How does the source voltage change with and without channel modulation? - Electrical Engineering Stack Exchange most recent 30 from electronics.stackexchange.com 2019-07-23T07:07:18Z https://electronics.stackexchange.com/feeds/question/251646 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://electronics.stackexchange.com/q/251646 0 How does the source voltage change with and without channel modulation? electronics https://electronics.stackexchange.com/users/117015 2016-08-10T13:22:47Z 2016-08-10T17:36:08Z <p><strong>The description here seems big, but it's only a two minute read. I'm stranded, desperately need help.</strong> </p> <p><a href="https://i.stack.imgur.com/qjIj7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qjIj7.png" alt="enter image description here"></a></p> <p>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. </p> <p>How I solved it:</p> <p>Vd = Vdd - IR => Vd = 2V</p> <p>Now, Vds = Vd - Vs = 2-Vs, Vgs= 2-Vs</p> <p>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&lt; 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. </p> <p><strong>MY QUESTION:</strong> 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. </p> <p><strong>How I solved it:</strong> </p> <p>For saturation to occur, Vds >= Vgs - Vt.</p> <p>Here, Vgs-Vt = 2-Vs-0.4 = 1.6-Vs</p> <p>Vds= 2-Vs</p> <p>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.)</p> <p>The final step: Taking ohms law: Considering Rsat as the finite resistance, Vds = I* Rsat (Inspired from the I-V characteristics curve)</p> <p><strong>Rsat = (Vd - Vs)/I</strong></p> <p>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, <em>Vs should increase to maintain the ratio constant.</em> This is my way of deriving it. </p> <p>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).</p>