The discussion focuses on calculating the transconductance (g_m) for a MOSFET while considering the body effect. Participants express confusion about the small signal gate-source voltage (vgs) and the implications of the body effect in their circuit analysis. Key equations for g_m are shared, including g_m = μ_n C_ox (W/L) V_OV and g_m = 2I_D/V_OV, which are essential for solving the problem. Clarifications are made regarding the connection of the body terminal in an NMOS, emphasizing that it should typically be tied to the source to avoid body effect complications. The conversation highlights the importance of understanding these concepts for accurate small signal analysis and amplifier design.
#1
icesalmon
270
13
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 ro = 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 vgs and i am not sure whether or not the problem actually is neglecting the body effect as this would include another VCCS gmb*vbs. I have calculated the DC bias drain current to be ID = 0.2mA and the DC bias drain-source voltage to be VDS = 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.
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.
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.
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.
#3
icesalmon
270
13
Joshy said:
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## + VSB) - 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.
Joshy said:
Hint: What happens to threshold voltage ##V_{th}## if ##V_B = V_S##?
vBS = vB - vS = 0
vOV = VBS - Vth = -Vth
This doesn't seem to make sense, having a ##negative## overdrive voltage.
Joshy said:
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.
#4
Jody
Gold Member
453
230
icesalmon said:
##V_t## = ##V_t##0 + ##\gamma##[sqrt(2##\phi_f## + VSB) - 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.
vBS = vB - vS = 0 vOV = VBS - Vth = -Vth
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.
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##
icesalmon said:
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.
#5
icesalmon
270
13
Joshy said:
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(2kn')*sqrt(W/L)*sqrt(ID)
##g_m## = 2ID/VOV
these are the ones I was looking for earlier.
#6
Jody
Gold Member
453
230
There we go :)
I think you solved for everything that you needed or are there anything else that is a little bit shaky?
Hi all,
I have a structural engineering book from 1979. I am trying to follow it as best as I can. I have come to a formula that calculates the rotations in radians at the rigid joint that requires an iterative procedure. This equation comes in the form of:
$$ x_i = \frac {Q_ih_i + Q_{i+1}h_{i+1}}{4K} + \frac {C}{K}x_{i-1} + \frac {C}{K}x_{i+1} $$
Where:
## Q ## is the horizontal storey shear
## h ## is the storey height
## K = (6G_i + C_i + C_{i+1}) ##
## G = \frac {I_g}{h} ##
## C...
