I am trying to understand the physical phenomenon that leads to increase in gain of an amplifier as the output impedance increases.
When we have resistive loads in a single stage amplifier, they convert the signal current change into voltage variation. Higher the value of load, more will be the conversion and hence the gain. In MOSFETs, since it is not necessary for the output impedance to be less, higher gain can be obtained by increasing the RD**(physical resistance connected to drain)** while ensuring that the transistor operates in saturation.
But how does increasing the rds(the internal drain-source resistance) help obtain higher gain? Since rds is not a physical resistor, how does it contribute to increase in gain?
By using long channel MOS, the rds could be increased. This would mean, lesser current variation in saturation region with change in Vds. When we model the effect of channel length modulation, we of course obtain a resistor of higher value.
Can anyone explain me how does the change in length could increase the gain?
I have edited the question and included correct notation for small signal dynamic resistance.
I understand that rds is not an actual resistor but a model that takes into account the effect of channel length modulation. But then how does it increase the gain of the amplifier?
Higher rds would mean small change in drain current for Vds variation i.e., a good current source. That is all I can gather from this model.
My question however is that how would this increase the intrinsic gain of the amplifier? I understand how gain increases as RD increases(which I have mentioned above). Am I correct?