Timeline for Why is the gain of this amplifier so small?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 4, 2016 at 17:35 | vote | accept | Georan | ||
| Apr 4, 2016 at 17:34 | comment | added | LvW | Yes , -1.6 is corect assuming gm=0.5mS and RD||Rload=3.3k. Perhaps the transconductance gm was larger in your experiment? By the way- up to now we didn`t consider the series resistor at the input (10k). We have a voltage division between 10k and the bias voltage divider. | |
| Apr 4, 2016 at 16:56 | comment | added | Georan | Right. And with the capacitor, it is -1.60 (if we ignore ro) which is greater than -1.07 without the capacitor. Does this make sense? I'm not sure about by answer because when I performed this experiment I obtained an output voltage of 2.70V when a 108 mV AC signal was applied, which is a gain of about ~25. | |
| Apr 4, 2016 at 16:50 | comment | added | LvW | Without bypassing the source resistor the gain is G=-3.2k/(2k+1k)=-1.066 (assuming gm=0.5mS and ignoring ro). | |
| Apr 4, 2016 at 16:46 | comment | added | LvW | The small-signal output resistance of the FET is ro=d(Vds)/d(Id). But this resistance should be listed in the data sheet. | |
| Apr 4, 2016 at 16:28 | comment | added | Georan | If I ignore ro in the pi-model, I obtain a gain of -1.60 . Is this the correct gain? | |
| Apr 4, 2016 at 16:06 | comment | added | Georan | "equivalent to the inverse slope of the Id=f(VDS) curves, normally in the higher kOhm range" can you please elaborate on this? how would i calculate the output resistance? thanks so much! edit: is it by doing ro = vds / id = 4.2933 V / 1.8074 mA = 2.38 kOhms (these are the experimental values i obtained)? | |
| Apr 4, 2016 at 16:00 | history | answered | LvW | CC BY-SA 3.0 |