Hi Team
As is known to many people, a common drain circuit can be used as a voltage shifter since it's gain~=1 and the input-output dc voltage is shifted by Vgs=Vth+Vdsat.
It came to me that what if Vgs is too big? If I only want to shift the DC level by 0.5*Vgs, is there a convenient circuit to use?
This is from Design of Analog CMOS Integrated Circuits by Razavi. The current sink, M2, can also be a resistor. It's not Vgs=Vth+vdsat but Vgs=Vth+sqrt(2Id/uncox(W/L)). From this the minimum difference between Vin and Vout is theoretically Vth.
Hi rmanalo
Thanks for the reply. How about the case that I need the level shifted less than that? For example, for a 1.2v Vdd process, vth is normally ~0.4v. This is quite a shift compare to 1.2v. What if I need only 0.2v shift, which is about 0.5 of Vth?
if you want to shift up (increase) signal voltage, then I recommend to use a common source (open drain) circuit. Mind that it inverts the signal logic.
With enough W and low enough Iload, you can see operation
at VT/2.
You could make a closed loop buffer controlling a common-
source PMOS, but this will cost you bandwidth (but gain
you accuracy / eliminate most voltage drop.
You could make a closed loop buffer controlling a common-
source PMOS, but this will cost you bandwidth (but gain
you accuracy / eliminate most voltage drop.
No, I cannot. I don't know what exactly dick_freebird talks about, probably he will explain.
But I show you an OPAmp solution:
<a href="https://imgur.com/IA7QKJ9"><img src="https://i.imgur.com/IA7QKJ9m.png" title="source: imgur.com" /></a>
The DC shift is IS1*R1, arbitrary. You can realize the IS1 with PMOS or NMOS current source, it depends on what you want, to shift up or down the DC level.