation in the applied AC bias signal. The ions may not migrate, but they will no longer be resonating to the electric field in the molecular structure as salt, where the capacitive energy is stored. So they are no longer contributing in the dielectric to store the energy. This means the applied AC bias field no longer has to use as much energy that it had to use to counteract the mini electric field within the salt. This in effect reduces the net capacitance, albeit miniscule on the individual event scale of things.

I will need a little time to work out your second question, I'm not so fast and have not thought of a problem like that, yet alone tried to calculate it. I'll try to give an answer by the end of today. Initially, I assume I would divide the liter in half, find how many atoms are in .5L using avagodro's number, and then look up the ionization potential required to pull an electron off Nitrogen and multiply that by how many charges there were when I found the number of atoms, and then divide that total energy by the mass of that ionized .5L of nitrogen.

Edit: The block diagram I attached in my first post shows how that modulation would be detected.

Last edited: Apr 9, 2012
16. Apr 9, 2012

### DragonPetter

To your second question, I will assume the nitrogen gas is in the form $N_2$

Calculating mass of nitrogen in .5L STP:
$\frac{14.0679 g}{mole} * \frac{1 mole}{6.022 x 10^{23} atoms}*\frac{6.022 x 10^{23} atoms}{22.4L}*{0.5L} = .314 grams$

And to calculate energy:
$.5L * \frac{6.022 x 10^{23} atoms}{22.4L} * \frac{1 ionization pair}{2 atoms} * \frac{15.5eV}{ionization pair} = 1.0418 x 10^{23}eV$

And to calculate coulombs:
$.5L * \frac{6.022 x 10^{23} atoms}{22.4L} * \frac{1 ionization pair}{2 atoms} * \frac{2 charged particles}{ionization pair} * \frac{1.602 x 10^{-19} coulombs}{charged particle} = 2153.4 coulombs$

$\frac{1.0418 x 10^{23}eV}{.314 gram} * \frac{1.6022 x 10^{-19}joules}{eV} *\frac{1000 grams}{kg} * \frac{100 rads}{gray} = 5.3166 x 10^{9} rads$

I see what you did there . . thats a lot of radiation and current. Assuming I calculated that correctly, which I'm not so sure. This was also assuming only one condition of ionization, where a single electron is stripped from a neutral nitrogen molecule.

But, first, the example about half a liter being ionized was purely to demonstrate the change in dielectric properties that I was trying to describe. I just am new to this so the numbers I used were very unrealistic. Secondly, the coulombs generated in radiation would not be what I'd try to measure with an AC capacitance measurement.

Last edited: Apr 9, 2012
17. Apr 10, 2012

### DragonPetter

Do those calculations look accurate?

18. Apr 12, 2012

### DragonPetter

I found at least one error in my calculation. I looked up $N_2$'s first ionization potential instead of it's W-value eV/ion pair number, which is roughly twice as much, 34.8eV. If I replace that number in then, the amount of energy needed would be $2.34x10^{23}$ eV, and so the rads would then be $11.93x10^{9}$ rads.

I'm a little confused why ionization potential is different from the w-value. I also have a feeling the true calculations would have to include other factors that empirical data would be used to find.

19. Apr 12, 2012

### Bob S

The empirical number is best. The Bethe Bloch dE/dx often excites non-ionizing atomic transitions. So the 100 ergs per gram (which is the definition of the Rad), accurately represents both the energy loss and the radiation dose of electrons, protons, neutrons etc. (with corrections for high LET (linear energy transfer) particles),

20. Apr 12, 2012

### DragonPetter

Oh, I've read the name of that equation before, but I'm very unfamiliar with what its about. I'll try to read about it and understand it, if possible.

Also, I read in a book by a guy named Knoll that historically the rad was referred to 100 ergs/gram but is gradually being replaced as 100 grays, which is in units of joule/kg, but the units work out to be equivalent. From what you said, it sounds like my attempt to calculate numbers to your original question is anything but accurate, but is it in the ballpark? I mean, that seems like a lot of dose, but I think that might be a reasonable amount to ionize so much gas.