Engineering Calculation of the Transconductance for a MOSFET

[icesalmon]

Homework Statement

Calculate the transconductance, gm, for the MOSFET shown.

Relevant Equations

gm = id/vgs

This problem was included as an exercise for a section on the small signal equivalent circuit for a MOSFET under the body effect. The problem states $λ =0$ so I believe the Early effect is negligible and r_o = 0. I don't see how the body effect would be negated so for part (b). I believe the circuit should look as the picture I've drawn and attached. My issues are the following: I don't really understand how to calculate the small signal gate-source voltage v_gs and i am not sure whether or not the problem actually is neglecting the body effect as this would include another VCCS g_mb*v_bs. I have calculated the DC bias drain current to be I_D = 0.2mA and the DC bias drain-source voltage to be V_DS = 3V which are correct according to the solutions, essentially when I switch to small signal analysis and redraw the equivalent circuit I am a bit lost on how to proceed.

 

[Jody]

What causes the body effect and is it applicable to your problem? I don't see where your body terminal is connected, but where would you assume it's connected. Maybe there's a body effect equation it might be able to tell you a little more about the body effect.

Spoiler

Hint: What happens to threshold voltage $V_{th}$ if $V_B = V_S$?

What are some equations for $g_m$? There are a few of them. I won't give you the full equation... should look for it in your notes or book.

$$\begin{align}& g_m = \mu_n C_{ox} \frac{W}{L}... \\
& g_m = \sqrt{2 \mu_n C_{ox}...} \\
& g_m = \frac{2I_D}{...}\end{align}$$

Can you draw on the small-signal model where $r_0$ would be and what that would be mean if it were $0 \Omega$... is it typically desirable for an amplifier?

They have a rule here that you have to make a attempt to solve problem before people can answer questions.

 

[icesalmon]

What causes the body effect and is it applicable to your problem? I don't see where your body terminal is connected, but where would you assume it's connected. Maybe there's a body effect equation it might be able to tell you a little more about the body effect.

$V_t$ = $V_t$₀ + $\gamma$[sqrt(2$\phi_f$ + V_SB) - sqrt(2$\phi_f$)]
I mostly forgot everything I learned about the body effect, so i'll have to read it over again as it was in a previous chapter I haven't seen in years.

Spoiler

Hint: What happens to threshold voltage $V_{th}$ if $V_B = V_S$?

v_BS = v_B - v_S = 0
v_OV = V_BS - V_th = -V_th
This doesn't seem to make sense, having a $negative$ overdrive voltage.

Can you draw on the small-signal model where $r_0$ would be and what that would be mean if it were $0 \Omega$... is it typically desirable for an amplifier?

In the picture I attached there is a portion where $r_o$ is and I wrote $\lambda = 0$ and $r_o = 0$
If $r_o = 0$ then the accuracy of the device goes down, but $|A_v|$ increases, this is a good thing for an amplifier. I would say this is desirable.

 

[Jody]

$V_t$ = $V_t$₀ + $\gamma$[sqrt(2$\phi_f$ + V_SB) - sqrt(2$\phi_f$)]
I mostly forgot everything I learned about the body effect, so i'll have to read it over again as it was in a previous chapter I haven't seen in years.

v_BS = v_B - v_S = 0
v_OV = V_BS - V_th = -V_th
This doesn't seem to make sense, having a $negative$ overdrive voltage.

Alright. Sorry I crossed some of it out on my own. You got some of it right. Good job. You got the right body effect equation too. If you see $V_{SB}$ go to zero, then those two square root terms they cancel out too and so $V_t=V_{t0}$; therefore: You have no body effect :). For an NMOS you'll want to tie your body to the source. What I've usually seen is if you don't draw the body terminal in an NMOS, then it's inferred tied to the ground reference. In your drawing your source is also tied to ground, and so you probably don't have to worry about body effect. This'll help you out a bit because now you don't have to worry about that second VCCS in your drawing.

Spoiler

For a PMOS the assumption if the body terminal isn't drawn, then it's tied to $VDD$. Your source is typically connected high $VDD$ and so that's usually what you'll want.

Some of this might change a little bit when you start looking at a common source with degeneration or cascoded amplifiers; however: different fabrication process might still allow you to tie the body to source. For example: A deep n-well process can help solve that problem.

Of course: It's always a good idea to clarify or double check with your professor so that you don't unnecessarily lose points.

By the way: $V_{OV} \neq V_{BS}-V_t$

In the picture I attached there is a portion where $r_o$ is and I wrote $\lambda = 0$ and $r_o = 0$
If $r_o = 0$ then the accuracy of the device goes down, but $|A_v|$ increases, this is a good thing for an amplifier. I would say this is desirable.

A professor use to say to me "hear what I mean; not what I say!" The resistance of an ideal wire is zero, and so the problem with saying that $r_0$ is $0\Omega$ is you've shorted your drain and source. In your drawing I don't see a short or a resistor (it's open circuit) and so I think what you meant to say is $r_0$ is $\infty$. I think I heard what you meant, but wanted to cover that. What that $\lambda=0$ means that you have no channel length modulation... the slope on your IV curve when it's in saturation is $g_{ds}$ which is $1/r_0$ you'll want the slope to be more horizontal or zero and so you want $r_0$ to be $\infty$.

A quick note: You might want to think of conductance $g_{ds}$ instead of $r_0$ in the future. I recommend practicing with that instead. A reminder that $g_{ds}$ is $1/r_0$. The reason this will be helpful in the future is you'll be dealing with a lot of parallel resistance. Equivalent resistance for parallel resistors can be a pain when there are a lot of resistors (lots of multiplication), but conductance you can simply add together.

Regarding the original question and sorry for straying a bit... you'll want those equations for $g_m$, which is not $i_d/v_{gs}$ as you put for relevant equations. Any chance you can fill in the blanks for post #2 regarding equations for $g_m$? I confess I have a very hard time memorizing these equations. While I took the class I had to spend a lot of time trying to memorize it and if I was allowed some notes those were definitely some equations I wrote down.

The problem is staging you for success by asking for you to solve for $I_D$. Once you've solved that it's given you everything else you can plug and chug into one of those three equations in #2.

 

[icesalmon]

What are some equations for $g_m$? There are a few of them. I won't give you the full equation... should look for it in your notes or book.
$$\begin{align}& g_m = \mu_n C_{ox} \frac{W}{L}... \\
& g_m = \sqrt{2 \mu_n C_{ox}...} \\
& g_m = \frac{2I_D}{...}\end{align}$$

$g_m = \mu_n C{ox} \frac{W}{L}V_{OV}$
$g_m$ = sqrt(2k_n')*sqrt(W/L)*sqrt(I_D)
$g_m$ = 2I_D/V_OV
these are the ones I was looking for earlier.

 

[Jody]

There we go :)

I think you solved for everything that you needed or are there anything else that is a little bit shaky?