Linear Power Supply Considerations

[UKRobotics]

I have designed and built a prototype linear powersupply for my cnc retrofit. It consists of a 240v - 110v 1kva transformer, a 35amp 400v Bridge rectifier, and 3 x and 4700uf 250v capacitors in parrallel. There is a 1.5ohm 2 watt bleeder resistor accross the caps and the final output from the supply is in the region of 140v DC

I'm going to have one of these for each axis so there will be four idential systems like this all neatly secured in one aluminium box when I make the final thing.

My question here is about inrush current and the best way to deal with it.

From speaking to various people there seems to be several schools of thought on this.

#1 - Do nothing, use big wires and 'D-type' breakers in your fuse box

#2 - Use a Negative Temperature Co-efficient (NTC) resistor to limit inrush current.

#3 - Use a standard resistor along with a relay to short it out after the initial surge period is over.

#4 Use a 2 sec time delay relay between each 1kva PSU

The first approach seems a little crude, and when powering up 4kva of torrodial transformer + caps at once seems a little risky.

Several people have suggested the second option as a good one to go for. Putting an NTC resistor in series with the + on the transformer input would limit inrush by providing a high resistance when cold, dropping to a very low resistance once it warmed up.

Potential disadvantages include the need to wait for it to cool after switching off before switching back on (and/or providing mechanism which prevents this condition from occouring) and the fact that power would be wasted through the resistor, though in my opinion I cant think that would be a huge ammount.

The third approach is fairly foolproof in theory, but the relay coil activation needs some thought. Do I initiate the shorting of the resistor by time delay relay, some sort of voltage/current sensing circuit or a different method alltogether.

A friend of mine suggested putting the relay coil accross the capacitor. The plan being that the resistor would cause the capacitor to fill up slowly, and by the time the voltage was high enough to trigger the relay coil, it would be the right time to short out the inrush resistor.

Would this theory work or is it flawed somewhere?

The forth option is a possibility but I think just one of these supplies alone might still be big enough to blow fuses if switched on at the wrong point in the mains cycle; which of course is impossible to predict.


I would really welcome any thoughts on this ..

Especially an idea of the best method to go for, and any shortcommings i've missed in my suggestions, along with any totally different alternatives i've missed out on.

[gar]

061122-2145n EST USA

UKRobotics:

First a 1.5 Ohm 2 W resistor across 140 V will explode, or at least burn up.

Your *1 approach makes no sense. If anything increase the resistance.

#2 will work and we have used this technique for many years.

But I favor your #3 and it is important that the resistor is in the primary. There are times, depends upon the transformer residual flux state at turn on, when the transformer, especially a toroid, will draw a large inrush current.

Operate your relay as a voltage threshold detector across the dc output. If you want tight voltage control then add electronics to the relay. Once this relay is energized you might use a seal contact to hold it in.

Why only 140 VDC from 110 VAC into a bridge rectifier with a capacitor input filter?

.

[Al_The_Man]

They must be some very hefty hi-torque motors to require a combined power rating of 4kva. Especially considering that in all probability this demand will not be seen due to all four motors seldom, if ever reaching maximum torque at the same time.
Depending on motor size and anticipated demand, it might be prudent to re-look at the power requirements, with a view to having a combined supply.
Al.

[kreutz]

You will get around 153 volt DC open circuit at the output with 240 volt AC at the input, and around 140 volt DC with 220 volt AC at the input. Depending on the load current and the winding resistances of the transformer your voltage will drop under load.

I suggest #3, resistor and relay contacts on the transformer's primary side. Switching the contactor on after a time delay of 200 mS to 250 mS. There are time delay contactors made for that purpose. Use only one contactor for the 3 power supplies.

1.5 ohm 2 Watt across the capacitors will burn immediately.

[UKRobotics]

Thanks for the comments guys.

Gar: I was very tired last night when posting , and unsuprisingly got a few numbers wrong. What I actually meant to say was the bleeder resistor is 1.5K 2.5w - does that sound more realistic?

I was told that the output will be higher when unloaded but would drop to the region of 140v under load.

Al: This is a rebuild of a Bridgeport Series II Interact 4 (The biggest and most powerful of the entire interact range). Each axis as a SEM Permenant Magnet 140v DC Servo Motor. Continious Stall current - 8.6amps, Current at peak torque - 50amps.

The original powersupply had a 3Kva transformer for the three axis machine. I want to add a 4th axis at a later date so It seemed sensible to add another 1kva to provide for that.

Kreutz: If I were to use a standard resistor + relay for inrush limiting, what values would you reccomend for the resistor. Also, would this item: http://www.tools-n-gizmos.com/store/...xm=on&ppinc=1a
be suitable for shorting out the resistor after the elapsed time ?

[audioandy1762]

Hi UKRobotics

Just had a look at the link, if you are to use it on the primary side 240v the unit on the link is for 120v. Just incase you had not noticed.

Andy

[UKRobotics]

I was going to put it on the secondary. Thx for the tip though.

I have just spent the last hour testing my power supply. Everything seems to work nicely except the bleed resistor which is still wrong (Assuming an orange glow with the smell of burnt toast is a bad thing ) I have re-looked at my calculations and worked out a more suitable rating.

Once I have the new resistor I'll do some load tests.

[kreutz]

What I actually meant to say was the bleeder resistor is 1.5K 2.5w - does that sound more realistic?

No, it will overheat too. You can use 15Kohm 3W across each capacitor.

Kreutz: If I were to use a standard resistor + relay for inrush limiting, what values would you recommend for the resistor.

10 Ohm 25W switched out (short-circuited by the contactor) at 300 to 500 mS.

You are near with that link. That particular one is not good, because of the 120 VAC rating and because of the delay time being too high. Look for a 3 phase rated contactor rated for approx. 3HP @ 240 Volt and a 240VAC rated time delay relay (adjustable from 0.1 to 3 sec), or use the contactor triggered by a homemade time delay circuit, in this case you will need an auxiliary power supply.

[gar]

061123-0837 EST USA

UKRobotics:

I assume a 240 to 110 transformer will be rated for 110 output at full rated load. Thus its no-load output voltage will be higher. Assume no-load output is 120. Then the peak output voltage is sq-root of 2 * 120, or 1.414 * 120, or 120/0.707 = 170 V. Subtract 2 diodes drops and you are still about 169 at no-load. If your input supply is high by 10%, then the open circuit voltage will be about 1.1 * 170 = 187 V.

Under full load at 240 V input your voltage may be in the range of 110*1.414 = 155 minus diode drops and other factors (such as ripple).

Your bleeder resistor design should be based on 200 V. At 1500 ohms this calculates to 40,000/1500 = 27 W. The time constant of 14,000 mfd and 1.5 K is 21 seconds or about 100 sec for a good discharge (about 1% of initial voltage). Instead I would suggest a relay or SCR controlled bleeder that only loads the supply at loss of AC line voltage. Then maybe use a 20 to 25 Ohm wire wound resistor of 25 W rating. At 200 V and 20 Ohms the inital current is 10 A. Short time (seconds) a wire wound resistor can easily handle 10 X overload. A 20 Ohm bleeder resistor will shorten the dischage time by a factor of 75 compared to 1500 Ohms. Using the same criteria this is about 100/75 = 1.3 seconds.

For AC line input current limiting consider limiting to a peak current of 20 Amps or 240/20 = 12 Ohms. 1 KVA / 240 = 4.2 Amps. Assuming failure of the time delayed shorting relay use P = 4.2 squared * 12 = 17.6 * 12 = 211 W. So use a 25 W resistor and assume the relay does not fail. Or peak limit at a higher current lower resistance.

Alternatively use the thermistor for surge limiting and short it with the time delayed relay. This will also eliminate the cool down time problem unless you cycle the supply on and off fast. This is probably the best approach.

At 4.2 A load a 12 Ohm resistor has a drop of about 50 V.

.

[kreutz]

I assume a 240 to 110 transformer will be rated for 110 output at full rated load. Thus its no-load output voltage will be higher. Assume no-load output is 120. Then the peak output voltage is sq-root of 2 * 120, or 1.414 * 120, or 120/0.707 = 170 V. Subtract 2 diodes drops and you are still about 169 at no-load. If your input supply is high by 10%, then the open circuit voltage will be about 1.1 * 170 = 187 V.

It all depends on the design of the transformer, 120 VAC no load / 110AC full load => Transformer regulation ~= 90% => effective winding resistance ~= 1 ohm (for 9 A full load current). I have seen better transformers in that power range. As an example to make a point it is a good example.

For AC line input current limiting consider limiting to a peak current of 20 Amps or 240/20 = 12 Ohms. 1 KVA / 240 = 4.2 Amps. Assuming failure of the time delayed shorting relay use P = 4.2 squared * 12 = 17.6 * 12 = 211 W. So use a 25 W resistor and assume the relay does not fail. Or peak limit at a higher current lower resistance.

If your time delayed relay fails you will notice immediately, by using a vitreous enameled resistor, it is going to fail after overheating, opening the circuit. No output voltage. I don't recommend other type of resistor for that function. You don't want to work under failure of the relay, so using the resistor as a fuse is going to assure that. Otherwise, you can use a thermal switch attached to each resistor, in series with the emergency-stop relay in order to shutdown power.

Use a power contactor designed for the rating of your load will be reliable for many years. Doing proper maintenance at least once a year will take care of normal wear on the contacts (a minimum wear since you are limiting surge current by means of the resistors)

In real life conditions anything could fail, of course, you want it to fail graciously, and not catastrophically.