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Very basic question here and I'm only looking for a generic over view, but is it a too high current or a too high voltage that will damage electronics? I assume it will depend on the component in question -

1) For example if you have a working circuit with a 10V battery, fixed 5 Ohms resistance and a current of 2A. If you then swap that battery to 20V, would it be the new current of 4A that does the damage or that fact that the voltage is now 20V? Both values are higher than they should be so which factor will do the damage, both?

2) Also, if when you swapped to a 20V battery, you also increased the resistance in the circuit to 10 Ohms, could this still damage components in the circuit? The current is now the same as it was in the original (2A), but the voltage has gone from 10V to 20V, could this do damage?

3) Finally, you could take the original circuit and lower the resistance to 2.5 Ohms, now the voltage is the same (10V) as it should be, but the current has risen to 4A, I assume this could cause damage depending on the components in the circuit?

Any help would be appreciated, thanks in advance.

Edit - I didn't make it very clear, I wasn't specifically talking about damaging the resistors, I mean't damage various components that could be in the circuit.

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    \$\begingroup\$ Heat, usually... \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 16 '15 at 1:00
  • \$\begingroup\$ Which is fancily called electrical overstress EOS. Now you know what to read/google. \$\endgroup\$ – Fizz Oct 16 '15 at 1:21
  • \$\begingroup\$ And if you want practical advice regarding how to select resistors: edn.com/design/components-and-packaging/4321070/… \$\endgroup\$ – Fizz Oct 16 '15 at 2:18
  • \$\begingroup\$ The different ratings (such as maximum voltage, maximum current, maximum power) are generally related to different failure mechanisms, so you need to keep all of them in mind in your design. For a similar question (specifically about connectors), see here. \$\endgroup\$ – The Photon Oct 16 '15 at 2:58
  • \$\begingroup\$ Resistors are hard to damage by voltage/current alone - if the stress is short enough (millisecons), you can apply hundreds of Volts (and tens of Amperes) on your 10 Ohms / 20 W resistor without any issues. If you keep the resistor under such stress, it will overheat and eventually burn out. \$\endgroup\$ – Dmitry Grigoryev Oct 16 '15 at 8:51
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A general answer is that electronic/electrical components are damaged when their electrical ratings are exceeded. Excessive current results in excessive heat which will destroy both passive and active components. Some passive components, such as capacitors have a max voltage rating, which if exceeded can result in failure of the dielectric (insulator) resulting in excessive current, and ultimately smoke. Generally, exceeding voltage ratings of passive compnents causes insulation failure. With active components, excessive voltage will cause a breakdown of the internal junctions of the diode, transistor, etc, which will also allow excessive current, heat and some smoke. However, in these cases the current will be quite a bit lower than when passive devices are overheated. My experience is that even a small spark on a transistors leads will destroy the component. The over-voltage condition breaks down the semi-conductor junction and it does not heal. The part is now just a lump.

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    \$\begingroup\$ Thanks @Wes, this is what I was looking for. To confirm, in a simple working circuit with various components, and I doubled the voltage, but also doubled the resistance so the current remained the same, this can cause damage? ..and vice versa, if I halved the resistance but kept the source voltage the same, this will double the current and can also cause damage. If I doubled the voltage but didn't increase the resistance, then both the voltage and current will be too high and will potentially do double the damage (and the power will also double)? In simple terms, something along those lines? \$\endgroup\$ – RJSmith92 Oct 16 '15 at 3:49
  • \$\begingroup\$ Also, when you say some components have a max voltage rating, does this mean that regardless of current, if you go above this value it will still potentially damage it? Even with a minuscule current in the circuit, a voltage higher than the components is rated for can cause damage/function incorrectly? Thanks. \$\endgroup\$ – RJSmith92 Oct 16 '15 at 7:27
  • \$\begingroup\$ Your comment regarding the relation ship of Voltage, Current and Resistance is fundamentally correct, simple Ohm's Law stuff. But the matter of what will cause damage is more related to the actual operational ratings of the components. If you increase the voltage applied to an operating circuits, you may see an increase in current, but not always. Some circuits are designed to self-protect and adjust to keep currents within safe values. Any change that increased current could result in damage due to excessive heat from the increased current. \$\endgroup\$ – Wes Oct 17 '15 at 4:21
  • \$\begingroup\$ The matter of exceeding voltage ratings is also related to current, but in a different way. When the absolute max rating of a device is exceeded, the internal materials become conductors when they should be insulators, and now current flows in places and amounts not intended. This is commonly referred to as "Letting the smoke out". It has really nothing to do with Ohms' law, this is just a catastrophic failure. \$\endgroup\$ – Wes Oct 17 '15 at 4:23
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Things like resistors will fail from excessive power dissipation- they get too hot and the materials they are made from suffer irreversible degradation. For example, the lacquer on the outside of a through-hole resistor may discolor or burn off, the resistance value change as the element oxidizes until finally it changes out of spec or opens up and starts arcing. Wires and PCB traces behave like resistors- too much current and the insulation burns off, the PCB delaminates or the trace opens up.

In low voltage circuits usually the voltage rating is not an issue- but if you were to take (say) an ordinary 0805 20M resistor and apply 2kV to it the power would be (in theory) only 200mW (which might be in spec or slightly outside of it) but the resistor could arc over and cause irreversible damage almost instantly. Similarly, you can have arcing between traces.

Things like capacitors and MOSFET gate oxide can fail when they are exposed to excessive potential which causes irreversible damage to the insulation. There will be some very localized heating (or more depending on what happens after the insulation is punctured) but that's not the main cause.

Things like diode and transistor junctions have breakdown voltages above which the current increases rapidly with voltage (sometimes they snap on with an avalanche/negative resistance characteristic). If the current is limited to that the heating is kept to a sensible amount (and doesn't increase too fast so that the heating is not localized to tiny areas) then this can be non-destructive. Otherwise the junctions can heat until they are not longer good semiconductor junctions any longer (in the hundreds of degrees C to destroy a silicon junction).

Getting back to your specific question about resistors- none of the voltages you mention is likely to run into a maximum voltage specification on the resistors (anything under about 25V you can forget about for resistors that are not an inhalation hazard).

So you're left with maximum power dissipation (and maybe maximum current if the resistance value is stupidly low, but let's ignore that). Here is a datasheet for a series of resistors, say we have a 10\$\Omega\$ resistor 0805 size. Rated power is shown as 0.125W and maximum working voltage as 150V. If you look at the "Power Derating Curve":

enter image description here

.. you can see that the rated power holds for ambient temperatures up to 70°C, but above that you must consider the rating to be less, according to the curve. Why does it level off at 70 degrees? Most likely the resistor would survive okay at >100% power if the ambient is kept cool, but the manufacturer does not want us testing that.

Recall that power dissipation of a resistor is $$P=I^2R$$ or $$P=V^2/R$$

(since power is $$V\cdot I $$ and Ohm's law).

In your first example, the resistance is fixed and you double the voltage- so the power should go up by 4:1. (from 20W to 80W) If your resistor is rated for 80W or more (under the conditions it sees in your box) then all will be fine. Otherwise, it may not be. Damage is caused by the heating, which is the product of voltage and current (obviously the current increases because the voltage is increased).

In your second example, you've doubled the resistance and the power is now 40W rather than 20W. If the resistor is rated for 40W then all will be fine.

The third example also results in 40W of dissipation. So if the resistor is good for 40W you're fine.

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  • \$\begingroup\$ Wow thanks for all that!, appreciate it. Was a bit more than I was asking for to be honest but thanks anyway! \$\endgroup\$ – RJSmith92 Oct 16 '15 at 7:32
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Too high a voltage tends to cause a catastrophic breakdown of a transistor. Once you apply over-voltage stress and the transistor breaks down, the pin will show short circuit (usually to ground). If you catch it, or limit the fault current some how, this type of failure will not be visible outside of the IC. It may be visible to microscope after exposing the die.

Of course, after a pin fails short, if the current is not limited, the component will probably get pretty hot and charred and show more obvious signs of destruction.

If you allow too much current to go through an IC, you will usually see smoke at some point. To be honest, I haven't had problems with this very often. A lot of regulators and such are protected against over-current. But I have seen ESD-type Zener diodes smoke after sustained exposure to high current.

This could also happen with a transistor that is dissipating more power than it is rated for. But, like I said, somehow I have not had to deal with that very often.

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  • \$\begingroup\$ Thanks @mkeith, Is the following correct, if you have a working circuit and you increase the source voltage, but at the same time increase the resistance so that the current remains the same, this could still cause damage (as you say to transistors)? Alternatively, if you keep the source voltage the same, but lower the resistance so the current increases, this may also cause damage to components in the circuit? Finally, If you just increase the source voltage, then that it turn increases the current, the damage could be double? (again depending on the components in the circuit) \$\endgroup\$ – RJSmith92 Oct 16 '15 at 1:36
  • \$\begingroup\$ In most circuits you can't really 'change the resistance'. And for many circuits, they don't behave like resistors anyway. \$\endgroup\$ – alex.forencich Oct 16 '15 at 1:40
  • \$\begingroup\$ Thanks @alex.forencich, I'm not asking this well tbh, what I mean is, if the power in a fairly simple circuit stayed the same (40W as Jrican shows), both a voltage increase or a Current increase could potentially cause damage (depending on the components in the circuit). I think mkeith has answered the question, if the original circuit had a transistor and I increased the voltage but lowered the resistance (so the power is the same as before), it could still do damage. The same applies if I just increase the current by lowering resistance (so the wattage is still 40W), this can also do damage. \$\endgroup\$ – RJSmith92 Oct 16 '15 at 1:50
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Ultimately, most things fail due to too much heat. If you are talking about a purely resistive load, such as a light bulb, this is pretty obvious. Exceed the voltage and current ratings, and the bulb burns out very quickly. In this case, you can't separate too much voltage from too much current as the resistance of the bulb is more or less constant so doubling the voltage will double the current and quadruple the power.

Things like diodes and transistors also generally fail from heat-related issues. Take a diode and apply a voltage that exceeds its reverse breakdown voltage. When the diode enters reverse breakdown, it will let current flow through. Current flowing through the diode times voltage drop across the diode equals power dissipated in the diode. Diodes can break down at a voltage of a few hundred volts, so this can dump heat into the diode at a rate of hundreds or thousands of watts, heating up the junction in the diode very quickly and causing it to melt. If it heats up fast enough, it can vaporize and you get a nice 'bang'. Same thing happens with transistors.

Dielectric breakdown is somewhat similar. Wires, connectors, MOSFET transistor gates, etc. can all be damaged by dielectric breakdown. When the voltage across an insulator gets too high, it is possible that the insulator will stop insulating and will instead start letting some current through. This current flow can cause damage. If voltages are high enough, dielectric breakdown can result in arcing, which can cause heating, pitting, etc.

In some cases, you can have issues with too LOW a voltage. Generally this is a problem when you have a poorly-designed step-up switching converter, such as a buck-boost or SEPIC, which tries to boost up the low input voltage and as a result generates a lot of heat by operating at a low efficiency.

One thing I should note: power ratings are generally related to operating temperature ratings. The maximum power will then be determined by the ability of the device to dissipate that power while remaining below the maximum operating temperature. It is possible to exceed rated powers for certain components under certain conditions. For example, a 5W resistor could actually dissipate 100W, so long as it only does that at a 5% duty cycle with a short enough on time that the resistor does not heat up enough to cause damage (i.e. 100W for 10s would probably cause it to fail, but 100W for 10us would probably be OK). It may also be possible to dissipate 100W in a 5W resistor continuously if you can build a system to pull heat out of the resistor fast enough to keep the temperature inside the resistor within its operating range (i.e. submerge it in liquid nitrogen).

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resisters are rated by the amount of power they can dissipate with out being damaged.

Power for a purely restive circuit is:

P = V * I

1) The Initial dissipated power is 20W (10V * 2A), then it changes to 80W (20V * 4A)

2) Now the dissipated power is 40W (20V * 2A)

3) Now the dissipated power is 40W (10V * 4A)

The damage is caused by the resistor dissipating more power than it is rated for, through heat.

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    \$\begingroup\$ But resistors are also rated by their (dielectric) breakdown voltage. \$\endgroup\$ – Fizz Oct 16 '15 at 2:21
  • \$\begingroup\$ Fair point. That would make the answer more complete, but probably beyond what a person asking this question needs or wants to know. \$\endgroup\$ – Jrican Oct 16 '15 at 2:27

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