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question about simulation

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h1_x2_r3

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Hi,everyone!
I have a quesition about Cadence:
I do not kown the difference between the component parameter sweep and model parameter sweep and how to use Cadence to test total Harmonic distortion(THD)?
thank you very much!
 

mike_bihan

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Question not well defined. Suggest polish the question before put it here.
 

DrLock

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component parameter sweep - This lets you change parameters of one or more components, such as the width of a transistor or the DC voltage of a voltage source. This changes any parameter that you can edit for a single component.

model parameter sweep - This changes something in your component models, such as your transistor model. For instance, you could change your nMOS transistor oxide thickness. All components that use that model (for instance all nMOS's) will change together.

I am not sure how you would use either in simulating THD. I assume that for THD you start with a transient simulation and compute the Fourier transform of the output, but am not sure beyond that.
 

present

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To compute the thd, you need to perform the following steps:
1. Choose thd in the Special Functions menu of the calculator.
The Total Harmonic Distortion form appears.
2. Set up an expression in the calculator buffer and click Get Buffer.
3. Specify the range and the number of samples.
4. Click OK.
5. Click Print to see the result.
The accuracy of the total harmonic distortion measurement depends on simulator options and the analysis parameters. For an accurate measurement set the following simulation options: Set the simulation timestep to be 1/100th of a cycle, and simulate for ten cycles. End the simulation slightly beyond the tenth cycle. When you use the calculator, measure during the tenth cycle by specifying the beginning of the cycle as the From time and the end as the To time.

Option Suggested Value
RELTOL 1e-5
ABSTOL 1e-13
VNTOL 3e-8
TRTOL 1
METHOD gear
MAXORD 3
 

h1_x2_r3

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thank you very much!:)
but i simulate a circuit in you way,the thd of the circuit is just -12dB,
but the requirement is -50dB,then,i simulate the ideal sine wave,
the thd of the ideal sine wave is just -51dB,
Does it mean the output of my circuit must close to ideal?
thank you very much!
 

pseudockb

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Well, not true. If it's an ideal THD, you should get -infinity theorectically. Try using a smaller time step to get the THD of an ideal sine wave. You will notice that the THD improves as the time step gets smaller. Try and see.
 

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