could some one help me to find out DC analysis for the following circuit for
examble, VDD= vdsat9 + vdsat10 +vdsat16 +vgs9 (correct me if i am mistake) but i need it for each branch
thanks in advanced
hi
you are mistake
M16 is current mirror so we have current of M16 so we have |Vgs9| and so we have PCS3 and vice versa if we have PCS3 we have ID9 and Vgs16.
regards
Thank you so much, now more clear for me
but i have 1 q more
that if we look to first branch frome the left side we cane see that
vdd=vdsat12+vdsat13+vdsat18
1.5=0.065+0.065+0.05
could u explain to me that
with my regards
Hi
assume
Id=0.5*beta*(Vgs-Vt)^2 (saturation)
so if draw Id vs. Vds we can find that for Vds> Vds(sat)
Id is Constant.
so in sat. region for specific Id and Vgs we can have any Vds that Vds> Vds(sat).
so we don't know Vds but we know Id and Vgs of each Transistor.
so
I know current of M18 (Id18) then I know Vgs12
=> Vds18=Vdd-|Vgs12|
Hi
I think that you have a basic mistake.
how you find Vds12 and Vds13?
you can not find Vds12 nad Vds13.
as I told in sat. region for specific Id and Vgs we can have any Vds that Vds> Vds(sat).
so you must just use Vgs to find each voltage.
for example Vds18=Vdd-|Vgs12|
but VICM=vdd-|Vds9|
in this case we must have Vds9 or we must use simulator or this equation:
Id=0.5*beta*(Vgs-Vt)^2 * (1+lambda*Vds)