Ideally they should be in the same package, but closely binding them to each other with a common heat sink is the next best way to go.

__________________

 

Your error calculation is wong

(38.9 - 26.8)/26.8 * 100 = 41.8%. I won't say that thermally coupling will eliminate the error. Eliminate as many variables as possible to study the operation of the circuit.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

When I come to think of it, the current through Q2 should have been constant even though we changed load since Q2 was saturated - meainng its in the constant current region.
(could it be that λn is that big?)

 

I doubt it. As you change Drain resistors, you require the transistor to dissapate different amounts of voltage. That in turn heats the device to different temperatures. The constant current of a saturated device depends on constant temperature. By thermal coupling, you affect a thermal feedback system.

You can look at the characteristics curves to see how Lamda will affect the output current. I'd do it for you, but then I'd rob you the richness of doing your own investigation.

Another thing you might try is simulating your circuit.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

Well, it seems that Lamda does have an effect on IDS, not that small an effect, but not that big also.

I think i'll first try to take one MOSFET and see how it behaves in its constant current region.
I'll also simulate it before i'll do that, its a great advice.

Are you saying that all amplifiers are thermally coupled so their bias point wont be effected by different loads?

I'm still not sure how to have it thermally coupled (one transistor for example).

Attached Thumbnails

 

 

No, I'm saying all current mirrors need to be thermally coupled. One transistor cannot be coupled, obviously. There are other methods of stabilizing a single transistor amplifier.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

I see, thanks.

So how can you enter the mosfet into its constant current region, without it changing its current due to different power dissipations due to different loads?

Thats exactly what happened here.
VGS of Q2 remained constant.
VDS2 changed.
IDS2 changed.
That shouldnt have happened, so how do i prevent this when working with a single MOSFET?

 

You need to read over the thread carfully. We've been over the requirement to design and construct your current mirror properly. Remember that transistors are not perfect current sources, and all effects must be considered and accounted for.

__________________
You don't need a quadraphonic Blaupunkt -- you need a curve
ball.

 

Yes I understand.

The thing is that i'm talking about having only a single mosfet, driving its gate with a constant voltage from a PSU to get it saturated, and changing its load, hoping to see a constant current for different loads.

I'm afraid that i wont see constant current for the same reason IDS2 was different for each load (while VGS2 remained constant).

So my question is, how do i get the current of a single saturated transistor not to change for different loads?

 

You can't. You need some form of active feedback to do that.

__________________
Carl
Curmudgeon Elektroniker

 

The reason is R2 is higher value then mirror resistor allowing Q2 drain to drop lower then its gate voltage. The variance will get greater as Q2 get closer to saturation.

Different FET's will be more or less sensitive to the reaction to lower Vds voltage on the current mirror.