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Hi

check the Model Definitions Functions rect() pulse (its smooth if you turn on smoothing)

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Good luck
Ivar

Hi Ivar

Thank you for the prompt reply. Currently, I tried to input the following command

if(t<=40e-9, (2/40e-9*t), if(t<=80e-9,2, if(t<=120e-9, (-2/40e-9*t),0)))

into under electric potential. However, Kept encountering the following error:

Repeated error test failures. May have reached a singularity.
Time : 8e-008
Last time step is not converged.

In what way, I can rectify this error? What might have went wrong? At first, I suspected is due to the "My Study Setting under Study was not set properly. And change to >> Times: range(0,1e-8,1.2e-7)" . But the problem still persist. Pleae advise.

Thank you.

Susan

Hi

even if you get a nice pulse like that you have discrete sharp (Dirac type) steps, the solver expect continuous functions to get the derivative to better control the solver steps, and with these steep slopes it fails. therefore the "rect()" function is proposed with "smoothing" Heavyside type functions. Just as for the step() function

use rather these

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Good luck
Ivar

Hi Ivar

Thank you for the prompt reply once again!! I really appreciate your helps.Please bear with me awhile more, I have tried the method that you've proposed by keying the Time waveform using the rect function. May I know how can I assign the rect function into the electrical potential?

The current function name: rect1()

Thank you.

Susan

Hi

use V0*rect(t[1/s]) it will give you a value, for each "t" (expressed in seconds) that your solver steps chooses. V0 is for a voltage of scalar amplitude "V0" adapt to whatever you need

to check your function plot it out (plot icon)

Another thing, if you use the default solver and time stepping, "dig" your way down to the
Study1 - Solver Configuration - Time-dependent Solver 1 and open the "Time Stepping tab"

change the Steps taken by the solver from "Free" to intermediate or strict, then be sure in the
Study 1 - Time dependent under Times that you define a range with a few steps on the rising and falling edge of your pulse

Else there is a fair chance that your solver will integrate over the steep rise and at least fall time regions
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Good luck
Ivar

Hi Ivar

Thank you so much!!!! By the way, how can I assign the rise and falling edge into Study 1 - Time dependent under Times?

For example, rise from 0s to 100ns, and stagnant from 100ns to 200ns and falls from 200ns to 300ns. How should I put it?

Please advise.

Thanks.

Susan

Hi Ivar,

Basically, I am trying to create a single voltage square pulse with time. For example, rise from time 0s to 2ns [from 0 to 2V] and pulse width of 100ns, follow by falling edge 5ns [2V to 0V]

I tried using rect function to key in the rise/fall time via size of transition zone to be 2.5e-9. and the width to be 100e-9 by entering defining it in the lower and upper limit.
Just like the following:
Under the Parameters tab ==> Lower limit: 0, Upper limit: 100e-9
Smoothing (Size of transition zone: 2.5e-9)

But an error msg pop up :

Repeated error test failures. May have reached a singularity.
Time : 1.0249995275321e-007
Last time step is not converged.

However, it was fine when I entered the rise/fall time (smoothing) to 10e-9.

How can I resolve this prob if I wants a steep rise and fall transition?

Alternatively, I have tried to use the step function
Step 1: Parameters >> Location: 0; From: 0; To: 1; Size transition zone: 5e-9
Step 2: Parameters >> Location: 100e-9; From 1; To: 0; Size transition zone 2.5e-9

And I tried to combined them into Analytic function: in the expression box: 2*(step1(t)+step3(t))

The following error message pop up when I was trying to see a preview of the plot
Syntax error
Expression: ()[1/ns]
Sub expression: ()[1/ns]
Position: 1

Please advise.

Thanks.

Susan

Hi

if you define the pulse as this (see file) it should be OK, but as you can define units in the analytical solution you do not need the units conversion when you call the analytical function, just use "t" as by default it's in seconds

But you pulse seems to be read from my understanding, so it's probably the solve thta has issues with the short and steep time series, you should use a time stepping of the type:

"intermediate" or "strict" (and not free, possibly try a gen alpha) and then use a range of the type

{range(0,0.1,2) range(5,5,95) range(100,0.1,102) range(105,5,150)}*1[ns]

to be sure you have enough points during your rise and fall times. It all depends on your model and equations, if you system is very dynamic you might need more points during the flat period, or you could use a power stepping to increase the time steps

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Good luck
Ivar

Hi Ivar,

Hmm..But I still have problem having to see the preview of the combined steps functions in analytic function. From the plot review I only manage to see a straight line.

What did you input in the plot parameters to allow you to see a nice analytic graph? How do you define the upper and lower limits for t (time)?
Thank you.

Susan

Hi Ivar,

Yes, I've got it!! Thank you so much for your help!! Really appreciated!!

Susan

Hi

Good to hear

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Have fun COMSOLing
Ivar

Hi Ivar,

After I executed the prog, I realised that the potential (v) across my model varies with my Voltage (set:1.8V) [file]

This is what I wrote in Vo under the electric potential.
1.8*an1(t)

However, if I just apply 1.8V into Vo it shows linear voltage increase across my geometry from 0 to1.8V (Expected)

How can ensure my model is receiving a maximum of 1.8V and not beyond or varies? Do I make change in the analytic function? how?


Thanks.

Susan

Hi

that's probably my error, I normalised to 2"V" but as the two step functions go from 0-1 and they add up you will see that the analytical amplitude, in my example is 4 and not 2 as expected.

Check carefully the amplitudes and remove the factor 2 (or divide by 2 depending on how you want to normalise), it should work better

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Good luck
Ivar

Hi Ivar,

I tried to divided the expression by 2 [file: an1]. but it still cannot be scale to 0 to 1 in the y axis. Please advise.

Hi

but your an1() it scales to one, but the output is now "Volts" as there are units defined to the analytical function.

If you set the amplitude in your equations then keep the v0=1.8[V] outside and remove the "V" for the units in the definition of an1(). Hence it gives a scalar function of "t" between 0-1

Note that COMSOL flag unit error by orange colour (these are warnings, no show-stopper to solve, but the results might be unexpected), while if you have any errors, such as missing (){} ... then the formula is flagged RED and you MUST resolve it to be able to solve

I cann't really say more, try to dig into the "model definitions probe values" or "plot while solving" to track better what is going on, then you might easier find out what is wrong

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Good luck
Ivar