Calculating gain of CG mosfet amplifier with potentiometer on source - Page 1
Forum section: electronics / beginners

author: igcamacho
author_rank: Newbie
timestamp: March 18, 2020, 04:50:38 am
content: Hi, in one of my courses we got an assignment and part of it is to calculate the gain of a common gate amplifier circuit (pic related), however, it's proved to be not so straightforward, as the circuit in question has a potentiometer at the source and the signal is injected through pin 2 with the potentiometer at 80%. I tried calculating the gain using the equation for the common gate with source resistance amplifier unsuccesfully, but I calculated the drain current to be about 5.3 mA and the gate to source voltage to be about 2.5V, but i just can't figure out the gain. Any help is greatly appreciated. Thank you very much in advance and sorry for asking noobie questions lmao.The assignment in question, M1 is a 2N7002 and Vin is 20mVpk@600Hz:Oh, btw, I'm making sure my answers are correct using NI Multisim 14.2, I simulate the circuit using a 2N7002BK as it's almost identical to the 2N7002, I'm not using the 2N7002 because it has two extra pins floating around and I can't really figure out why they're there and drain current and gate to source voltage don't match at all with my calculations.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 18, 2020, 05:13:39 am
content: Huh. The extra 2N7002 pins aren't for thermal simulation, are they?Anyway, assuming mid frequency conditions, the capacitors go away, and we see that the bottom half of RS is shorted into Vin. We don't care about that current flow, so we can ignore it entirely. That reduces the circuit to a variable resistor between M1-S and Vin.We calculate the DC bias condition with the 82k + 68k gate divider, Vgs(th) and K_n, and RS (the total, since C3 is open at DC). From this we get the small-signal equivalent, which will have an effective source input resistance approx. 1/Gm and drain output resistance something something lambda.We calculate the AC equivalent circuit, where the capacitors are effectively shorted and the resistances and gain are as found in the previous step.Tim

author: igcamacho
author_rank: Newbie
timestamp: March 18, 2020, 06:15:06 am
content: Tim, thank you very much for your kind response. I shall see if those pins are used for thermal simulation, it makes a lot of sense. I have indeed calculated the bias just the way you mentioned, but I'm having a bit of trouble with the ac equivalent. You mentioned using the impedances found in the second step, should I replace the resistances in the circuit with the calculated input and output resistances when analyzing the ac equivalent? How should I place them in the circuit? Sorry, I'm not that good when it comes to transistors and i still have some trouble when it comes to ac equivalents, so please be patient with me here.Thank you very much in advance.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 18, 2020, 06:19:33 am
content: You use the equivalents when calculating by hand.You can simulate the equivalent using dependent sources. The MOSFET small-signal AC equivalent should be in your textbook.Otherwise, that's the purpose of the 2N7002 model, SPICE does this for you automatically in the AC steady-state analysis. Just measure the input and output voltages, and take their ratio.Tim

author: igcamacho
author_rank: Newbie
timestamp: March 18, 2020, 11:51:24 pm
content: Hi Tim, good evening, I tried doing the ac model and while this time everything is closer to reality, I get double what I should for some reason. I assumed lambda=0 for the moment being to make things easier, I'll attach a picture of my calculations,in case you or anyone else can spot some mistake that would be responsible for my doubled results. Any ideas in where this error might be coming from? Thank you very much in advance.