Trying to understand a MOSFET

[robzy]

Hey guys, ive read a few different webpages on MOSFETs but it seems as if they either go into too much detail, or too little.

Here's what i know so far:
- Essentially infinate impedance on the gate (for DC)
- Transconductance refers to amp per volt

But thats as much as i know, and as for its actual electrical behaviour i get very confused. Take this circuit for example:

(Assume transconductance = 1)

[edit]: Can we assume that the MOSFET will actually be "happy" with 1volt at its gate? Just for theory?

If i apply 1 volt to the gate, that means that the current through the mosfet would be 1 amp, correct?

This would mean that one amp would be flowing through the resistor, and the voltage drop accross the resistor would be its resistance (as per ohm's law) - correct?

And then if i apply 2 volts to the gate, that means that the current through the mosfet would be 2 amp. Those two amps would then be flowing through the resistor, and the voltage drop accross the resistor would be its resistance multiplied by two in volts (as per ohm's law) - correct?

From the theory i've read that would all be correct. Although i think i am missing something, as in the simple simulation i ran (through an, albeit, very dodgy program) the resistance of the resistor ended up influencing the current through the MOSFET in quite a major way. So is there something i am missing? Or is it a bad simulation.

Thanks a lot,
Rob.

[edit]: I forgot one of the most important things, the voltage from Gate to Source has to be over 3, doesnt it? What happens if it is under? Will the MOSTFET not conduct?

 

[peterpan188]

First of all, MOSFET does not have any current passing through the gate, so lets modify what you said about "current through the MOSFET".

IMHO, its better to read up about BJT first, it will be easier than to read on about MOSFET. That's how I learnt it.

Regard,
Peter

PS: Sorry for not going deep into this, really sleepy now. When I get a chane tomorrow I will come back to this.

 

[robzy]

Sorry, when i say "current through the mosfet" i mean Ids. Since, in my mind at least, for DC applications Igs=0 making that simplification should be more or less okay - right?

As for a BJT, if you replace the mosfet in that circuit with a BJT i would fully understand what would happen. But if it were a BJT then i know that the value of the resistor would not change Ice, where the simulation i was running was telling me that the resistor would change Ids.

Rob.

 

[amb]

If you already know how a BJT works, then a MOSFET is only slightly different. For a BJT, Ic = Ib * hfe, where Ic = collector current, Ib = base current, and hfe = DC current gain of the BJT. Ie (the emitter current) is the sum of Ic and Ib. Because the Ib "drives" Ic, the BJT is a current-to-current amplification device.

For a FET (both MOSFET and JFET), the drain, source and gate are the conceptual equivalent of a BJT's collector, emitter and base, respectively. The main difference is that the FET's gate input is not a current, but a voltage, so it's a voltage-to-current amplification device (much like a vacuum tube). The FET gate has a very high input impedance, which for practical purposes can be treated as infinite for DC analysis. For a MOSFET, Id = Vgs * gm, where Id is the drain current, Vgs is the gate-to-source voltage, and gm is the transconductance. Since there is essentially no gate current, Id = Is.

Note that Vgs is the gate-to-source voltage, not gate-to-ground. In your schematic, as Id increases, the voltage across R1 will also increase, which causes the voltage at the source pin (relative to ground) to go up too. Unlike silicon BJTs, where Vbe is fairly constant at around 0.6V-0.7V due to the "diode" at the base-emitter junction, a MOSFETs' Vgs will vary due to the nature of its voltage input. Virtually all MOSFETs are of the "enhancement" type, which means that when Vgs = 0, Id = 0. The threshold Vgs where the MOSFET begins to conduct varies between different devices.

Contrast that to "depletion" type devices like JFETs, which self-bias when Vgs = 0 and Id in that scenario is the maximum the device will flow (this is the Idss parameter). To reduce a JFET's Id, Vgs must go negative (assuming N-channel device), until we reach the cutoff threshold where the device stops conducting. This cutoff is also device-dependent.

Note also that the BJT's hfe is not a constant, but vary with Ic, Vce, temperature and other parameters. Similarly, the FET's gm also vary with the Id and Vds, etc. When the device is to be used in a linear fashion (rather than as a switch), the bias should be carefully chosen to minimize distortion prior to application of negative feedback.

These descriptions ignore the input capacitance of all device types which must also be considered when designing an amplfier for AC signals. A MOSFET's input is typically much more capacitive than a BJT's.

 

[robzy]

Quote:

Originally Posted by amb Note that Vgs is the gate-to-source voltage, not gate-to-ground. In your schematic, as Id increases, the voltage across R1 will also increase, which causes the voltage at the source pin (relative to ground) to go up too.

That is exactly what i had wrong. While i knew that - i forgot to take it into account with my calculations.

Something that seemed kinda interesting to me:

Assuming:
- R = 1ohm
- Transconductance = 1

Vgs = Vg - Vs
Ids = Vgs
Vr = Ids
.: Vr = Ids = Vgs = Vg - Vs = Vs
.: Vg - Vs = Vs
.: Vg = 2Vs

Which means that if i apply X amounts of volts to the gate, twice (edit2: actually, make that half) that voltage will apear at the source.

edit: Actually, that doesnt make sense. After checking a common drain has only current gain. What have i done wrong? :S

edit2: Ah, i see what. What I've actually done is made a voltage gain that is less than one (a la Szekeres). I mucked up the fractions

Rob.

 

[robzy]

Hey guys, i come baring another question:

If you have a MOSFET setup as a common drain amplifier (is that the correct term?) but replace the source resistor with a CCS, how does that affect things?

To me that just doesnt make sense - at Vg = x the MOSFET will want to pass maybe 100-500mA, but the CCS only wants to pass 200mA (For example), how does the "fight" end up getting resolved?

Thanks a lot,
Rob.

 

[amb]

Quote:

Originally Posted by robzy If you have a MOSFET setup as a common drain amplifier (is that the correct term?) but replace the source resistor with a CCS, how does that affect things?

The CCS will have a much higher impedance than the resistor, but being an active device, it will want to pass the amount of current that it's supposed to pass (provided that you bias the MOSFET to allow at least that much Id to flow).

Quote:

To me that just doesnt make sense - at Vg = x the MOSFET will want to pass maybe 100-500mA, but the CCS only wants to pass 200mA (For example), how does the "fight" end up getting resolved?

As you increase Vgs from the turn-on threshold, Id will start to go up, but will stop increasing when the "programmed" CCS current is reached. In this case the CCS works like a current limiter.

 

[robzy]

Quote:

Originally Posted by amb As you increase Vgs from the turn-on threshold, Id will start to go up, but will stop increasing when the "programmed" CCS current is reached. In this case the CCS works like a current limiter.

Hrm, i think i have now ended up a touch confused about CCS's that are on the source pin.

Taking an example: If i have a n-channel-mosfet set up with 20v on the drain, 8v on the gate and a CCS set to 700mA on the source going to ground - what will actually be happening? (Assuming the MOSFET has a transconductance of .1) (What would the Ids be, and what will the voltage drop across the MOSFET and CCS be?)

Thanks a lot,
Rob

[edit]: My problem can be put down to not fully understanding current sources. Wikipedia linked me to http://www.st-andrews.ac.uk/~www_pa/...rt1/page3.html though and i now understand whats happening.