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Registered Member #1571
Joined: Wed Jul 02 2008, 03:26AM
Location: Bendigo Victoria Australia
Posts: 44
Hi there,
I am hoping someone might shed some light on this for me.
I understand that if you have an air transformer say that has 10 turns on the primary and 100 turns on the secondary the increase in voltage from the primary to the secondary is 10 times. Thats fine but how do you calculate the voltage increase if the primary is driven with a capacitor in series and the frequency is such that its running at resonance?
Further on from that, how do you calculate the voltage increase if its the secondary resonating instead.
Finally and this is what I am aiming to do... if both the primary and the secondary each have a capacitor in series such that the one frequency is the resonant point for both primary and secondary??? how do you calculate the voltage increase over and above the normal increase based on the turns ratio..
Are there formulas etc I can use?
Hope someone might be able to shed some light on this for me....
Registered Member #162
Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141
The first assumption is wrong because air coils have leakage inductance, ... not all of the magnetic flux of one coil is linked to the other. This will cause a voltage transfer of less than the turns ratio when the non-driven coil is open circuit
If there is a load on the non-driven coil the output will be much less than the turns ratio indicates.
Two resonant circuits of identical resonant frequency magnetically loosely coupled will have two frequencies of peak voltage transfer.
If the coupling is loose and the non-driven coil is lightly loaded (high 'Q') then the output voltage may be many times the turns ratio predicts.
Registered Member #1571
Joined: Wed Jul 02 2008, 03:26AM
Location: Bendigo Victoria Australia
Posts: 44
I am making a Cockcroft Walton multiplier. The transformer will be have an air core which Inca indicates a k of 0.7... The multiplier has an equivalent capacitance..
Would the Cockcroft Walton multiplier qualify as a "light load"?
Is there a way of calculating the voltage on the secondary based on this?
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
I think there's a flaw in the reasoning here that makes the analysis hard.
Normally a CW multiplier has capacitors that are "large". That means they have negligible reactance at the operating frequency, hence a small ripple voltage.
But if the first capacitor is resonant with the output transformer, this can't be true, so the classical analysis of the CW won't work.
The usual approach to driving a CW is to use a power supply that is happy with a "short circuit" on the transformer secondary, such as the SLR converter. The transformer primary can be resonant, but the secondary isn't: the CW with its "large" capacitors presents a heavy load that damps the resonance.
A SLR converter with some beefy IGBTs might be a good solution in this instance. It could easily handle the 200 amp peaks we calculated, and pulse density modulation could be used to keep the average power under control.
Another solution I've seen is to make the transformer parallel resonant with its own leakage inductance and secondary self-capacitance, and the CW not be a "heavy load" compared to this. The result is an easy-to-wind transformer and small, cheap CW capacitors.
Registered Member #2901
Joined: Thu Jun 03 2010, 01:25PM
Location:
Posts: 837
Isn't an unloaded CW simply 0 reactance (or infinite impedance) regardless of capacitors? Even with a load you can't represent it as a capacitance AFAICS. AFAICS it only pulls current at the tops/bottoms of the sines.
The only way to run it at a load independent resonance frequency is to have an additional secondary tank cap >> N*capacitors you use in the multiplier, so most of the circulating current is in the tank even when the multiplier gets nearly short circuited.
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A question about transformers.
