Yup. A single SCR on a the DC side of bridge rectifier with the AC side in series with the primary would assure symmetry.
Best regards, Spehro Pefhany
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This is assuming there is not mains ripple getting into the triggering circuit. It only ensures that it is as symmetrical as the triggering is. A triac running from that same trigger signal would be about as good.
We could put a large capacitor in series to ensure no DC component. It leads to some fun waveforms and interesting shopping trips at these current levels, however.
how about two back to back FETs or one FET and a diode bridge, then you can you can also chop anyway you like.
-Lasse
Yeah, and that the diodes are pretty much the same.
Triacs, especially those of the sensitive-gate persuasion, don't like to commutate when there is much dv/dt. Alternistors are better, SCRs are the best thyristor.
I don't think it's an issue with a reasonable design. although toroidal xfmrs are supposed to be worse in that regard than E-I lam types.
Heck, if you want to get fancy, why not detect any miniscule DC component and actively null it?
Best regards, Spehro Pefhany
-- "it\'s the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
At the voltages involved, we would have to use some strange diodes to make enough mismatch to matter.
Even an SCR would likely need a snubber in this circuit.
Neither really do I. At worst we will have a small fraction of the peak current as an average.
I think that is a somewhat silly idea so lets do it :)
We know that the current stops when the SCR is off. We can use a current transformer and restore the zero when the current stops. I assume we are tripping the SCR with a tranformer so we can have the whole circuit isolated from the mains.
The current sense transformer would drive a circuit like this:
-----/\\/\\--------- ( ! ( ! ( ! ! --/\\/\\---+ -GND ! ! ! ! ---/\\/\\-+-!-\\ ! C1 ( ! >----+---!!--+---!+\\ To DC servo ( GND-!+/ ! ! >--+--- ( / -!-/ ! ---GND SW1 O ! ! ! ------ GND
The first opamp lets us use a cheap design for the current transformer and get a big enough signal to drive C1.
C1 is snapped to ground by SW1 when the SCR is off to reset the zero.
Perhaps no need to use a semiconductor. This could be an opportunity to investigate the design of the old GE theatre light dimmer, based on the saturable reactor. There should be no net DC component from that, and relatively soft waveform modulation.
La ACin-----------/////////---------+ ========= | DC Control---////////------------- DC Control ========= | ACout----------/////////---------+ Lb
The saturable reactor can be wound on E-I laminations or on two toroids stacked together, with the control winding wound on the stacked assembly.
Effective inductance control is just a matter of a variable DC constant current source driving the control winding.
-- Tony Williams.
Just take two transformers and wire the primaries in series and the secondaries in series bucking as the control. It worked fine some 40 years ago to control a light bulb so a bigger one should control a heater quite nicely. DC in the secondaries makes the cores saturate.
If you want to be a purist about it, you need to place a large capacitor across the control winding. When you feed DC into the control some AC currents flow in the control windings. The capacitor keeps this out of the control circuit.
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Phase Control - I.M.O.
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Boat Anchor ;-) But .... In a switcher I can testify that "the old way" is really Neat!
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Well, since we are into plumbing supplies anyway, one might as well run cooling water through the "wire" - i.e. make it from household copper tubing ;-)
Brick Outhouse more probably.
-- Tony Williams.
One transformer with two secondaries, wired in series bucking, wouldn't work, I suppose, since the dc fields would cancel.
Yes, you need to have a net field in the core(s) from the control winding.
I'd appreciate a response on this. The saturable reactor is appealing to me because it's a linear way of dealing with an ac-transformer power-control problem. One thing that slows me down is considering the two transformers needed to create a saturable reactor - how big do they need to be? Thinking about this, with no DC current, they create an open circuit, no current flows and no power is dissipated. Alternately, with maximum DC current the two transformers are saturated, and simplified, look together like a length of copper wire. So the intermediate condition provides the greatest stress. That should be the region to evaluate.
Another issue, how much power will be involved generating the DC-current to control say 1.0kVA? Will it be a similar amount, say 500W dc? Ahem.
I was interested in Ken's idea, so sketched a possible circuit, as below.
Each transformer's primary winding has to be able to take the final max load current. So if you had a 230V/2500VA load then you would possibly use a pair of 115V/1250VA transformers.
I suspect that if the transformers were designed for this application then they would be smaller because more window area could be allotted to the AC side.
As a first pass, the load current, Iload = Vin/Z.
Where Z^2 = (w.(2Lp + Lleak))^2 + (2Rpri + Rload)^2.
The effective '2Lp' is the variable that is controlled by the DC current, by varying the incremental permeability. DC control will probably be non-linear, more or less following the shape of the B-H loop.
I suspect that the control power will be far less than the normal rated output power of the secondaries.
The B-H loop will show what DC field strength is required to push the flux density into saturation. An old graph that I have shows that a sample of 0.014" Silicon Steel required a polarising field strength of about 5 Oersteds to reduce the incremental permeability from over 2000 to about 100.
Polarising field strength, H = 4.pi.N.I/10.le Oersteds.
So the required N.I can be estimated. It would seem useful to buy unpotted toroidal transformers so that a few turns can be wound on, to get some idea of the turns/volt the secondaries were wound at.
-- Tony Williams.
Actually, why can't the two transformer primary windings be in parallel? That'd reduce the copper resistance. And L(leak) for that matter, right? Editing Tony's drawing:
AC high + Pri - Load current --+-----------////////-------+--->---, | T1 ======== | | | DC --+---////////---, | |( | | + Sec - | | |( L(leak) | +_|_ | | |( | --- | | | | | + Sec - | | / Win's Load. | DC --+---////////---' | \\ Rload | T2 ======== | / '-----------////////-------' | + Pri - AC low
OK, now I see that Tony took advantage of a series primary connection to use transformers with half the primary voltage rating. Hah, it appears there's no free lunch.
A nice bit of work, there, Tony!
For quick estimates, could we guess that the DC current would be 3 to 5 times higher than the peak AC current at the VA rating of transformer? Given that the DC winding resistance of the transformers is quite low, the DC- control requirement doesn't look too bad. A fair amount of DC control current, sure, but not too much voltage...
I don't think it is even as high as that Win.
If you were to leave the Pri open circuit, connect a voltage source to the Sec, and turn it up until the current waveform had nice wide flat tops then the measured Ipeak could be a reasonable estimate of the polarising DC amps needed to saturate the core.
3 to 5 times the secondary magnetising current peak is probably much lower than 3 to 5 times the peak secondary load current.The control power would then be the DC amps times the winding resistances.... but see below on what could buggerate that low control power.
There could be a little ambush in the two-transformer saturable reactor. Those two series'd secondaries may not cancel completely, so there could be a residual AC voltage at their terminals. I suspect that the DC current source must have a high enough compliance/supply voltage such that no AC current flows in the secondaries.
The same (or worse) AC current flow could occur with the circulating currents that would arise from the paralleled windings in your previous post.
This is getting interesting.......
-- Tony Williams.
The Lleak in my sketch is to allow for the possible leakage inductance in your transformer Win. It could be significant.
-- Tony Williams.
Oh bum! Wrong way round.
Drive the Sec from a resistive source until the voltage is seen to have nice wide flat tops. Then measure the peak of the current waveform.
-- Tony Williams.
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