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Frequency Locked SLR Inverter?

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Turkey9

Tue Dec 28 2010, 09:19AM

Registered Member #1451 Joined: Wed Apr 23 2008, 03:48AM
Location: Boulder, Co
Posts: 661

I have had some issues with having an SLR resonant frequency change as a capacitor charges. How would you go about using a microcontroller to lock the frequency onto resonant? Specifically, how would a micro be used to detect the zero cross of the current waveform. I'm thinking that if that can be detected, a pulse can be sent out on every other zero cross. The pulse can also be the same length of the time between zero crosses. The length of the pulse could be actively changed while the pulse train is created.

GeordieBoy

Tue Dec 28 2010, 10:33AM

Registered Member #1232 Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881

The load current in an SLR converter smoothly commutates from the switch to it's co-packaged free-wheel diode automatically when it goes through zero.

As long as the switch is kept on for the first half-cycle of the resonant waveform, the remaining half-cycle will always automatically flow in the free-wheel diode. So you don't need to match the switch's gate drive to the resonant frequency of the tank circuit exactly. In short the gate drive width for each switch should be between 50% and 100% of the period of the resonant load network. In other words you can switch off the gate drive to each IGBT any time after the current has naturally commutated to the free-wheel diode. When set up correctly, nothing actually happens at the falling edge of the gate drive because the current has already left the active part of the IGBT die.

I hope this helps,

-Richie,

Turkey9

Thu Dec 30 2010, 06:09AM

Registered Member #1451 Joined: Wed Apr 23 2008, 03:48AM
Location: Boulder, Co
Posts: 661

Oh interesting, Makes sense. That will help with getting the drive waveform tuned.

The PIC I'll most likely use has an interrupt on change pin. Would this work to detect the zero volt? Or am I just going to have to sample the adc to find the zero?

GeordieBoy

Thu Dec 30 2010, 12:02PM

Registered Member #1232 Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881

You really don't need to spend lots of time tuning anything. Just set the on time of the two gate pulses to something in the region of 75% of the period of the SLR load network's resonant frequency. That way the gate drive for each IGBT (or pair of IGBTs) will be turned off mid-way through the reverse half-cycle of the load current. You shouldn't see any glitches or steps or whatever in the waveforms at this point though, because the IGBT is already reverse biased and the current load current has transferred to the free-wheel diodes long ago.

As long as the IGBTs are firmly off before the resonant load current would start to swing positive again for the beginning of a new cycle then all should be well.

If you really want to detect the point when the load current smoothly commutates from IGBT channel to co-packed free-wheel diode then you could look at the voltage across the device. You will see a small positive Vce drop from collector to emitter when the device is on and carrying forward current. When the current goes through zero and reverses this voltage will change to a small negative voltage equal to the voltage drop of the free-wheel diode as it carries the negative half cycle. Once you see this reverse in Vce you can switch off the device any time onwards from this point, but definitely before the voltage goes positive again.

-Richie,

Turkey9

Thu Dec 30 2010, 11:28PM

Registered Member #1451 Joined: Wed Apr 23 2008, 03:48AM
Location: Boulder, Co
Posts: 661

When charging a large capacitor, 320uF, to 1,100V the SLR load network resonant frequency shouldn't change, correct? In the past I had issues with the charger going out of tune as the voltage on the cap climbed. This is the reason of wanting to have an auto tuned driver.

GeordieBoy

Fri Dec 31 2010, 05:04PM

Registered Member #1232 Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881

The resonant frequency of the SLR load network shouldn't change significantly. (It may change by a couple of percent due to thermal expansion as the tank capacitor warms up or the windings on the transformer move slightly. But there shouldn't be anywhere near enough frequency drift to make it necessary to adjust the drive waveforms because there is a lot of tolerance on the point at which the IGBTs turn-off due to the zero-current soft-switching behaviour that I already described.)

As the output load voltage at the secondary increases the transformer reflects this back to the primary as a voltage-drop (or a loss) in the resonant SLR network so the apparent Q-factor will fall significantly. I can't remember what the critical Q is for the SLR inverter but there will be a Q-factor below which the current waveform becomes so damped that it no longer rings back in the opposite direction for a free-wheeling negative half-cycle after each driven positive half-cycle. Essentially all of the energy will be dissipated leaving little circulating energy to free-wheel through the diodes when the IGBTs are off.

If this is the case then you probably need to increase the loaded Q of the SLR load network by either:

1. Changing the transformer ratio to support a higher output voltage without reflecting as much voltage drop to the primary side SLR network, or...
2. Increasing the leakage inductance of the transformer to raise the resonant inductance, usually combined with...
3. Decreasing the tank capacitance to lower the resonant capacitance. (Thereby keeping the actual resonant frequency the same.)

You essentially need to increase the Q by increasing the characteristic impedance of the resonant network, so I hope I got the last two options the right way around. It's a lot of years since I used to work on this stuff.

Basically there are an infinite number of combinations of transformer leakage inductance and tank capacitance that will give the same resonant frequency. However the ratio of inductance to capacitance determines the characteristic impedance, and when combined with the load's resistance, determines the loaded Q or damping of the SLR network. (To an approximation output current is constant, but voltage increases, this can be modelled as a resistance. Since R=V/I, a gradually increasing V and constant I implies an increasing load resistance and therefore deteriorating Q factor.)

Designing an SLR inverter is always a compromise between switching losses and conduction losses. A higher Q factor maintains soft switching longer, so minimises switching losses but the peak currents are larger so the conduction losses are larger.

If you have a storage scope you should be able to look at the "buttock waveform" inverter current signal with a CT. Then compare with that which others have shown on this site and decide if the loaded Q is too low near the end of the charging cycle.

I hope this helps,

...and best wishes for 2011 to everyone!

-Richie,

Turkey9

Fri Dec 31 2010, 11:20PM

Registered Member #1451 Joined: Wed Apr 23 2008, 03:48AM
Location: Boulder, Co
Posts: 661

I think that is exactly what was happening with my last SLR charger. I had the bridge powered by a variac. The only time I got the nice double hump current waveform was when the input to the bridge was low. As I turned up the power the waveform disappeared and then went crazy.

Just to check what I think I know, the total voltage that the inverter can charge to is still determined by the turns ratio of the transformer, correct? The SLR topology is a constant current source, Correct? In order to charge to a high voltage a step-up ratio on the transformer would be needed. A 1:1 ration wouldn't work, right? Sorry for all the questions, but you your post have already taught me a lot about SLR topology! Thanks!

GeordieBoy

Sat Jan 01 2011, 01:54PM

Registered Member #1232 Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881

> I think that is exactly what was happening with my last SLR charger. I had the bridge powered by a variac. The only time I got the nice double hump current waveform was when the input to the bridge was low. As I turned up the power the waveform disappeared and then went crazy.

You should get the best looking double-humped current waveform with the output short-circuited. The SLR inverter will then drive an constant average current through the s/c with minimal losses. Q-factor is highest at this point because there is no voltage drop at the output to be reflected back to the resonant circuit by the transformer. Therefore both cheeks of the buttock waveform should be of almost the same amplitude.

> Just to check what I think I know, the total voltage that the inverter can charge to is still determined by the turns ratio of the transformer, correct?

Yes. There is no steady state continuous resonance taking place in the SLR inverter like you would get in say a Tesla Coil. The resonance is discontinuous in an SLR inverter. It essentially rings for one complete cycle, then stops dead, then rings for another complete cycle (albeit in the opposite polarity,) then stops dead again. This process repeats indefinitely. The polarity reversals are put in there to stop the transformer from experiencing flux walk and saturation due to any DC current component as the loaded Q falls. If you keep periodically reversing the direction of current flow then there can be no average DC component to saturate the transformer even if one buttock is smaller than the other!

> In order to charge to a high voltage a step-up ratio on the transformer would be needed.

Yes, the resonant action does not increase the voltage significantly. In order to get a huge output voltage you need to have a transformer in there. Typically this transformer is designed with some leakage inductance that provides part or all of the necessary resonant inductance for SLR operation.

> The SLR topology is a constant current source, Correct?

Yes, but only up to a point. It is easiest for the inverter to maintain constant average current into a short-circuit because this presents no load. As the load resistance rises this causes a voltage to be dropped across the output. This represents power being dissipation and an increasing load on the supply. A given design can only maintain constant current up to a certain load resistance or output terminal voltage, then the current will typically start to fall off. If you think about it in the limit, you can't drive constant current through an open circuit with infinite resistance, so something has to give.

> A 1:1 ration wouldn't work, right?

A 1:1 transformer ratio will work fine provided you only want to charge the output capacitor up to some fraction of the DC bus supply voltage to the inverter. A 1:1 transformer will reflect all of the output voltage back to the resonant load network so if the output voltage gets near the drive voltage the losses will be huge and the Q factor dives into the dirt. So if you want constant current compliance range up to an output voltage much higher than the DC bus voltage to the inverter, you need to put a step up transformer in there. That way you can charge the output capacitor to a high voltage without reflecting too large a voltage drop back into the SLR network and killing the resonant action (lowering the Q factor.)

> Sorry for all the questions, but you your post have already taught me a lot about SLR topology! Thanks!

No worries, there isn't too much info about this topology online. I'm sure others will chip in after the holiday period is over. As I said it's a while since I did this stuff so it will be interesting to hear other people's more recent experiences and corrections!

Cheers,

-Richie,

Ash Small

Sat Jan 01 2011, 04:01PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Any chance of putting something on SLR topology in the Wiki?

Turkey9

Sat Jan 01 2011, 06:19PM

Registered Member #1451 Joined: Wed Apr 23 2008, 03:48AM
Location: Boulder, Co
Posts: 661

I agree about the wiki. When tuning and adjusting an SLR inverter, would it be alright to rectify the output and put it to a high power resistor? This way there is a constant load. I read somewhere that if the output isn't fed through a fullwave rectifier and into a load that the inverter would be damaged.

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