Eddy Current Heating

[TinMan]

Dear Brad.

I have 7 W LED lamps in use all over the house. Our house is ancient, 316 years old! No ceiling roses, wall lights. I had to open one to direct the light more downward.

They are SMD LED's with a bridge and a storage cap, plus a few dropping resistors.

I can provide a photo if you want one.

Cheers Grum.

That would be great Grum.

So just a FWBR,cap and resistors to drop the current?
So how then is this pumping up the current like some sort of tank circuit,as i assume that the cap is after the FWBR,to smooth out the pulses the LEDs would see.


Brad

[ion]

Food for thought.

The nature of a reactive circuit such as a tank circuit is that you can produce thousands of volts and thousands of amperes with just a tiny excitation, depending on the "Q" of the circuit. Theoretically, in an ideal L-C circuit the voltage and current would go to infinity, but in the real world this is not possible, as the dielectrics would break over long before that.

It does take many input impulses for the energy to accumulate in the reactive circuit, but once it does build to it's maximum value, it will stay at that high value, limited only by the resistive and radiative losses of the tank circuit. These tend to bleed off real power which must be resupplied by the source of excitation.

Thus far all "known" attempts to get at that reactive power and have it do real work by becoming real power, have AFAIK failed, but it is an area ripe for investigation.

I have postulated that the key to OU may be in using high Q circuits to advantage by not directly bleeding off the circulating energy, thus lowering the Q, but letting it build such that it can act as a catalyst in a reaction which releases energy from "some other potential source". A catalyst is typically "not used up" in the reaction, but acts as a trigger.

If you could capture the instant of connection of your LED  lamp to the stepper output with your scope you would probably catch the first impulses from the stepper motor pumping up the tank circuit. It would look like an oscillation that builds in amplitude with each cycle.

So for each cycle a max of 76 mA is pushed into the tank until it accumulates it's maximum, which is limited by the stepper winding resistance and the power dissipated in the LED.

If you try tuning the output of the stepper motor with a capacitor of the right value, instead of the LED lamp, you should get even higher values of voltage and current circulating in the tank.

You asked:

But first a question-->and this may sound silly,but
When measuring AC current and voltage,how do we know which way-or,from what device(source) the bulk of the power is coming from?.
To make my question more clear,lets say we have two black boxes,and we have a common or neutral wire from one box to another,and a live (or hot) wire with a 1 ohm CVR in series on the hot wire to measure the current flowing from one to the other black box. Lets assume that it's an AC current flow. So we place one channel of our scope across the neutral/common and live wires to measure voltage. We then place the second channel of the scope across the CVR to measure current. Without knowing which black box is the power source,and which is the sink,how do we know which way the power is flowing ?.

One way to do this is by measuring the voltage across the CVR with respect to ground, this should tell you which direction it is flowing. If it is flowing out of the source into the DUT, it will be higher on the left and lower on the right side of the CVR, and vice versa.

Also if you compare the phase of the current across the CVR WRT to the voltage input you can tell which way it is flowing if it is a pure resistve load. If it is reactive, it gets trickier.

[ion]

That would be great Grum.

So just a FWBR,cap and resistors to drop the current?
So how then is this pumping up the current like some sort of tank circuit,as i assume that the cap is after the FWBR,to smooth out the pulses the LEDs would see.


Brad

Typically the cap responsible for resonance in your stepper setup is in series with the bridge and the mains (220nF), acting as a dissipation-less resistor dropper at, in your case, 50Hz. (as shown in the attachment). The other cap (4.7uF) is the smoothing cap.

You can calculate the effective resistance or impedance at 50 Hz if you know the value of the cap. It is this cap that is resonant with the stepper motor winding (inductance). The rest of the components in the lamp are lossy items that lower the Q of the resonant circuit.

That's why I suggested just using a cap tuned for peak resonance on the output of your stepper motor.

If you draw out the equivalent circuit you will see it.

These are the dimmable LED lamps, other types with a switching circuit are supposedly not dimmable.

[Grumage]

That would be great Grum.

So just a FWBR,cap and resistors to drop the current?
So how then is this pumping up the current like some sort of tank circuit,as i assume that the cap is after the FWBR,to smooth out the pulses the LEDs would see.


Brad

No problem......

[TinMan]

No problem......

Ah--different type of bulb.
Mine may be different inside ?.


Brad

[TinMan]

Here is the one i have,pictured below.

As you can see,i have already tried to break into it lol,without success 

[Grumage]

Hi Brad.

You might be correct however the front end of my lamp has a low value ( capacitance Unknown ) high voltage cap in series with an SMD resistor directly across the incoming 240 V input. Tank circuit?   

Cheers Grum.

[TinMan]

Typically the cap responsible for resonance in your stepper setup is in series with the bridge and the mains (220nF), acting as a dissipation-less resistor dropper at, in your case, 50Hz. (as shown in the attachment). The other cap (4.7uF) is the smoothing cap.

You can calculate the effective resistance or impedance at 50 Hz if you know the value of the cap. It is this cap that is resonant with the stepper motor winding (inductance). The rest of the components in the lamp are lossy items that lower the Q of the resonant circuit.

That's why I suggested just using a cap tuned for peak resonance on the output of your stepper motor.

If you draw out the equivalent circuit you will see it.

These are the dimmable LED lamps, other types with a switching circuit are supposedly not dimmable.

Thanks ION

I will get my box of caps out,and see if i can find one that will get the resonant tank going,as with the LED bulb.
I did try a couple of microwave oven cap's,but it would seem they were far to large a value.

It would be good if we could design a variable cap with a value range of say between 50nF and 1uF 


Brad

[TinMan]

@ ION

Did you read my post 23?

What i would like to know how to do,is be able to shift the voltage/current phase angle.
I want to try and get the current and voltage out by exactly 90*,and see what happens.
This is in relationship with that question posed by Poynt--can you produce real power when the current and voltage are out by 90*. Is there some way of being able to shift the phase angle on the fly ?-some sort of !inline! circuit i could make up ?.


Thanks

Brad

[Grumage]

Dear Brad.

A large ranged variable capacitor would be a wonderful asset in our game!

As I see it the only way would be to use variable inductance. In the late 70's a lot of car radios used variable inductance tuning with a slug of ferite that moved in a solenoid coil. Perhaps you could engineer something on that line.

Cheers Grum.

[ion]

@ ION

Did you read my post 23?

What i would like to know how to do,is be able to shift the voltage/current phase angle.
I want to try and get the current and voltage out by exactly 90*,and see what happens.
This is in relationship with that question posed by Poynt--can you produce real power when the current and voltage are out by 90*. Is there some way of being able to shift the phase angle on the fly ?-some sort of !inline! circuit i could make up ?.


Thanks

Brad

Any high quality inductor or capacitor in series with an AC source will shift the current away from voltage by 90 deg.

A capacitor will cause the current to lead the voltage and for an inductor the current will lag the voltage.

If you require more than 90 deg, it can be done with additional components.