Question Regarding Soft-Start

J.R.Freeman

Disabled Account
2008-08-11 1:15 pm
For a relatively large transformer, must the soft-start circuitry be on the primary side? Is the current in-rush required to charge the magnetic field of the transformer the primary issue, or is it the load seen on the secondary side, as presented by a bank of uncharged capacitors? Or is it that both are of equal concern?

Jim
 

J.R.Freeman

Disabled Account
2008-08-11 1:15 pm
Hi Frank,

Thank you for your reply. Yeah, I think you are right. I've been considering a few soft-start methodologies for an upcoming project. It would be nice to have it on the secondary side - that way I could have the secondaries go straight from the transformer into a PCB containing the soft-start, and there's no messing with mains voltages. But I suppose that would be neglecting the in-rush current of the transformer.

I've been toying with the idea of making a time-delay relay, using a thermistor. I'll post up a schematic or two when I have something that might work :p

Jim
 
When you connect a transformer primary to the mains it may draw a DC current of up to the value of the initial AC current - it depends on where in the AC cycle the switch closes. This DC current then decays, according to the L/R time constant of the transformer. It can cause transformer saturation, which can often be heard.

Except for small items, I now routinely include an NTC inrush limiter in the mains circuit. This handles both the capacitor inrush and the DC, by temporarily increasing the resistance. Inrush limiting on the secondary side will help by limiting the initial AC current, but it won't quicken the decay of the primary DC.
 
I always put the soft start on the primary side as it both covers inrush and filter capacitor charging. For a system with filter capacitors use a relay or contactor with a coil voltage equal to any convenient winding on the transformer usually the primary. wire the coil straight across that winding and use the contacts to short out a heavy duty resistor in series with the primary, Avoid the white ceramic block type wirewound resistors they blow operating into a heavy load. I have used this technique on power supplies up to 40 Kw NTC devices are fine for small power supplies I have found they give trouble once you get in the Kw range and restart after momentary power loss can be quite a problem.
 
could you give some (or any) example of what not to do?
Inrush current limiting on a 35 Kva transformer 415V supply 6 x 2.2 ohm 5W ceramic resistors per phase to get one ohm, 1 second delay to switch out of circuit, found one or more resistors failing from blown wires every few starts.

The wire doesn't have enough thermal mass to absorb the current pulse, switching to 1 ohm 50W circular wirewounds with heavy wire fixed the problem, pulse rated monolithic resistors would be ideal. Most circular wirewound resistors have quite robust wire, the ceramic block type have very thin wire, crack one open and have a look.

As for NTC resistors in the Kw range I had a spate of VFD failures where the NTCs would fail with low resistance causing the bridge rectifier to fail short circuit discharging the partially charged filter caps and vaporising plenty of pcb track in the process it could just be that brand of NTC
 
Inrush current limiting on a 35 Kva transformer 415V supply 6 x 2.2 ohm 5W ceramic resistors per phase to get one ohm, 1 second delay to switch out of circuit,
415V through 1r0 of resistors is peaking out at near 415A. That is a recipe for blown resistors.
Holding then in-line for 1000ms sounds like torture.

Working back from 415V and 35kVA I would select ~15r as the current limiting resistors in each phase and bypass them in 200ms.

The 15r resistors could be a series string of 15 off 1r0 5W resistors.
That's my first guess. But it's no more than a guess because I have never worked with anything approaching 35kVA
 
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If 1R0 gives 415A then omitting it would give infinite amps! This does not happen because there are other impedances present: transformer winding resistance, transformed rectifier and capacitor series resistances etc. The initial peak lasts for only a short time as capacitors charge.

My wild guess on inrush current-limiting resistor value is that it should be somewhere around 10-50% of the normal load impedance. For 35kVA at 415V this is 0.5-2.5 ohms, so 1R is bang in the middle of my wild guess range. 15R is way too big!

I have heard that white ceramic block wirewound resistors tend to explode when seriously overloaded, while some other types just get very very hot and hang in there safely.
 
415V through 1r0 of resistors is peaking out at near 415A. That is a recipe for blown resistors.
Holding then in-line for 1000ms sounds like torture.
It works out about 300A peak because the effective phase to phase resistance is sqrt3 x the individual phase resistance and the peak is 1.414 x the RMS. The 1 second delay does not cook the resistors because the current settles down after a few cycles to about 10A making the I^2R x T losses less than with a higher resistance though the delay could probably be reduced to about 10 cycles like you suggest. I got into the habit of using low resistances when my first power supply (3Kw 4Kv) would not start under full load with a 25 ohm resistor, switching to 4 ohms solved the problem and had no adverse effects. I forgot to mention if starting under load you need to get about 80% normal load voltage before shorting the resistor so a resistor 1/5 of the supply impedance or less works well, most 3 phase motors have a locked rotor impedance of 1/7th of the full load impedance and this gets a bit hard on the supply with larger motors but they tend to have longer start times than power supplies.
 
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J.R.Freeman

Disabled Account
2008-08-11 1:15 pm
Hi all,

Thank you to everyone for their contributions. It seems there are lots of solutions out there. I'm trying to come up with a simplistic method that could be applied generally.

Concerning the use of NTC thermistors as a one-component solution: my biggest concern is the temperature of the thermistor during steady-state operation. Is this a valid concern? Should the device be thermally insulated?

Lately I've been thinking of a method that has a small part count. What about a small NTC thermistor in series with the coil of a '120 volt' relay - one that uses mains voltages to operate the coil. This would create a time delay on the relay closing after mains voltage is applied. Once closed, the relay could be looped back on itself so that it pulls its own coil up to main levels thus latching it. During the delay time, an alternate path could be provided - this would be the soft start path - for the current. The alternate path could be another NTC thermistor, or power resistor, or even an incandescent lamp was a thought that came to mind.

As an aside, an incandescent lamp might make a good power resistor, as it is already setup to handle heat dissipation.

Jim
 

compex

Member
2010-09-18 11:35 pm
Consider these, when i have to "plug in" a big toroidal trafo i use 18–20 W ballast for fluorescent lamps in series with trafo. Balast will limit inrush current to 0,37A, time delay relay outputs (NO - or normally open side) are connected parallel with balast so when time "runs out" (3-5 sec depending on how big trafo is) it will short the balast. ;)

:)
 
I've been thinking of a method that has a small part count
The method suggested in my first post only uses 2 parts a relay and a resistor and it uses the filter capacitors albeit indirectly to provide the delay Unlike an NTC it will not hard start after brief power outage. No need for anything extravagant a 50W heavy duty wire wound resistor had no problem charging 500J of filter capacitors.
 

ernsttt

Member
2009-07-04 4:55 pm
use a relay with a coil voltage equal to any convenient winding on the transformer usually the primary. wire the coil straight across that winding and use the contacts to short out a heavy duty resistor in series with the primary

so how does this work?? is it that when the coil is sucking huge power at startup, the relay does not activate, sending the current through a resistor, then when everything settles down the relay gets current, activates and removes the resistor from the circuit. ?