# MOSFET switch question

Woo-Woo Discussion in 'Twilight Zone' started by Nikša, May 7, 2018.

1. ### Nikša

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Hi, i am just learning how to use MOSFET as a switch.

I am watching this video

So in the video, Vdd is 9V and Vin is +5V using a separate power source. Now this is confusing me. In BJT base has to be 0.6V above emitter. How does this translate to MOSFET? In other words, if Vdd was not 9V but 1000V, would it still be enough to put +5V on the gate to make MOSFET conduct or would the gate have to be at 1000V + 5V?

Nikša, May 7, 2018
2. ### BobK

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Depending on the specific MOSFET, the gate must be anywhere from about 2V to 10V above the source. This is for an N-channel device. 5V might or might not be enough, you need a logic level MOSFET to turn on at 5V.

Bob

BobK, May 7, 2018
3. ### Nikša

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Thank you, i know threshold voltage varies etc. What's confusing me is does the gate have to be brought up to Vdd + those 5V or whatever or. For example if Vdd was 1000V, would the gate then have to be at 1000V + ~5V or? I can not understand this.

Last edited: May 7, 2018
Nikša, May 7, 2018
4. ### AnalogKid

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As Bob said, it depends on the FET. Ont he datasheet is a quantity called Vgs(th) - the Threshold Voltage from gate to source. This is the voltage when the FET just begins to conduct, equivalent to about 0.45 V to 0.5 V for a bipolar transistor. And like the BJT whose base-emitter voltage will increase to 0.6V to 1.0 V at full "on" status, a higher Vgs makes the FET more "on". The datasheet will say the minimum gate voltage for this. It usually is around 10 V for an n-channel power MOSFET, and 5 V for a "logic-level" part designed to operate at lower Vgs values.

ak

AnalogKid, May 7, 2018
5. ### Nikša

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Thank you, i know about threshold voltage and how it affects the width of the channel making FET more "ON" etc..that is not what's confusing me, but the biasing. Let's suppose you want to run a high voltage MOSFET as a switch, let's say Vdd is 4kV. Let's say you use common source configuration. Now, how do you bias such switch so that pulse circuit doesn't get damaged by high voltage Vdd. Would you use a high resistance load to bring the source voltage all the way down and voltage divider network on the gate side? Example would be much appreciated.

PS. I apologize if my question is unclear, i am a noob and just learning.

Last edited: May 7, 2018
Nikša, May 7, 2018
6. ### Bluejets

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The gate is insulated.

Bluejets, May 7, 2018
7. ### Nikša

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Some simple example of high voltage MOSFET switch circuit would help the most. I tried googling it, nothing really simple and helpful.

Nikša, May 7, 2018
8. ### Nikša

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Ok, i'm getting no responses, so a simpler question that will also help. In the common emmiter BJT amp example below, does the 1V signal source see the 5V battery?

Nikša, May 7, 2018
9. ### AnalogKid

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I did a quick search to find something close enough:

As long as the input signal and output +Supply share a common GND, it is no problem to have a 5 or 10 V signal at the gate and a 1000 V power source for the load. The MO in MOSFET is basically a thin sheet of glass rated for the peak drain voltage.

ak

AnalogKid, May 8, 2018
10. ### Nikša

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Thank you! Like i suspected but wasn't sure. Now, i want to interupt discharge from a HV cap and i wasn't planning on grounding it. Let's just say device might have to operate where ground connection isn't available.

Isn't the - of the cap enough of "ground". Does ground have to be physical connection to earth?

Nikša, May 8, 2018
11. ### Audioguru

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The video is wrong to show an IRF540 Mosfet with a gate to source voltge of only 5V. The datasheet for an IRF540 Mosfet shows a maximum threshold voltage of 4V when some of them barely conduct and shows 10V when it conducts well. The video should have shown an IRL540 (see the "L"?) that conducts well with a gate to source voltage of only 5V.

Ground usually has nothing to do with earth. It is simply a common point in a circuit that usually has no signal.

Audioguru, May 8, 2018
12. ### Nikša

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Aha.. I wonder why then is ground symbol so often used in addition to + and - of a power supply. More so confusing is the fact that the "ground" side is actually the source of electrons. This is cause "convetionional current" (to hell with it) follows the flow of imaginary positive charge, not the electrons.

Last edited: May 8, 2018
Nikša, May 8, 2018
13. ### Bluejets

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Probably because there are supplies that run positive and negative with respect to ground.(chassis)

i.e. split power supply.

Bluejets, May 8, 2018
14. ### hevans1944Hop - AC8NS

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Yes. This is an enhancement mode MOSFET with an silicon dioxide (glass) insulated gate, so +5V or more on the gate terminal with respect to the source terminal should be enough to "turn on" the drain-to-source circuit if the MOSFET is rated to operate with 1000V between drain-terminal and source-terminal.

If you had paid attention to the video, you would have noticed that an IRF 540 MOSFET was used. This particular device has a maximum drain-to-source voltage of 100 V according to this datasheet. So raising the voltage from 9 V to 1000 V would likely destroy the device. Most of the specs for the IRF 540 specify a gate-to-source voltage of +10 V to turn the device fully on, but the allowable maximum gate-to-source voltage is ±20 V.
"Ground" is NOT the source of electrons. Electrons are everywhere, surrounding the nucleus of atoms and balancing out the positive charge of the nucleus. That is why almost everything has zero net electrical charge in the world about us. You are actually supported by "electron clouds" pushing against each other.

To get electrons to produce a current in a circuit, it is only necessary to get them moving from one place to another place along a conductor such as a copper wire. "Ground" is totally irrelevant to the process. "Ground," in terms of circuit analysis, is simply a convenient zero-potential reference point against which to measure circuit voltages. By definition, all potentials between "ground" symbols are zero potentials. If you do measure a potential between two different points labeled as "grounds" in an equipment, there is something creating that potential, and neither "ground" should be considered a "true" ground. Even the Earth itself can exhibit differences in potential between two copper rods driven into the ground.

Consider the common dry-cell connected to an LED with a series current-limiting resistor. The dry-cell produces a charge separation at its two terminals by means of chemical action within the cell. One terminal has a negative charge with respect to the other positive terminal. The converse is also true: one terminal has a positive charge with respect to the other negative terminal. The dry-cell, considered as a whole, has zero charge. What it does have is electromotive force (emf) or voltage potential between its two terminals: the ability to move electrons through a circuit externally connected to the dry-cell terminals. The ability to generate an emf and produce an electrical current in a resistive load requires the expenditure of energy, in this case chemical energy produced in the dry-cell. Nowhere is a "ground" ever necessary or required.

Why don't you try purchasing some parts to breadboard a circuit similar to the one shown on the video? Although computer simulation may eventually lead to some understanding, simulation is not reality. You need some "hands on" experience with real parts, making real measurements with real test equipment, such as a digital multimeter, to gain practical knowledge. A multimeter is essential for electronics experimentation. It doesn't have to be expensive to measure AC and DC voltages and currents as well as resistance to three significant figures. Asian products are cheap and will do the job until you find something with better "bells and whistles."

hevans1944, May 8, 2018
15. ### Audioguru

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If a power supply has a positive voltage and a negative voltage then its common (called ground) is usually 0V.

Audioguru, May 8, 2018
16. ### Nikša

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Yes, as AnalogKid said, "As long as the input signal and output +Supply share a common GND, it is no problem".

That video was just an example. I won't use that MOSFET but this one or a similar one rated for high voltage.

I referred to ground being a source of electrons as in the circuit AnalogKid posted above (reproduced below), "ground" is negative terminal of the power supply which IS a source of electrons. In reality, electrons barely move - 10cm to 1m per hour, it is EM wave and it's Poynting Vector that move and carry the energy. Of course true "ground" is not required, only potential difference to make the energy flow through a conductor. Yes, electrons are everywhere. Very fabric of space is densly packed with virtual electron-positron pairs in eternal jitter due to fundamental imbalance of their electric and magnetic components - these little bubbles of energy literally make up all-there-is, forces and "matter".

I possess a multimeter.

Nikša, May 8, 2018
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17. ### AnalogKid

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Often that symbol refers to earth or chassis ground, as opposed to the lower potential of a two-connection floating output.

ak

AnalogKid, May 8, 2018
18. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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@Nikša I have reverted your earlier posts.

Editing them away like you did is impolite. It means that others can't follow the thread of conversation.

(*steve*), May 8, 2018
19. ### hevans1944Hop - AC8NS

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@Nikša: Although you have self-identified as
in your post #5, your later response in post #16 seems to indicate some understanding of physics with a willingness to add electronics to your repertoire. Good for you. And having a multimeter is a good start. You can use it to rustle up some free electrons, among other things.

The problem of driving the gate-to-source at low voltages, in the presence of a large voltage on the source with respect to power supply common, comes up frequently., especially in high-side H-bridge drivers for large motors. H-bridge drivers usually solve the problem with a "charge pump" circuit for the high-side drivers.

It appears you want to place a conducting MOSFET in series with a high-voltage capacitor and then turn the MOSFET off to interrupt the discharge of the capacitor into some sort of circuit. A typical use might be to control the exposure from a xenon gas-discharge arc lamp after a certain level of light exposure is reached. The problem is the source terminal is operating at the same potential as the drain terminal when the MOSFET is on. You need some means to isolate the gate-to-source low-level drive signal from this voltage. Two means come readily to mind: optical isolators and pulse transformers. Neither circuit is particularly simple. And you may still need a charge-pump circuit to obtain the desired bias levels.

It would help us to help you if you just explained what it is you are trying to do.

Oh, and on electron motion... most of us responding here are aware that those of us in good physical shape can easily outrun an electron traversing a wire. I like to use the analogy of people trying to run through a crowded tunnel or tube and having to bump into each other, dodge and weave to find a path. The speed of an individual electron in a conductor has nothing to do with current, which by definition is measured in units of charge-per-second passing a given point. One coulomb of charge moving past a point in one second represents a current of one ampere, no matter how fast individual electrons are moseying along. A coulomb of charge is represented by a fairly large number of electrons, about 6.2415 × 1018, so moving that many electrons past a given point on a wire every second represents one ampere of current, no matter how fast or slow individual electrons may require to "cross the finish line". Take a look at electrolysis for some practical applications of moving electrons from one place to another.

And just in case nobody has mentioned it yet: Welcome to Electronics Point!

hevans1944, May 8, 2018

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