i am using gpdk045 technology and i want to get the value of the capacitance

Dear Hagar Hendy,

Hagar Hendy said:

yes you understood my question,

Great - at least I did that much right anyway!

Hagar Hendy said:

what I want to show you if i understand the steps that you said or not ,and i am doing right or not ?

i am trying to do a small signal analysis: for the current source I choose the dc current source and I put 1 A  at the AC value as shown in the screenshot.  For the resistor when i  put << 1 ohm, it gives an error so I chose a value less than 1 for example 0.8,

You set the current source correctly.

However, perhaps there is some confusion on how to set the resistor values. You should not get an error when setting its DC value to something << 1. As an example, Figure 4 shows how to set the DC resistance to 1 milliohm and the AC resistance (circled in red) to 1e12. Setting the DC resistance to 1 milliohm does not result in an error. Did you set the AC resistance to something large in your simulation?

Figure 4.

Hagar Hendy said:

For AC setup is shown below, please check it  if the setup is right 

It appears you are using ADE-L. This is no longer supported by Cadence. I would seriously consider migrating to ADE Assembler/Explorer. Explorer is very similar to the interface for ADE-L and has far more features and capabilities than ADE-L. The migration process is very easy to convert your state file to the new maestro view that is shared by Assembler and Explorer.

For your AC simulation GUI, it appears you are attempting to simulate the impedance at 1 MHz and sweeping the value of design variable "var". Is there a reason you only want to see the impedance at 1 MHz? It can be instructive to see how the impedance varies with frequency. Nevertheless, your basic AC setup appears fine.

Hagar Hendy said:

and this the graph that results, which i didn't understand, i think i am doing something wrong 

I do not know your test bench nor the impedance you are simulating. Hence, I cannot tell you if what you plotted is "correct" or even what it means. Is /net1 the positive node of your AC current source? If /net1 is the positive node of your current source, then the plot is the magnitude of the impedance.  If so, the plot suggests is that as you swept variable "var" from 0.0 to 1.0, the magnitude the impedance varies from about 0.25e6 at var = 0 to 7.5e6 at var = 1.

To determine the real and imaginary parts, you need to plot the real and imaginary parts of the response at node /net1 - not the magnitude. Use the Results Browser to send output node /net1 to the Calculator and make use of the real() and imaginary functions of the Calculator to plot each. If you are interested in the effective capacitance, you will need to compute it from the imaginary part of the impedance using the 1 MHz simulation frequency. This can all be automated as you can create an output that contains the real impedance and a second output that contains the effective capacitance.

Shawn

Dear Hagar Hendy,

Unfortunately, my response was characterized as "spam". Hopefully, it will be released soon. However, I thought you ay also be interested in a prior Forum post where I included expressions for the real and equivalent capacitance for an impedance element using a similar test bench. At the time, to my knowledge, an AC resistor was not available so I used a large inductor to isolate the DC and AC frequencies. However, I recommend you use an AC resistor. An updated version of the test bench I used to determine the impedance of an arbitrary network is also attached.

https://community.cadence.com/cadence_technology_forums/f/rf-design/38996/capacitance-vs-bias-voltage-curve-for-ferroelectric-varactor