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Re: [Amps] Re: Step-start calculation

To: "amps@contesting.com" <amps@contesting.com>
Subject: Re: [Amps] Re: Step-start calculation
From: Will Matney <craxd1@ezwv.com>
Date: Sun, 21 Nov 2004 19:28:09 -0500
List-post: <mailto:amps@contesting.com>
Jeff,

I think what they're talking about is the core saturating during the first few 1/2 cycles. When this happens, there is little inductance and thus no impedance except the DC resistance of the winding. I think their example was a primary with a 1.5 ohm DC primary at 120 volts. 120 / 1.5 is 80 amperes for a 150 VA transformer which is pretty darn small. The ZSC SSR would need to be able to withstand this surge for the first few 1/2 cycles until the core settles down to normal. Their thinking is it would be better to apply full line voltage at the peak instead of at zero voltage. below is a quote from the texts;

"A “zero-crossover” SSR does not always turn on at precisely zero voltage.
It takes perhaps a millisecond or more for the circuitry to react. Therefore,
the load switch may not be fully on until load voltage is perhaps 15 to 20
volts. In this event, surge current isn’t as great, but it is still potentially
destructive. Also, a random turn-on SSR may, at times, turn on at or near
zero cross-over. The best method of turning on transformers and other
saturable, highly-inductive loads is by use of a peak voltage turn-on device.
Turn-on at peak voltage results in minimal surge, if indeed any surge is
present at all."


In other words, because of the time it takes for them to switch, 0 current draw can't be acheived. 1/4 cycle ( the time for the wave to peak) is .00417 sec. or 4.17 milliseconds. From this one can see there's not much time for the ZSC SSR to switch on before the voltage rises to a good amount if the time is 1-2 milliseconds. The only way for it to switch 0 current would have a 0 sec. switching time.

In Richs original example, he was using a transformer way larger than the above example. I forget what the primaries DC resistance was but it was very little, 0.14 ohm maybe at 240 volts? At least that's the way I read it.

Best,

Will Matney



jeff millar wrote:

Will...

Wow...that's a scary paper. But, I just don't understand it. ...And the calculations don't make sense. For example, the paper insists that zero voltage switching will cause horrible problems and computes the current as

I = E/R = 120V / 1.5A = 80 Amps

...which reads to me as 120V not 0V. So where does the 120V come from if the line is crossing through zero????

Maybe they're talking about phase control SCR's or something. Maybe they're talking about some extreme power system with very high power factor. But a zero crossing switch turns on once in a condition with the transformer at idle and the line a zero volts. There's no residual magnetism or stray currents running around. There's no DC in the windings.

Also there a reference to switching an SSR at 90 degrees and it shows some scope pictures with high currents in a small transformer. But 90 degrees amounts to switching the SSR on at the peak of the line cycle, not a zero volts.

I'm very confused. Do you have any other references to this effect?

jeff, wa1hco

Will Matney wrote:

I thought I ought to give a link to the following PDF file by Potter & Brumfield about the effects of using a zero-crossover switch with transformers. From this, I think most will see how and why surge currents act. Also, see the references (footnotes) on page 2.

Link; http://relays.tycoelectronics.com/app_pdfs/13c3206.pdf

Will Matney


Dennis12Amplify@aol.com wrote:


In a message dated 11/21/04 1:11:18 PM Central Standard Time, wa1hco@adelphia.net writes:
Dennis...


A transformer has inductance when open circuit, but when something's connected to its secondary, that impedance get transferred to the primary by the square of the turns ratio. So, when the transformer runs open circuit, a nearly infinite impedance gets transferred to the primary and it draw little current. When a transformer has a a load such as a resistor or discharged capacitor, then the primary impedance depends on the winding resistances, turns ratio and coupling coeficient. For most transformers one can model the coupling as perfect.

jeff, wa1hco
***Jeff,

***Thanks for the education.

***I still find it very hard to believe that there can be an almost instantaneous change in the secondary current when the secondary inductance is probably measured in multiples of HENRYS, and I know it's been a long time since I worked in that area, but when I did, I rarely saw 'M' values greater than .85 for a high voltage step up transformer.
The turns ratio is a big part of the problem in this case, because the ratio is so high, it is very hard to get a coefficient of coupling much higher than that. Maybe the technology has changed and that is no longer the case.


***Now if that same transformer was bifilar wound with a 1 to 1 or even a 4 to 1 ratio I would understand completely and absolutely agree with you.

***If what you are saying is true, then wouldn't the reciprocal also be true and allow me to greatly extend the frequency range of my tube stereo output transformer by making my speaker load impedance as low as possible; like paralleling ten 4 ohm speakers for an output load of only .4 ohms on the secondary of the output transformer? I understand that the efficiency would be poor unless the turns ratio was readjusted, but the audio bandwidth should be HUGE!

***I guess the best way to test this concept would be to run a 'significant', (like 1 ampere for a high power linear amplifier transformer), amount of DC current through the secondary winding, then short the primary and open the secondary circuit. If what you say is true, there should be no high voltage spike when the secondary discharges because of the shorted primary and the perfect coupling. I, personally, would not want to be holding both ends of those secondary leads while they were being open circuited......

***Regards,

***Dennis W. Ostrowski
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