Fet voltage regulator: Why/how does it work

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Hi all,
:ashamed: I fear this to be a darn silly question, but I can't help myself here. :ashamed:
Take a FET voltage regulator, e.g. the borbely thing here

Q7 is a average source follower, but the reference voltage is not set by a simple zener or similar, rather by the diff-pair that controls the voltage through feedback path via c7/R7.

The part I don't understand, and I think there might be something fundamental I do not get: Isn't the left side gate of Q1 positive biased in respect to the source? I was under the impression that a N-jFET can't work with positive gate voltages above say 0,6V :dead:
How does it work?

thanks,
Rüdiger
 
D5 and D6 set the reference voltages into the diff-amp. The diff-amp is comparing the reference voltage with a sample of the output voltage. Q7 is the VAS, feeding Q9 which is the power source follower. It's a negative feedback loop that controls the output voltage.

The JFETs self-bias, with the voltage at the source pins rising until it is held at whatever the gate voltage is at.

Note that the supply as shown has no current-limiting mechanism.
 
The JFETs self-bias, with the voltage at the source pins rising until it is held at whatever the gate voltage is at.

Thanks for that explanation.
Note that the supply as shown has no current-limiting mechanism.
Could a current source before the regulator cure that?

At the expense to say something silly again, couldn't we relativly easy tweak that circuit to be a shunt regulator, with Q9's source going to ground, a low value resistor in the drain, leaving C7 and clamping R7 to the positive rail? Or is that totally ***not so good***? If not, i guess the noise reduction wouldn't be so stiff?

Sorry in case I bore you,

Rüdiger
 
The circuit as shown can only be used as a series regulator. A shunt regulator is quite easy to design, have a look at some of the app-notes showing how to add circuitry around a TL431 to give you some ideas.


The usual way to provide current limiting is to sense the current through a low-value resistor, which turns on the Vbe junction of a transistor when the voltage across the resistor reaches 0.6V or so. I've not studied this regulator circuit to see how to add this though.
 
I tried to compile a somewhat simpler version of that regulator with parts out of the box intending to get 9V out of 12V.
example: http://www.borbelyaudio.com/pics/EB-604-415-regulator-web.jpg

While I succeeded to setup the differential pair regulating the voltage at the gate of Q5 (from about 2V to 12V), the output of the mosfet (in my case irf9160) did not change but giving only the full 12V. Do I need another part with more gm (like the 2sj148) or is something else wrong here?

Rüdiger
 
If you can change the voltage on the gate of the P-channel pass transistor (Q5) then the output voltage should change.

Did you have a load resistor on the regulator and (I know this may sound silly), but are you certain that you had the D and S of Q5 connected the right way round?
 
No, I'm not sure if I have the correct pinout cause the datasheet does not show it. I assumend it to be the same as its N-Type complement.
However, would the device pass the voltage if it died?
I once had an jfet that dissipated the right amount of IDSS but did not work in the circuit any more.
thanks for your helping,
Rüdiger
 
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