Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.
lavitaebelle is right. The circuit as you have drawn it will have no DC path from Vdd to Gnd since both transistors are in cutoff, so the output will be floating.
As lavitaebelle also mentioned, you can diode connecting both transistors to get a voltage divider. However, you must be aware of the body effect. Certainly, you can get the output to be half the supply, but to do it with two identical devices, you need to connect the source of each transistor to their respective bulks. Otherwise, you can adjust the sizing until you get the output you want.
However, note that this is by no means a good way to get a DC voltage reference. Not only does the output change with supply, process, and temperature variation, it wastes a lot of power (you're basically putting two relatively small resistors from Vdd to Gnd).
cicer,you are right. and I also agree with gevy's viewpoint.
the two Pmos are cutoff,and the leak current is weakly,almost zero,and the impedance of these pmos are very large(infinity),if they have the same sizes,this circuit would be a voltage divider by 2.
other way,we can regard cotoff pmos as capacitor(Cgd),then the result is the same.
but this circut can not work accurately.