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Registered Member #1784
Joined: Tue Oct 28 2008, 02:30AM
Location: Rochester, MN, USA
Posts: 14
Okay, another quick question or 2 for the forums (I'm waiting on some parts to arrive, so I've got plenty of time to plan this out):
With regards to the matching coil (apparently labeled "Tlumivka" in Czech), to match the impedance of the LC tank to the inverter output, is there any reason (other than size) than an air-core coil wouldn't work? At these frequencies, it would be a large-ish thing -- several inches in diameter and several inches long. My concern with a solid ferrite core is that it would saturate, given the large currents (I plan to just use rectified AC, so 120 V DC going into the half bridge, with 5 A giving about 600 watts (assuming a good power factor). On the other hand, an air core with a long wire will have more loss due to simple wire-resistance than a solid core coil with less wire.
Also, any recommendations on powering pin 1 of the IC in the schematic above?:
I was thinking of just attaching it to the +120 VDC rail with a 5W, 22k ohm resistor and letting the resistor drop "ballast" the chip. That *should* give it the voltage it needs at the current it should draw, but it's totally unregulated, of course. How would you attack this problem differently? A 78xx-series regulator can't handle 120V input.
Registered Member #95
Joined: Thu Feb 09 2006, 04:57PM
Location: Norway
Posts: 1308
Air core coils leak an enormous magnetic field, which means it may affect other objects around it, but most importantly the coil's inductance will change if it's brought close to magnetic objects. None of that is much of a problem if you give the coil some space however, but the wire resistance argument still holds true. As for ferrite cored, just place an air-gap in the core.
The safest way to power the IC is with a small wart and regulator.
Registered Member #1784
Joined: Tue Oct 28 2008, 02:30AM
Location: Rochester, MN, USA
Posts: 14
Uzzors wrote ...
The safest way to power the IC is with a small wart and regulator.
I thought about this, but wasn't sure if it would be "okay" to tie the main ground of the circuit to the wall wart's ground. If that arrangement is acceptable, it would be a lot safer for the IC than a simple resistor.
My crude ballast-resistor idea is from the manufacturer's application note for the chip, actually, but I think that they intended the chip for use in equipment like electronic fluorescent lamp ballasts where cost and size are at a premium and the conditions are a little more forgiving and predictable...
Registered Member #152
Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384
jshine wrote ...
I was thinking of just attaching it to the +120 VDC rail with a 5W, 22k ohm resistor and letting the resistor drop "ballast" the chip. That *should* give it the voltage it needs at the current it should draw, but it's totally unregulated, of course. How would you attack this problem differently? A 78xx-series regulator can't handle 120V input.
Thanks! -Jon
Zener diode. The IC has a built-in one but AFAIK it is good only to 25mA. I've had bad experience with powering such chips by dropping resistor, the voltage can sag in some conditions, exploding your bridge, or even your bridge can be destroyed at start- up when the voltage on the IC slowly ramps through the resistor (low gate voltage> linear region> less current handling capacity and huge dissipation)
Registered Member #1784
Joined: Tue Oct 28 2008, 02:30AM
Location: Rochester, MN, USA
Posts: 14
Dr. Kilovolt wrote ... Zener diode.
You mean from pin 1 to ground? That is probably a good idea. I bought 10 of the IC's off of eBay (since they are cheap and are being shipped from Hong Kong, I didn't want to wait for more if I blow one or two). I'll probably use this idea first, and only resort to more complex solutions if I loose a few chips.
Registered Member #1232
Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881
> I've had bad experience with powering such chips by dropping resistor, the voltage can sag in some conditions, exploding your bridge, or even your bridge can be destroyed at start- up when the voltage on the IC slowly ramps through the resistor (low gate voltage> linear region> less current handling capacity and huge dissipation)
Thats why UNDER-VOLTAGE LOCKOUT is important in power electronics.
Registered Member #1784
Joined: Tue Oct 28 2008, 02:30AM
Location: Rochester, MN, USA
Posts: 14
Well, the IR2153 driver chip seems to be working well. I got a bunch of spare power mosfets off of a friend (200 V, 25 A in a TO-3 can) and hooked up a quick oscillator. This is what it looks like running off of 14.5 volts with no load (2 ms/div horiz, ~5 volts/div vertical). It sure does "ring" a lot, but I suppose those will damp out when I hook it up to a load.
Registered Member #1784
Joined: Tue Oct 28 2008, 02:30AM
Location: Rochester, MN, USA
Posts: 14
Well, while I gather components to build this thing (almost done!), I've been working on a theoretical analysis of the circuit in Matlab. If anyone is interested, I thought I'd post the .m file here. It uses the LCLR model as described on Richie Burnett's page, and implemented in Matlab, since Matlab handles complex variables so well. Bear in mind that I'm not an electrical engineer, so I can't guarantee that this is correct! Also, the way this is implemented, it implicitly assumes that the voltage source is a sine wave, which, of course, it's not. How much difference that makes, I'm not sure.
Here's an example analysis of a system with the following parameters:
Inverter: 170 V sine wave (from 210 kHz to 216 kHz) Work Coil: 1.5 uH Tank Capacitor: 0.376 uF Virtual Resistance: 300 ohms (a heavy load) Matching Inductor: 85 uH
Registered Member #1232
Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881
Nice work! You used MATLAB, and I used Microsim, but we both arrived at the same results.
Those are a good set of plots for the LCLR workcoil arrangement. The important conclusions to draw from them are:
1. Maximum real power occurs at around 213.7 kHz as shown by the peak in the green line in the first plot, and the maximum tank voltage (blue line) in the second plot.
2. Inverter load current in the last plot (blue line) does not make a good control variable because it's phase is neither 0, 90 or 180 at 213.7kHz. (It is actually load dependent.)
3. Tank circuit voltage in the last plot (green line) passes through 90 degrees at 213.7kHz (the maximum power point) and makes a good control variable.
4. Tank circuit current in the last plot (red line) passes through 180 degrees at 213.7kHz (the maximum power point) and also makes a suitable control variable.
A PLL (or some intelligent controller) that controls the drive frequency in order to achieve this phase condition is what is required to track the maximum power point. Heating power can then be decreased from "full power" by either detuning the drive frequency on the high side of resonance, or by reducing the DC link voltage using a thyristor controlled bridge ahead of the IH inverter.
You may also find these two Induction Heating papers interesting:
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