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Input Capacitance of cmos inverter chain

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nickwillnam

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Hi guys,how to find the input capacitance of inverter chain (inverter connected in series) using cadence? I need to find input capacitance so that I can size each inverter stage using logic effort.I am referring to this slide page86 formula. **broken link removed**

I am thinking to use Cin=I/(dv/dt) where I measure the Iin at input node using .measure tran iavg avg I(vin1)
Below are some part of the stimulus:
vin1 vin1 0 pwl (0 0 10n 1.8 20n 0)
.tran 20p 20n
.measure tran iavg avg I(vin1)
.measure tran Cin param = '-iavg/(1.8/20n)'

Is my steps to measure the Cin correct?
 

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Apply a pulse current of 1pA to the input then your input voltage starts to increase like a ramp.

check your voltage it should be around VDD/2.

so as you said Cin=I(dv/dt)
 

Hi Ata_sa16,
I apply 1pA current source by using
Iin1 vin1 gnd pulse(0 1pA 0 10n 10n 0 20n) $is that correct?
I am getting a 'V' shape for the I graph and the input voltage not seemed like a ramp function.

Actually, why 1pA is selected? Is that any other value of current source also can be accepted?

And for my previous steps that using a voltage source, can it be accepted?
 

nickwillnam said:
Hi Ata_sa16,
I apply 1pA current source by using
Iin1 vin1 gnd pulse(0 1pA 0 10n 10n 0 20n) $is that correct?
I am getting a 'V' shape for the I graph and the input voltage not seemed like a ramp function.
Click to expand...

thats true since u apply pulse not a step. But its the same, if you increase your time period your Vin will increase more.

nickwillnam said:
Actually, why 1pA is selected? Is that any other value of current source also can be accepted?
Click to expand...


because of ur capacitance value if u apply 1 mA for example your voltage will increase so much to hundreds of volts.

I=C(dv/dt)

- - - Updated - - -

Just find the slope of Vin around VDD/2 and calculate this

Cin=I/(dv/dt)

Cin = 1p / slope
 

You can step voltage and use a cccs referred to the input
voltage source to integrate the charge onto a (say) 1pF
capacitor, then voltage is the charge in picocoulombs.
There may be more realism when you apply a sensible
(i.e. circuit-realistic) rise (or fall) time to the front end.
You may need to apply a reset (ideal) switch to the cap
if there's either leakage or just gmin conducted charging
in the DC solution.
 

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