# For the same simulation, does Gear method requires more points than Trap method

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#### Rich_HSV

##### Newbie
I was using NGSPICE ver32 to simulate a full-chip design, and I encountered a very peculiar number -141 Volts generated between 2 pmos. Here is my post in the NGSPICE forum. https://sourceforge.net/p/ngspice/discussion/120973/thread/5722f02293/

And I tested a few more cases and I found a trend that using Method Gear tends to use shorter time step, (which generates more points) and usually better results than Method Trap does. So can I say for a simulation with various time steps(we want the step to be as large as possible while maintaining the accuracy, so I set the maxStep to be a big number, (Tstop - Tstart)/20 instead of the default step given by the command) method Gear tends to use a more conservative step size than Method Trap does? Does anyone have literature/proof for my conjecture?

Check chapter about transient analysis from this book

Check chapter about transient analysis from this book
I read that book again after you mentioned it.
"For a fixed time step, the most accurate of these methods in a local sense is the trapezoidal rule, followed by Gear2. ... but the trapezoidal rule would typically allow larger time steps and so require a shorter run time. ...
thus it is not usually the best choice when reltol is loose."
First, I understand that it is hard to say the default reltol(1e-3) is tight or loose without having the circuit in hand, but I assume the default one for a full-chip setting is loose.
"The higher-order backward-difference formulas are efficient when tolerances are tight or when computing very smooth waveforms."
Usually, Gear2 is the go to (not a higher order) and I would not say it is a higher-order method.

I tried to derive the relationship from the formula standpoint.

Gear2:$X^{n+1} = \frac{4}{3}X^n - \frac{1}{3}X^{n-1} + \frac{2}{3}h^n \dot{X}^{n+1}$
Trap: $X^{n+1} = X^n + \frac{h^n}{2}(\dot{X}^{n+1}+\dot{X}^n)$

F(t) = Gear2 - Trap with the assumption that $X^n, X^{n-1}$ are the same timept under 2 methods, aiming to check who has a bigger $X^{n+1}$ under different assumption.

$F(t) =/frac{1}{3} X^n -\frac{1}{3}X^{n-1} + \frac{2}{3}h^n \dot{X}^{n+1} - \frac{h^n}{2}(\dot{X}^{n+1}+\dot{X}^n)$

Where can I go from here ?

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