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Registered Member #190
Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567
Capacitance on the HV DC supply will drop the power factor. How low should one go on the smoothing capacitors? What is the downside to really low capacitance other than a larger ripple with regard to its use powering an induction heater?
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
If the capacitors are too small, they'll overheat from ripple current and explode.
Also, as the ripple gets bigger, the power output of your induction heater decreases, if it was limited by the peak current.
In the limit, with no smoothing capacitors at all, the average output power is only 1/4 of the peak, so you have to size your inverter to handle 4 times the peak power that it would if it were running off smooth DC.
If you're using IGBTs this may not be a problem, they laugh at high peak currents.
Registered Member #162
Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140
If the main ac supply has a low impedance (at the induction heater frequency) in theory you need no capacitance at all. The output would look like the switching frequency 100% modulated by the rectified mains voltage.
Since the actual ac supply has resistance and inductance, a capacitor will be required. The capacitor is required to minimise the drop in dc voltage during 1/2 of one rf cycle. The value of capacitance will be determined by the output power required. e.g. 110 Vac in, 1.1 kW out, at 100% efficiency input current = 10 Arms = 14 Apk At say 100 kHz 1/2-cycle = 5 us, 5 us x 14 A = 70 uJ. Since dQ = C.dV , C = dQ/dV so for 1 V 'ripple' on the dc supply you would need 70 uF Polypropylene 'lighting' (pfc correction for fluorescent lamps with inductive ballasts) capacitors are quite cheap and easily available, or a motor run capacitor. Or you could use an 'invertor grade' (low esr) electrolytic.
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