Cannot get the VTC curve of inverter in cadence for my ckt

Hi all,

Please refer my inverter ckt in the attachment (scematics.png). i am designing the floating Gate programmable inverter.So by panning thru a lot of IEEE papers i have been able to design this inverter.

So in my inverter i am using two capacities Cin =1pF and Ctun=1pF. Nmos anf PMOS are of same sizes for 180nm technology.
NOTE:Ctun and Vtun is used to tunnel out the electrons that are trapped between the two gates..i.e tunneling of electrons concept





see test_schematic attachment for the test ckt. Wherein Vpulse is the Vin to the inverter with v1=0v, v2=1.8v , td=0 ,tr=tf=1ns and ton=10n T=20n ,,vdc=1.8v ,,,Vtun=5V..

I have done the trans and DC analyses of the above ckt with trans of stop time of 200ns and DC for sweep range of 0 to 1.8




see the attachment output for the ADE window...here you can see the trans and DC outputwaveforms ....



in TRANS analysis the input is properly inverted to give the output signal ...

BUT the DC analysis fails to give the expected VTC curve of the inverter( i am getting straight line of 1.8v)..WHY??

How can i get the VTC curve ...where i am going wrong in the analysis please point it out....

Thanks in advance!!

The simulations are correct. For the inverter will work only with transient inputs.
At DC, all caps are open. Running DC is like injecting your input, wait for an infinite time for all the caps to charge till its equilibrium point, which is defined purely by the leakage gate currents of the nmos and pmos. Normally, the simulator may even fail with a DC, as it sees the gate as a high impedance node with no path to ground.