Any Voltage through any material - Page 1
Forum section: electronics / beginners

author: challie2
author_rank: Contributor
timestamp: October 06, 2016, 10:12:47 pm
content: Is it possible too put an unlimited voltage through almost any electronic component without severe damage so long as the amperage is low enough? I get with diode breakdowns and insulation leakage but anything else maybe passive components, or a single peice of 2.5mm wire rated at 250v and putting 10kv through it with a ultra low amperage would it not melt? stupid question maybe but it would help me with my theory in a weird way. Thanks

author: cowana
author_rank: Frequent Contributor
timestamp: October 06, 2016, 10:17:06 pm
content: Ohms law: the current through an item will always be equal to the voltage applied divided by the resistance.(Okay, lets not go into complex impedances and AC effects here...)That means for a length of wire with 1 ohm resistance, the current flow will *always* equal the voltage - if you put 10kv across the two ends, there is no way to stop 10kA from flowing. The rate that the wire heats up will depend on the current flowing. The total temperature rise will depend on the current and time the current is applied for - while a 5A fuse might melt in 50ms with 10A flowing through it, it might take 10 seconds to melt with just 6A flowing.Limiting the current will in turn reduce the voltage - there is no way to adjust them independently.

author: rs20
author_rank: Super Contributor
timestamp: October 06, 2016, 10:19:31 pm
content: You're asking an impossible question. Consider a piece of copper wire with a resistance of 1 ohm -- if you put 1 microamp through the copper wire, the voltage across itwill be1 microvolt (V = IR). If you put 10kV across that copper wire, the current through the wirewill be10,000 Amps (at least until it vaporises). Hence, it is meaningless/impossible to speculate that would happen if you put 10kV "with 1 microamp" across the wire.In particular, if you had a 10kV source and put 10 gigaohm current limiting resistor on it to limit the current, you would have (in some sense) a "10kV 1 microamp power supply". But crucially, if we connect our copper wire to this power supply, the copper wire only sees 1 microvolt; the current limiting resistor is the part that has to handle the 10kV.

author: Brumby
author_rank: Unknown
timestamp: October 06, 2016, 11:35:44 pm
content: Quote from: challie2 on October 06, 2016, 10:12:47 pmThis is where fundamental terminology is absolutely critical.Generally speaking- you putcurrentthrough something- you put voltageacrosssomethingIn the case of a simple conductor (or resistor), there is a simple parameter - resistance - that characterises it. This is where Ohm's Law comes in:1) With this resistance - and the current passing through it, you can calculate the voltage that will appear between the two ends of the conductor (resistor).2) With this resistance - and the voltage that appears between the two ends of the conductor (resistor) you can calculate the current passing through it.Note that the voltage discussed here is the voltage between the two ends of the conductor (resistor) and hasabsolutely nothingto do with the voltage with respect to Earth.

author: amspire
author_rank: Super Contributor
timestamp: October 07, 2016, 12:17:42 am
content: Quote from: challie2 on October 06, 2016, 10:12:47 pmThe simple answer is no.If you apply enough voltage to, say, a mosfet gate, it will reach a voltage where it punches a hole through the gate insulation. Even if you try and limit the current, the gate has capacitance which stores a charge so when the breakdown happens, all the energy stored in the gate capacitance will be added to the applied current. This means there is nothing you can do to limit the actual current that causes the damage.Once the gate has a tiny hole in the insulation, the device may even appear to work fine, but it will probably now fail. The failure can be immediate, next week, next month or next year but the device is on the road to complete failure. I was once with a company that initially had poor electrostatic practices. They were getting a lot of failures in test, and a lot of returns within the first 12 months. They implemented proper ESD practices and the test failures and returns dropped to almost nothing. If a component is damaged to the extent that it is not longer useable in a product, I would call that severe damage even if it appears to work immediately after the overvoltage.Basically you can device parts into two groups - those that it is possible to apply a high voltage to with a very low current supply and those you cannot due to their low impedance. Many parts in the first group will be damaged by an overvoltage. In all cases there will be a limit to the voltage you can supply. So if you have a 15v zener, the maximum you can apply is 15v and at low currents, it will survive a breakdown without any problem. The Mosfet gate will breakdown at something like 30V and will not survive.

author: sleemanj
author_rank: Super Contributor
timestamp: October 07, 2016, 02:45:42 am
content: Quote from: amspire on October 07, 2016, 12:17:42 amI think it is important to pick this nit and expand your statement toIf you apply enough voltage to, say, a mosfet gate,relative to the source or drain, it will reach a voltage where it punches a hole through the gate insulation.I pick the nit because I suspect that "voltage is relative" may not have been grasped quite by the original poster. I appreciate that most of the readers here would know what you meant.

author: Brumby
author_rank: Unknown
timestamp: October 07, 2016, 04:23:42 am
content: Quote from: sleemanj on October 07, 2016, 02:45:42 amThis is another expression of exactly the point I was making.Putting it simply ...Dear OPIn order to answer your question, we need to know where the voltmeter probes are placed in relation to the component under consideration.

author: max_torque
author_rank: Super Contributor
timestamp: October 07, 2016, 12:30:19 pm
content: To expand i simple terms, imagine the following:A battery with two terminals + and -A 1 ohm resistor.Let say the battery is, to make things easy, 1volt (ie the chemisty i the battery pushes the +ve terminal to an electrical potential that is 1volt higher than it's negative terminal)Now put the 1ohm resistor across the battery + and -ve terminals. Ohms law tells us that 1 amp will flow (V=IR) because one end of the resistor is at 1v, and the other at 0vwe also know that the resistor must disipate some energy, i this case, 1Watt (W= V X I)If we now pick a battery with a higher potential difference, say 10V, we have 10v across our 1 Ohm resistor, 10Amps flow, and the resistor disipates 100W as heatAt some point, the resisitor burns out, because it can't disipate the heat without actually going overtemperature and melting!However, there is another case, and this time, we simply connect one end of our 1 ohm resistor to the +ve battery terminal, and leave the other end unconnected, in free air.Here, there is no currently flow, and both ends of the resistor have 1v on them. In fact of course, we still have a complete circuit, it's just that part of the circuit is made of air, and not copper. As air has a relatively high resistance, very little current flows through it. However, if we increase the battery voltage, at some point (in the thousands of volts range) even this air gap breaks down and current flows. Those mechanisms are more to do with other factors such as the ionisation threshold of the air, and shape of the conductors, and the impurities (such as dust, or water vapour) carried in the air.

author: CraigHB
author_rank: Regular Contributor
timestamp: October 07, 2016, 04:07:47 pm
content: The question is asking if the impossible is possible. If a component can handle a high voltage then it has to have a high resistance. There can not be high source voltage and low current flow without high resistance somewhere in the circuit. It's possible to limit current, but then there has to be an active device somewhere in the circuit taking the required voltage drop. That puts a lower voltage across the component which does not comply with the question.Possibly what the OP was considering is if a current limiting device such as a benchtop DC supply can protect components when voltages are high. In most cases it can, but as pointed out the energy stored in the gate charge of a FET can result in enough current flow to do damage even when source current is limited. However a lot of chips have protection diodes on transistors which will avalanche well before gates are compromised. In that case a current limit can provide protection, though it's still possible to blow protection diodes which of course will destroy the circuit. In my own experience setting a reasonable limit on my benchtop supply has saved me a number of times from blowing circuits.