MOSFET question

Hi all. With mosfets, such as the one at this link here …click here…

This question relates to using a mosfet as a relay switch … if there’s no voltage across drain and source, then the drain-to-source path (for the N-channel mosfet) can’t be just open-circuited or short-circuited simply by controlling the gate-to-source voltage, right? In other words, the mosfet isn’t the ‘same’ as an open-circuit/short-circuit relay switch, right?

Eg… if the drain is left open-circuited (not connected to anything), then will applying a gate-to-source voltage (and removing that voltage) repeatedly cause the resistance between drain and source to go low and high repeatedly?

Thanks all!

The open-drain MOSFET needs to connect to something. With the source connected to Arduino Gnd, then the drain can e treated as an open circuit with no voltage on the base, and as a very low resistor to Gnd with voltage applied to the base.

So no, not quite the same as an open or closed relay, but similar to one if one side of the relay is connected to Gnd.

CrossRoads:
The open-drain MOSFET needs to connect to something. With the source connected to Arduino Gnd, then the drain can e treated as an open circuit with no voltage on the base, and as a very low resistor to Gnd with voltage applied to the base.

So no, not quite the same as an open or closed relay, but similar to one if one side of the relay is connected to Gnd.

Thanks very much crossroads! Thanks for your time for helping me out with my question. Very much appreciated.

Southpark:
Eg.... if the drain is left open-circuited (not connected to anything), then will applying a gate-to-source voltage (and removing that voltage) repeatedly cause the resistance between drain and source to go low and high repeatedly?

Yes, sure, the gate voltage creates a strong electric field across the gate oxide which causes
a conducting channel to form between drain and source.

MarkT:
Yes, sure, the gate voltage creates a strong electric field across the gate oxide which causes
a conducting channel to form between drain and source.

Thanks MarkT!

Southpark:
Eg… if the drain is left open-circuited (not connected to anything), then will applying a gate-to-source voltage (and removing that voltage) repeatedly cause the resistance between drain and source to go low and high repeatedly?

If the drain is unconnected, the answer to this question doesn’t matter since the drain-source channel is not part of a circuit. Any possible answer to this question is of literally 0 practical value and is totally meaningless.

What do you intend to do with this information? If you are asking such a strange question, there is a high chance that you will misinterpret one of the answers here and try to do something that is doomed to failure.

Jiggy-Ninja:
If the drain is unconnected, the answer to this question doesn’t matter since the drain-source channel is not part of a circuit. Any possible answer to this question is of literally 0 practical value and is totally meaningless.

My question was simply this - if the drain is left unconnected, then is the path between drain and source effectively a ‘short-circuit’ when an appropriate voltage for the N-channel MOSFET is applied across gate and source.

What do you intend to do with this information? If you are asking such a strange question, there is a high chance that you will misinterpret one of the answers here and try to do something that is doomed to failure.

It is just for understanding. For example, an N-channel mosfet with open-drain (not connected). If an adequately large voltage is applied across gate and source (but not so large as to damage anything), then will the drain and source become electrically shorted together? (like a relay switch that can connect two terminals together - regardless of whatever is connected to the pair of terminals).

Crossroads taught that if the source of the mosfet is connected to a ground (such as an arduino ground) then the drain and source will be shorted when a gate-source voltage is applied.

Its worth knowing the chain of causality:

gate-source voltage causes channel formation
drain-source voltage AND channel existing means current flows in drain-source circuit.

Southpark:
It is just for understanding.

Understanding what, though? What point is there in asking what really happens when you flip a switch that isn't connected to anything?

Suppose the answer is "the drain is high impedance until something is connected to the drain".

Suppose the answer is "the drain is low impedance even without anything connected to the drain".

What difference does either answer make when the drain is unconnected? How would either answer increase your understanding of MOSFET operation? I'm more confused about what train of thought would lead you to think that this is a useful question than about any of the answers. I have nothing against hypothetical or theoretical questions, but this one just doesn't make sense. You might as well be asking "where on earth is the end of a rainbow".

Have a look at the data sheet for an STP120N4F6 from ST microelectronics.
Figure 4 shows a graph of ID (drain current) against Vds (Drain- source voltage).

For Vgs=10V, if the Vds is 0volt then ID will be 0 amps (inconclusive).

If now you increase Vds by even a small amount then ID suddenly becomes enormous i.e 150amps at 0.5 volts. Since R=V/I then the resistance is 0.5/150=3.3mOhms.

So you can't have resistance without Vds, it just doesn't make sense.

I have just tried measuring the resistance of the channel of the above device with an ohmmeter with a gate voltage of 7.5 volts. The ohmmeter showed a short (near enough) but then the ohmmeter supplies current.

So there is no point in having a conducting channel without any current to flow, i.e a Drain source voltage.

In theory, if there is a gate voltage then there is a channel that can conduct between the source and the drain.

Does that answer your question?

Dead_Ard

Dead_Ard:
So you can't have resistance without Vds, it just doesn't make sense.

That's just not true. The resistance is perfectly well defined in the limit as Vds goes to zero.

What you are saying is that the bag of 100 ohm resistors on my desk right now can't really be 100 ohms because as yet I haven't placed them in any circuit. They are just loose in a bag with no voltage and no current. So you tell me that this means their resistance is undefined.

Ok, so perhaps I should return them all to the shop because I payed for 100 ohm resistor not "undefined" ones.

Resistance is a physical property of the device. It doesn't suddenly became undefined the moment that you stop measuring it. In the same way my weight doesn't suddenly become undefined the instant that I step off the scales.

And there are ways to measure resistance purely passively, such as measuring thermal noise...

An FET is a non linear device so you cannot compare it to a resistor.
What I meant to say is there is no point in having a resistor in a circuit if you are not going to apply voltage to it.
Thermal noise, get real, this is practical electronics not advanced physics.
Does Schrodingers cat exist if you can't observe it?

Dead_Ard

A MOSFET that is switched on(*) is very much like a resistor, that's why the parameter is called Rds(on).

(*) Assuming the Vds isn't very large, which is the normal case for a switching application.

Thermal noise not practical electronics? Clearly you've not done much RF circuitry or even
low-level audio preamp design...

Dead_Ard:
An FET is a non linear device so you cannot compare it to a resistor.

A fet (in the "on" state) is at its most linear operating point precisely at Vds=0. In fact when Vgs>>Vtheshold and Vds close to zero the fet is very close to a linear resistance!

Again I repeat. Resistance is a physical property of the mosfet. It is well defined (can be calculated from the physical geometry, the oxide layer dielectric constant and thickness, the doping levels and fundamental physical constants). It is both well defined and stable, it definitely doesn't mysteriously vanish or become undefined just because we don't apply a voltage.

The original question in this thread was whether or not the mosfet still had well defined high impedance and low impedance states if the the drain is left open circuit. It does.

I see the link provided in the original post refers to Open Drain GPIO

When thinking of how the pins on an Arduino could be configured and used, its possible to use the pins themselves as a switch, providing you stay within the pin specifications (5V, 20mA for UNO).

Open Drain:

pimMode(pin, OUTPUT); // Switch ON (output LOW)
pimMode(pin, INPUT);  // Switch OFF (output HIGH-Z)

Open Source:

pimMode(pin, INPUT_PULLUP); // pin as output will default HIGH 
pimMode(pin, OUTPUT);       // Switch ON (output HIGH) 
pimMode(pin, INPUT);        // Switch OFF (output HIGH-Z)

Weak Open Source:

pimMode(pin, INPUT_PULLUP); //  Switch ON (output pulled HIGH with approx 30K) 
pimMode(pin, INPUT);        // Switch OFF (output HIGH-Z)

Note: All code above is open source.

That's not the point. Of course it has a resistance but its not a lot of use if there is no voltage. I agree that the FET is most linear at low Ids with suitable gate drive but a resistor without a voltage is impractical. If there is no voltage there is no current so it doesn't matter what the resistance is. Do you get my point.
The FET is non linear when the gate drive is not sufficient to pass the required amount of current, so in the example I quoted ( STP120N4F6) when the gate-source voltage is only 4volts then the max current is approx. 8 amps so the 'on' resistance is 0.5 ohm so it is a non linear device. So stuartO, what is the resistance of the current channel when no gate voltage is applied?
I always tried to keep out of the noise when designing amps, it was always a problem with early transistors, and by reducing the bandwidth then you reduced the noise, thank goodness transistors have improved but measuring resistance by measuring noise definitely is not a method that I would use. Anyway, resistive noise is not the only noise in a device, Shot noise is prevalent in transistors which is proportional to junction current so how do you measure resistance by measuring noise. Then there is generation - recombination noise.
Another question - what limits the current through the drain-source channel at a fixed Vgs when the load is increased?
PS RF design is and was not my forte. RF designers use pipes not wires.

Dead-Ard(uino)

southpark, dlloyd

I see what you are getting at. sorry that I got distracted. Yes the lower FET drive can be configured as an open drain device (just like an open collector BPJT) so yes it is almost like a set of relay contacts (except the lower FET will have an 'ON' resistance, even a relay has a contact resistance).

Dead_Ard