# How does the source voltage change with and without channel modulation?

The description here seems big, but it's only a two minute read. I'm stranded, desperately need help. The variable resistance R in the figure is set to 10K ohms. The first question was to find the voltage Vs (Source voltage). It is a short channel device with Vdsat=0.6V(Minimum drain voltage for saturation). Threshold voltage Vt=0.4V. Assume zero body-effect and zero channel length modulation.

How I solved it:

Vd = Vdd - IR => Vd = 2V

Now, Vds = Vd - Vs = 2-Vs, Vgs= 2-Vs

As, gate and drain voltages are equal, the device should be in saturation, with or without velocity saturation. (That's what I think, because with velocity saturation the device saturates with Vds< Vgs. Hence, it easily qualifies this condition.) Now, I calculated Vs using the Id (drain current for saturation) formula. I ended up with Vs = 1.3V, which according to the key is right.

MY QUESTION: The follow-up question was if Vs would increase or decrease if channel length modulation was considered finite or not equal to zero? It asks for a qualitative explanation.

How I solved it:

For saturation to occur, Vds >= Vgs - Vt.

Here, Vgs-Vt = 2-Vs-0.4 = 1.6-Vs

Vds= 2-Vs

So, for any value of Vs, Vds >= Vgs -Vt. A more important point is the Vds, here, is higher than the voltage required for saturation, which points to the fact that the current should increase linearly with increase in Vds (once saturation is attained, an increase in Vds increases drain current, as we are considering channel length modulation; else the current would've been constant.)

The final step: Taking ohms law: Considering Rsat as the finite resistance, Vds = I* Rsat (Inspired from the I-V characteristics curve)

Rsat = (Vd - Vs)/I

Here, as mentioned, we observe that Vd has increased more than the minimum saturation voltage, hence if Vs was constant, 'I' should increase too, to keep the value Rsat constant. But, since 'I' is constant here, due to the constant current source, Vs should increase to maintain the ratio constant. This is my way of deriving it.

Is my solution right? If not, please enlighten and if it is right, please provide your approach to the question, to create a deeper insight of the scenario, both qualitatively and quantitatively(if possible).

• imgur link seems broken. – Brian Drummond Aug 10 '16 at 13:43
• The source behavior should be independent of the drain condition. The only thing that changes is that the charge sharing at the source edge will change due to the bulk condition, but this sort of question is more advanced than you'd probably need. (most books assume the source and bulk are tied) Fix your image link and we'll probably be able to give you an answer for sure. – b degnan Aug 10 '16 at 14:25
• @BrianDrummond Please recheck the link now. – electronics Aug 10 '16 at 17:22
• @bdegnan Please recheck the link now. – electronics Aug 10 '16 at 17:23