Power Supply Upgrade

17 07 2012

All photos(1/2)
All photos

During the first trial, our langmuir probe told us that our electron beam intensity was fluctuating at 60 Hz.

This is a problem, because one of the main things we are trying to study is the way changes in beam intensity affect potential well depth, so we want a steady intensity. The frequency of the fluctuations suggest that the AC-powered hot cathode is to blame.  I don’t totally understand the details of how a hot cathode running on AC 120v 60Hz translates to this waveform:

data from the langmuir probe displayed on the oscilloscop

The important thing is to prevent it. To do that, I put a full wave bridge rectifier in the power supply. It converts the AC coming from the wall to DC


It has three essential components.

1) The bridge rectifier


This change the AC sine wave into a waveform expressed by the function abs(sin(x)):

Better, but still not steady DC.

2) The filter capacitor


This gets rid of the ripple. you could compare the capacitor to a bucket with a hole in the bottom. Even if I vary the rate at which putting water into the bucket, the rate at which it come out is always going to be more or less the same, provided that it is sufficiently large compared to the volume of water going in.

However, its impossible to get an absolutely perfect DC output with this setup, because the ammount of charge on the capacitor does affect the voltage at which the current comes out.

This 680 uF capacitor takes away enough of the ripple for our purposes:

the output of the power supply when hooked up to a light bulb

3) An isolation transformer


Usually, the diode bridge and the capacitor would be enough, but our AC isn’t coming from the wall, its coming from a grounded auto transformer. this is a problem because the rectifier only works if the AC input is floating. A transformer with an equal number of primary and secondary wingdings accomplishes this without stepping the voltage up or down.

Nest step is to test it in the chamber.

Domenick Bauer

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4 responses

18 07 2012

that waveform doesn’t seem very abs(sin(x))
Sure you’re measuring from the right point?

18 07 2012

bottom troughs are pointer than humps.

18 07 2012

The way this used to be solved in valve (tube) audio equipment was to centre tap the heater transformer and ground the centre tap so drive was balanced. Or have a pot across it with a grounded wiper to null out the hum.

I am not sure how well this would work with a direct heated cathode though. Most valves use indirect heating.

19 07 2012
Remy Dyer

With it isolated like this, the whole heater circuit can float at the cathode negative bias. No need to centre tap the transformer, unless you want the electron beam energy to spread either side of the total acceleration voltage, rather than just one side…

This is because the filament itself will have a voltage gradient across it due to ohm’s law through it. If the negative side of the filament is connected to the adjustable negative bias, the effective electron energy will be (Vanode – (Vfilament bias + integral [from 0 to ~ +6.3 rectified from the negative most end of the fillament to the other])).

This spread is why most valves use an indirectly heated cathode. It’s much less of an issue with the EBM cathode though, due to the much lower voltage across that one.

If the filament is biased, say at -20 V, and the anode is at +200, and the filament drive ends up at a fairly steady +7V, then the electrons as they enter the (grounded) magrid will have energies from 20 eV to 13 eV, depending on where on the filament they came from…
If the magrid *is* the anode, then they’ll retain the full 220 to 213 eV energy while inside it.

Exactly what coil current will be required to reach beta=1 (WB mode) for this geometry and electron beam current + energy, I have as yet no idea. I’m still working on coming to terms with the math… The beta = 1 condition depend on the electron number density, which depends on the total well depth voltage achieved and the volume – maybe reasonable to set it to ~80% drive or about 160 V… I get around 4.7 e-6 Coulomb / cubic metres, or about 30e+12 electrons / cubic metre in your 30mm radius wiffleball to sustain that voltage… Applying the beta=1 condition (with electron energy ~200 eV) gets a magnetic flux of only ~ 0.1257 T… so coils would need, what? 6 thousand Amp – turns? Can you hit this? If so the potential well should get down to about 160 V around here, and it’ll probably be the minimum. (or there will be one nearby).

And the above is dependant on getting enough electron beam current into the thing to get that electron number density up… And this last part I think will be the hard part… The original Sydney experiment sure didn’t manage it… They were certainly running in mirror confinement mode – far from beta =1, since there wasn’t enough electron beam current to keep the wiffleball inflated. IMHO all the *interesting* physics happens in WB mode..

Don’t take this as golden, I’m an amateur at this…

As always, good luck!

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