Your channel is really good. Please make more videos. However, could you please explain if this effect can be ignored if we short body with source (which is the typical case)?
Yes, the only reason the body effect exists is because the channel is at a different potential than the bulk (body). Since the channel is continuous with the source in any mode of operation except cutoff), because it’s just a bunch of electrons floating around, this means whenever the body and source are shorted together we can ignore this effect.
Thank you for this video, I would like to ask you, what is the advantage and disadvantage of the body effect, and why we need a connection between the source and drain ?
For the body effect what we care about is the difference in voltage between the body and all the other terminals. We could use Vsb instead of Vgb if we wanted, the math would just be different.
Thanks for the lecture! However, I don't understand why we don't need to account the change of phi_s when we calculate the change of threshold voltage. When the depletion region gets wider from V_sb > 0, should phi_s also increase from 2phi_FP to 2phi_FP + V_sb?
Great question. Phi_s is the surface potential (or the difference between the semiconductor bulk and the semiconductor interface with the oxide). This can only get so large, because silicon's band only has so much room to bend. So regardless of the V_bs you apply, only some of it gets dropped across the semiconductor itself in the form of Phi_s.
@@JordanEdmundsEECS Thank you so much for the quick response! Sorry, I am still a little confused. When we have V_sb > 0 and Phi_s can only get so large(Phi_s < 2phi_FP + V_sb), should that also limit the total charge Q we use to compute V_ox? But we plugged in Phi_s = 2phi_FP + V_sb to compute total charge Q. Thanks again!
if VSB is more negative than 2phiF then Vth becomes imaginary, explain? If we apply Modulus then for sweeping VSB across negative range we still get VSB + 2phif as an Increasing function due to modulus,same as VSB in positive range. Ran into this problem while solving Razavi Cmos Analog design problems
you emphasize that body effect is just an artifact because we are appling Vgs instead of Vgb (a matter of where we reference the voltage to). I think this is not correct. physically speaking, even if you keep the absolute Vg voltage constant (Vgb constant), as you increase the absolute value of source voltage Vs (hence changing Vsb), the required absolute Vg (or Vgb since bulk is ground) voltage to get inversion will still increase, hence altering the definition of device Vth. therefore, there is in fact a "real" physical change that happens to Vth of the device when we change the absolute value of Vs, hence the name, "body effect"
Hi, really nice explanation about the body effect. I have a question abouth how threshold voltage(Vth) changes with temperature. When body effect is not present Vth is complementary to temperature, thus Vth becomes smaller when temperature rizes. Does this changes when body effect effect is present? Does it has to do with Φs which contains thermal voltage(Vt) on it? Could Vth become proportional to temperature if we apply a Vsb in a device? Thanks in advance🙂
Thank you so much! You helped simplify and clarify two 50 mintute+ lectures about parasitic diodes and the transistor body effect!
Thanks! I wish our lecturers could explains things like that
Your channel is really good. Please make more videos. However, could you please explain if this effect can be ignored if we short body with source (which is the typical case)?
Yes, the only reason the body effect exists is because the channel is at a different potential than the bulk (body). Since the channel is continuous with the source in any mode of operation except cutoff), because it’s just a bunch of electrons floating around, this means whenever the body and source are shorted together we can ignore this effect.
Thank you for this video, I would like to ask you, what is the advantage and disadvantage of the body effect, and why we need a connection between the source and drain ?
Useful when understanding t-gate on resistance modulation
Sorry if you mentioned, I just wondering what is “Phi”?
awesome explanation! thank you so much
Thank you :)
You're really good. Thank you a lot for a nice explanation!
Very clear and nice explanation. Thank you very much!
Thank you for the video, but can you explain why we don't just use Vgb instead of Vgs?
For the body effect what we care about is the difference in voltage between the body and all the other terminals. We could use Vsb instead of Vgb if we wanted, the math would just be different.
Thanks for the lecture! However, I don't understand why we don't need to account the change of phi_s when we calculate the change of threshold voltage. When the depletion region gets wider from V_sb > 0, should phi_s also increase from 2phi_FP to 2phi_FP + V_sb?
Great question. Phi_s is the surface potential (or the difference between the semiconductor bulk and the semiconductor interface with the oxide). This can only get so large, because silicon's band only has so much room to bend. So regardless of the V_bs you apply, only some of it gets dropped across the semiconductor itself in the form of Phi_s.
@@JordanEdmundsEECS Thank you so much for the quick response! Sorry, I am still a little confused. When we have V_sb > 0 and Phi_s can only get so large(Phi_s < 2phi_FP + V_sb), should that also limit the total charge Q we use to compute V_ox? But we plugged in Phi_s = 2phi_FP + V_sb to compute total charge Q. Thanks again!
if VSB is more negative than 2phiF then Vth becomes imaginary, explain? If we apply Modulus then for sweeping VSB across negative range we still get VSB + 2phif as an Increasing function due to modulus,same as VSB in positive range. Ran into this problem while solving Razavi Cmos Analog design problems
you emphasize that body effect is just an artifact because we are appling Vgs instead of Vgb (a matter of where we reference the voltage to).
I think this is not correct.
physically speaking, even if you keep the absolute Vg voltage constant (Vgb constant), as you increase the absolute value of source voltage Vs (hence changing Vsb), the required absolute Vg (or Vgb since bulk is ground) voltage to get inversion will still increase, hence altering the definition of device Vth.
therefore, there is in fact a "real" physical change that happens to Vth of the device when we change the absolute value of Vs, hence the name, "body effect"
What will happen to the equation for pmos devices? will it be the same? 9:48
Hi, really nice explanation about the body effect. I have a question abouth how threshold voltage(Vth) changes with temperature. When body effect is not present Vth is complementary to temperature, thus Vth becomes smaller when temperature rizes. Does this changes when body effect effect is present? Does it has to do with Φs which contains thermal voltage(Vt) on it? Could Vth become proportional to temperature if we apply a Vsb in a device?
Thanks in advance🙂
Thank you very much! really helped!! :D
can vsb > Vth ? what is the limitation of VSB
So is this the body diode?
Thank you ❤️
Thank you sir 🥰
omg!! the video didn't star with Hello friens whit desk top windows with lots of icons.
Please teach EECS170 A-E
God I would love to 😍
Very clear explanation !! Very useful !! (●°u°●) 」