Bit of steel water pipe bent into a coil?
Frank
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.
... be required and have parasitic inductances that would raise the 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.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..
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.
Bit of steel water pipe bent into a coil?
Frank
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.Hi,
If a regular coil (non bifilar): then inserting a movable short circuit ring causing eddy current should lower Q....and it is adjustable.
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.
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
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.
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.
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
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.
When we tried that, we couldn't get the Q to be less than 1.
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.
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