Current mirror.

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Hey guys,

I got back to the lab and I wanted to see if i indeed built a 18-19mA current source.
So Instead of 150ohm load (R2), I connected a 75ohm load (R2), and the current I got was IDS2 = 26.5mA.

Do you have any idea whats wrong here?
I think that Q2 might not be saturated (I measured VGS = 2.2V), i'll check its VTN and post it up here.

 

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According to the **broken link removed** of the BSS670S2L L6327, VTn is 1.2V-2V.
The VGS in the circuit is 2.2V

For R2 = 150ohm (18.7mA), VDS = 1.89V > VGS - VTn.
For R2 = 75ohm (26.5mA), VDS = 2.7V > VGS - VTn.

I dont understand why I got two different currents for two different loads
 
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Yes its strange.

I conducted the experiment again here are the results:

R1 = 100ohm, IR1 = 26.85mA, ***VCC = 5V.
VGS = 2.3V

Each time I used another load:
R2 = 150ohm, IDS2 = 20.5mA, VDS2 = 1.99V.
R2 = 75ohm, IDS2 = 27.6mA, VDS2 = 2.92V.
R2 = 25ohm, IDS2 = 38.9mA, VDS2 = 4.02V.

I really dont understand whats wrong here.

--
Edit:
I added an important detail that VCC = 5V.
 
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R2 = 150 ohm; (4.7 - 1.99)/150 = 18mA
R2 = 75 ohm; (4.7 - 2.92)/75 = 23mA
R2 = 25 ohm; (4.7 - 4.02)/25 = 27mA

The current is tighter than you think. Have you thermally coupled T1 and T2? If not, then do so.
 
BrownOut said:
R2 = 150 ohm; (4.7 - 1.99)/150 = 18mA
R2 = 75 ohm; (4.7 - 2.92)/75 = 23mA
R2 = 25 ohm; (4.7 - 4.02)/25 = 27mA

The current is tighter than you think. Have you thermally coupled T1 and T2? If not, then do so.
Click to expand...

Hey, i'm sorry for not updating you about it.
I changed VCC from 4.7V to 5V.
Again, sorry, I thought only to give you these details.

Anyways, this is the only difference (the VCC).

About thermally coupling Q1 and Q2, I only soldered them as close as i could get.
The gates are touching each other, but the drains are 1cm away from each other.
The sources are 0.5cm away from each other.
 
It is critical in current mirrors that the transistors be in thermal contact. Otherwise, the output won't track the diode connected transistor's current.
 
Oh I see.

Do you think that its that crucial? (an error of 38.9mA/26.8mA * 100 = 145%).

How do you thermally couple the transistors please?

Thank you for your help and guiding.
 
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.
 
alphacat said:
Oh I see.

Do you think that its that crucial? (an error of 38.9mA/26.8mA * 100 = 145%).

How do you thermally couple the transistors please?

Thank you for your help and guiding.
Click to expand...

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.
 
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?)
 
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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.
 
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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).

 

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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).
Click to expand...

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