Differentiation for relative sensitivity calculation

2020 10 21
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更新时间：2020 10 21
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I need to differentiate xN with respect to ln (Nd) and plot the result for Nd varying between 10^14/ cm^3 to 10^18/cm^3. Can any one please help. Heartfelt thanks...
K=1.38e-23;
T=300;
q=1.6e-19;
esi=11.8;
e0=8.854e-14;
Na=1e19;
ni=1.1e10;
syms Nd
v=(K*T)/q;
x=((Na.*Nd)./(ni^2));
Vbi=(v.*log(x));
xN=sqrt((2*esi*e0/q)*((Na*Vbi)./(Nd.*(Na+Nd))));

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回答（1 个）

Stephan
Stephan 2020-10-21

0 个投票

D_xN = diff(xN,Nd)

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SWASTIK SAHOO
SWASTIK SAHOO 2020-10-21
It's coming all 0.. and how can I plot the differentiation result???
Stephan
Stephan 2020-10-21
编辑：Stephan 2020-10-21
Run the complete code:
K=1.38e-23;
T=300;
q=1.6e-19;
esi=11.8;
e0=8.854e-14;
Na=1e19;
ni=1.1e10;
syms Nd
v=(K*T)/q;
x=((Na.*Nd)./(ni^2));
Vbi=(v.*log(x));
xN=sqrt((2*esi*e0/q)*((Na*Vbi)./(Nd.*(Na+Nd))));
D_xN = diff(xN,Nd)
To plot:
fplot(D_xN,[-1e4 1e7])
Stephan
Stephan 2020-10-21
Also you can use:
>> pretty(xN)
/ / 10 Nd \ \
| log| ----- | |
| \ 121 / |
sqrt(6007439) sqrt| ------------------------------ | 750000000
\ Nd (Nd + 10000000000000000000) /
and for the derivative:
>> pretty(D_xN)
/ / 10 Nd \ / 10 Nd \ \
| log| ----- | log| ----- | |
| \ 121 / 1 \ 121 / |
sqrt(6007439) | ------------------------------- - ------------------------------- + ------------------------------- | 375000000
| 2 2 2 |
\ Nd (Nd + 10000000000000000000) Nd (Nd + 10000000000000000000) Nd (Nd + 10000000000000000000) /
- -------------------------------------------------------------------------------------------------------------------------------
/ / 10 Nd \ \
| log| ----- | |
| \ 121 / |
sqrt| ------------------------------ |
\ Nd (Nd + 10000000000000000000) /
SWASTIK SAHOO
SWASTIK SAHOO 2020-10-21
I want to do differentiation between xN and log(Nd) and plot that so that I can see the change in xN with respect to change in Nd. That I am not able to do..Could you please help?? Thank you so much

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