Need to discipate 1A. - Materials Question

[CircularLogic]

This is pretty simple actually. I have a PLC that has a "light out" (LO) circuit that checks the filament of an incandescent bulb by intermittently pulsing the output. If there is no current draw, it sets a LO bit in the program.

I simulate the LO condition with an LED and a switch. I use a 15W 8ohm resistor to draw the required current.

My question is this...is there a discrete component that I can use to draw an amp (1A) that is smaller than a 15w resistor and/or produces less heat?

Any feedback is greatly appreciated.

 

[IRstuff]

Power = voltage * current = current^2 * resistance = voltage^2 / resistance

TTFN

FAQ731-376

 

[itsmoked]

What IR is correctly stating is that if you want to draw an amp thru something with a voltage across it you will have power dissipation.

The only other suggestions would be make the load something different.

A motor with a fan? The fan will help disperse the heat.

An LED - then some of the power will leave as light.

A buzzer - then a wee bit will leave as noise.

A battery charger - as then more than half will go into storage, at least until the battery is charged.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[IRstuff]

The next obvious question is why are you pulsing the load so long that it actually gets hot? Or, why do you need to pulse it with so much current?

If the load is present, 10 mA would be enough to tell that it's still there. And you don't really need to pulse it much longer than a millisecond.

TTFN

FAQ731-376

 

[OperaHouse]

I imagine a LED is used in place of a lamp. If just pulsed in a momentary test, a much smaller power resistor could be used. A lamp test shouldn't require more than 0.1A, who uses 20W panel lamp. Not that familiar with PLC's, but this request doesn't add up.

 

[CircularLogic]

The LED circuit is part of a simulator that tests the internal programming of the PLC. Under normal conditions the lamp could be lit constantly, but if the PLC detects anything less than an amp +/- 10% it will flag the lamp as defective. The "pulsing" that is does is only done when the lamp is not activated as a "cold filament check" and is a very low current.

My issue is that my simulator is heavy as hell and has a piece of aluminum that is covered with about 50 of these 15W resistors. In an effort to reduce the overall weight, I was trying to figure a way to reduce the physical size of the resistors while maintaining my 1A current draw. I know I am limited by physics here, but was hoping for some insight into some different components/materials that could perform the same task.

Thanks for all your help!


————————————————————————————————
James W. Hudgins, III
Southwest Signal Engineering Co.

http://www.swsignal.com

 

[IRstuff]

Had you posted THAT as your OP, better answers would have been possible:

Bigger fins and fan. Component size is about heat removal.

TTFN

FAQ731-376

 

[itsmoked]

If these are pulse tests they are very short periods. Why aren't you using very small resistors. Keep the average power below the resistor's power rating.

You could shift to some terminal strips with resistors between screws.

Other info desired would be frequency of the "cold tests".
Duration of each test.
Current peak of each test.

This could all be found by using a scope across one of the existing resistors. (careful of your grounds!)



Keith Cress
kcress -

http://www.flaminsystems.com

 

[CircularLogic]

So the REAL answer to the question is NO. There doesn't appear to be any device that is smaller and produces less heat than what I already have. I will post any new information that I may find elsewhere.

Thanks.

 

[IRstuff]

Again, if you want to use smaller devices, you could, but you'd need to provide adequate heat removal.

Alternately, since this is an emulator, there's no reason to actually run an amp through the resistors; you just have to fudge the PLC output to drive a lower current, and the feedback to the PLC to report the current current.

TTFN

FAQ731-376

 

[BrianG]

CircularLogic:
Some areas where you have not provided data:

1) Is the normal output voltage for the lamps 8V (hence 8 ohm resistor) or is that a test voltage?
2) Is the output current limited at 1A /does the PLC check for a short circuit?

Depending on the exact test conditions you may be able to reduce the real load resistors to small electronic loads.

 

[IRstuff]

As emulator, it only needs to LOOK REAL, not be real, to the PLC.

If you really wanted, you could have a PC fake the inputs to your PLC.

TTFN

FAQ731-376

 

[OperaHouse]

How long is the test? I have taken 2 & 5 watt resistors and submitted them to hig currents for short periods of time Suspend them in air with a couple light computer fans and weight could be reduced. Everyone thinks aluminum is light, but it has the same specific gravity as granite!

 

[itsmoked]

This is a "field simulator". It may need to run for hours(?)

He's running the resistors as the lamps. That means if a lamp happens to be ON the resistor will be getting a steady state ON and so needs to dump the X watts for a while. It's not just the brief "cold test". The "cold test" is just the part of the requirement that forces the actual current demand of the actual lamp loads. Otherwise he could use some 10mA LED stand-ins to just show the lamp activity.

There are no easy work-arounds that I can see. He was just hoping to cheat thermodynamics somehow. IR's use a fan/s and crowd the resistors together is about the only way to shrink this mess. I think the best solution for this test stand is to just mount it on a 19" vertical, single bay, rack mount and roll it around not carry it around. Then its weight wouldn't be an issue.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[CircularLogic]

I will post some pics shortly.

————————————————————————————————
James W. Hudgins, III
Union Switch & Signal

 

[biff44]

Well, you could make this very small indeed. If I were going to do it, I would pick a Sn-Pb plated heat sink, and directly mount ALN chip resistors to it. We are talking pretty small chips here--0805 size to dissipate 25 watts each at 100 deg C baseplate. So the issue is to size a heat sink to dissipate 50 * 8 watts = 400 watts, and remain under say 80 deg C. A fairly easy project.

 

[biff44]

If you use an avid heatsink 477000L00000, and fifty ALN chip resistors, you could have a load box that is 12 x 5 x 4 inches. It will work at a worst case surface temperature of 88 deg C WITHOUT a fan. With a fan it would be much lower.

 

[itsmoked]

biff44; Those are pretty cool.. er Hot!

But tell me what is the typical way you would mount these to a heatsink?

How would you connect to them??

Keith Cress
kcress -

http://www.flaminsystems.com

 

[biff44]

I would probably solder the resistors directly to a 0.1 inch thick copper plate. I would then bolt the plate to the heat sink, and bolt the distribution PCB board to the heat sink too. Would hand solder ribbon from the pcb to the resistor pads.

 

[IRstuff]

All great ideas, but this thing is only required to exercise the logic of the PLC, duplication of the physical behavior of the loads really isn't necessary.

The PLC can only receive voltages representing the behavior of the loads; it doesn't know where they come from or how they're generated. All you need to do is to program another PLC to generate the appropriate voltages at the appropriate times. This has the added advantage of being programmable for other scenarios or changes in the system.

TTFN

FAQ731-376