Electronic – High Current or High Voltage that damages components

basiccurrentdamageresistancevoltage

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.

Best Answer

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.

Related Solutions

Electronic – Voltage distribution between components

Yes there is a general method, but its generally not useful.

If you go back to first principles, there is a equation for each component relating voltage and current. Let's simplify the problem to where one is a direct function of the other, like with your LED and resistor example. Put another way, there is only one degree of freedom to decide what each component is doing.

The problem then is a set of simultaneous equations. The general answer is to solve these simultaneous equations. Usually this is rather complicated and the extra level of accuracy is not necessary and even misleading because it exceeds the tolerance to which other values can be known.

In your specific LED and resistor example, you could solve this simple set of simultaneous equations visually with a graph. Plot the resistor and LED current separately, each as a function of the voltage of the node between them. You know one additional constraint, which is that the current thru both is the same. Visually, this means you look for the place the two curves intersect, because that is the point at which both have the same current.

Hopefully you can see that it gets a lot more complicated for anything but a simple contrived example.

Electronic – Current flow in series circuit with different (unknown) resistances

Perhaps a worked example would make things clearer.

Suppose I had three lamps - each has a different working voltage and takes a different current. What would happen if I connected all three in series and connected them to a 9V battery.

enter image description here

For the purposes of illustration only (keeping it simple) I am assuming the lamps work as pure resistances that don't change their values.

First we calculate the individual resistances of the lamps. In this case 6,24 and 15 ohms.

The total circuit resistance will be 6 + 24 + 15 = 45 ohms

The current that will flow around the circuit will be

                               9/45  = 0.2 Amps

This is much less than any of the rated values because there is MORE resistance in the circuit

Each lamp will 'drop' a different voltage

lamp1 will drop 0.2 x 6 = 1.2V

lamp2 will drop 0.2 x 24 = 4.8V

lamp3 will drop 0.2 x 15 = 3.0V

If we add up all the voltage drops we get 1.2 + 4.8 + 3.0 = 9V

In other words the sum of the voltage drops around the circuit is equal to the supply voltage (kirchoff's voltage rule)

Again note that these voltages are not related to the ratings of the devices.

enter image description here

Best Answer

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Electronic – Voltage distribution between components

Electronic – Current flow in series circuit with different (unknown) resistances

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