/sci/ - Science & Math » Thread #11152460

9KiB, 310x270, nmoist.png

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Anonymous Sat 16 Nov 20:10:53 2019 No.11152460 View Reply Original Report

Quoted By: >>11152563 >>11152597 >>11152602

biasing an nmos amplifier

what do i do for the source and the drain resistance values????????????

if I modify the source resistance it starts to screw with the quiescent drain current, which affects the voltage across both drain and source resistances and the vds. this in turn affects the vgs, which affects what you need to bias the transistor at. also, i am stuck using a standard load resistance which affects the small signal gain. anyone please help????????????????????????

is this even possible?

Anonymous

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Anonymous Sat 16 Nov 2019 21:07:13 No.11152563 Report

Quoted By: >>11152575 >>11152582 >>11152599

>>11152460
i had to do this for a class like years ago

iirc basically the problem is feedback from the source resistor changes both the gain and q point which in turn changes overdrive voltage. but guess you already said that.

i can't actually exactly wtf we did except that it took all night.
but when we built the circuit IRL during the lab period, it worked. thank god.

so we made a spreadsheet (pic related) and then put in values.

you can see which the "inputs": Vs (i.e. voltage measured at the source terminal of the nmos), Rs, Rg1 (R1 in your picture), and Vd
then the rest are calculated
this way, at least you can just punch in numbers and see what you get. some spreadsheets can even do this for you (you tell it what output you want and it'll brute force the inputs for you)

going off what my past self said in his lab report years ago:
>Vdd was only 10V, so for Vds to have a nice linear swing but never fall below Vov, we wanted Vg to be relatively low. Therefore Vs must also be low.
>Vgs at the DC bias point needs to have enough margin so the transistor stays in saturation on input swings
>Common source amplifier gain = -(Rd/Rs).
We selected 200mV and 470 ohm as initial values of Vs and Rs.
>Id=Vs/Rs. (ohm's law)
> $I_D=0.5\cdot k_n (V_{ov})^2 \implies V_{ov}=\sqrt{\frac{2I_D}{k_n}}$
>transconductance value $g_m=\frac{2I_D}{V_{ov}}$
> $V_G=V_{ov}+V_t+V_S$
> $R_{G2}=\frac{V_G}{I_{G2}}=\frac{V_G}{\frac{V_{DD}-V_G}{R_{G1}}}$
Choosing Vd: it can't exceed Vdd, we needed a 2V swing (based on the parameters of the assignment, we needed gain of 20 on a 0.1v signal), so it has to be 2V beneath Vdd and 2V above Vov to keep the transistor on.
Then
> $R_D=\frac{V_{DD}-V_D}{I_D}$
theoretical volt per volt gain is $A_v=-\frac{R_D \| R_L}{\frac{1}{g_m}+R_S}$

my past self sure was a smart guy, sure wasn't a lazy brainlet like me. i'll ask him for more info if i get in touch with him.

hope it helps

Anonymous

Anonymous Sat 16 Nov 2019 21:11:08 No.11152575 Report

Quoted By: >>11152599

>>11152563
$I_D=0.5 \cdot k_n (V_{ov})^2 \implies V_{ov} = \sqrt{\frac{2I_D}{k_n}}$
transconductance value $g_m=\frac{2I_D}{V_{ov}}$
$R_D=\frac{V_{DD}-V_D}{I_D}$
not sure why these didnt render up there

Anonymous

Anonymous Sat 16 Nov 2019 21:12:28 No.11152582 Report

Quoted By: >>11152599

>>11152563
transconductance value $g_m = \frac{2I_D}{V_{ov}}$
$R_D = \frac{V_{DD}-V_D}{I_D}$

Anonymous

Anonymous Sat 16 Nov 2019 21:15:16 No.11152597 Report

Quoted By: >>11152823

>>11152460
Why does your schematic look like a classic Class-A bipolar transistor amplifer that you just pasted a FET into?

FETs are voltage amplifiers; bipolar transistors are current amplifiers, why are you treating a FET like it's a bipolar transistor?

Anonymous

Anonymous Sat 16 Nov 2019 21:15:42 No.11152599 Report

Quoted By:

>>11152582
>>11152563
>>11152575
tldr: Don't choose the drain resistance value, choose drain voltage instead, and let Rd follow from that.

Anonymous

Anonymous Sat 16 Nov 2019 21:16:35 No.11152602 Report

Quoted By:

>>11152460
how does one go into engineering and not end up just doing physics? you seem to always struggle with the mechanical theory behind these applications and the solution would be to just double major or minor in engineering no?

Anonymous

Anonymous Sat 16 Nov 2019 22:24:24 No.11152823 Report

Quoted By:

>>11152597
>implying you can't use a FET as an amplifier

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