resistance fail

I feel like I am starting to get a better grasp on general electronics, but I still have no quite wrapped my mind around some of the V=IR level basics.

I understand giving something too much power and blowing it up. I understand what it means to short circuit. However with the analogies I have floating around my brain, I do not quite get how this all comes together.

Start with an example of a laptop power brick. Usually what happens for power-range failure is not that the laptop gets blown up by too big a power adaptor, but rather that the too small power adaptor blows up trying to keep pace with the bigger laptop. I guess that makes me feel like it is power consumers that hurt power suppliers. However we all know that jamming too much power blows things up too. Anyone have a better explanation to help me get my brain around?

Considering how resistors and other power consumers need more and more power as they get bigger, why is it that short circuits even happen? How can I figure out how much resistance I need to put on the line so that GPIO pins or whatever else does not get blown up?

Thanks for any help!

a single UNO-pin can handle max 40mA without getting hurt.
If u use 5 volts only…then, as u mentioned yourself, R=V/I => 5/0,04 = 125 Ohm
Lesson. Keep all resistors >=220 Ohm and U’ll be safe
(this is the truth, but not the whole truth)

Damage is caused by heat. If you exceed the wattage (volts x amps) then it overheats and melts/dies.

Not just heat, Damage is also caused by excessive voltage. Exceed the breakdown voltage of the semiconductor and it dies.

fungus:
Damage is caused by heat. If you exceed the wattage (volts x amps) then it overheats and melts/dies.

Further information is that as you increase the heat (Power) in a semi-conductor, the resistance increases, which increases the power dissipated by the semi-conductor, which increases the heat... and so on. It can cascade fail. Then it melts and shorts things out or burns them open usually. Also, over voltage will arc through the insides of the part and cause damage inside. This is why you worry about your power dissipation.

Also be careful with the "40mA" per pin. One pin can handle that much current, but if you feed that much into all of them, the whole thing overheats. I think the chip can handle 100mA total (this is off the top of my head, but that's a guess). I'd recommend 5-10mA per pin max to be safe, or less if you can get away with it. Personally I only attach up to maybe 3-4 LEDs at 10mA max each to the IC directly, and if I need more, I just use something like the ULN2003A.

mirith:
Some semiconductors have negative temperature coefficients and some have positive. They do not all react the same.

KeithRB:
Exceed the breakdown voltage of the semiconductor and it dies.

Isn't that just excessive localized heat?

For avalanche breakdown, yes, but you also have gate rupture. Also, if you drive an emitter base junction to breakdown it will lower the Hfe, I am not sure if that is thermal or not.

First of all, it is good to redefine the equation V=IR as I=V/R....if you apply a voltage across a load(a resistor), it draws a current I depending on the voltage V. This means that the current depends on the resistance, also often called the load, and the voltage applied across this load. Now in the case of a laptop adapter, the voltage of the adapter is constrained within the acceptable limits....now as the laptop works it will draw a current from the adapter...but unlike a resistor, it is going to draw a varying current because the system is going to need more current for processing bigger things and lesser current for processing lighter things. Now, the adapter has a power rating. This is the maximum power that the adapter will be able to supply. Power=V*I. This means that when the output voltage of adapter is V, the maximum current it will be able to provide is I. Attempting to draw a larger current i.e a large laptop blows up the adapter....not the laptop.......the adapter or any voltage source can supply a specified amount of power....the load draws the necessary power.....trying to draw much more power kills the voltage source.
I hope this helps.

I want to have a minor objection to a statement you made

I understand giving something too much power and blowing it up

Power = Current * Voltage and it is a measure of "work".

I wish to clarify that providing too much power will not make something fail... specifically.

You can provide too much voltage with a power supply and get bad things to happen if the load can't regulate that voltage . You can provide too small a voltage with a power supply and very little will happen. You can provide too little current with a power supply and likely burn up the power supply... but you cannot provide too much current. That is why I object to your statement about giving too much power to a load. A load will only draw as much current as it needs.

A short circuit is a form of failure where the power source is asked to supply an unrestricted amount of current between the two supply connections, namely Power and GND. As if a wire or component of high conductivity is place between the connections. The unrestricted request for current will usually exceed the capacity of many parts, causing some rather catastrophic failures of at the very least... a fuse.

KeithRB:
mirith:
Some semiconductors have negative temperature coefficients and some have positive. They do not all react the same.

Usually intrinsic semiconductors have a large negative coefficient as the number of charge
carriers rises steeply with temperature - this is how thermistors work.

Metals and heavily doped semiconductors have a more-or-less constant number of charge
carriers, in which case the temp. co. is +ve as temperature increases the rate of interaction
between charge carriers and crystal lattice (which shortens mean-free-path).

A well designed power supply never blows up, it shuts down on overload.

Considering how resistors and other power consumers need more and more power as they get bigger, why is it that short circuits even happen? How can I figure out how much resistance I need to put on the line so that GPIO pins or whatever else does not get blown up?

Short circuits happen when a circuit is shorted out - the resistance is just the wiring which
is probably a few tens of milli-ohms. For, say, a 12V supply and 20 milliohms of wiring the
current that can flow (given a powerful enough supply) is 600A, enough to vaporize the
wires rapidly. A car battery is perfectly able to do this which is why you must not wear
metal wrist watch or jewelry working on them, as an accidental short could melt it and
burn your wrist to the bone.

Fuses are used that are designed to melt/vaporize faster than the wiring, thus saving the
wiring from damage.

Maybe "semiconductors" was a bit generalized. I was thinking more of bipolars vs DMOS Fet's where the collector current tends to rise with temperature - leading to hot spots, vs the drain current going down with temperature which promotes better sharing in large devices.

A long time ago I did this chart… it might shed some light on what is being discussed.

pwillard:
I want to have a minor objection to a statement you made

I understand giving something too much power and blowing it up

Power = Current * Voltage and it is a measure of “work”.

I wish to clarify that providing too much power will not make something fail… specifically.

You can provide too much voltage with a power supply and get bad things to happen if the load can’t regulate that voltage . You can provide too small a voltage with a power supply and very little will happen. You can provide too little current with a power supply and likely burn up the power supply… but you cannot provide too much current. That is why I object to your statement about giving too much power to a load. A load will only draw as much current as it needs.

A short circuit is a form of failure where the power source is asked to supply an unrestricted amount of current between the two supply connections, namely Power and GND. As if a wire or component of high conductivity is place between the connections. The unrestricted request for current will usually exceed the capacity of many parts, causing some rather catastrophic failures of at the very least… a fuse.

Can’t agree more. On one of our Guitar forums I am getting similar misunderstands about Electric power. Many think that giving more power will put more pressure on a component. And I keep saying that a circuit requests power based on setting of the components. A short circuit asks lots of power and if the power supply can deliver most of the power will be turned into heat frying the short. If the power supply cannot deliver the power supply will either stop delivering at all (short circuit protection) or it will fry trying to deliver the power requirements.

nicoverduin:
Many think that giving more power will put more pressure on a component.

Have you pointed out to them that their houses are connected to megawatt power grids but they seem to survive?

fungus:

nicoverduin:
Many think that giving more power will put more pressure on a component.

Have you pointed out to them that their houses are connected to megawatt power grids but they seem to survive?

Nah... these ar Dutch musicians.... it's all magic :grin: