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Some Bifilar coil experiments

Started by evostars, April 11, 2017, 04:31:06 PM

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evostars

Quote from: dieter on April 12, 2017, 07:40:41 PM
evostars, you need to get the loading time into the calculation. Your caps cannot deliver 900vdc constantly. This only how high the voltage goes at max. It's the noload voltage.


Again, 900*900=81'000*1.6uF=~128milli *0.5=64 millijoule. How do you get 1.8 Joule?


Anyway, by adding a load such as a 100 ohm resistor, you can calculate the amps based on voltage. If the voltage were 200vdc at the load, then it were 200 vdc / 100 ohm= 2 ampere. And hat would be 200 vdc * 2 amps = 400 Watt.


Then again, your yellow multimeter looks like it has amp section.


Well, it may be relatively easy to autoadjust the frequency to resonance, the sam way tv and radio did it to follow a drifting station. I think it was termed AFC? Raher simple circuit.
oops 900x900=810000 misted a zero
uF is micro farad.  micro is 1/1000000 so 0.0000016F
now calculate again
maybe buy a new keyboard ;)

indeed its unloaded. but it shows the energy potential.and the rise of energy related to the rise in input voltage.


evostars

Quote from: Dog-One on April 12, 2017, 10:40:25 PM
Quite interesting considering the turns ratio is one to one.  At first glance with two coils, one would think 24 volts is about all you would get out of there.  A rather novel form of boost converter you have assembled there Evo.  I'll bet getting that 180 degree phase relationship was tricky to do.  Does it move around when the storage capacitors are charging?

So these coils produce a nice sine wave even though you are injecting a short duration pulse?  What's the approximate duty cycle?


Thanks dog one,
The 180 degree phase shift, happend. when I reversed the coil, and retuned it. I saw the voltage rise in my caps. Then I kept tuning, until i reach maximum voltage. Then the phase was 180 degrees.

yes, a sine wave, from being pulsed.

The power factor is a big question. What is the input power, what is the output power?
In this low voltage setup I'm rather convinced that the input is larger then the output power.

The relation between the input power and output power isnt linear. It needs to reach a certain voltage before it becomes really interesting. Below that voltage, the input is bigger, above that voltage, the output is bigger, due to its realtion to the square of the voltage.

Grumage

Dear evostars.

You could get one of these for your input measurements.

http://www.ebay.co.uk/itm/Digital-LCD-60V-100A-Balance-Voltage-RC-Battery-Power-Analyzer-DC-Watt-Meter-CI-/272291431181?hash=item3f65d5830d

I recently bought one but have not used it in any practical way yet.

Measuring the output will be a whole different " ball game " though.

Cheers Graham.

AlienGrey

Interesting device but no good for RF ;) but ok for the power supply I suppose

cheers anyway AG

evostars

Quote from: Grumage on April 13, 2017, 10:54:58 AM
Dear evostars.

You could get one of these for your input measurements.

http://www.ebay.co.uk/itm/Digital-LCD-60V-100A-Balance-Voltage-RC-Battery-Power-Analyzer-DC-Watt-Meter-CI-/272291431181?hash=item3f65d5830d

I recently bought one but have not used it in any practical way yet.

Measuring the output will be a whole different " ball game " though.

Cheers Graham.
Thanks for the advice, looks like a nice piece of gear. but its restricted to 60Vdc
I found a 750uOhm shunt on my igbt driver. maybe i will use that.
But In the end, I think i need a different kind of pulse generator.
Maybe one that can charge a capacitor, and then discharge the capcitor at both ends by switching both ends of the capacitor.
My IGBT driver, now only switches one end (to ground)