EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: Meshka7 on February 14, 2015, 07:03:21 pm
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Let's say I walk over the carpet and get charged to 10KV. Then I touch a MOSFET.
Is it correct to assume that the MOSFET won't be damaged unless it has a path to ground?
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It will have very small capacitances from the other leads and the case to ground, so there is a good chance it will be damaged even though it looks like it is floating and has no ground connection. Charging one electrode to 10kV at a very fast rate will likely damage it.
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Is it correct to assume that the MOSFET won't be damaged unless it has a path to ground?
No. The MOSFET has capacitance. When you touch the MOSFET it will try to equalize its potential to your potential. To equalize potentials current must flow, and if that current flows through the wrong bit of the MOSFET it will be damaged.
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Let's say I walk over the carpet and get charged to 10KV. Then I touch a MOSFET.
Is it correct to assume that the MOSFET won't be damaged unless it has a path to ground?
This is a very simplified explanation to get you thinking in the right direction.
The mosfet will have a capacitance to ground, with the mosfet being one "plate" in that capacitor.
The discharge path will always hit one terminal first, and there will be a flow of charge across the device towards the terminal that is nearest ground, until all terminals are at the same voltage. How much transient voltage there is between the two terminals will depend on the direction of charge flow and the mosfet junctions between the terminals; in general it will be dangerously high.
The current will cease when the mosfet (i.e. the capacitor's plate) reaches its final voltage.
Will that current and voltage damage the mosfet? How lucky are you feeling today?
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So if the MOSFET was sitting on some insulating material, it won't have any capacitance to ground. In this case it won't get damaged?
Alternatively, if it does have capacitance to ground, but the person who is charged is insulated from ground, would it get damaged?
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So if the MOSFET was sitting on some insulating material, it won't have any capacitance to ground. In this case it won't get damaged?
The definition of a capacitor is something conductive sitting on some insulating material. So yes, it will have capacitance to ground.
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Yes, but if it's well insulated, the capacitance is practically zero
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CV = Q
for constant Q, lim(C->0) V = inf
"well isolated" means if you plop some charge onto it the voltage will be even higher... it will probably be less likely to be damaged by the initial discharge into it, but now it's packed with charge that really wants to get out.
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Yes, but if it's well insulated, the capacitance is practically zero
Crudely speaking, the capacitance actually goes up if you use a better insulator.
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CV = Q
for constant Q, lim(C->0) V = inf
"well isolated" means if you plop some charge onto it the voltage will be even higher... it will probably be less likely to be damaged by the initial discharge into it, but now it's packed with charge that really wants to get out.
Or alternatively, the MOSFET will just become at the same potential (10kV) and nothing will happen
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Or alternatively, the MOSFET will just become at the same potential (10kV) and nothing will happen
...except possible damage to the MOSFET. Go back and re-read the 2nd, 3rd and 4th posts in the thread.
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Right. The point is that the MOSFET is now at 10kV, meaning it has become a serious ESD hazard (as soon as you get something at a lower voltage near it, it'll discharge and be damaged). The initial discharge has not harmed it, but now it's full of potential.
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Right. The point is that the MOSFET is now at 10kV, meaning it has become a serious ESD hazard (as soon as you get something at a lower voltage near it, it'll discharge and be damaged). The initial discharge has not harmed it, but now it's full of potential.
Being at 10kV doesn't mean it has any charge. The gate will have 10kV and the source will be at 10kV. dV=0 -> Q=0. No charge has moved, no energy needed to be expended. All that happened that the voltage at both the gate and the source rose to 10kV. Once you stop touching the MOSFET, the gate and source go back to zero.
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Crudely speaking, the capacitance actually goes up if you use a better insulator.
Ok assume you put the MOSFET over a piece of rubber 10 inches think or a wooden desk
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Right. The point is that the MOSFET is now at 10kV, meaning it has become a serious ESD hazard (as soon as you get something at a lower voltage near it, it'll discharge and be damaged). The initial discharge has not harmed it, but now it's full of potential.
Being at 10kV doesn't mean it has any charge. The gate will have 10kV and the source will be at 10kV. dV=0 -> Q=0. No charge has moved, no energy needed to be expended. All that happened that the voltage at both the gate and the source rose to 10kV. Once you stop touching the MOSFET, the gate and source go back to zero.
no they won't
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Right. The point is that the MOSFET is now at 10kV, meaning it has become a serious ESD hazard (as soon as you get something at a lower voltage near it, it'll discharge and be damaged). The initial discharge has not harmed it, but now it's full of potential.
Being at 10kV doesn't mean it has any charge. The gate will have 10kV and the source will be at 10kV. dV=0 -> Q=0. No charge has moved, no energy needed to be expended. All that happened that the voltage at both the gate and the source rose to 10kV. Once you stop touching the MOSFET, the gate and source go back to zero.
...thats...not how charges work. Once something becomes charged, it retains that charge until it is removed. By shorting to ground when you set the mosfet down, for example.
But lets think of it this way. A mosfet is not all connected inside. The gate is insulated from the gate and source. This makes it a capacitor, and why you need to pull the gate down if you want to turn a mosfet off - it holds a charge. If you charge the gate to 10kv, you get an opposing charge on the other side of that insulator. What's the breakdown voltage of a mosfet gate insulator? Not much. Mosfets are one of the most ESD sensitive components you'll find in a circuit, because of that gate insulator.
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no they won't
Try this: Charge a cap to 10V, then touch a piece of wire to it, then remove the wire. Does that make the piece of wire automagically at 10V?
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You didn't pay attention in physics class, did you?
If you want magic, explain where the static charge in your mosfet goes when you stop touching it. It must go somewhere, there has to be a discharge path. You are claiming it magically vanishes. Explain your work. While you're trying to figure it out, ponder how you, a giant sack of highly conductive salt water, can retain a static charge but other conductors can't. And why a Van de Graaff generator is impossible.
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Try this: Charge a cap to 10V, then touch a piece of wire to it, then remove the wire. Does that make the piece of wire automagically at 10V?
no
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You didn't pay attention in physics class, did you?
If you want magic, explain where the static charge in your mosfet goes when you stop touching it. It must go somewhere, there has to be a discharge path.
My point is that the MOSFET never got charged to begin with. Just connecting it to a 10kV charged capacitor doesn't charge the MOSFET. to charge the MOSFET, current has to flow through its internal capacitor, and current won't flow unless the circuit is closed. If the MOSFET is isolated from any conducting path, the circuit is not complete, no charge will move, and the MOSFET will retain 0 charge before and after you connect the 10kV charged cap to it.
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My point is that the MOSFET never got charged to begin with. Just connecting it to a 10kV charged capacitor doesn't charge the MOSFET. to charge the MOSFET, current has to flow through its internal capacitor, and current won't flow unless the circuit is closed. If the MOSFET is isolated from any conducting path, the circuit is not complete, no charge will move, and the MOSFET will retain 0 charge before and after you connect the 10kV charged cap to it.
Your approach here is one of denying knowledge, insisting you must be right, and refusing to learn. Not only that, you have come here and are trying to teach experts that everything they know is wrong. Why do you think that will be successful?
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no they won't
Try this: Charge a cap to 10V, then touch a piece of wire to it, then remove the wire. Does that make the piece of wire automagically at 10V?
If the cap is ground referenced, yep. buy nice expensive electrometer and some teflon supports that are clean enough and you can measure it yourself.
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Your approach here is one of denying knowledge, insisting you must be right, and refusing to learn. Not only that, you have come here and are trying to teach experts that everything they know is wrong. Why do you think that will be successful?
So you want me to believe everything the experts say without being convinced? Is that how you learned, by not questioning anything? Can you respond to my argument rather than commenting on my approach?
In fact, I just tried it and I do think you are wrong. I charged a cap to 10V, grounded one terminal, then connected the gate of a MOSFET to the other terminal. According to the experts here, I should see the capacitor discharging quickly. It didn't happen.
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In fact, I just tried it and I do think you are wrong. I charged a cap to 10V, grounded one terminal, then connected the gate of a MOSFET to the other terminal. According to the experts here, I should see the capacitor discharging quickly. It didn't happen.
No-one here has said the capacitor will discharge if you do that.
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So you want me to believe everything the experts say without being convinced? Is that how you learned, by not questioning anything? Can you respond to my argument rather than commenting on my approach?
Look at the picture on the left next to my name.
The Van de Graaff generator has a dome supported on an insulator isolated from earth. It is charged to a high voltage. The girl is standing on a rubber mat and touching the generator. She has also become charged to a high voltage. You can tell this because her hair is standing on end. Her hair is acting as an electrometer (a voltage detector).
In order for the girl to become charged up an electric current has flowed between the generator and her body through her hand. We know an electric current has flowed because if not her body would not be charged up above ground potential. A change in potential requires a flow of charge, and a flow of charge is an electric current.
Now imagine the Van de Graaf generator is a highly charged finger, and imagine the girl is a MOSFET sitting on a rubber mat. When the finger touches the MOSFET the MOSFET will be charged up to the same potential as the finger. In order for this to happen an electric current must flow between the finger and the MOSFET. If the current flows through the gate insulation layer inside the MOSFET while equalizing then the MOSFET will likely suffer damage.
The lesson from this is to keep the MOSFET grounded on a static dissipative surface, and to keep your body grounded through a static protection wrist strap. That way when you touch the MOSFET your finger and the MOSFET will be at the same potential and no current will flow between your finger and the MOSFET.
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Ok. Let's take this a bit further. Assume the girl touched a ground rod before touching the Van de Graaff generator. When she touches the generator, she is charged to high voltage and you say that it is caused by transfer of charge. How come the girl doesn't die?
You also say that a change of potential requires flow of charge. However if you look at a voltage doubler circuit. One of the caps jumps from V to 2V without any transfer of charge. Just by the act of opening a switch and closing another the voltage on one plate of the cap doubled. As long as the voltage difference between the cap plates remains the same, no charges should flow.
One more thing, if you touch a resistor to a cap charged with 10V. In no time does the other end of the resistor go to 10V. Did that require any flow of charge? Could the girl here be acting like a large resistor? her potential changes without any current flow.
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Here's an actual demo. The point is the amount of current.
https://www.youtube.com/watch?v=ubZuSZYVBng (https://www.youtube.com/watch?v=ubZuSZYVBng)
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Ok. Let's take this a bit further. Assume the girl touched a ground rod before touching the Van de Graaff generator. When she touches the generator, she is charged to high voltage and you say that it is caused by transfer of charge. How come the girl doesn't die?
It is the volts that jolt, but the amps that kill.
You also say that a change of potential requires flow of charge. However if you look at a voltage doubler circuit. One of the caps jumps from V to 2V without any transfer of charge. Just by the act of opening a switch and closing another the voltage on one plate of the cap doubled. As long as the voltage difference between the cap plates remains the same, no charges should flow.
Think before writing. Use simple examples, not complex ones.
One more thing, if you touch a resistor to a cap charged with 10V. In no time does the other end of the resistor go to 10V. Did that require any flow of charge? Could the girl here be acting like a large resistor? her potential changes without any current flow.
You need to understand very basic physics. Start with understanding potential difference, charge, current, resistance and capacitance. I suggest a physics textbook aimed at 11-15 year olds. There's no short cut; you have to put in the effort.
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Imagine yourself at 30,000 ft. Would you prefer to be in an airplane or on snow skis at the top of Mount Everest? Height is gravitational potential. Volts is electrical potential. You are safe provided you are not free to move or electrons are not free to move.
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I have a strange feeling of Deja vu with this thread. It strongly reminds me of Mrwing in the free energy thread. :scared:
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Let's say I walk over the carpet and get charged to 10KV. Then I touch a MOSFET.
Is it correct to assume that the MOSFET won't be damaged unless it has a path to ground?
No, unless you short all the pins together and then touch it.
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All right, let's try a different approach.
Let's say I walk over the carpet and get charged to 10KV. Then I touch a MOSFET.
Is it correct to assume that the MOSFET won't be damaged unless it has a path to ground?
Forget ground. It isn't important. Neither you nor the MOSFET needs to be grounded, in order for the MOSFET to get zapped.
Suppose the MOSFET is sitting on an insulated rubber mat. You walk along across a woolly carpet wearing nylon clothing and rubber soles on a dry day. Your voltage with respect to the MOSFET is now 10kV. Suppose you touch the gate terminal of the MOSFET. The MOSFET is very likely to be damaged because the gate terminal is now at 10kV with respect to the drain and source and its maximum rated voltage is only 20V. A spark jumps across the the thin gate oxide layer and destroys it.
It's potential difference which is important not ground. Here's another example:
Suppose there are two people, both standing on rubber mats and have previously shaken hands to equalise the charges i.e. they have a potential difference near zero. Now one of them puts their hand on a Van de Graff generator and develops a positive potential of 50kV. Now if they shake hands with the other person standing on a rubber mat they'll both receive a shock because a current will flow between them to equalise the potential difference.
This also applies to aircraft which need to be refuelled in mid flight. Fuel is flammable and any spark could cause an explosion. Before the fuel line is connected between the aircraft carrying the fuel and the aircraft which is being refuelled, a piece of electrical cable is connected between them, to discharge them, so they both have the same potential difference.
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So what is the explanation of the fact that when I connect one terminal of a 1000uF cap charged with 10V to another 1000uF cap that is discharged (only via one terminal so circuit is not complete), the other terminal of the discharged cap has a voltage of 10V with respect to the ground terminal of the charged cap. Once I disconnected the caps, the voltage difference across the second is zero so zero or very little charged transferred from the charged cap to the discharged despite an initial potential difference.
Is a charged cap not a good model for static electricity?
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Another one. You talk about potential difference.
How is that when I charge two identical 1000uF caps to 10V, then connect one pair of positive and negative terminals together then measure the potential difference across one of the caps, it doesn't change from the original 10V. Shouldn't connecting the two terminals equalize potential? and it should be like half the original voltage or something?
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So what is the explanation of the fact that when I connect one terminal of a 1000uF cap charged with 10V to another 1000uF cap that is discharged (only via one terminal so circuit is not complete), the other terminal of the discharged cap has a voltage of 10V with respect to the ground terminal of the charged cap. Once I disconnected the caps, the voltage difference across the second is zero so zero or very little charged transferred from the charged cap to the discharged despite an initial potential difference.
Is a charged cap not a good model for static electricity?
Another one. You talk about potential difference.
How is that when I charge two identical 1000uF caps to 10V, then connect one pair of positive and negative terminals together then measure the potential difference across one of the caps, it doesn't change from the original 10V. Shouldn't connecting the two terminals equalize potential? and it should be like half the original voltage or something?
Normally when we talk about capacitance, we really mean mutual capacitance. i.e. the capacitance between two conductors, whether they be wires or the plates of a capacitor.
There's also self-capacitance, which doesn't rely on having more than one conductor. All objects have self capacitance which is due to the fact that removing or adding electrons to an object will alter its potential difference.
https://en.wikipedia.org/wiki/Capacitance#Self-capacitance
A 1000uF capacitor has 1000uF of capacitance between its plates but its self capacitance is tiny, as is the capacitive coupling between two 1000uF capacitance sitting next to one another.
If you charge a 1000uF capacitor to 10V, then connect one of the terminals to another discharged 1000uF capacitor it will discharge a little but the amount of charge transferred will be tiny because the capacitive coupling between the two capacitors will be several orders of magnitude smaller, say <50pF. It would be like connecting a <50pF capacitor in parallel with it.
Self capacitance also becomes noticeable at higher frequencies. If you have an oscillator connected to a battery and touch the output with your finger, a small AC current will flow into you, as the oscillator pushes and pulls electrons back and fourth between you and the battery. The current can be increased by attaching one of the battery terminals to a large sheet of metal (ground plane) and by sitting on a ground plane yourself. Even if the two ground planes are not electrically connected to one another and are a significant distance apart, they still have self capacitance which is increased by having larger ground planes. Larger ground planes have a higher self capacitance and lower impedance because electrons repel one another and having more of them makes it easier to add/remove electrons from them. This is useful when building transmitters: a ground plane is used as a source/sink of electrons for the oscillator to move in and out of the antenna.
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Meshka7,
I like your question. The answers given by others above, that the MOSFET will die if its gate is raised to 10kV and the other 2 ports left floating, is correct, as we know from experience, or from reading the absolute maximum DC ratings of the MOSFET vendor's datasheet, or from the fact that a legion of ESD engineers have been gainfully employed for the last 50 years to protect this from happening. But I know you agree with the answer, it is the reasoning you are interested in. Let me give you what I understand as the reason.
As you pointed out, the gate (G) is one terminal of a capacitor with, in this case, the other terminal (T) floating. G is brought to a potential of 10kV. This means there is an excess charge on G, enough to raise it to 10kV. Now T is floating, which means we really don't know how much excess charge is on T, but suppose it's none so that T is at potential 0V. Your arguments above have rested on the fact that T at 0V is not the same as T at GND since GND is a supply that can source current whereas T cannot source current. I agree with this and the fact that there is no current between G and T. Then you conclude that since there is no current there can be no damage to the cap. This I disagree with since you are forgetting the large E field created on G. The E field will cause the dialectric between G and T to breakdown and kill the cap. Remember, the E field is a force field and the dialectric is matter, ie electrons and protons. I believe Nerull made this argument above.
However, there is a case where you are correct and that is if the cap is made in a vacuum. By this I mean G and T are metal terminals but the dielectric between them is empty space or nothing. In this case there is no dielectric to break down and so the cap will survive.
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So what is the explanation of the fact that when I connect one terminal of a 1000uF cap charged with 10V to another 1000uF cap that is discharged (only via one terminal so circuit is not complete), the other terminal of the discharged cap has a voltage of 10V with respect to the ground terminal of the charged cap. Once I disconnected the caps, the voltage difference across the second is zero so zero or very little charged transferred from the charged cap to the discharged despite an initial potential difference.
Is a charged cap not a good model for static electricity?
Another one. You talk about potential difference.
How is that when I charge two identical 1000uF caps to 10V, then connect one pair of positive and negative terminals together then measure the potential difference across one of the caps, it doesn't change from the original 10V. Shouldn't connecting the two terminals equalize potential? and it should be like half the original voltage or something?
Normally when we talk about capacitance, we really mean mutual capacitance. i.e. the capacitance between two conductors, whether they be wires or the plates of a capacitor.
There's also self-capacitance, which doesn't rely on having more than one conductor. All objects have self capacitance which is due to the fact that removing or adding electrons to an object will alter its potential difference.
https://en.wikipedia.org/wiki/Capacitance#Self-capacitance
A 1000uF capacitor has 1000uF of capacitance between its plates but its self capacitance is tiny, as is the capacitive coupling between two 1000uF capacitance sitting next to one another.
If you charge a 1000uF capacitor to 10V, then connect one of the terminals to another discharged 1000uF capacitor it will discharge a little but the amount of charge transferred will be tiny because the capacitive coupling between the two capacitors will be several orders of magnitude smaller, say <50pF. It would be like connecting a <50pF capacitor in parallel with it.
Self capacitance also becomes noticeable at higher frequencies. If you have an oscillator connected to a battery and touch the output with your finger, a small AC current will flow into you, as the oscillator pushes and pulls electrons back and fourth between you and the battery. The current can be increased by attaching one of the battery terminals to a large sheet of metal (ground plane) and by sitting on a ground plane yourself. Even if the two ground planes are not electrically connected to one another and are a significant distance apart, they still have self capacitance which is increased by having larger ground planes. Larger ground planes have a higher self capacitance and lower impedance because electrons repel one another and having more of them makes it easier to add/remove electrons from them. This is useful when building transmitters: a ground plane is used as a source/sink of electrons for the oscillator to move in and out of the antenna.
Self capacitance IS mutual capacitance with second conductor or "plate" being earth. Atleast that's what I was taught.
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Being at 10kV doesn't mean it has any charge.
EVERYTHING that is not at 0V relative to ground has a charge. If the FET's effective capacitance to ground from whatever surface it is resting on is 1pF, then Q=1e-12 x 10e3 = 10nC.
When you stop touching it, it still has that 10nC charge unless leakage is draining it away or you discharged yourself before breaking contact. That's what a dissipative anti-static mat connected to ground is good for: you put an unknown static charge on components by handling them and drop them on the dissipative mat to safely bring them to the same local ground voltage as everything else on it before bringing parts together.
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Self capacitance IS mutual capacitance with second conductor or "plate" being earth. Atleast that's what I was taught.
It's an oversimplification.
Self capacitance doesn't have anything to do with the distance of an object from earth or the dielectric separating it from the earth.
The earth has a self capacitance of the earth is 710 µF which is much larger than the self capacitance of any object on it so it's a very handy source/sink of electrons for large objects on the earth such as radio antennas.