# MOSFET as voltage buffer - is this correct?

Discussion in 'The Projects Forum' started by Dyslexicbloke, Sep 8, 2010.

1. ### Dyslexicbloke Thread Starter Well-Known Member

Sep 4, 2010
528
34
Hi folks,
Trying to break down some stuff I don't yet fully understand into manageable chunks.
What I know about electronics is self taught and I am increasingly realising that I lack understanding of some basic circuit elements.
(Mostly because of things I am learning here right now)

Is the attached going to work 'like an emitter follower' (I think) and maintain the Source at VG + VGS(th) circa 12V.

VGS max = +/- 30V
VGS(th) min = 2v ( Dose this imply a range that corresponds to )
VGS(th) max = 4v ( fully Off @<2v and Fully On @>4v )
VDS = 400v
W max = 125W

If I am not correct, what am I missing?

(The range of input voltage is because R Load, the circuit I want to supply modifies the incoming AC once it is active)

Thanks
Al

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2. ### beenthere Retired Moderator

Apr 20, 2004
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That is not correct. The threshold voltage is some voltage difference between the source and the gate that just starts the device to conduct -
In this case, it is from 2 to 4 volts. Full conduction need a Vgs of at least 10 volts.

Part of the fun here is that any current that gets passed through the FET will cause a voltage across the load resistance. That will affect Vgs in a way that causes the FET to turn off. Your maximum conduction might only be 15 - 40 ma.

FETs make poor linear devices, unlike transistors.

3. ### Dyslexicbloke Thread Starter Well-Known Member

Sep 4, 2010
528
34
Ah OK that makes sense given that I keep seeing references to much higher gate drive voltages than range above.

I get the whole 'source comes up reducing VGS' once the device starts to conduct, that is exactly what I was looking for.

Obviously I am not understanding the spec sheets, is there some way to extract a satisfactory gate voltage, device fully on i mean, from the spec?
Max VGS cant be right can it?

Is the region between VGS(th) and whatever gate voltage is required to fully turn on the device likly to have a linier relationship to the RDS(ON)?

Sorry if I am way out here but I learn quickly.
Thanks for the help thus far
Al

4. ### Ghar Active Member

Mar 8, 2010
655
73
FET current is proportional to (Vgs - Vth)^2

Since the gate voltage is fixed it is going to conduct a current such that the source voltage increases to a point where current through the load resistor balances the operating point of the FET.

Ids = k (Vg - Vth - Ids*Rload)^2

k isn't really given on datasheets and won't be very linear, you can sort of figure this out using the Ids and Vgs curves on the datasheet.

Here is an example of how you could try to figure out the operating point from the datasheet of this part:

(my writing on the bottom)

So, with R = 130 ohms you see that Vgs will stay just above the threshold, so Vs will be pulled up to essentially Vg - Vth = 6V
(The load line I drew is very inaccurate, since 50 mA is 100 times smaller than the 5A division)

With R = 1.3 ohms Vgs will increase, meaning Vs will stay much lower.
You get Vgs = 5 V, and Vs = Vg - Vgs, or Vs = 5V.

This will vary quite a lot based on the FET's exact transfer characteristic as you can see by how much the curves vary with temperature even.

Also take a look at the datasheet's plot of Rds-on vs Vgs... it's not as abysmal as people make it sound.
The Vgs = 10 V is simply the point where they guarantee the spec.
It's listed as the test condition under the Rds-on spec.

The relationship of Rds-on to Vgs under the assumption of being well on (Vgs - Vth >> Vds) is approximately:
$R_{ds-on} \alpha \frac{1}{Vgs - Vth}$

Check out all the curves on the datasheet, they show quite a bit.

Last edited: Sep 8, 2010
5. ### Dyslexicbloke Thread Starter Well-Known Member

Sep 4, 2010
528
34
Thats a fantastic explanation thanks .... There are newances of it that I wll have to study but I get the gist of it ....

cant believe I added TGS(th) instead of subtracting it, that was the only bit I thought I did understand ....
Ah well loads to learn, starting with fully understanding the figurs you provided.

Anyway the concept is basically sound right?, I can use the fet as a variable power resistor with a fixed gate and floating source.
The output will vary a little with load and temperature and if I want to be more precise and find the actual values and practical limmits I need to test it.

It sounds like the sort of thing that dosnt get done in practice because there are better ways of achieving the same goal.
Should I be abandoning this and looking at another way to go?

I had sort of envisiged this being used to bleed off anything over 30V and allow me to use an LM317 as the final reg.
Al

6. ### Dyslexicbloke Thread Starter Well-Known Member

Sep 4, 2010
528
34
Whilst trying to understand the answer I recieved earlier a little better I have looked at all sorts of stuff, turns out I still had it wrong .... well partially.

It turns out that the circuit is [Wikipedia] 'a common-drain configuration also known as a source follower, it is analogous to a BJT common-collector amplifier.'

Anyway found the attached, and much basic concepts here:-
http://www.softwareforeducation.com/electronics/notes/AS/mosfet/sourcefollower.php

Thought somone else my find it as useful as I did, it gave me even more to look up IE other 'standard' configurations and their uses.

Thanks again Ghar
Al

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7. ### shortbus AAC Fanatic!

Sep 30, 2009
4,549
2,588
Al, I'm in the same boat as you. Self taught and learning that what I thought I knew isn't always correct. But the guy's here have been great ,helping out.

The schematic from the link you posted isn't correct as I now understand things, I don't think. From what I understand, when a N channel mosfet is used like shown, it won't turn on unless the gate is 10V higher than the source. As the source starts to rise in this circuit it is going to put the supply voltage back into the gate voltage frying what ever is supplying the gate voltage.

The way this is drawn is called a "high side switch" For a high side switch you need a mosfet driver.

If I'm wrong in this, I hope someone will chime in and explain more!! Hope I didn't hijack your thread too much. I just thought we could both learn.

8. ### Dyslexicbloke Thread Starter Well-Known Member

Sep 4, 2010
528
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The source is isolated from the gate, in fact every thing is isolated from the gate.
In a FET N or P the channel is opened, enhancement mode or closed, depletion mode by the Field around the gate, internal to the device channel obviously.

I am not saying that current isnt required to charge and discharge the gate but exactly like a capacator no current path exists between the gate and anywhere.

Precisly how BJT's DONT work ... I am fairly sure that is right ...
maby I will get corrected, it happens a lot.

The first time I used a FET was a couple of weeks ago, now that was a steep learning curve involving huge amounts of blue smoke and alarming flashes, You should have seen my first attempt at a totem-pole driver detinate !!!!
I got it totally wrong and dead shorted about 6600μF bank (3) of 3 big electrolitics charged to around 30v.

Possibly not as spectacular as the first generator field control circuit bursting into flames ... but I was primed and ready for that.

Cap discharge desulfater now working and no flames at least from the genny.
Progress
Al

Al

9. ### Ghar Active Member

Mar 8, 2010
655
73
If you apply a positive voltage to the gate above threshold and there is some sort of source connection, such as a resistor, a short, whatever, it WILL turn on. Always. The only question is how much current you'll get (if saturated) or how low the resistance will be (if triode, aka 'fully on')

With very light loads (high resistance, capacitor, open circuit) the FET will pull the source up to gate minus threshold, Vs = Vg - Vth.

With heavy loads (low resistance, short circuit) the FET can't pull up appreciably without significant current. Since current is related to (Vgs-Vth) you must have Vgs increase.
This means Vs will rise but Vs < Vg - Vth because Vgs is constrained by the I - Vgs relationship.
In the case of a short Vs won't be pulled up at all, you simply have Vgs = Vg

Again, the 10V is simply a specification test condition. Look at the plot of current vs Vgs or Rds-on vs Vgs and you'll see that it can conduct appreciable current as long as Vgs >> Vth

10. ### Dyslexicbloke Thread Starter Well-Known Member

Sep 4, 2010
528
34
Yep ... not suprised at all that you are bob on ... see I had to go test that!

400V fet I have lying around.
25VDC on the rails / 470R S > gnd / 470K G > gnd (just so I could disconnect it)

VG from my bench supply > comon gnd

One can easily adjust the temperature of the resistor (1/2W)
The fet didnt apriciably warm up.
S tracks G with an increesing, but aparamtly linier, negative offset that is about 2.2V with VG @ 10V.

Now to try with 120VDC rails ... perhaps i'l do some calcs first
Al

Last edited: Sep 9, 2010
11. ### Dyslexicbloke Thread Starter Well-Known Member

Sep 4, 2010
528
34
Given that VD is not playing a part here if the upper rail is anywhere above VGS(th) the FET will be on, at least partially.

Adding a voltage reg with its supply connected to S should be fine.

If I use an LM317T, set for 10V and its input is 1.2v or more above that but less than 30V that too should be fine too right?

Dose anyone know what would happen if VS and hence the reg input is below the set point of the reg. will it pass what it has or shut down?

I know power disipation will be important but I am only looking for 10mA or so total output at 10 V.

I think i worked out it was good for about 40mA whilst disipating 25% of the spec for the FET (W that is).

That said I have never got my head arround doing this sort of thing propily and taking thermal conductivity into account.

Comments anyone .. all are welcome as always and thanks for all the help so far.

Thanks
Al

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Last edited: Sep 9, 2010
12. ### Ghar Active Member

Mar 8, 2010
655
73
The power dissipation specs are always given for infinitely large heatsinks (case temperature of 25C) and the numbers are never achievable.
You need to do a thermal calculation.

13. ### shortbus AAC Fanatic!

Sep 30, 2009
4,549
2,588
Al and Ghar - I have been told numerous times on this and other forums that a N fet that is not connected source to ground is considered high side. Are the others not correct? Any time a load is connected to source it is high side right?

If the circuit shown will work why do they make high side driver chips? If the Vin shown is from a logic gate or other circuit won't the higher =supply voltage feed back into that logic gate or circuit?

14. ### Potato Pudding Well-Known Member

Jun 11, 2010
684
92
SMPS. Switch Mode Power Supplies.

Please tell me you have at least looked at them.

As interesting as it is to look at using a FET as a linear regulator there are so many reasons why you almost never see this done and you see so many SMPS.

You are talking about Power dissipation when the FET should be switching instead and averaging less than a quarter watt.

Power equals Voltage X Current

Switch between Some/High current with no voltage and Some/High voltage with no current.

Both states multiply by nothing so you get no power loss in the transistor (idealized - milliwatts possible in practice).

The voltage current drop is all across your coil which recovers power by transformer action.

Linear needs 10 Amps input to get 10 amps output.

A switcher that drops 50 volts to 5 Volts needs maybe just over 1 Amp at 50 Volt input to get 10 Amps at 5 Volts.

Please give up on linear regulation for anything more than 10mAmps or at most 100mAmps.

15. ### Ghar Active Member

Mar 8, 2010
655
73
It is high side you're right but that's just a name.

The circuit does something useful but it's not the general way of using a FET. Notice how I explained how Vgs depends on the load, and available current / on resistance depends on Vgs.
It's not a situation you want to be in usually.

With a switching supply you need the FET on hard and fast with a predictable Vgs which turns it on completely and minimizes on resistance.
To do this with a high side FET the gate voltage needs to be higher than the drain voltage (usually the highest input voltage available).
Among other things a high side driver for an NMOS will generate a voltage higher than the input to do this.

In this application he just wants the FET to drop some voltage, he's using it in a linear way. It doesn't really matter how 'on' it is. [it won't work for all loads so it still matters to some degree]

The supply voltage won't leak through the gate assuming the voltage isn't too high. The constraint for this FET is that is that |Vgs| < 30V
If you exceed that then the FET will be destroyed.
The other limit is that Vds < 400V
As long as you stay within those limits the FET won't breakdown.

16. ### Dyslexicbloke Thread Starter Well-Known Member

Sep 4, 2010
528
34
Thanks for the input guys ... I know this isnt 'normal' but then the application isnt normal either.
The rails here are from the secondary stator in a genset.
The circuit supplya a switching, PWM, circuit that controls the genny field with a fet in a 'standard' switching role.

What this circuit needs to do is supply the field control circuit when the genny first starts up and dosnt yet have any field current.
It starts the oscilator and switches the PWM FET on within two cycles the voltage will from the secondary winding builds to over 100 and the PWM circuit, still being supplied from the circuit, starts backing the field off.
Obviously when its all running a standard switching supply would make more sense, rather than this linier arrangement I mean, but it wouldnt work at startup.
I supose I could use this to suppply rails to a switching reg with a much higher potential current capability and effectivly have the switching supply ' take over' once the voltage had come up but as I dont need any more current than this will give me I dont see the point.
-- Unless I am missing something. --
Anyone know about how the 317 will handle to low a supply voltage?,
I havnt tested that yet.

Al

17. ### Potato Pudding Well-Known Member

Jun 11, 2010
684
92
Ok, I just wanted to know that you had looked at switching supplies.

I don't think the 317 has a low voltage shutdown. I think it just saturates and tries to not drop too much of the remaining voltage.

I haven't used one in a while, so I am not much help. I would find a datasheet and check to see what it says.

18. ### Dyslexicbloke Thread Starter Well-Known Member

Sep 4, 2010
528
34
Hay ...
As I am always saying, any and all comments greatfully accepted.

I had considderted a SMPS buit you wernt to know that and the comments were not only valid but sencible and conventional to boot.

Thanks for taking the timre to reply
Al