## Pulsed fusor #23 How to design a linac/ with example

### Pulsed fusor #23 How to design a linac/ with example

Warning: The following post contains procedures that should not be attempted with out a thorough and COMPLETE understanding of the radiation hazard involved.

High vacuum is required to make this method work.

Remote operation procedure HAVE to be in place from the beginning.

Appropriated warning signs with a flashing warning light are necessary.

High voltage safety will be stringent due to the extra high potentials involved.

The radiation is short range but very intense close to the linac and the experintal area

Proper safety interlocks with keys are a must.

The local radiation will be on the order of 100 Rads or above.

It exceeds the lifetime permissible dose many fold.

How to build a linac.

********** NOTICE******************

Unless you just really want to know read this BUT....

The DC linac is where it's at

3L

These are truly simple machines to design.

If you look at the books you see calculus…with some head scratching it is not necessary.

The design is very old ,it dates to the mid thirties.

R. Wideroe of Germany devised the first system.

In a linear accelerator a particle is given successive pushes until it reaches the desired energy.

The accelerations happen at the gap between two tubes.

The voltage must alternate between plus and minus voltages because

the front and back of the tubes are used as accelerating gaps.

The particle is shielded during the voltage change by being inside the tube at that time.

The particle drifts along the inside of the tube until it reaches the business end, hence the name drift tubes.

The mass plays a role because in order to know how long a tube has to be in order to function the kinetic energy is known…just work backwards on the kinetic energy formula.

You know KE=1/2 M V^2

Units

V=Velocity (m/sec,cm/sec)

M=mass in kg

KE=joules

Rearranging we get:

Units

100cm = 3ft

A deuteron weighs 3.24 x 10^-27

D=drift tube length (m,cm)

T=time sec

F=frequency

C=speed of light (3x10^8m/sec,3 x10^10cm/sec)

WL=wavelength

Vt =Volts

A=amps

V=(KE/2M)^.5

V = D / T

T= D/V

F= C/WL

W=Vt x A

So lets take what has been derived and apply it.

Example: I want to build a 100 kv deuteron accelerator. I would like to use a small 10 kv per tube power supply to run it. I want it to provide 500 ma of deuterons. The device runs at 1 megacycles.

Find the number of drift tubes.

Specify the length of each tube.

Define the amount of power in watts the linac will draw.

You have thirty minutes … show all work.

Sorry… I miss physics teaching sometimes.

Step one:

Determine the time of a full cycle.

1 megacycles =1 x 10^ 6/sec

Time per cycle = 1x 10^-6 sec

Step two: Determine the number of tubes.

Simply divide the desired energy by the accelerating voltage.

In this case …. 100 kv /10 kv/tube = 10 tubes The kv divide out leaving tubes.

Step three: Determine the length of each tube.

The trick to not going mad at this stage is to simply get the velocity at 10 kv then just remember to multiply by the tube number.

>>>>> See you like physics right?<<<<<<<

Tube 1 @ 10 kv /particle

10 kv =1x 10 ^4 eV

KE =1.60x10^-19ev/J x 1x10^4 ev=1.60 x10^-15 J

The mass of the deuteron is 3.24 x 10^-27

V=square root of (1.60 x 10^-15 j/(2x3.24 x10^-27 kg)

So V = 5.04 x 10 ^ 5 m/sec

D=VT=(5.04 x 10 ^5 m/sec)(1x10^-6)=5.04 x10-1 m=.5 m or 50 cm

Tube 2 @ 20 kv /particle

V=square root of (2x1.60 x 10^-15 j/(2x3.24 x10^-27 kg)

D=VT=(7.02 x 10 ^5 m/sec)(1x10^-6)=7.02 x10-1 m=.70 m or 70 cm

Tube 3 @ 30 kv /particle

V=square root of (3x1.60 x 10^-15 j/(2x3.24 x10^-27 kg)

D=VT=(8.60 x 10 ^5 m/sec)(1x10^-6)=8.6 x10-1 m=.86 m or 86 cm

Tube 4 @ 40 kv /particle

V=square root of (4x1.60 x 10^-15 j/(2x3.24 x10^-27 kg)

D=VT=(9.93 x 10 ^5 m/sec)(1x10^-6)=9.93 x10-1 m=.54 m or 93 cm

Tube 5 @ 50 kv /particle

V=square root of (5x1.60 x 10^-15 j/(2x3.24 x10^-27 kg)

D=VT=(1,11 x 10 ^6 m/sec)(1x10^-6)=1.11 m or 110 cm

Tube 6 @ 60 kv /particle

V=square root of (6x1.60 x 10^-15 j/(2x3.24 x10^-27 kg)

D=VT=(1.21 x 10 ^6 m/sec)(1x10^-6)=1.21 m or 121 cm

Tube 7 @ 70 kv /particle

V=square root of (7x1.60 x 10^-15 j/(2x3.24 x10^-27 kg)

D=VT=(1.31 x 10 ^6 m/sec)(1x10^-6)=1.31 m or 131 cm

Tube 8 @ 80 kv /particle

V=square root of (8x1.60 x 10^-15 j/(2x3.24 x10^-27 kg)

D=VT=(1.41 x 10 ^6 m/sec)(1x10^-6)=1.41 m or 141 cm

Tube 9 @ 90 kv /particle

V=square root of (9x1.60 x 10^-15 j/(2x3.24 x10^-27 kg)

D=VT=(1.49 x 10 ^6 m/sec)(1x10^-6)=1.49 m or 149 cm

Extra Credit: Why can tube 10 be ring and still function?

If you increase the frequency what happens to the drift tube length?

(Take tube #9... 1 MHZ 149 cm ,10 MHZ 14.9 cm,20 mhz 7.45 cm ,27 MHZ (Break 19 Good buddy) 5.5 cm)

Oh Yeah by the law of conservation of energy ,the energy imparted to the deuterons times the amperage of said particles equals the watts you have to feed the wee beasty of a linac.

100 kv x 500 ma = 50 kw (1.00x10^5V)x(5x10^-1A)

Fusion is Fun!

Larry Leins

Fusion Tech

High vacuum is required to make this method work.

Remote operation procedure HAVE to be in place from the beginning.

Appropriated warning signs with a flashing warning light are necessary.

High voltage safety will be stringent due to the extra high potentials involved.

The radiation is short range but very intense close to the linac and the experintal area

Proper safety interlocks with keys are a must.

The local radiation will be on the order of 100 Rads or above.

It exceeds the lifetime permissible dose many fold.

How to build a linac.

********** NOTICE******************

Unless you just really want to know read this BUT....

The DC linac is where it's at

3L

These are truly simple machines to design.

If you look at the books you see calculus…with some head scratching it is not necessary.

The design is very old ,it dates to the mid thirties.

R. Wideroe of Germany devised the first system.

In a linear accelerator a particle is given successive pushes until it reaches the desired energy.

The accelerations happen at the gap between two tubes.

The voltage must alternate between plus and minus voltages because

the front and back of the tubes are used as accelerating gaps.

The particle is shielded during the voltage change by being inside the tube at that time.

The particle drifts along the inside of the tube until it reaches the business end, hence the name drift tubes.

The mass plays a role because in order to know how long a tube has to be in order to function the kinetic energy is known…just work backwards on the kinetic energy formula.

You know KE=1/2 M V^2

Units

V=Velocity (m/sec,cm/sec)

M=mass in kg

KE=joules

Rearranging we get:

Units

100cm = 3ft

A deuteron weighs 3.24 x 10^-27

D=drift tube length (m,cm)

T=time sec

F=frequency

C=speed of light (3x10^8m/sec,3 x10^10cm/sec)

WL=wavelength

Vt =Volts

A=amps

V=(KE/2M)^.5

V = D / T

T= D/V

F= C/WL

W=Vt x A

So lets take what has been derived and apply it.

Example: I want to build a 100 kv deuteron accelerator. I would like to use a small 10 kv per tube power supply to run it. I want it to provide 500 ma of deuterons. The device runs at 1 megacycles.

Find the number of drift tubes.

Specify the length of each tube.

Define the amount of power in watts the linac will draw.

You have thirty minutes … show all work.

Sorry… I miss physics teaching sometimes.

Step one:

Determine the time of a full cycle.

1 megacycles =1 x 10^ 6/sec

Time per cycle = 1x 10^-6 sec

Step two: Determine the number of tubes.

Simply divide the desired energy by the accelerating voltage.

In this case …. 100 kv /10 kv/tube = 10 tubes The kv divide out leaving tubes.

Step three: Determine the length of each tube.

The trick to not going mad at this stage is to simply get the velocity at 10 kv then just remember to multiply by the tube number.

>>>>> See you like physics right?<<<<<<<

Tube 1 @ 10 kv /particle

10 kv =1x 10 ^4 eV

KE =1.60x10^-19ev/J x 1x10^4 ev=1.60 x10^-15 J

The mass of the deuteron is 3.24 x 10^-27

V=square root of (1.60 x 10^-15 j/(2x3.24 x10^-27 kg)

So V = 5.04 x 10 ^ 5 m/sec

D=VT=(5.04 x 10 ^5 m/sec)(1x10^-6)=5.04 x10-1 m=.5 m or 50 cm

Tube 2 @ 20 kv /particle

V=square root of (2x1.60 x 10^-15 j/(2x3.24 x10^-27 kg)

D=VT=(7.02 x 10 ^5 m/sec)(1x10^-6)=7.02 x10-1 m=.70 m or 70 cm

Tube 3 @ 30 kv /particle

V=square root of (3x1.60 x 10^-15 j/(2x3.24 x10^-27 kg)

D=VT=(8.60 x 10 ^5 m/sec)(1x10^-6)=8.6 x10-1 m=.86 m or 86 cm

Tube 4 @ 40 kv /particle

V=square root of (4x1.60 x 10^-15 j/(2x3.24 x10^-27 kg)

D=VT=(9.93 x 10 ^5 m/sec)(1x10^-6)=9.93 x10-1 m=.54 m or 93 cm

Tube 5 @ 50 kv /particle

V=square root of (5x1.60 x 10^-15 j/(2x3.24 x10^-27 kg)

D=VT=(1,11 x 10 ^6 m/sec)(1x10^-6)=1.11 m or 110 cm

Tube 6 @ 60 kv /particle

V=square root of (6x1.60 x 10^-15 j/(2x3.24 x10^-27 kg)

D=VT=(1.21 x 10 ^6 m/sec)(1x10^-6)=1.21 m or 121 cm

Tube 7 @ 70 kv /particle

V=square root of (7x1.60 x 10^-15 j/(2x3.24 x10^-27 kg)

D=VT=(1.31 x 10 ^6 m/sec)(1x10^-6)=1.31 m or 131 cm

Tube 8 @ 80 kv /particle

V=square root of (8x1.60 x 10^-15 j/(2x3.24 x10^-27 kg)

D=VT=(1.41 x 10 ^6 m/sec)(1x10^-6)=1.41 m or 141 cm

Tube 9 @ 90 kv /particle

V=square root of (9x1.60 x 10^-15 j/(2x3.24 x10^-27 kg)

D=VT=(1.49 x 10 ^6 m/sec)(1x10^-6)=1.49 m or 149 cm

Extra Credit: Why can tube 10 be ring and still function?

If you increase the frequency what happens to the drift tube length?

(Take tube #9... 1 MHZ 149 cm ,10 MHZ 14.9 cm,20 mhz 7.45 cm ,27 MHZ (Break 19 Good buddy) 5.5 cm)

Oh Yeah by the law of conservation of energy ,the energy imparted to the deuterons times the amperage of said particles equals the watts you have to feed the wee beasty of a linac.

100 kv x 500 ma = 50 kw (1.00x10^5V)x(5x10^-1A)

Fusion is Fun!

Larry Leins

Fusion Tech

### Re: Pulsed fusor #23 How to design a linac/ with example

This reminds me of some stuff I've been reading in the book Accelerators: Machines for Nuclear Physics.

Another idea for Linacs, instead of drift tubes, is the corrugated tube, or a tube with open-centered discs connecting to the rest of the tube. When you pump UHF high voltage down the tube the eletricity flows in waves of high positive and negative voltages (remember, the electrons dont travel all the way, its the electron compression wave). This wave would need to be tuned to the distances, but the advantage of this is that theres less possibility for arcing (because its already shorted, so to speak) and it doesnt require drifts within a large tube, the whole tube is solid drifts.

This design requires microwave frequencies, however, so I don't know how well a transistor/tube amp would work to power it. You might need a klystron, maybe. But you can get much higher output energies because the number of charging discs (or cavities as any two are called) is great. If the power input is 15KV and you have 20-30 cavities, you've got 450KV. And the cavity size can be damned small (2.45GHz wavelength is 12.5cm so the tube is a mere 3.75 metres long)

And whats good is, the higher the frequency the smaller the distance, so 10GHz input = 3cm cavity and a 0.9 metre tube. But this is all advanced accelerator shit, and you'd need steadily increasing frequencies because of acceleration. And you don't need these kinds of energies for the reactor tubes. Infact, without magnetic confinment you couldnt use drift tubes or microwave frequency tubes because you'd have no way of confining the fuel to the center of the tube.

Unless, ofcourse, you're using a non-evacuated main tube (possibly filled with high pressure inert gas) and you have a glass beam tube within the center of the main tube. But the pressure gradient across the glass might be too great.

Really, all you need is a few (~40) grids within a pipe to get good acceleration. You'd still need appropriately oscillating electricity, however.

Another idea for Linacs, instead of drift tubes, is the corrugated tube, or a tube with open-centered discs connecting to the rest of the tube. When you pump UHF high voltage down the tube the eletricity flows in waves of high positive and negative voltages (remember, the electrons dont travel all the way, its the electron compression wave). This wave would need to be tuned to the distances, but the advantage of this is that theres less possibility for arcing (because its already shorted, so to speak) and it doesnt require drifts within a large tube, the whole tube is solid drifts.

This design requires microwave frequencies, however, so I don't know how well a transistor/tube amp would work to power it. You might need a klystron, maybe. But you can get much higher output energies because the number of charging discs (or cavities as any two are called) is great. If the power input is 15KV and you have 20-30 cavities, you've got 450KV. And the cavity size can be damned small (2.45GHz wavelength is 12.5cm so the tube is a mere 3.75 metres long)

And whats good is, the higher the frequency the smaller the distance, so 10GHz input = 3cm cavity and a 0.9 metre tube. But this is all advanced accelerator shit, and you'd need steadily increasing frequencies because of acceleration. And you don't need these kinds of energies for the reactor tubes. Infact, without magnetic confinment you couldnt use drift tubes or microwave frequency tubes because you'd have no way of confining the fuel to the center of the tube.

Unless, ofcourse, you're using a non-evacuated main tube (possibly filled with high pressure inert gas) and you have a glass beam tube within the center of the main tube. But the pressure gradient across the glass might be too great.

Really, all you need is a few (~40) grids within a pipe to get good acceleration. You'd still need appropriately oscillating electricity, however.

### Re: Pulsed fusor #23 How to design a linac/ with example

Humm... the object of the linac post was to put foward an easy solution that would not cost an arm and a leg. Microwave design takes time and a thorough understanding of the rf design and engineering. I was a radar tech as one of my duties when I was in the Air Force.

You would need circulators at the particular frequencies you speak of. Are you versed in the Smith Chart? Waveguide theory?

Microwave measurement? The work can get very hairy as any

EE will tell you. Designs on paper will have to be debuged...

serious time and effort. But if it floats your boat ...knock yourself out. I tend to like simplistic solutions where possible.

Larry Leins

Fusion Tech

You would need circulators at the particular frequencies you speak of. Are you versed in the Smith Chart? Waveguide theory?

Microwave measurement? The work can get very hairy as any

EE will tell you. Designs on paper will have to be debuged...

serious time and effort. But if it floats your boat ...knock yourself out. I tend to like simplistic solutions where possible.

Larry Leins

Fusion Tech

### Re: Pulsed fusor #23 How to design a linac/ with example

teehee :p

even with drift tubes you need some form of confinement, otherwise the fuel will disperse as its attracted to the wide-spread negative field in front of it. Without confinement it'll get messed up seriously. Tho confinement isn't terribly hard.

you're right tho, I got too advanced. heh Best bet is to just use grids instead of drift tubes unless you want to deal with magnetic confinement. However, using drift tubes DOES leave you with a gridless design that has no losses from fuel-grid collisions, so theres no HARD XRAYS.

even with drift tubes you need some form of confinement, otherwise the fuel will disperse as its attracted to the wide-spread negative field in front of it. Without confinement it'll get messed up seriously. Tho confinement isn't terribly hard.

you're right tho, I got too advanced. heh Best bet is to just use grids instead of drift tubes unless you want to deal with magnetic confinement. However, using drift tubes DOES leave you with a gridless design that has no losses from fuel-grid collisions, so theres no HARD XRAYS.

### Re: Pulsed fusor #23 How to design a linac/ with example

True enough and Ace Hardware carries the raw stock for the cheapo linac with Ebay picking up the slack, I'm figuring in the 150 dollar range counting the feedtrus ,Vacuum fittings and a NST for power and a two tube oscillator that is crystal controlled.

All a person would need is a chop saw to make it happen with a little welding.

Fusion is fun!

Larry Leins

Fusion Tech

All a person would need is a chop saw to make it happen with a little welding.

Fusion is fun!

Larry Leins

Fusion Tech

- Richard Hull
- Moderator
**Posts:**12431**Joined:**Fri Jun 15, 2001 1:44 pm**Real name:**Richard Hull

### Re: Pulsed fusor #23 How to design a linac/ with example

At 100kev, which is a mouse level linac, I would go with straight DC acceleration! Zero complication. Tube length would be about 24" or less if working at 10e-5 torr or better. No drift tubes, no oscillator, maybe some ion optics, but that is it.

Richard Hull

Richard Hull

Progress may have been a good thing once, but it just went on too long. - Yogi Berra

Fusion is the energy of the future....and it always will be

Retired now...Doing only what I want and not what I should...every day is a saturday.

Fusion is the energy of the future....and it always will be

Retired now...Doing only what I want and not what I should...every day is a saturday.

### Re: Pulsed fusor #23 How to design a linac/ with example

I know these are small powers.

I thought it might be fun to raise the amperage on one.

Besides there seems to be a shortage of xray tranis at the moment.

Also I wanted to try other fuels at levels as high as 2 mev.

Larry Leins

Fusion Tech

I thought it might be fun to raise the amperage on one.

Besides there seems to be a shortage of xray tranis at the moment.

Also I wanted to try other fuels at levels as high as 2 mev.

Larry Leins

Fusion Tech

### Re: Pulsed fusor #23 How to design a linac/ with example

2 MeV? You're insane. I will build a temple to you if you complete it, tho.

### Re: Pulsed fusor #23 How to design a linac/ with example

The company I worked for in the seventies had a 20 mev linear.

They used it to test hardened electronics for missile use.

It all fit in a small area. You could load it in your car.

Larry Leins

Fusion Tech

They used it to test hardened electronics for missile use.

It all fit in a small area. You could load it in your car.

Larry Leins

Fusion Tech

- Richard Hull
- Moderator
**Posts:**12431**Joined:**Fri Jun 15, 2001 1:44 pm**Real name:**Richard Hull

### Re: Pulsed fusor #23 How to design a linac/ with example

There is a big difference between the discussion and th' doin'. Even at the liniac whimpy level of 100kev, and, no matter how you construct it, you would be lucky to survive the X-ray blast. I find that glibley thrown out figures over 60kev are pretty much pie in the sky numbers for amateurs. Operation anywhere above 100kev is a dreamworld. If brought to any form of fruition such projects would demand a special room for operation.

Of course, random musings are always nice. I often imagine myself a lotto winner, inspite of never buying a ticket.

Richard Hull

Of course, random musings are always nice. I often imagine myself a lotto winner, inspite of never buying a ticket.

Richard Hull

Progress may have been a good thing once, but it just went on too long. - Yogi Berra

Fusion is the energy of the future....and it always will be

Retired now...Doing only what I want and not what I should...every day is a saturday.

Fusion is the energy of the future....and it always will be

Retired now...Doing only what I want and not what I should...every day is a saturday.