Source follower noob question

[mayor]

Hello,

based on the attached simplified schematic, and the Gate to Source voltage to Drain to Source current chart, I am a little bit stumped as to how a source follower can possibly produce a high amount of current, since as I understand it, Vgs will always be pretty much == the gate threshold voltage?

Can anyone point out what needs to be changed to source more current with such a setup?

Tx

[w2aew]

The op amp is going to drive the Gate until the voltage at the source (also inverting input of op amp) is equal to the voltage at the non-inverting input.  The amount of current provided by the FET depends on the load.  The heavier the load, the more current (and the larger the resulting VGS).

[mayor]

The op amp is going to drive the Gate until the voltage at the source (also inverting input of op amp) is equal to the voltage at the non-inverting input.  The amount of current provided by the FET depends on the load.  The heavier the load, the more current (and the larger the resulting VGS).

See, that's where I'm not entirely clear. You say Vgs will increase, but since it's a source follower, how will it do so? The gs voltage drop in the mosfet increases?

[w2aew]

The op amp is going to drive the Gate until the voltage at the source (also inverting input of op amp) is equal to the voltage at the non-inverting input.  The amount of current provided by the FET depends on the load.  The heavier the load, the more current (and the larger the resulting VGS).

See, that's where I'm not entirely clear. You say Vgs will increase, but since it's a source follower, how will it do so? The gs voltage drop in the mosfet increases?

Yes.  The MOSFET's drain current is a function of Vgs - larger Vgs give a larger drain current.  The closed loop with the op amp causes the output of the op amp to supply enough voltage at the gate to ensure that the drain current * load resistance equals the voltage at the non-inverting input.

[mayor]

Yes.  The MOSFET's drain current is a function of Vgs - larger Vgs give a larger drain current.  The closed loop with the op amp causes the output of the op amp to supply enough voltage at the gate to ensure that the drain current * load resistance equals the voltage at the non-inverting input.

Thanks. I understand the drain current is related to Vgs. But with the feedback in the follower configuration, Vg and Vs are tied to one another. So, I guess I am trying to understand how they are allowed to diverge so much. Does the internal resistance of the mosfet increase enough to explain it, etc?

[Benta]

First, learn how to draw a schematic. What you are presenting is unreadable. Paper and pencil might help, before putting something onto a printout.

Now, decrypting your "schematic", we have what is a MOSFET source follower.
Due to the feedback, the opamp will try to keep the voltage difference at the inputs at 0.
The noninverting input is at 3.5 V.
The inverting input should also be at 3.5 V, which means that 35 mA has to flow through the load resistor RL.
The opamp will drive the MOSFET gate voltage to a point, where this condition is fulfilled.
You do not need to think about gate cutoff voltage, unless it moves outside supply/opamp output voltage limits.

Simple, no?

[mayor]

First, maybe I don't draw great schematics, but I at least can be civil in every day life, and be respectful of others.
Second, no one is forcing you to answer.
Third, yeah let's not worry about it. Why try to understand, right?
Fourth, you're not even answering the question.
Fifth, no need to reply back, thanks very much.

[KerryW]

A FET is not like a BJT, the Gate to Source voltage can be any value (within limits).  In a BJT emitter follower, the emitter voltage will always be ~0.6V less than the Base voltage (thus a 'follower').  In a FET circuit this is not true, and it should probably be called a 'common drain' instead of a 'source follower'.

[Simon]

First, learn how to draw a schematic. What you are presenting is unreadable. Paper and pencil might help, before putting something onto a printout.

Now, decrypting your "schematic", we have what is a MOSFET source follower.
Due to the feedback, the opamp will try to keep the voltage difference at the inputs at 0.
The noninverting input is at 3.5 V.
The inverting input should also be at 3.5 V, which means that 35 mA has to flow through the load resistor RL.
The opamp will drive the MOSFET gate voltage to a point, where this condition is fulfilled.
You do not need to think about gate cutoff voltage, unless it moves outside supply/opamp output voltage limits.

Simple, no?

Moderator note: No need to be so aggressive.

[danadak]

There is a range of Vbe vs Ic in a BJT, albeit fairly narrow.

https://www.google.com/search?q=bjt+vbe+vs+ic&tbm=isch&imgil=844UH3rGMDGsZM%253A%253BrH1nQo4KXG4YWM%253Bhttp%25253A%25252F%25252Fwww.zen22142.zen.co.uk%25252FDesign%25252Fbjtbias.htm&source=iu&pf=m&fir=844UH3rGMDGsZM%253A%252CrH1nQo4KXG4YWM%252C_&usg=__jka3nuisdT4_PFgiK86Vr29OIqY%3D&biw=1517&bih=665&ved=0ahUKEwjEnZn_kIDRAhWIOSYKHdMcDqkQyjcILg&ei=u8FXWITUOojzmAHTubjICg#imgrc=ASKt1PWPvsPmuM%3A


Regards, Dana.

[salbayeng]

A follower is a follower, here is an analogy:

Consider a bull with a ring through its nose, that is wearing a harness (connected to a sled).

With the bull standing still the nose ring is 1metre from the harness attachment
When you pull on the nose ring , the distance to the harness attachment is just a little more than a metre.
Wherever the nose ring goes the harness follows. The legs do what has to be done to keep that 1m seperation.

So nose = gate  ,  harness = source ,  legs= drain,  force = current , distance =voltage

Wherever the nose ring gate goes the harness source follows. The legs drain does what has to be done to keep that 1m seperation Vgs =1v , in the process you can pull a heavy load current with very little effort.

(The analogy is better with a BJT)

[KerryW]

To extend the analogy a bit:  With a BJT, the harness is leather and can stretch a little so under heavy load the distance might increase to 1.1 meters.  With a FET, the harness is a bungee cord and can stretch to 3 or 4 meters.  The sled will follow (eventually), but you can't really predict the position of the sled from the position of the ring.
 

[mayor]

To extend the analogy a bit:  With a BJT, the harness is leather and can stretch a little so under heavy load the distance might increase to 1.1 meters.  With a FET, the harness is a bungee cord and can stretch to 3 or 4 meters.  The sled will follow (eventually), but you can't really predict the position of the sled from the position of the ring.

Great analogy, thanks! The elasticity of the harness is what I was struggling with, so this is great. So I guess the cord will extend far enough so that Vgs is large enough to allow required amounts of current to flow between D and S.

Thank you again for taking the time.