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Registered Member #2099
Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716
The cap can definitely source more current then a mot.
What's the basis of that statement? Imagine a nice big MOT after removing the HV secondary and the core shunts. I bet we can run it a bit over 1 volt per turn.
Now give it a 1-turn secondary made of rectangular copper bar, say 1x4 cm in cross section and 20 cm effective length. With rho_Cu of 1.7e-6 ohm-cm, the new secondary would have a resistance of about 8.5 microohms (vs. 700 microohms for the boostcap under discussion). You want to match voltages, make it 2 turns (34 uO) or 3 turns (77 uO).
For each turns count, let's triple the effective resistance to account for the primary winding resistance reflected to the secondary. (the copper-filling factor of mains voltage primary is not as good as the solid copper we're figuring for the secondary). They will all still be much less than the rated ESR of the boostcap, so short circuit current will be higher. And the MOT windings won't get appreciably hot in the time it would take the boostcap to be discharged.
Registered Member #2431
Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639
klugesmith wrote ...
And the MOT windings won't get appreciably hot in the time it would take the boostcap to be discharged.
Im certain the secondary and core would not get warm or hot in this short time, but what about the primary? The primary will see alot of current through its gaggle of turns. Surely that would cause some temp rise?
Registered Member #2099
Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716
Patrick wrote ...
Im certain the secondary and core would not get warm or hot in this short time, but what about the primary? The primary will see alot of current through its gaggle of turns. Surely that would cause some temp rise? or do others say different?
Excellent question. Not counting the magnetizing current, the primary and secondary will have about the same number of ampere-turns. Their relative current densities will be inversely proportional to the total cross-sectional area of copper in all the turns. My previous resistance estimate implied that the primary has half as much total copper area as the secondary (2 cm^2 vs 4 cm^2). So [edit] twice the current density in amps/cm^2. Within the "inertial cooling" time frame, which is measured in minutes for things as massive as MOTs, it will have four times as much temperature rise.
Registered Member #2431
Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639
klugesmith wrote ...
Patrick wrote ...
Im certain the secondary and core would not get warm or hot in this short time, but what about the primary? The primary will see alot of current through its gaggle of turns. Surely that would cause some temp rise? or do others say different?
Excellent question. When heavily loaded, the primary and secondary have practically the same number of ampere-turns. Relative current densities are inversely proportional to the total cross-sectional area of copper in all the turns. My previous resistance estimate implied that the primary has half as much total copper area as the secondary (2 cm^2 vs 4 cm^2). So four times the current density in amps/cm^2. Within the "inertial cooling" time frame, which is measured in minutes for things as massive as MOTs, it will have four times as much temperature rise.
I see...let me ponder that which you have said for someminutes.
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What to do with a Boostcap?
