# Common drain / source follower stage circuit analysis

I have been recently working on different types of MOS amplifiers and the following are my main doubts,

1. When we have a high gain stage such as the common source stage we use the output of this stage as an input to a common drain stage. What makes the common drain stage so special? I know that they charge capacitors in a much quicker rate (slew rate) compared to the common source stage but is there an intuitive way to understand what makes the common drain stage good for this?

2. In terms of slew rate is the push pull stage (with the distortion effect corrected) a better option compared to the common drain stage as the final output stage?

Common drain stage <----------> Push-Pull stage

If you make the assumption that the devices are large so the threshold currents match, the result is: $$I_{th}e^\left(\frac{\Phi_{sc1}}{U_t}\right)=I_{th}e^\left(\frac{\Phi_{sc2}}{U_t}\right)$$ This then can be modified with voltages based on the schematic above: $$I_{th}e^\left(\frac{\kappa V_{bias} -0}{U_t}\right)=I_{th}e^\left(\frac{\kappa V_{in} -V_{out}}{U_t}\right)$$ You can then push through the math and you get: $$\Delta V_{out} = \kappa V_{in} -\kappa V_{bias}$$
• So, common-drain stage is commonly in subthreshold for high gain, but it's not necessary. Above threshold is fine, but the gain is significantly lower. Vbias fixes the flux in the channel, that means that Vout must respond for whatever condition is set by Vin. This means that $\Phi_{DC}$ will have to accommodate whatever flux is in the channel to make the net flux equal between the devices. This, of course, assumes that no charge is lost through Vout. Apr 23, 2016 at 11:31