Now you are starting to scare me. What does your tutor think ?
With that quality/functionality ethic, perhaps you are in the wrong career branch. I'd suggest a switch from hardware design, to working for Microsoft, where your talents would be a better fit ?
-jg
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J
Jim Granville
No, *WE* do not know that, You presume that.
WRONG. A little knowledge is a dangerous thing.
Google surge and resistor, and you might learn something. Failure modes are anything but simple.
-jg
H
Hans-Bernhard Bröker
There's no nice way to put this: you're delusional.
Please help spare readers prolonged therapy against post-traumatic shock by immediately and publicly declaring you're *never* going to work on cars. Ever.
H
Hans-Bernhard Bröker
Then why do you insist on ignoring every single design rule and hard-learned experienced advice that comes your way?
Do you honestly believe parts makers specify all those limits in their data sheets just to get on your nerves?
You aren't? Hmm, then why are you trying so hard to achieve the impression that that's exactly what you're doing?
In your dreams that may be so. In reality it's not.
... within a specified tolerance, and assuming you're not too far from the specified temperature.
Only as long as you stay within specified limits of all other parameters.
No. You think so. We know better.
No, it's not. Other failure modes exist, even for resistors. E.g. some may shred themselves by electromagnetic forces if you apply an allowable voltage, but at very high frequency.
Only as long as the pulse isn't short enough that it's HF components trigger the above-mentioned mechanical failure.
L
linnix
That's not enough. The OP needs to be barred from any mission or life critical devices. Would the OP please tell us what country you are from? We have to avoid any electronic device from there.
C
CBFalconer
... snip ...
A LED is not a common resistor. Doesn't work that way.
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]:
Try the download section.
** Posted from http://www.teranews.com **
M
mng
Dude, you've obviously never read an LED datasheet. I don't know how you can argue so confidently.
You should design things conservatively when starting out, for a variety of reasons.
D
David Brown
As others have said, it's not nearly that simple. Resistors are not necessarily "just a plain old lump of material", and excessive current can break them in many ways other than just averaged global heating. Local heating will cause damage quickly for high current pulses. Passing high currents means using high voltages, which can cause breakdowns of the insulation and other materials in the resistor.
A resistor spec'ed to 100 W will probably handle a bit more than that without problem, especially for short durations. But it will *not* handle pulses of much higher than 223 mA, even if the average power is under 100W.
High power resistors are specified for different power ratings under different usage, such as different frequencies of signals and different cooling arrangements.
You do understand, incidentally, that if you use a 50% duty cycle and twice the current, then you double the power dissipated by the resistor?
No, it's not acceptable in a product - at least, not a product that needs to be reliable over time. There are certainly plenty of products for which reliability is not a priority, and failure means you just through the thing out and buy a new one. But until you understand how to make *good* designs, you should not be trying to "cheat" - you have no concept of when it is appropriate, and how it can be done safely.
Write on the blackboard 100 times:
"I must not exceed the specifications given in the datasheets".
That's a precondition for an acceptable product design.
G
Guy Macon
You would need to test enough units to get a good statistical sample and you would need to test them at the extreme ranges of temperature, voltage, etc., and even *then* all that you would know is that this run of ICs, LEDs, etc. works in this circuit.
In other words, you are going down the wrong path. What you are doing is called kludging / kludging, while what you should be doing is called engineering / designing.
I give you my personal promise that if you read the following four webpages, you will end up thanking me. Note: the first three URLs make the point by hitting you over the head with a hammer, but the last one in the list is subtle and requires thought, but if you "get it," it will be the most valuable of all.
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Guy Macon Guy Macon
Guy Macon Guy Macon
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Guy Macon
I saw a situation like that a couple of weeks ago. A product that had been in production for many years had a zener diode that was getting a lot less current than the spec sheet called for, and the knee was soft enough to impact performance of the circuit. When I pointed this out to the original designer, he told me that when it was designed all zeners from all manufacturers worked great at the lower current, but that the zeners they are making now don't. As he correctly pointed out, depending on testing instead meeting the dayasheet requirements was a mistake. He had put in ECOs fixing all the places where the mistake was made, but there were still a few boards floating around that were old enough to not have gotten the fix but young enough to have the newer zeners.
Guy Macon
G
Guy Macon
Really? *That* simple? So a 1W carbon composition resistor, a
1W thin-film resistor and a 1W wire-wound resistor will all have the exact same ability to survive short-term over-current?
If the resistor datasheet specifies that it can take that overload for that amount of time, you are fine. If it doesn't, you are assuming that all future shipments of that resistor will meet a specification that is not guaranteed.
You are dead wrong if the resistor datasheet does not specify that it can take that overload for that amount of time. Not specifying that parameter means that the manufacturer can change his process or even his basic resistor technology in such a way that the behavior under overload changes wildly.
I should add that what you are making can also be important. At Mattel, we often ignored the datasheet specs if doing so would save a fraction of a cent. At a production quantity of
100,000 units per hour and a 2% failure rate in the field being acceptable, it was often worth the risk. Someone making a manned aircraft would be fired for doing something like that.
Guy Macon Guy Macon
Guy Macon Guy Macon
Guy Macon Guy Macon
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G
Guy Macon
Experience is a harsh teacher, but some people refuse to learn from any other.
Guy Macon
A
avitzur
About the last link, thanks to a three month delay in shipping the original PPC hardware, Graphing Calculator 1.0 had the luxury of four months of QA, during which we added no features and did an exhaustive code review. Being the only substantial PowerPC native application, everyone with prototype hardware used it, resulting in more thorough QA than any product I had worked on before or since. With no management or marketing pressure on features, we could focus solely on usability and stability.
As a result, for the ten years it shipped, Apple support told customers experiencing unexplained system problems to run the Graphing Calculator Demo overnight, and if it crashed, they classified that as a *hardware* failure. I like to think of that as the theoretical limit of software robustness.
J
Jim Granville
Did they also change suppliers ? If pushing specs, we often lock down the supplier, and also choose one like NXP.
Of course suppliers are not perfect, I recall a design done many years ago that failed to margin. Testing revealed the Philips devices did not meed their specs; Philips response was a shrug, and a change of the data sheet!
We use Zeners in 'tight' places, and tested 5% and 2% ones to see if 'same reel' had good matching. (often happens). What we saw was a distinct hollow in the 5% stats curve, where all the 'good' ones had gone into the 2% bin!
The 2% ones had a higher MOQ and longer led time (more of an issue than the price dlta ) but they got specified.
-jg
R
Robert Adsett
Another cautionary tale of overlooking data sheet parameters. I saw a design that had an RC delay between a pair of schmidt trigger inverters. It had been in production for some time when it started failing. The inverters had been substituted with another manufacturer's part (it was on the approved list). That part turned out to have a lower drive current than the original part. Both parts met the same specifications but the first had exceeded the specifications more in practice. Needless to say the RC was redesigned to work within the parts actual specifications.
Robert
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R
rickman
Actually, if you are working with amplifiers, the power dissipated in the amp is not as simple as the resistor because the voltage and current are not the same function. In particular, the power dissipated in an amp gets very hard to calculate if the output clips. In my current design I needed to calculate the instantaneous power as well because the frequency of the signal was fairly low, in theory it could be as low as 20 Hz in a software failure mode.
The simple resistor assumption the OP is using does not hold very often. As others have pointed out, pulsed current is the same as steady state current only under the conditions where it is the same. Yes, I know that is circular. I don't know off the top of my head all of the conditions where the two are not the same. But that does not mean they are always the same.
That is why I asked the OP to think about this himself. If he really wants to *learn* something, he has to figure it out for himself. Rather than continuing to stare at a problem and only seeing one side of it, he needs to get up and walk around it and *learn* to see it from other perspectives. I think even an undergraduate student should be able to figure out some differences between the effects of pulsed current and steady state current instead of making an ***assumption*** that they are the same in his "lump of material" model. Heck, he has been given many indications of how pulsed current affects material differently from steady state current. Instead of trying to learn, he just argues.
R
rickman
l
he
Ok, so you are *NOT* putting 0 volts on the base. You are pulling the base low with an MCU pin. This is totally different. The MCU is using a MOSFET to pull the base low and the MOSFET has limited drive capability. The B-E junction is very voltage limited. So you have maybe 0.8 volts across the B-E junction and 4.1 volts across the I/O pin with a relatively high current through it. This means it is dissipating a *lot* of power relative to what it is designed for. Does the MCU maker give a max current rating on the I/O pin? I am sure it is much lower than the current through your B-E junction and the I/O pin. Whether multiplexing will proportionally increase the max current rating on the MCU pin is doubtful. If you try to get the maker to tell you it is ok, you will find they won't do it. There are too many things that can go wrong. Since you don't even have an idea of what the current is, there is no way to say it is ok.
At this point, I am not as worried about the LEDs as I am the MCU. However, the same thing that is happening at the I/O pin is happening with your transistor. It is potentially dissipating more power than it is designed for. Unless you know the current, you don't know the power in it. Also, the current will vary greatly as the power supply voltage changes and the parts change with process, and let's not forget temperature. Those are the big three variables in semiconductor design, voltage, temperature and process. Your design needs to accommodate all of them separately and together.
You didn't measure the voltage on the base, because I *KNOW* it is not
0 volts if the emitter is 5 volts. I asked for these voltages in order to know more about the circuit and to show you how badly you are treating the parts.
I am pretty sure that even a 9 volt battery can blow out a transistor or LED. The problem I am having is that you don't seem to understand that you are operating all of the parts outside of their spec so that you don't know the current and voltages on the parts. So you don't have *any* idea of how far you are outside the specs. You have to measure something. You can't just assume all the parts are working the same as normal.
Yes, this is what I wanted you to think about. Now make some measurements and *find out* where the 4.3 volts is being dropped. This is the sort of learning that will stick with you forever.
Yup, you are starting to catch on. You still need to tell me *where* the excess voltage is being dropped.
A lot of diodes are rated for pulsed operation. They even publish curves showing how the max power (or current) varies with the duty cycle. The instantaneous power is the same as the steady state rating at wide pulse widths and rises with lower duty cycles. Obviously it can't rise forever until it becomes an impulse with infinite current and zero width. Instead it levels off at some point that depends on the internal construction of the diode. Dig around for diodes that aren't LEDs and find one with this curve.
Ok, give this a try. Connect your B-E junction across a current limited supply and tell me how much current flows at 1 volt output. You won't be able to because the junction won't support 1 volt at any reasonable current. The supply will become "non-ideal" before that point. My point is that the transistor will be very non-ideal before your LM7805 does.
No, you have already said that the MCU is driving the base. You are not putting 5 volts across the B-E junction. Got that?
Sure, this can work, sort of, because the current can be kept well within ratings of the I/O pin. But the problem you will have (even if you don't see it on the test bench) is that the slowly rising and falling edge of the reset signal can be seen as a set of pulses, potentially some with *very* narrow width. Narrow pulses on the reset pin can disrupt the device. You *will* find a minimum pulse width on the reset in the data sheet. If you violate this spec, you can introduce errors due to metastability. If you don't know what metastability is, you need to take an afternoon and read up on it. There are any number of sources on the Internet and you should read
*many* sources because I have never seen any that give you the full picture.
o
How does a capacitance limit the current? A capacitance *supplies* current and limits (or slows actually) voltage changes. Inductance slows current changes and the inductance of your I/O pins won't have an effect on anything wider than 10's of ns (actually it is closer to
1's of ns). We are talking about the LED and transistor having problems well above that time frame.
Yes, the LED *will* take more current when multiplexed. Most are even rated for multiplexing (possibly not with Nx current) and your 16:1 ratio may not be outside the spec. But you don't know the current in your circuit and you don't know the voltages either. So how can you tell if the LED and transistor are being operated within the multiplexed spec?
If you really want to tell if something works by testing, you need to use very high margins. If the part fails with current 10x the rating, I would not multiplex it at all. Find the multiplexed current where it does fail (Nx current, 1/N duty cycle, between 1 and 10 kHz rate). I would then be confident using it at a pulsed current 1/10th of this level, maybe even 1/5th. But I would not use it at half the failure level and I would not use a pulse width more than 1 mS. This is all in lieu of actually finding a proper spec for the parts.
Rick
L
linnix
I already fried some AVRs like this, driving three colors LEDs without resistors. They shine very bright, but last very short (days).
And 5 volt can blow out a uC.
N
Nils
Jim Granville schrieb:
A little story about that:
I've fixed my guitar amp a year ago. There was a huge chasis mount 10W
10K resistor running at a constant voltage of roughly 370V.
My first thought was just a big WTF? Why would someone puts such a overrated resistor into a commercial product. The resistor was expensive and I bet the labour work for mouting the resistor and hand-soldering the cables wasn't cheap either.
It turned out that this very resistor burned out. More than 10 times overrated but still it turned into a piece of ash. I traced it back to some arcing at the tube-sockets. That - together with the inductivity of the power-transformer - caused spikes that burned away the resistive wire inside the heatsink.
I know that this is a completely different world for those who deal with
3.3 or 5V supplies. Arcing is more common in these devices than you may think. It happeds often if you cut-off the power supply or do other nasty things. Overall it tought me a lesson:
Your job is not done if you put voltage, resistance and current into the equation and find out that you don't exceed the power-rating in the
*usual* case. You have to take the corner-cases into account as well.
In my tube-amp I measured thousands of voltages comming out from the transformers in the switch-off moments. I did some math and research (inductance is crazy!), found out that this had to be expected in the case and then understood why the original designer used a 10W resistor at that place.
Crazy stuff, these tubes. So simple and elegant on the one side, but so difficult to make a reliable product on the other.
Nils
Btw, comment to the OT: Every minute you torture an innocent tranny, microcontroller output-pin or LED god kills a kitten :-)
T
Tomás Ó hÉilidhe
Thanks for that, I appreciate your help. It's late here now and I've an exam in the tomorrow so I'll read over it again tomorrow.
And by the way I'm taking everyone's suggestions into account regarding spec limits and so forth. At the moment I've got my LED multiplex circuit made up on a patchboard; I'm going to order in some of the 3-pin LED's and see how bright they light with 20 mA running thru them multiplexed. If they're good enough then I won't bother messing around with higher currents.
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