Simple current source - but what doea the mosfet do?

[FriedMule]

I have this simple current source with a added pot for adjusting the current / dimming the LED
I have also added a switch across the mosfet and I see no change if the mosfet is shortet put or not, why is the mosfet there?

http://tinyurl.com/y8qbs72e

[Jwillis]

It looks like it's using the transconductance of the JFET to sink some current to ground. Increase the beta or transconductance and it sinks more current .

[FriedMule]

Thanks, unfortunately have I not found much that explain transconductance but I'll try to understand that when I have learned more of the basic stuff.

But as I understand the N-ch JFET, is it always closed in this circuit, just like if you had a jumper or just some copper trace.
Why use a permanently open JFET instead of a copper trace?

[Jwillis]

I'm still learning to so I could be wrong. Maybe one of the experts could explain it better.

[Dave]

A transistor is like a valve on a faucet. There are states between fully open and fully closed where it is just partially open.

But as I understand the N-ch JFET, is it always closed in this circuit, just like if you had a jumper or just some copper trace.

That is incorrect.

[FriedMule]

Thank you both!
So the JFET is like driving at full speed but with the possibility to throttle down if the JFET "sense" it has to?
But the ground can never go below zero and therefore never close, only limit a small degree?

[Benta]

I don't understand the 1.5 kohm resistor, and I don't understand the switch.

Removing those, it's a perfectly normal JFET adjustable current sink. Built millions of times and works wonderfully.

It's NOT a MOSFET.

And please: disable the 70s disco-light animations.

[FriedMule]

The 1.5k is to allow current to go to zero. The switch is to illustrate what I mean by the LED functioning perfectly no matter if I have a JFET or just a piece of wire.
What I am trying to learn is why the JFET is there, what it do, when it is always closed and never open. A piece of wire is also never open but always closed. So why put a component in the circuit that never get triggered?

[Dave]

The current through the LED actually changes dramatically when you short the transistor.

Also the terminology with transistors is the opposite of relays and switches - when a transistor is open, it conducts current, when it's closed, it doesn't.

[JackJones]

Here's a little modification to hopefully make it clear: http://tinyurl.com/yaz4oeo2

I added some ripple on top of that DC voltage. If you close the switch and short out the JFET the current follows that ripple. (it's a lot higher too)

But with the switch open the JFET is in circuit and it creates a nice and steady DC current, despite the ripple. That's the reason for it. It acts like a current sink, keeping the current constant even with varying voltages.

[Manul]

I guess you have misunderstanding about different FETs.

N-channel mosfet as you know typicaly has a gate threshold voltage somewhere from +2 to +5V. That means if gate voltage relative to source is above that, the mosfet starts to turn on.

Now N-channel JFET has a threshold voltage (actually it is more common to say pinch-off voltage) in the negative region. Like for example -4V to -1V relative to source. So JFET is conducting and is in fact almost fully on when gate is at the same potencial as source. To turn it off, you need to drive gate to lets say -4V (4V below source potential).

In this circuit there is voltage drop resistor on the source. So when current increases, the source is becoming more and more positive. But gate stays at zero. So gate is becoming negative relative to source. As gate becoming more and more negative, JFET starts closing, so current decreases. So it effectively regulates (balances) current.

[Benta]

Just as Manul and I said:
you don't understand the operation of a JFET. Remove the 1.5 kohm resistor and the switch and do your disco-simulation again.
A JFET lets through a constant current from drain to source depending on the gate voltage.
Shorting the JFET gate to source is what's known as a "current regulator diode". The variable resistor in the source-ground limb allows you to regulate this current.

Now, if you're doing a simulation using a MOSFET, it'll never work. Period.

Try simulating with a 2SK170 (a very famous N-JFET). This should give you ~12 mA if the variable source-ground resistance is zero. The current will drop as you increase the resistance.

After you've done that little task, we can discuss how to completely cut off the current (your dubious 1.5 kohm resistor).

[FriedMule]

The current through the LED actually changes dramatically when you short the transistor.

Also the terminology with transistors is the opposite of relays and switches - when a transistor is open, it conducts current, when it's closed, it doesn't.

Sorry my terminology is wrong, what I meant were open=nothing goes trough, extremely high resistance.

Here's a little modification to hopefully make it clear: http://tinyurl.com/yaz4oeo2

I added some ripple on top of that DC voltage. If you close the switch and short out the JFET the current follows that ripple. (it's a lot higher too)

But with the switch open the JFET is in circuit and it creates a nice and steady DC current, despite the ripple. That's the reason for it. It acts like a current sink, keeping the current constant even with varying voltages.

Wow thanks, what a great illustration, thanks for that, it learned me a lot!! :-)
It looks like the JFET do act as a throttle to keep the current constant.

I guess you have misunderstanding about different FETs.

N-channel mosfet as you know typicaly has a gate threshold voltage somewhere from +2 to +5V. That means if gate voltage relative to source is above that, the mosfet starts to turn on.

Now N-channel JFET has a threshold voltage (actually it is more common to say pinch-off voltage) in the negative region. Like for example -4V to -1V relative to source. So JFET is conducting and is in fact almost fully on when gate is at the same potencial as source. To turn it off, you need to drive gate to lets say -4V (4V below source potential).

In this circuit there is voltage drop resistor on the source. So when current increases, the source is becoming more and more positive. But gate stays at zero. So gate is becoming negative relative to source. As gate becoming more and more negative, JFET starts closing, so current decreases. So it effectively regulates (balances) current.

Oh yes I see, I just thought that "connected to earth" clamped it to ground current and Voltage, but I can see it's not what is happening.
But do the JFET's model / data not have a lot to do with the amount of current it allows to go trough?
If so, how do I select the right JFET? It may be a big difference if you need 200mA or 5A?

Just as Manul and I said:
you don't understand the operation of a JFET. Remove the 1.5 kohm resistor and the switch and do your disco-simulation again.
A JFET lets through a constant current from drain to source depending on the gate voltage.
Shorting the JFET gate to source is what's known as a "current regulator diode". The variable resistor in the source-ground limb allows you to regulate this current.

Now, if you're doing a simulation using a MOSFET, it'll never work. Period.

Try simulating with a 2SK170 (a very famous N-JFET). This should give you ~12 mA if the variable source-ground resistance is zero. The current will drop as you increase the resistance.

After you've done that little task, we can discuss how to completely cut off the current (your dubious 1.5 kohm resistor).

No you are right, I did not understand the circuit right. :-) Did I say MOSFET earlier? If so, am I sorry, sinse yes a MOSFET, BJT and JFET is nothing like each other.
My reason for the 1.5k is I was recommended to use it, if I would let it be a precision dimmer, down to zero.

It's thanks to you guys that I learn so much, I am super grateful!! :-)

[Manul]

Also you need to know about JFET saturation. In general for some fixed gate voltage, JFET conducts some amount of current and this current is not changing a lot with a changing drain to source voltage. Source resistor makes this effect even sharper, because as current increases gate gets more negative. So it is negative feedback going on. And of course it sets the saturation current itself, so it becomes adjustable constant current sink - a path, where current stays more or less constant while voltage changes.

[FriedMule]

About using the current source, I can of cause make a LED glow as much I want. But lets say I want several LED's to light up with exactly same lumen and they are specified to 20mA typical and 50mA max. I imagine it means that every single LED, typical will need maybe 15mA-25mA to glow with the same lumen?

Except building a current source for every LED and adjust the pot for every single LED, is there not a simpler way to make them all glow with same lumen?

[Manul]

Powering multiple indentical LEDs in series should give close to equal lumens. How equal do you need? Also do not forget that there will be optical variantions between LEDs. So multiple LEDs may not have identical pattern and intesity even if you drive them equally.

[FriedMule]

What I mean is the ability to adjust the lumen on every single LED. I know that I can just build 10 current sources with 10 pots and adjust them separate, but there has to be an simpler solution?