Class D Commutation of a SCR

Hi everyone,

I recently ran into two variants of a circuit for class D commutation of a SCR. (This circuit is also known as the Jones Chopper.)

I have uploaded two files that capture these two variants:

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I could understand the first variant which has a single commutation inductor.

The second variant uses a coupled inductor (shown as L1, L2 in the image) in place of the single commutation inductor. This coupled inductor is also referred to as a "auto transformer" in some texts!

I am unable to understand why the coupled inductor is required (in place of a single inductor) and does it really work as a "auto transformer"?

Regards, Anand

Reply to
Anand P. Paralkar
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Your quote marks around auto transformer and your exclamation mark indicate surprise. Is this because the term "auto transformer" is new to you, or are you just surprised that it's in that circuit?

Well, clearly it's not _required_, or the circuit without it wouldn't have been presented. I suspect that it enhances the turn-off behavior of the circuit somehow. A simulation may be a good idea here.

Note that I _absolutely positively_ have not gotten my head wrapped around that circuit -- any time I try to think of an SCR in a DC circuit my imagination fills up with so much stinky smoke that it boils over into my intellect and I can't think about it until I douse the fire with a big mug of beer or something -- and then I just can't think.

I'll stick to devices that don't need help to turn off, thank you very much.

--
Tim Wescott 
Wescott Design Services 
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Reply to
Tim Wescott

The lower SCR turns off the upper by shorting it. A.C. coupling ensures the lower SCR turns off, eventually--there's no d.c. current possible.

The capacitor has to be charged for the turn-off scheme to work. For the auto-transformer version, on main SCR turn-on, the auto-transformer produces a positive pulse that charges C through D, leaving C ready for turn-off duty.

I don't see how the first version charges C. Looks faulty--I don't see how the "commutation inductor" ever has current. L must be the secondary of a pulse transformer?

Cheers, James Arthur

Reply to
dagmargoodboat

The question is not so much what charges C, but what discharges it. Or are you thinking that C needs to be charged in a way that helps to turn off the upper SCR?

When the lower SCR is fired current is passed through the lower SCR which may or may not be sufficient to drop the current in the upper SCR below it's holding current and turn it off. Assuming that happens the lower SCR will turn off when C charges sufficiently and the current drops below it's holding level. C now has a charge and no discharge path to drain it... until the upper SCR turns on again and current from C passes through the diode and inductor. Ah, now I see it. Once the current builds in the inductor and C has drained its charge, the field in the inductor continues the current flow until the voltage from it's field is less than the sum of the voltage on C and the diode drop leaving a charge of opposite polarity on C that will help to shunt the current in the upper SCR when it is needed to turn off.

That explains the purpose of the auto transformer. The first circuit doesn't get the reverse polarity charge on C the first time the upper SCR fires. So it seems to me this circuit could have problems getting started. The auto transformer enhances the reverse voltage on C using the current flowing to the load which should be significantly larger and happens on the first and every cycle.

--

Rick
Reply to
rickman

Oh come on, what's the worst that could happen? :-)

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This one actually works pretty well up to a kHz or so. It does have the worrisome latch state, and since the only SCRs are the ones doing conduction, there's no "out". Not very safe. Sometimes, that series resonant tank is fired by a third, and the tank is kept charged with diodes and resistors from the bus, rather than the switching side, so it's always available for commutation.

I've seen SCR fired induction heaters up to 50kHz (those ones are 3rd or maybe 5th harmonic resonant). They're apparently very well regarded: robust and reliable. The adjustable range on them is awful, though (good luck using them for anything but dull production line work).

I think the ultimate safety factor on those type units comes from the AC line: they normally use massive inductors (choke input filter) and film capacitors (of the boxed variety), so there isn't terribly much energy storage on the bus. And if the whole damn thing latches up, end to end, the worst that can happen is shorting 1/2 of an AC line cycle. And if that draws fault current, semiconductor fuses open up in just a few ms, enough to save the SCRs (something that's impossible to do with MOSFETs or IGBTs, which do the opposite, they go open (constant current region that is, but still conducting heavily), then fail shorted :) ).

The hardest thing about using SCRs is, like toobs, they are ideally suited to constant current mode inverters, and suddenly everything looks all topsy-turvy and confusing.

Why? You can perform a voltage-to-current, series-to-parallel transformation on the half bridge inverter circuit, and get something that looks suspiciously like a push-pull circuit -- except instead of a CT'd transformer winding, the left and right side inductors don't need to be coupled, and the supply is a constant current source. And, if you started with switches having antiparallel (e.g. MOSFET body) diodes, then after the transformation, you have series diodes (prevent voltage in reverse direction --> prevent current in reverse direction). Thus, the switches have to be unidirectional -- SCRs or tubes.

And analogously, switching behavior is inverted, so instead of dead time between pulses, shoot-through is a required feature!

Often, the constant current requirement is faked with a generous inductor (see the circuit at the beginnning), but ultimately, using a constant voltage DC bus is clearly a dangerous thing!

Tim

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Seven Transistor Labs 
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Reply to
Tim Williams

Thanks everyone for your replies. I think the self start explanation from Rick makes sense. Will need go over it again to understand it better.

Tim (Tim Wescott), yes, I was surprised to see an auto transformer. I thought auto transformers are useful only for stepping down (or up) ac mains voltages. Wouldn't we call this application a coupled inductor?

Anyways, if it works, it works no matter what we call it! :)

Regards, Anand

Reply to
Anand P. Paralkar

I've seen the term used in other places than AC power applications.

I used to try to clean up what I saw as inconsistencies in language between technical disciplines. Now I just smile and add them to a mental list.

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Tim Wescott 
Wescott Design Services 
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Reply to
Tim Wescott

Actually the first circuit should not have startup problems, but it requires the proper startup sequence to do it.

For a reliable start, Saux must be triggered FIRST, then it will precharge C and turn itself off when C is charged.

A side effect will be an unavoidable initial pulse through the load when C is first charged.

Afterwards, Smain can be triggered to start it and Saux can be triggered to stop it again.

When Smain is triggered, the L+D+C circuit will complete a resonant half-cycle which will flip the polarity of C, charging it to almost the negative supply voltage (less some ESR and diode losses).

This will make C ready for "extinguishing" duty of Smain.

When Saux is than triggered, it will connect the reverse-charged C across Smain, thereby for a short time reversing Smain's voltage and forcing it to commutate off.

Another side effect will be an unavoidable short-time doubling of the supply voltage across the load (the other half coming from C). The load must withstand this pulse and NOT clamp it.

This will also recharge C again because Saux will not stop conduction until C has been fully charged to the (positive) supply voltage and the charging current has ceased.

The circuit is than ready for another cycle.

Note that this circuit has limitations concerning both the on-time and the off-time of Smain:

  1. Smain has a minimum ON-time limit. If Saux is triggered too early, before the L+D+C resonant half-cycle has completed, C will lose all its charge and it won't be able to turn off Smain. Smain stays latched forever and C cannot be recharged any more.

There is no workaround that avoids this condition, so the minimum on-time must always be met.

  1. Smain has a maximum ON-time limit. If Saux is triggered too late, C will have self-discharged to the point that it won't be able to reverse Smain's polarity long enough while also carrying load current. An attempt to trigger Saux will just drain C and leave Smain on forever.

There is no workaround that avoids this condition, so the maximum on-time must always be met also.

  1. Smain has a maximum OFF-time limit. If Smain is NOT triggered too long, C will have self-discharged too far. Triggering Smain will flip C's polarity, but (because of the too low remaining charge) the voltage on C, after reversal, won't be enough to ever force Smain off again.

There IS a workaround that avoids this condition. The maximum off-time can be extended indefinitely by periodically retriggering Saux in order to keep C topped up to full charge.

This has the side effect of the charge-up pulses going through the load.

Note also that the load must always be connected and must be mostly resistive or inductive. It cannot ever have an opening switch or relay (this could disturb C's charge-reverse cycle at the wrong moment and leave C critically discharged) and it also must not be capacitive (C relies on the load impedance being non-capacitive, to NOT arrest the voltage doubling pulse, otherwise Smain would not turn off).

Regards Dimitrij

Reply to
Dimitrij Klingbeil

What makes it an autotransformer is that it is one coil rather than isolated coils. There is no isolation between the two coils.

I see some others have jumped in to explain the first circuit more fully with proper terminology and a lot of detail on running conditions. I wonder what the real advantages of the second circuit are then? From what the others have said I suppose it eliminates some of the operational conditions the first circuit has.

--

Rick
Reply to
rickman

This all looks like a 1980s RCA horizontal deflection circuit. Or was that late 1970s ? They used two SCRs, each turned off the other.

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Reply to
jurb6006

I think both SCR and GTO's were used..

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Reply to
Maynard A. Philbrook Jr.

Not an unreasonable analysis. The commutated SCR circuits i have seen have always taken the load from the anode which makes it easier to understand and build. Just the same, here is an LTSpice sim yall can play around with:

Mind the wrap!

Version 4 SHEET 1 2276 1104 WIRE -448 320 -528 320 WIRE -368 320 -448 320 WIRE -240 320 -304 320 WIRE 0 320 -240 320 WIRE 128 320 0 320 WIRE 304 320 128 320 WIRE 304 336 304 320 WIRE -448 352 -448 320 WIRE -192 368 -304 368 WIRE -528 400 -528 320 WIRE -192 464 -192 368 WIRE -128 464 -192 464 WIRE 0 464 0 320 WIRE 0 464 -48 464 WIRE -448 528 -448 416 WIRE -368 528 -448 528 WIRE -240 528 -240 320 WIRE -240 528 -304 528 WIRE -192 576 -304 576 WIRE 0 576 0 464 WIRE 0 576 -112 576 WIRE 304 624 304 416 WIRE -448 640 -448 528 WIRE -416 640 -448 640 WIRE -336 640 -352 640 WIRE -240 640 -240 528 WIRE -240 640 -256 640 WIRE 128 720 128 320 WIRE -528 832 -528 480 WIRE -496 832 -528 832 WIRE 128 832 128 784 WIRE 128 832 -496 832 WIRE 304 832 304 704 WIRE 304 832 128 832 WIRE -496 880 -496 832 FLAG -496 880 0 SYMBOL Misc\\SCR -368 336 R270 SYMATTR InstName U1 SYMATTR SpiceModel SCR SYMBOL Misc\\SCR -368 544 R270 SYMATTR InstName U2 SYMATTR SpiceModel SCR SYMBOL ind -240 624 R90 WINDOW 0 5 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName L1 SYMATTR Value 1u SYMATTR SpiceLine Ipk=20 Rser=0.03 SYMBOL ind 320 432 R180 WINDOW 0 36 80 Left 2 WINDOW 3 36 40 Left 2 SYMATTR InstName L2 SYMATTR Value 187u SYMATTR SpiceLine Ipk=100 SYMBOL res 320 720 R180 WINDOW 0 36 76 Left 2 WINDOW 3 36 40 Left 2 SYMATTR InstName R1 SYMATTR Value 4.6 SYMBOL cap -464 352 R0 SYMATTR InstName C1 SYMATTR Value 1u SYMATTR SpiceLine V=100 SYMBOL Misc\\battery -528 384 R0 WINDOW 123 0 0 Left 2 WINDOW 39 24 132 Left 2 SYMATTR SpiceLine Rser=.01 SYMATTR InstName V1 SYMATTR Value 48 SYMBOL schottky -352 624 R90 WINDOW 0 0 32 VBottom 2 WINDOW 3 32 32 VTop 2 SYMATTR InstName D1 SYMATTR Value MBR20100CT SYMATTR Description Diode SYMATTR Type diode SYMBOL schottky 144 784 R180 WINDOW 0 24 64 Left 2 WINDOW 3 24 0 Left 2 SYMATTR InstName D2 SYMATTR Value MBR20100CT SYMATTR Description Diode SYMATTR Type diode SYMBOL voltage -144 464 R270 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V2 SYMATTR Value PULSE(0 3 0.1m 1u 1u 10u 120u) SYMBOL voltage -208 576 M90 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V3 SYMATTR Value PULSE(0 3 0.18m 1u 1u 10u 120u) TEXT 648 96 Left 2 !.tran 0 10m 0 10u TEXT 352 24 Left 2 ;SCR Commutation circuit No 1 for Anand P. Paralkar TEXT 1000 16 Left 2 !..subckt SCR 1 3 2 * *Models used by the SCR Model: ..model STH VSwitch (ROn=0.1, VOff=0.5) ..model DLG D (Rs =25m) * *We assume that the SCR in the on-state is equivalent to 0.1 ohm: SSCR 1 4 5 0 STH VSCR 4 2 0V * *Trigger to cathode impedance is a 1k ohm resistance - * for simplification: RTrig 3 2 1k * *The following circuit locks in the on-state when a 1mA current flows *through the gate to the cathode with a positive anode to *cathode voltage. This circuit locks in the off-state when the *current through the SCR tends to become negative: VPlus 6 0 2V SiTrig 6 5 3 2 STH CState 5 0 1uf IC=0 DLogic 5 8 DLG HCur 8 0 poly(1) VSCR 1 10000 * *The potential of node 5 reflects the state of the SCR. *It is less than 2Vfor the off-state and *between 1v and 2V in the on-state *SCR Extracted from PONT_DIPH.CIR file from EDN website. *I renamed that file as SCR_CKT.CIR * Anode * | Cathode * | | Trigger * | | | * 1 3 2 ..ends

Reply to
josephkk

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