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WHAT AND WHY: I've finally built a isolated supply with a current limiting bulb for my bench. The transformer is a Triad Magnetics VPT230-4350 torrid, 8.5 amps max on the output, 1:1 ratio (home page, datasheet). I went with the 'bigger is better' and hopefully I don't ever need more. Started with a 4A fuse on the output, and 8A on the input.

PROBLEM: So it wasn't a surprise that it will sometimes pop the 8A primary fuse even with the output open. It was a surprise that replacing it with a 15A slow blow will sometimes blow. So inrush current is way higher than I expected, some research shows torrids have higher inrush current.

I can go with a 20A slow blow to match the 20 breaker it's wired to in the breaker box, now I'm starting to worry about the 20A light switch I'm using to turn it one and off. Seems like I should do something better.

I can't get my head around using an inrush limiting ICL for 2 reasons. First is that my load is going to vary greatly. From a 10W solid state guitar amp, to large tune amps. So how does one size the ICL for somewhere between 200mA and 8A? Doesn't seem possible. Next is that the ICL needs to cool between cycling ON->OFF->ON.

Here is my proposed circuit design.

schematic

simulate this circuit – Schematic created using CircuitLab

So my proposed idea is to use a ICL to soft start the transformer a double pole relay to bypass the ICL and connect the output when the transformer is up to voltage, or at least the relay voltage. Also adding a thermal switch to the box. Then I can stay with small fuses to protect the transformer. Thinking that a ICL with a starting resistance of 60-120 ohms would be about right.

Good Idea? Bad idea? Missed Something? Thanks for looking Cory

EDIT Dec 24th. The link posted by Andrew Morton to an article reconfirmed my concerns about using an ICL and that bypassing it with a relay is a good idea. Since it reduces recycle time and eliminates the effects of varied load on the ICL device. I believe that the relay also switching the output should eliminate most of the concerns with the relay dropping out and the full load being across the ICL and blowing it, except in case of a failure of the transformer primary on start up. But being able to use a smaller fuse with this circuit should help that. If anything really can help with that lol.

Only question I have left is the issue of having the relay pull in slowly mentioned in the article. I'm not sure if this is an issue, or not, in this case. The problem with the time delay in the example from the article using a resistor to charge a capacitor up to the voltage required to pull in a relay. In that case I think the voltage to the relay will rise much slower than this circuit.

What I don't know now: 1)Is how the output of the transformer responds during the inrush period. Does the output spike or dip during the inrush? Closing the relay sooner or later? I'd be pretty sure the output will dip if the windings were separated like on a EI core transformer, but they look interleaved on this torrid. Mmmm, so do the windings couple before the core is magnetized on a torrid?

2)How long it takes for a large relay close. Will this be long enough to get a solid enough voltage to the coil?

Without any further input, I'm going to build the circuit and see if the relay chatters or arcs as a sign that it's pulling in too fast or too weak.

EDIT Dec 28th: After some more reading and experimenting. I ordered a 30A dual pole contactor (relay in case contactor isn't a widely used term) to build the proposed circuit. I also ordered a 400ohm 200Ma max ICL(highest starting ohms I could find) to pre-charge the transformer. But after experimenting with the transformer's output open I have found that a 1K ohm 1/8 watt resistor is nearly enough to fully energize the transformer and provided 90 of the 120V on the output. I don't have a way to test inrush current accurately. But both a 100ohm and 400ohm resistor show to have a 50mA current. Making it pretty clear the resistor is the better option, as the link provided by Andrew Morton concluded.

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    \$\begingroup\$ Soft-Start Circuit For Power Amps looks like it could be useful to you. \$\endgroup\$ – Andrew Morton Dec 24 '17 at 0:40
  • \$\begingroup\$ Thank you, this was useful information. Pretty much confirmed the problems I had foreseen using just an ICL, and that a bypass relay was the answer to both problems. Adding an edit to the OP to address the one question I have left. \$\endgroup\$ – Cory Lytle Dec 24 '17 at 18:03
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    \$\begingroup\$ (1) Cory, get into a habit of adding datasheet links to your questions. (2) If somebody else does you a small favor by adding datasheet links, don't edit it out. \$\endgroup\$ – Nick Alexeev Dec 24 '17 at 20:46
  • \$\begingroup\$ I didn't intentionally, it's possible that I was editing at the same time. Maybe that deleted it, sorry. And good point! \$\endgroup\$ – Cory Lytle Dec 24 '17 at 21:00
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Have you missed the simplest fix?

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Placing the lamp in the primary gives the required soft-start on the transformer and provides the required current limiting on the secondary too.

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  • \$\begingroup\$ That is a simple fix. The downside is that I'd always have to cycle both switches to start and stop. The current limit switch is mounted out of reach, mainly so the I can't accidentally flip it off instead of the power. Also means that every time I started something I would have to start in current limit. Not sure that would be good for certain circuits. Mainly for soft starting tube circuits. \$\endgroup\$ – Cory Lytle Dec 23 '17 at 18:33
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The only cause of inrush current of idle transformer is core saturation due to inappropriate moment (phase) of connecion to mains. It's better to prevent saturation at all, than mitigate it's consequence, I guess. So, you can build a circuit, that connects primary to mains via triac at the best moment (phase) - when voltage is at it's extremum (max or min peak). Also, you can later make this circuit act as a fuse!

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  • \$\begingroup\$ I am still trying to get my head around why the maximum voltage it the correct time to connect the transformer. I know it's the correct answer, just still working through it in my head. I was editing the OP when you posted. Wait, now I think I get it. If you turn on at max voltage there is less time before current reversal and the core's field to reduce current? Right? \$\endgroup\$ – Cory Lytle Dec 24 '17 at 19:13
  • \$\begingroup\$ Yes! Current in inductor is proportional to integral of voltage (square under voltage-time curve). So, when started at max, current increases from zero to max only while 1/4 of period and then it starts to decrease. 1/4 of time it decreases back to zero and then increases down to -max. But if started at zero, current increases for 1/2 of period and can be as large as 2*max. In fact, at values less than 2*max, core saturation happends, so current quickly becomes much larger than 2*max, blowing almost any fuse:) \$\endgroup\$ – Eugene K Dec 24 '17 at 21:54
  • \$\begingroup\$ Now that makes sense. Still thinking that initially limiting the current has the same effect of keeping the transformer out of saturation. Using a SCR the circuit is more complicated, but with less questions left unanswered than the relay circuit. Also if i want to add an 'adjustable fuse function' the SCR should be way faster shutting down than a relay, never more than half a cycle from being off. \$\endgroup\$ – Cory Lytle Dec 25 '17 at 2:19
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I once used an "inrush current limiter", a kind of thermistor. Ametherm sells them on Digi-Key. enter image description here

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    \$\begingroup\$ Yes, is an "inrush current limiter" or ICL usable in a circuit with such wide range of current loads? Also that leaves the problem of the required reset time. I have one in my proposed circuit, but it gets switched out after startup. Which removes the load variation, and starts the reset time immediately after startup. If the ICL even gets warm enough to lower its resistance to a point where a 2nd start is a problem. \$\endgroup\$ – Cory Lytle Dec 23 '17 at 19:04
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I built the proposed circuit. The ICL with 400 ohm starting resistance caused the relay to chatter intermittently on start up. So I went with a 100 ohm resistor. I only had a 1/4 watt resistors available. I know initial wattage is extremely high, but there is no secondary load until the relay closes and then the resistor is bypassed.

I tested it with no load and a 1/4 amp(250mA) fast blow fuse holds up just fine. I didn't experience any chatter when starting a 1000 watt heater either.

I'm calling this solved. Probably order a better resistor next time I get parts. Thanks for the education

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    \$\begingroup\$ "a 1/4 watt fast blow fuse" You don't mean that - fuses are not rated in watts. I don't know if you mean a 1/4 A (250mA) fuse, or if you were trying to refer to both 1A and 4A fuses, or if you meant something else. Please can you edit your answer to clarify? Thanks. \$\endgroup\$ – SamGibson Dec 30 '17 at 22:50

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