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Registered Member #1911
Joined: Mon Jan 05 2009, 06:30PM
Location: Salem, Oregon, USA
Posts: 165
Hello everyone,
I recently bought a ferrite core from a forum member and I didn't realize that I would need to build a driver rather than just drive it off of mains, since I didn't realize that ferrite was anything different than the iron plates. So, now, here I am with a transformer that exceeds my abilities to drive it. I was given a reference to Steve Ward's CCPS and to Uzzor's CCPS, but I was unsure of how to lay it all out and was wondering if somebody could tell me a lower power driver that uses simple, common components to drive IGBTs at relatively high energy levels. I would also like to know at what voltages IGBTs operate at maximum performance.
At the moment I have this core -
and I have a few 2KV .15µF (150nF) CDE capacitors and a few rated at 2KV and .33µF.
I have 30N60 IGBTs (eight of them, and I'm hoping for either a half-bridge or full-bridge).
Well, I suppose that's about it. I'm looking forward to any responses you have.
Registered Member #1107
Joined: Thu Nov 08 2007, 10:09PM
Location:
Posts: 792
There is really no "easy" way to get high energy levels. Usually your basic inverter consists of oscillator---> gate driver---> gate drive transformer---> full bridge of igbt's---> hf ferrite transformer. For the oscillator i recommend using a 555 "easier" or tl494 "greater control over voltage and current" fed into a ucc37322/21 pair driven by 15v. Assuming you know somewhat how bridges work, you feed the ucc outputs into a gate drive transformer and put the 30n60 igbt's in a full bridge configuration. Make sure the igbt's have internal freewheeling diodes or you will have to add your own "mur860 works well". I will not tell you how to hook up the full bridge as it has been covered a billion times already, just use the search function. There are other easier ways of driving it like with a zvs driver but it would not be ideal based on the igbt's you have at your disposal.
Registered Member #152
Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384
There is no advantage of using fullbridge over halfbridge, at least with MOSFETs. If you connect 4 FETs in a fullbridge or in a half bridge (parallel each two), the heating and power handling will be almost the same, with easier gate drive.
With IGBT's this also holds true as long as they are well matched.
Registered Member #1232
Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881
The full-bridge has a number of advantages at high powers. It develops twice the peak-to-peak voltage across the primary of the transformer. For the same power throughput this means that the current is halved. Resistive losses are proportional to the square of the current so they fall by a factor of four. There are however always two switches carrying current at any point in time in the full-bridge so conduction and switching losses will double due to this. Therefore the net gain is only a factor of two, but this is significant at power levels of several kW. The spreading of losses between four devices is also desirable when dissipation starts to approach the limits of package dissipation.
The full-bridge also gets rid of the DC blocking capacitor(s) which start to become bulky and expensive when required to carry lots of amps of AC load current.
Registered Member #152
Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384
The losses will be the same. Lets take a fullbridge of mosfets with R of turn-on resistance switching current I. The conduction losses will be 2*I*I*R on all devices, because two are conducting at a time.
Lets connect them up to half-bridge, (2 per leg). They are now switching current 2I and the on-resistance of each leg is R/2. But only one "leg" is conducting at a time, so conduction losses are 2I*2I*R/2 = 2*I*I*R.
The same applies to IGBTs with voltage drop being half with halfbridge.
Registered Member #1819
Joined: Thu Nov 20 2008, 04:05PM
Location:
Posts: 137
GeordieBoy wrote ...
The full-bridge has a number of advantages at high powers. It develops twice the peak-to-peak voltage across the primary of the transformer. For the same power throughput this means that the current is halved. Resistive losses are proportional to the square of the current so they fall by a factor of four. There are however always two switches carrying current at any point in time in the full-bridge so conduction and switching losses will double due to this. Therefore the net gain is only a factor of two, but this is significant at power levels of several kW. The spreading of losses between four devices is also desirable when dissipation starts to approach the limits of package dissipation.
The full-bridge also gets rid of the DC blocking capacitor(s) which start to become bulky and expensive when required to carry lots of amps of AC load current.
-Richie,
Good point. I did hear of an alternative to the splitter capacitors; a double-ended supply could be used. However, the use of one of these supplies on a half-bridge results in bus pumping, where the supply rails receive current and increase their voltage, causing regulation and feedback control problems. I would always recommend building a full-bridge, as the expense of the additional components usually is balanced by less problems.
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GIANT ferrite core driver
