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CCCS circuit for ADS simulation

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I simulated the Mosfet circuit with try and error method and draw real part of input impedance vs frequency but I did not get the desired response yet (it isnt negarive and is positive for all frequency)

It seems this circuit doesn't work as a NIC, probably because the additional currents needed to create the correct reference points (i.e. due to the multiple current mirroring). I must admit I didn't know in the very beginning of this thread that you need a NIC circuit.

I tried the circuit with supposed technology currents and Kp values for your process and a DC input current of 10mA (it works between 1..100mA).
To get reasonable operation points, I had to scale up your resistor values by a factor of 1000 (and still had to use ridiculous large MOSFET widths), s. this PDF: View attachment CCCS3.pdf
With the given DC input current of 10mA, the pos. input impedance is 7.7Ω .
 
It seems this circuit doesn't work as a NIC, probably because the additional currents needed to create the correct reference points (i.e. due to the multiple current mirroring). I must admit I didn't know in the very beginning of this thread that you need a NIC circuit.

I tried the circuit with supposed technology currents and Kp values for your process and a DC input current of 10mA (it works between 1..100mA).
To get reasonable operation points, I had to scale up your resistor values by a factor of 1000 (and still had to use ridiculous large MOSFET widths), s. this PDF: View attachment 86422
With the given DC input current of 10mA, the pos. input impedance is 7.7Ω .
Thank you for the time you spent helping me
now dont you have any idea for replacing ideal CCCS at first circuit and it remain and acts as a NIC?
 

... my problem isnt NIC , I have problem with circuitary model that replace to ideal CCCS

Just here this part of my circuit is a good example: CCCS3.png

The current injected into T3 (≈2µA) is repeated via T4, T10, T11, T12 (all the same current, but now referred to a different node), then translated via T13 with the required transmission ratio of 1.5 ((W13/L13) / (W12/L12)), now ≈3µA to the load resistor.

Isn't this what you wanted?
 
Just here this part of my circuit is a good example: View attachment 86479

The current injected into T3 (≈2µA) is repeated via T4, T10, T11, T12 (all the same current, but now referred to a different node), then translated via T13 with the required transmission ratio of 1.5 ((W13/L13) / (W12/L12)), now ≈3µA to the load resistor.

Isn't this what you wanted?
Hello erikl
as clear in below schematic:
3517581400_1360252733.jpg

the current to the load resistor is (A'+1)i2 that is (1.5+1)*2um, and it seem good in your circuit, maybe the problem is on the first CCCS configuration beacuse its clear that the above circuit should act as a NIC and the simulation with ideal CCCS prove it
 

3517581400_1360252733.jpg

... beacuse its clear that the above circuit should act as a NIC and the simulation with ideal CCCS prove it

Yes, and it's also clear that my MOSFET circuit will not act as a NIC, because it has to transfer the initial currents (from T1 & T3, s. my CCCS3.pdf from above) to different reference nodes (which is no problem with CCCS), and by this creates additional currents which contribute to i1.
 

Yes, and it's also clear that my MOSFET circuit will not act as a NIC, because it has to transfer the initial currents (from T1 & T3, s. my CCCS3.pdf from above) to different reference nodes (which is no problem with CCCS), and by this creates additional currents which contribute to i1.
ok dear erikl, where is the problem you think? and now how I can replace ideal CCCS with circuitary models that the final two-port act as a NIC? do you have any idea for this implementation?
 


You would know if you had read my above explanation thoroughly.
OK, but your proposed CCCS dont work like Ai1 with 2 port, if all of circuitary model is like the ideal form then why the operation of them are different
Sorry, no, this isn't my concern, it's your business. I already told you above where to go on searching. Good bye and good luck!
yes, I know, but at first I searched in google and beacuse of I could not find anything I post my question here
 

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