[SOLVED] Current Source of 10 uA

I did a lab experiment to design and implement a 10 µA current source .These details of the current source that I used is given below,
1) Widlar Current Source using a 3V DC supply
2) The range of load resistance is calculated to be from 0kΩ to 258 kΩ.

This is what I observed, from the theoretical range of load resistance the practical range of load resistance was obtained as follows,
1) 50-70 kΩ yielded exact current of 10 µA.
2) The voltage supply was found to be 3.109 V as compared to the supply reading of 3 V in the dc supply kit.The error is around 3 % approx. and the shift is observed as per this error.
3) The range of load resistance changes for every 20 kΩ by 0.1 µA.(negative slope)
4)The resistors used in this circuit showed the exact standard values despite tolerance errors of 5-15 %
The measurement of current and resistances and the voltage supply used in this circuit is made using Multimeter.

My question is why is there a deviation in the range of load resistance obtained between theoretical and practical values?
Is it because of Early Effect but this effect sets in only for large values of V_ce...

Reference :https://en.wikipedia.org/wiki/Widlar_current_source

My thoughts..
There are a great many constant current source op-amp circuits out there, even temperature compensated ones.
https://en.wikipedia.org/wiki/Current_source

The key reference part is the sampling resistor. It needs to be known, or trimmed, or made up so that it's value is *accurately* known.

There are also IC's devoted to this kind of instrumentation. If I recall, Maxim, and Analog Devices, and Linear Technologies have application notes that show example circuits.

While the Widlar circuit is conveniently small, exploiting the very high gain and low offset voltage of a modern op-amp is still low parts count, and gives a very predictable result.

Hi Adi,

Have you considered mismatch between the two transistors? It is highly unlikely that the two Vce of the transistors will also be equal. The Early voltage also plays a role, yes.

Regards Ken.

Thanks a lot,Kenneth..