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Mosfet as a switch.

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alphacat

New Member
Hey,

When I was using a BJT as a switch, then I saw to it that the BJT will be in the saturation region, by having IB at least 1/10 of IC, for a given IC.
I wanted the BJT to be saturated to have a minimal VCE voltage drop.

When using a Mosfet as a switch, do we want it to be in the linear region, where VDS is minimal?
The problem is that for it to be in the linear region (for a given current), then if the current IDS is too large, we would need a very large VGS (and/or very low VTn) in order to keep the Mosfet in the linear region.
 
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Hero999

Banned
I can't remember the technical term for a MOSFET but if you want to use it as a switch, you just need to ensure the gate voltage is high enough for the desired on resistance at the desired current. 10V is enough for all MOSFETs and only 5V is required for logic level MOSFETs.
 

alphacat

New Member
Thanks for the info guys.

Nigel,
I assume that you define not linear by having VDS barely changed when IDS changes, right? (which is what happens when the Mosef is saturated).

But for the same reason you would want a BJT switch to be forwared biased and not saturated (in order for VCE to be barely influenced by changes in IC).
 

Hero999

Banned
But for the same reason you would want a BJT switch to be forwared biased and not saturated (in order for VCE to be barely influenced by changes in IC).
Actually, when a MOSFET is saturated, it's operating in its linear region.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Thanks for the info guys.

Nigel,
I assume that you define not linear by having VDS barely changed when IDS changes, right? (which is what happens when the Mosef is saturated).
I would define the linear region as where you can adjust the base/gate up and down, and the collector/drain current increases accordingly.

But for the same reason you would want a BJT switch to be forwared biased and not saturated (in order for VCE to be barely influenced by changes in IC).
If it's a switch you want it ON and ON as much as possible, why would you want it less than fully ON?.
 

MrAl

Well-Known Member
Most Helpful Member
Hey,

When I was using a BJT as a switch, then I saw to it that the BJT will be in the saturation region, by having IB at least 1/10 of IC, for a given IC.
I wanted the BJT to be saturated to have a minimal VCE voltage drop.

When using a Mosfet as a switch, do we want it to be in the linear region, where VDS is minimal?
The problem is that for it to be in the linear region (for a given current), then if the current IDS is too large, we would need a very large VGS (and/or very low VTn) in order to keep the Mosfet in the linear region.

Hi there,

MOSFETs are mostly operated so that they are fully turned on when
used as a switch, however this isnt always the case with BJT's.

With BJT's sometimes we want to keep them slightly out of saturation
so we can avoid the lengthly storage time of the transistor which can
be quite long. This allows us to switch the transistor at a much higher
frequency than if we allowed it to go into saturation.
Thus, although BJT's are more often driven into saturation for switching
applications this isnt *always* the case.
 
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smanches

New Member
You always want a mosfet in the linear region for almost all applications. WHERE it is being operated in the linear region depends on the application. For switching, you want the Rds to be nearly as low as possible, which means using the recommended gate voltage from the datasheets. Some amplifiers run them in the lower linear region to take advantage of a more "linear" part of the curve.

The saturation region of a mosfet is where the DRAIN is saturated and cannot pass any more current. At this point, the mosfet is a constant current limiter. No matter how much higher you make Vdd, it will not push any more current through the device. This is the main difference between the linear region (where gate voltage affects Rds, and Vdd directly affects Ids) and the saturation region (where gate voltage affects Ids directly, regardless of Vdd)

Btw, there was just a long thread about this very same question a few weeks ago.
 
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q5101997

New Member
Running a MOSFET in the linear region disipates most power (controlled resistance). For high current, low speed switching (motor speed controller), its best to drive the gate as high as possable with as much current as possable, so it charges quickly and turns Vds on and off quickly.
 

picbits

Well-Known Member
Running a MOSFET in the linear region disipates most power (controlled resistance). For high current, low speed switching (motor speed controller), its best to drive the gate as high as possable with as much current as possable, so it charges quickly and turns Vds on and off quickly.
This is how I've always driven them.

The linear region is where they dissipate the most power due to internal resistance hence either driving them hard on or hard off.
 

shambhukumar

New Member
i think it will work...
why does LED lamp glow.it is because in BJT base is forward biased with respect to emitter.so current goes from collector to emitter. if we make base -emitter reversed biased then there will be no current from collector to emitter and LED will not glow.
just apply some voltage at base of BJT to make initially LED on.now connect the emitter with the sensor (heat sensor) output.i.e voltage produced by the heat sensor.initially LED will be glowing. when there will be too much heat then sensor voltage will increase and at some time it will be more than base voltage. at that situation base emitter will be reversed biased and there will be no more current coming from collector to emitter. thus LED will be off. i hope it will work.if you want to stick to BJT.
 

RCinFLA

Well-Known Member
From vacuum tube lingo, it's called triode region, as opposed to constant current region where Ids flattens out versus Vds.
 

Roff

Well-Known Member
i think it will work...
why does LED lamp glow.it is because in BJT base is forward biased with respect to emitter.so current goes from collector to emitter. if we make base -emitter reversed biased then there will be no current from collector to emitter and LED will not glow.
just apply some voltage at base of BJT to make initially LED on.now connect the emitter with the sensor (heat sensor) output.i.e voltage produced by the heat sensor.initially LED will be glowing. when there will be too much heat then sensor voltage will increase and at some time it will be more than base voltage. at that situation base emitter will be reversed biased and there will be no more current coming from collector to emitter. thus LED will be off. i hope it will work.if you want to stick to BJT.
What are you doing? You made this exact same post (#22, here) about a totally different topic.
 
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