Power Output vs. Power Dissipation of ULN2803A

The dissipation of the device* is the current controlled by each Darlington pair times the voltage drop through the Darlington, which may be a few tenths of a volt. The load power is this same current times the voltage drop across the load which, for your application is 24V but could be as high as the maximum voltage this device is capable of controlling (I don't have the data sheet, so I don't know hat that is).

• plus a little bit of power that it consumes internally.

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Paul Hovnanian     mailto:Paul@Hovnanian.com
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You are seeing a bit of specmanship, here. Each of 8 outputs is individually rated for 500 mA, continuous or 600 mA peak. With a full

50 volt supply, and all 8 outputs passing 600 mA momentarily, the peak power being controlled is 8*0.6*50=240 watts. The power dumped into the chip is the load current times the on state voltage.

The on state voltage is not specified any higher than 350 mA, but even that low, that saturation drop can be as high as 1.6 volts. So at that current times 8 outputs, the chip power would be 8*0.35*1.6=4.61 watts. So it is pretty obvious that no where near 24 0watts can be continuously powered through this chip.

It looks like your steady state load current is about 5.4W/24V=0.23A. It is safe to assume that since this is lower than the 0.35 amps that the 1.6 volt saturation voltage is specified, you can safely assume that the chip dissipation is less than 0.23A*1.6V=.37 watts. This is well below the 1 watt power rating for a single output in a 25 C ambient. In fact, you might get away with using 6 of the 8 outputs this way, before you exceed the total power rating of 2.25 watts.

Of course, if you need only one driver, why not use a single darlington or mosfet, instead.

The switches are rated to 95V.

Obviously they won't control the same power in a 24V system.

RL

I have looked at the TI data sheet, which states 500mA(single output) and

50V. Nowhere anything about 260W. There are specs for combined current, and combined dissipation, which you do not seem to have understood.

Because the package can not dissipate more, and even then only with extra cooling measures.

You seem to argue like a digital designer, who has never build anything. You should learn how to read the data sheet. You will find everything inside. But even without reading it, common sense will tell you that it is not possible to simply multiply and add the maximal values for a single channel.

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ciao Ban
Bordighera, Italy

Thanks to everyone for all the help!

I guess I should have been more specific about some things:

The 260W power was at 95V, and was not for the 2803A, but rather another version. The 260W power figure was given by an Allegro data sheet for a now-discontinued version of the device.

I need to control 80 valves, and will be using a 96-channel PCI digital I/O card, so I will need several devices, not just one channel.

To my perceptive friend in Italy: You are correct, I do argue like a digital designer who has never built anything. I am a mechanical engineer by trade, and although I've used lots of off-the-shelf industrial control equipment, I've never tried to build any.

Thanks to all for the help.

greenman_11 snipped-for-privacy@yahoo.com wrote:

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If your solenoids are DC, then that's 225mA/coil. (If they are AC (!) you'll likely need something different to drive them-- check this point because wattage ratings are more typical with AC coils).

I'm not going to look at the data sheet for you, but I'd guess that you won't be able to drive more than one or two of these safely, allowing for reasonable ambient rise, chip package and so on. That chip is a darlington array so it has a *lot* of voltage drop, maybe over a volt typically. A logic-level MOSFET and 1N400x catch diode per channel whould allow you to drive this level of current very safely and they can be designed to run cold, because of the low voltage drop.

Best regards, Spehro Pefhany

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Good morning. Keeping it simple, the power dissipation in any given transistor, regular or darlington, is approximately the voltage across the transistor (Vce) * the current going through the transistor (Ic). (This ignores base current.) Device power dissipation doesn't have anything to do with load power dissipation -- two different things.

The darlington transistors in the ULN2803 will typically have a saturation voltage of about 1.0V to 1.2V. Each transistor that is on will then typically dissipate about 0.225 to 0.27 watts when on, with your 0.225 amp load. If you multiply that by the eight darlingtons in the package, that will give you 1.8 watts to 2.16 watts of power dissipation in the IC (which is different than load power). Given that the data sheets specs a maximum Vce(sat) of 1.6V, that _will_ exceed total maximum power dissipation.

As a practical matter, that's way too close to the absolute maximum power dissipation. In addition, if you consider variances in coil resistance, power supply (current through the coil will go up 10% if your power supply goes up 10%), and variations in Vce(sat) (which will probably be a little higher than 1.0V), it looks like a mistake to use this IC.

However, you may have an application that only requires one or a couple of these solenoids to be on at one time. In that case, the ULN2803 might be an ideal solution. Personally, I would worry about going much over 1.0 watt (4 of your loads) on these ICs. (N.B.: By the way, make sure to connect the diode pin of the ULN2803 to your +24V supply, or the diode protection won't work and your chips will smoke).

So, you have several potential solutions here:

• If you can guarantee that you're not going to be driving all of the solenoids at once, you might be OK with your setup. In fact, if you can guarantee you're not going to be lighting up more than four at one time, it's a great solution.

• If you have to do all eight at once, you might want to consider doubling up on your ULN2803s, and only using half the outputs on each IC. That will guarantee you will not go much above 1 watt power dissipation for each IC.

• You might want to take a tip from the programmable logic controller people, and use your ULN2803s to drive eight small relays capable of handling the loads. If you do this, be sure to place 1N4002 diodes across the load to avoid arcing at the relay contacts when they open.

• Instead of a "one IC solution", you might want to consider using discrete darlington transistors to do the job. Assuming you have a TTL signal, you might want to try something like this (view in fixed font or M$ Notepad):

' VCC VCC ' + + ' | | ' 1N4002| C| ' - C| ' ^ C| ' | | Solenoid ' | | ' | | ' | | ' '-----o ' | ' | ' ___ |/ 'TTL o-|___|- -|TIP120 or equiv. ' 10K | |> (NPN Darl.) ' .-. | ' 10K| | | ' | | === ' '-' GND ' | ' | ' === ' GND created by Andy´s ASCII-Circuit v1.24.140803 Beta

This is something like what you'll see with a programmable controller with a "high current" DC solid State output module. If you pop one open, you'll see the discrete darlington transistors all in a row, just like this.

Unfortunately, there aren't any slick answers here. Your load is right at the point where you have to start thinking about alternative, higher power control solutions than the trusty, ever-popular ULN2803.

A lot of the digital control of "real world" stuff like solenoids, heaters, and other stuff isn't that complicated. Look around at the stuff that's out there, learn from it. Take the time to look at websites, schematics, and learn from what you see. And take the time to carefully read all data sheets. They're made to sell the product as well as accurately characterize it. Most IC manufacturers make a special effort to make their data sheets as clear as possible to a wide range of potential customers.

By the way, questions of this type usually are asked at sci.electronics.basics or sci.engr.control.

Good luck with your project. Chris

I see that. Actually, I've decided to use the ULN2803A to drive some small relays to control the solenoids. Thanks to all for the help.

Really? I have not done the calculations, but I find it hard to imagine that an inefficient Darlington switch that can switch in the microsecond time range would get that much more inefficient switching every 50 or 200 msec or whatever speed a solenoid valve could keep up with.

I think the DC power dissipation is going to dominate. Place 8 chips side-by-side with tinfoil on top. Apply 64 loads for a few minutes or until the oil begins to sizzle. Add the beaten egg with a bit of butter, chopped chives and cilantro. Turn lightly and serve.

Best regards, Spehro Pefhany

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Thanks for all the straightforward help, Chris. I've done exactly as you suggested; I've looked at a lot of data sheets and schematics, and found the best solution (after getting all the great advice here) is to use the 2803 to operate some small relays. Thanks for the tip on using the diode across the load.

Thanks again to all who've helped with this question.

gm

Very badly.

Graham

On 9 Apr 2005 06:36:42 -0700, greenman_11 snipped-for-privacy@yahoo.com wroth:

Also take into account the fact that chip dissipation will increase momentarily during the time that the valves are being turned on or off. In other words, If many rapid on/off cycles are anticipated, the dissipation could be much higher than the DC or steady-state predictions.

I suggest getting one chip and connecting eight solenoids to it. Exercise all the outputs at a speed somewhat greater than you expect in the final product. Adjust the supply voltage to a bit higher than you expect the normal supply voltage to be. Let the test chip run for ten minutes and if you can then hold a finger on top of it for longer than a few seconds, you'll be reasonably assured that the completed systen will be OK.

Jim "empirical designs-r-us" Meyer

But with 225mA*1.6V=360mW dissipation for each channel there is no way driving 8 solenoids at the same time from 1 package. Here is a better advice: You can parallel two drivers and have guaranteed some headroom left. It will be too difficult for a beginner to design drivers from scratch, so the choice of an array is a good one. There are also all protection circuits incorporated, nevertheless I would still solder some 1A schottky-diodes directly across the coil connections with the cathode(where the stripe is) to the 24V line. Fairchild sells the SB150, which would be appropriate for this application. Unfortunatly this means double the driver ICs, but now you do not need to think about cooling, if you leave enough space between the devices. I hope you have some helper with a bit of knowledge to look over your work before you order circuit boards. In case you need some support, drop me a mail.

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ciao Ban
Bordighera, Italy

On 9 Apr 2005 09:19:25 -0700, "Chris" wroth:

Quite correct. This news group, sci.electronics.design, is more properly reserved for posts concerning the death of fishes, lame jokes, and celebrity trials.

Jim