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18th January 2018, 12:42 #1
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Cascode LNA output impedance in cadence
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

18th January 2018, 12:42

18th January 2018, 13:28 #2
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Re: Cascode LNA output impedance in cadence
The gain of cascode is gm*(output impedance)^2. I want to increase the gain so I like to increase the output impedance.
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.

18th January 2018, 13:37 #3
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18th January 2018, 13:37

18th January 2018, 14:11 #4
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Re: Cascode LNA output impedance 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 multistage amplifier.

18th January 2018, 14:43 #5
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Re: Cascode LNA output impedance in cadence
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.

13th March 2018, 18:08 #6
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Re: Cascode LNA output impedance in cadence
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

13th March 2018, 19:45 #7
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Re: Cascode LNA output impedance in cadence
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 nonlinear 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.
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13th March 2018, 19:45

16th March 2018, 13:36 #8
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Re: Cascode LNA output impedance in cadence
This is fine, I understood. We need to remove the load while finding the resistance of the circuit.
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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