Duty cycle and current capacity relation

How is the duty cycle and current capacity generally related? linear, exponential, logarithmic? (approximate ofcourse)

Maybe if they specified an exponential approximation Ae^(-Bx) + C then only

3 parameters would need to be given(which is just one more) for a description of an arbitrary(within reason) of any duty cycle.

This assumes that a linear approximation is not very good.

Right now I'm trying to figure out if a linear interpolation is good enough.

1/10 duty has a current of 100mA and duty of 1(average) is 25mA.

This gives the linear equation -83.3x + 108.3 where x is the duty cycle.

So at 50% duty I would expect that I could have a current capacity of 66mA. This seems aweful high though? (for some reason I guess I'm expecting to double it from 25mA to 50mA since I cut the duty in half)

It also seems that there would then be an optimal duty cycle for a given device. If f(x) is the current capacity given duty cycle x then x*f(x) is the average current. If f(x) is approximately linear then this has a maximum.

For my example above it occurs at a duty cycle of .65 and gives a maximum current capacity of 54mA and average current of 35mA. An increase of 15mA(a

40% increase)!!!

Also, since at the maximum it is somewhat "flat" the duty doesn't have to be perfect. A variation of 10%, i.e. 55% to 75% duty, gives a variation of <

1mA in the average current capacity.

I expect though that the curve is not linear but exponential... can anyone verify this?

Reply to
Jon Slaughter
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Also, if the component is an LED, then one can bring in the brightness and maximize brightness vs duty cycle. (i.e., the would maximize efficiency)

Reply to
Jon Slaughter

"Jon Slaughter" schreef in bericht news:5BNel.11628$ snipped-for-privacy@flpi146.ffdc.sbc.com...

Mostly it is about heat. The average heat output at a 10% duty cycle is 10% of that which you'd get at a 100% duty cycle, and that's a linear relationship.

If you want to try for more current when the device is on, you have to work out how much the voltage drop across the device will rise when you raise the current, and that is a lot trickier.

The only sensible answer is that it depends on the device, and you need to experiment to find out what actually happens.

Buy quire a few devices before you start experimenting - they tend to blow up.

Imagining possible mathematical relationships isn't a particularly useful way of spending your time.

--
Bill Sloman, Nijmegen
Reply to
Bill Sloman

what the heck is that suppose to mean? There are only so many possible relationships... (just a handleful... like 3... It's not going to be some random function and we don't need a perfect model)

I say blowing up devices just to test them is much worse.

if the device dissipates the duty cycle less heat and it is linear and the heat dissipation is the main factor then my analysis is correct...

And hence an optimal duty cycle can be found for giving the most current capacity. That is pretty damn significant IMO. You may not care too much about mathematics but history has proven it to be much more useful than trial an error.

I suppose you didn't understand what I was doing because I definitely wasn't "imagining" possibilities....

Reply to
Jon Slaughter

You still haven't told us what kind of device you plan on misusing.

If the critical element was a bipolar transistor acting as a saturating switch, with just enough base drive to keep it in saturation with 100mA going through it, none of your "three possible relationships" would be useful in modelling the relationship between heat dissipation and peak current.

Unfortunately, one needs an educated imagination to create useful computer models, and experimenting is pretty much the only way of educating the imagination.

If you understand what you are doing, you won't blow them up.

Unlikely.

That is a thoroughly wrong-headed proposition. In fact, history proves that computer modelling is a very useful technique for making sense of experimental results. Unvalidated computer models are an invitation to disaster.

I understand all too well what you are trying to do. Too many of my junior engineers had tried to substitute guesswork for systematic measurement, and spent a couple of days devising what they think is a plausible computer model, and rather longer on doing an elaborate design that could never work before I got around to getting them to take some measurments.

-- Bill Sloman, Nijmegen

Reply to
bill.sloman

Fortunately I don't like your condescending attitude...

Reply to
Jon Slaughter

I wondered how long it would take before the pent up belligerence started seeping out. Stop arguing and experiment, that's how it's done. You can't change reality thru argument, but the reciprocal is entirely possible.

Reply to
Anthony Fremont

I knew you would reply... didn't think it was going to be so quick. You and Bill must have came from the same bad apple.

Anyways... I guess you will go back on the ignore list... That won't stop you from replying though.

Reply to
Jon Slaughter

It's Sunday, not much going on today. At any rate you're welcome for the other input I gave you about using the 24 channel LED driver.

I wish I knew as much as Bill. I don't agree with Bill politically, but that doesn't make him wrong when it comes to electronics.

Of course it won't; people can't be controlled thru a killfile. OTOH, you can certainly continue protecting your ignorance by using one.

Reply to
Anthony Fremont

I believe that in general the total thermal resistance would just be a sum of individual thermal resistances which would be constant excluding temperature. The capacity would be dependent on the power dissipation which is a direct consequence of the thermal resistances.

Hence the relation I am looking for should be linear(approximately at least). Hene the analysis I gave for optimality would hold and there is an optimal duty cycle for maximizing current capacity.

Reply to
Jon Slaughter

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Duty cycle should be based on the circuit requirements. Once you know the operation of the circuit, you size up the components accordingly. At that point, you let things like average current due to duty cycle come into play.

Basically, I think your approach is backwards.

While I mentioned the 1ms rule for pulsed electromigration, it occurred to me that diodes used in power line rectification run a lot longer than 1ms. Granted, the current flow is not a pulse but a piece of a cycle, but still that makes me think the 1ms rule is really guard banded.

Reply to
miso

x-no-archive:

the point you miss is that the heat produced in the device may not be (probably isn't) linearly related to the current through the device

Mark

Reply to
makolber

The point I missed? That was my question.. duh... maybe you should read the OP?

Now you say "probably isn't" but no proof? The thermal resistance is not constant with respect to temperature or linearly related although it is linear with small changes... see the temperature coefficients.

It seems you guys like to do a lot of guessing instead of relying on fundamental physical prinicples. You claim I'm "guessing" when in fact I was asking then turn around and guess that I'm wrong.

In fact it seems you are the one who is missing the point.

Reply to
Jon Slaughter

Depends on the current vs voltage relationship of the device, and which specified limit you hit first.

Parts have continuous and peak current limits, maximum junction temperatures, thermal impedances and time constants.

A pure resistance (like a fet) will exhibit a peak to average penalty that a fixed forward voltage drop device (like a thyristor or diode) will not. This is one reason why cuurrent ratings for fets are often a waste of print and why rds dominates their spec.

I believe you're dealing with leds. These respond to average current like most diodes, though some have more resistive tendencies than others. Thyristors, diodes and bipolar devices are typically supplied with pulsed duty nomograms, in their specification.

You neglect the first problem - actual required luminous intensity in the application. I understand that this intensity is maintained in leds at lower average current, when pulsed duty is employed. This is useful.

RL

Reply to
legg

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