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Cascode LNA output impedance in cadence

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circuitking

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Can anybody tell the step by step procedure in cadence to find the output impedance(attached image). The gain of cascode is gm*(output impedance)^2. I want to increase the gain so I like to increase the output impedance
 

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The gain of cascode is gm*(output impedance)^2. I want to increase the gain so I like to increase the output impedance.
Don't confuse cascode output impedance with output impedance of a single MOSFET in common source configuration. The quoted formula is referring to the latter, and only valid for the low frequency range, not for typical LNA frequencies.

Apart from this, the consideration is wrong because the gain is actually determined by Zout || ZL. In most LNA applications, ZL is smaller or maximal in the same range as Zout.
 

Don't confuse cascode output impedance with output impedance of a single MOSFET in common source configuration. The quoted formula is referring to the latter, and only valid for the low frequency range, not for typical LNA frequencies.

So what are the ways to increase gain

Apart from this, the consideration is wrong because the gain is actually determined by Zout || ZL. In most LNA applications, ZL is smaller or maximal in the same range as Zout.
is this ZL equal to 50 ohm? and how to find the Zout in cadence?
 

A simple estimation says that gain can't be larger than gm*ZL. If ZL is fixed to e.g. 50 ohms, you need to increase gm (respectively Id). Or move to a multi-stage amplifier.
 

How to measure |Zout| in cadence?
1. Add a relatively large capacitance (like 1mF) to the output of the cascode and place a signal generator series with it (vdc from analogLib is good for it).
2. Set AC magnitude of the generator to 1V. Be cautious that other generator's AC magnitude is 0V.
3. Open ADE L -> Analysis tab -> choose ac, set frequency range, click on OK
4. In ADE L click on "Setup Outputs" -> From Design -> click on any pin of the added vdc source or capacitor to save the current of them.
5. At "Setup Outputs" type an expression: 1/mag(IF("/C0/PLUS")). ("1" in the nominator represents the 1V magnitude of the generator with 0° phase, mag(IF("/C0/PLUS")) in the denominator is the magnitude of output complex current. /C0/PLUS is your placed capacitor's PLUS pin, you can choose the generator's of course if you save that) Click on OK.
6. Check and Save the test schematic, then click on "Netlist and Run".
7. Plot the expression.
 

How to measure |Zout| in cadence?
1. Add a relatively large capacitance (like 1mF) to the output of the cascode and place a signal generator series with it (vdc from analogLib is good for it).

I have few questions from your reply.
1.Can I use 50ohm resistor because that what my circuit is matched to.
2.Could you tell me why did you tell me to use large capacitor
3.When we are using Vdc (Dc source),Why are giving AC amplitude,isn't it supposed to be DC Voltage value that should be provided. I want to know the difference between DC voltage field and AC magnitude field in the properties of Vdc
 

1. Where do you want to use 50Ohm? In |Zout| simulation? Not recommended. I don't understand this question actually.
2. When you measure |Zout| of your circuit you actually measue |Zout + Xc|, where Xc is the added capacitor's reactance. To keep this Xc part low, especially at low frequencies, you have to use large capacitor because Xc = 1/(2*pi*f*C).
3. When you simulate in DC a non-linear circuit the AC field of the vdc source is not important. But if you simulate AC, you want to know how is your circuit AC transfer function in a DC operating point. vdc source has got an AC field, because in AC simulation we don't want to change any source, just type in an AC magnitude where it is necessary.
 
1. Where do you want to use 50Ohm? In |Zout| simulation? Not recommended. I don't understand this question actually.

This is fine, I understood. We need to remove the load while finding the resistance of the circuit.

2. When you measure |Zout| of your circuit you actually measue |Zout + Xc|, where Xc is the added capacitor's reactance. To keep this Xc part low, especially at low frequencies, you have to use large capacitor because Xc = 1/(2*pi*f*C).

what if my circuit is operating at large frequencies,say 60GHz. How should be the value of this capacitor. But if we look back into our comments we have not given anywhere the circuit frequency of operation.Don't we need to provide that
 

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