[SOLVED] How do you make a Q=0.7 inductor, very high power?

[ionp]

I have a non-communications requirement to build a test load that is around 2uH with a Q=0.7 at 30kHz. To make matters worse, I have to pump 100kW or more into it!. Water cooling is OK.

It is turning out to be extremely difficult to get a Q that low at high power.
Everything has been cut and try, and it is difficult fabricating the attempts.

Any thoughts?

Thanks.

 

[FvM]

I don't understand the specification. I would expect a specification of Isat, Irms, Rdc, ESR at 30 kHz, possibly core losses separately as everyone does it. What does Q=0.7 mean for you? Very high ESR? Why not add a separate resistor? What's the application.

 

[ionp]

This is not a component in a product design.
The intent is to attach this to an induction heater to simulate a particular heavily loaded coil.
Resistors and their interconnects have inductance, and at this power level a bunch of power resistors would be required and have parasitic inductances that would raise the Q.
There need not be a core, so Isat should not come into play There is no size constraint.
Rdc depends on if all losses are due to conductivity. With hysteresis losses on steel components, for example, Rdc could be less than 0.5 ohms but there would be no minimum.

Irms would be around 225A, although a higher capability would be better.

Thanks for thinking about it.

 

[chuckey]

Bit of steel water pipe bent into a coil?
Frank

 

[ionp]

Bit of steel water pipe bent into a coil?
Frank

When we tried that, we couldn't get the Q to be less than 1.

Thanks, tho.
Your thoughts are appreciated.

 

[BradtheRad]

Suppose you introduce a capacitor (or combination of capacitors) somewhere, to create a center frequency with the inductor? It doesn't necessarily have to create a resonant action, but its Q would be at a different frequency than the one you are operating on. Therefore it could have the effect of reducing your inductor's Q at your operating frequency.

I would not be able to say what value of C, or its position, etc., could work. It is a different concept using a component other than resistors.

 

[ionp]

Suppose you introduce a capacitor (or combination of capacitors) somewhere, to create a center frequency with the inductor? It doesn't necessarily have to create a resonant action, but its Q would be at a different frequency than the one you are operating on. Therefore it could have the effect of reducing your inductor's Q at your operating frequency.

I would not be able to say what value of C, or its position, etc., could work. It is a different concept using a component other than resistors.

Interesting!
For testing at fixed frequency, a series cap would let me have a small effective reactance by cancelling most of the inductance, leaving the losses intact giving me a low inductive impedance to resistance ratio (low Q).

That could be part of a solution, but the system finds its actual operating frequency by resonating the induction heating coil in parallel with a cap already and uses the zero phase point as part of this. In other words, the tuner couldn't be tested if there was another resonance in band.

An impressive idea tho, thank you!

 

[FvM]

Still think inductance and resistance/conductance should be implemented separately. I guess parallel circuit is preferable, about 0.3 ohm in case of a hypothetical lossless inductor. Could be implemented as massive parallel circuit of multiple low inductance resistors.

 

[KlausST]

Hi,

... be required and have parasitic inductances that would raise the Q

Am I wrong?
I'd say connecting a resistor in parallel or serial to an inductance generates heat, it dissipates power, this is power loss.
Any power loss means lowering Q..

Klaus

 

[FvM]

I'd say connecting a resistor in parallel or serial to an inductance generates heat, it dissipates power, this is power loss.
Any power loss means lowering Q..

It depends on the resistor design. Standard tubular power resistors or rheostats might have higher Q than 0.7 at 30 kHz, aluminium case brick or planar power resistor have considerably lower Q also suitably designed metal sheet resistors.

 

[Warpspeed]

Could a water cooled copper tube inductor be used to generate an initially high Q inductor.

Then something like a stainless steel shorted turn, or solid slug (also suitably water cooled !) moved into its proximity to selectively adjust the Q as desired.

 

[c_mitra]

Could a water cooled copper tube inductor be used to generate an initially high Q inductor.

Then something like a stainless steel shorted turn, or solid slug (also suitably water cooled !) moved into its proximity to selectively adjust the Q as desired.

The clamps (metal ones) to hold the turns in place would sufficiently reduce the Q: getting 0.7 should not be a problem at 30 kHz.

 

[KlausST]

Hi,

Bit of steel water pipe bent into a coil?
Frank

If this construction gives a too high inductance, then one should consider to bend the water pipes in bifilar style.
Stainless steel has higher resistance than normal steel, this may improve the effect even more.

Klaus

Added:
If a regular coil (non bifilar): then inserting a movable short circuit ring causing eddy current should lower Q....and it is adjustable.

 

[chuckey]

If you have your water pipe coil to hand, try putting sodium carbonate in the water, enough to make a super saturated solution. It will make the water more conductive so you will have a resistor in parallel with the coil. I am not sure that you will get a big enough effect.
If your device is meant to represent a slug of metal in a RF heater, use a tank through which you pump water the same shape as the slug. If you have washers inside the tank you can get the same RF circulating currents.
Frank

 

[Warpspeed]

Hi,

If a regular coil (non bifilar): then inserting a movable short circuit ring causing eddy current should lower Q....and it is adjustable.

I feel adjustability would be a very desirable feature here, because it would be very difficult estimating the coupling coefficient into an eddy current load.
Its all pretty much beyond any precise calculation from first principles.

It should be possible to measure the Q readily enough by the half power bandwidth, assuming you have a suitable low loss induction heater type of capacitor to test it with.

 

[FvM]

If this construction gives a too high inductance, then one should consider to bend the water pipes in bifilar style.
Stainless steel has higher resistance than normal steel, this may improve the effect even more.

I feel adjustability would be a very desirable feature here, because it would be very difficult estimating the coupling coefficient into an eddy current load.

I agree with the adjustability point, both for L and Q parameter. A copper pipe inductor coil with movable tap and a bifilar stainless steel pipe coil (or similar low inductance geometry, e.g. meander) in parallel or series circuit would be my first attempt for a water cooled design.

 

[Ata_sa16]

Q is lw/R so of u have 30khz and 2uH
Then ur resistance should be 0.5 ohm
If u have higher Q it means that u obtain lower resistance. U have to increase resistance by making the inductor metal thinner which also incease the L and therefore Q there should be trade off between this. You can change the metal which are u using.

Try to increase ur R while keeping ur L constant

 

[ionp]

Q is lw/R so of u have 30khz and 2uH
Then ur resistance should be 0.5 ohm
If u have higher Q it means that u obtain lower resistance. U have to increase resistance by making the inductor metal thinner which also incease the L and therefore Q there should be trade off between this. You can change the metal which are u using.

Try to increase ur R while keeping ur L constant

Thank you for your suggestion.
The problem is that even non-inductive resistors tend to have as much inductance as a piece of wire, and by the time you have enough strung together to handle 100kW, the high Q stray L on the hookups becomes very significant!

I may get back to this kind of technique, including circulating oil bath and water to oil heat exchanger for cooling, but that and the cost of the resistor bank vs the risk that it may not work puts that at the bottom of the list.

As far as the metals used for inductor elements, most have been steel alloys to take advantage of hysteresis losses. But even making the lowest inductance up-and-back using iron pipe only gets the Q down to 1.

If I could find a source of nichrome tubing in 6 foot lengths, that might help, or I might find that magnetic steel is lossier due to hysteresis losses that the nichrome would not have.

- - - Updated - - -

The clamps (metal ones) to hold the turns in place would sufficiently reduce the Q: getting 0.7 should not be a problem at 30 kHz.

We have been making inductances entirely out of various metal pipes (pipe is preferred for cooling water flow), and the Q to beat is below 1.

- - - Updated - - -

Hi,



If this construction gives a too high inductance, then one should consider to bend the water pipes in bifilar style.
Stainless steel has higher resistance than normal steel, this may improve the effect even more.

Klaus

Added:
If a regular coil (non bifilar): then inserting a movable short circuit ring causing eddy current should lower Q....and it is adjustable.

We have been doing bifilar (up and back lengths of pipe, as close together as possible) and can only get down to a Q of 1 with black steel gas piping.

- - - Updated - - -

If you have your water pipe coil to hand, try putting sodium carbonate in the water, enough to make a super saturated solution. It will make the water more conductive so you will have a resistor in parallel with the coil. I am not sure that you will get a big enough effect.
If your device is meant to represent a slug of metal in a RF heater, use a tank through which you pump water the same shape as the slug. If you have washers inside the tank you can get the same RF circulating currents.
Frank

When we tried ball bearings in a similar configuration, the induced currents caused lots of arcs amongst the bearings, and that line was abandoned.

The solution of which you speak would be in parallel with a steel conductor, and almost no current would flow.
I am not sure if the ionic conduction would be effective at 30 kHz, in any case.

- - - Updated - - -

I feel adjustability would be a very desirable feature here, because it would be very difficult estimating the coupling coefficient into an eddy current load.
Its all pretty much beyond any precise calculation from first principles.

It should be possible to measure the Q readily enough by the half power bandwidth, assuming you have a suitable low loss induction heater type of capacitor to test it with.

Adjustability would be nice, but right now I would be happy just getting close.
I have a meter that can measure Q that low, but I do have water cooled capacitors at hand if need be.

 

[_Eduardo_]

Quote Originally Posted by chuckey View Post
Bit of steel water pipe bent into a coil?
Frank

When we tried that, we couldn't get the Q to be less than 1.

Try same winding dimensions but less diameter pipe.

 

[ionp]

Quote Originally Posted by chuckey View Post
Bit of steel water pipe bent into a coil?
Frank


Try same winding dimensions but less diameter pipe.

Going straight up and back (nearly bifilar) seems to always be lower Q than a coil or ring, and with various diameters and types of steel or copper has never resulted in a Q less than 1.