C ↑↓ 1 s ↑ 2 s ↑ 2 p x ↑ 2 p y ↑ 2 p z {\displaystyle C\quad {\frac {\uparrow \downarrow }{1s}}\;{\frac {\uparrow \,}{2s}}\;{\frac {\uparrow \,}{2p_{x}}}\;{\frac {\uparrow \,}{2p_{y}}}\;{\frac {\uparrow \,}{2p_{z}}}}
V B = V 0 + V {\displaystyle \mathbf {V_{B}=V_{0}+V} }
S Q = V B {\displaystyle \mathbf {S_{Q}=V_{B}} }
s u m l = 0 m − 1 {\displaystyle \mathbf {sum_{l=0}^{m}-1} }
S q = ( N ) ( 1 ) {\displaystyle \mathbf {S_{q}={\frac {(N)}{(1)}}} }
z ( x k x l ) {\displaystyle \mathbf {z(x_{k}x_{l})} }
S_{o} = \frac{1}{M N} \cdot sum_{l=0}^m-1
T ( K ) = T 0 1 + [ T 0 ⋅ l n ( R ( T ) R ( 298 ) ) ⋅ 1 B ] → T ( K ) = 298 1 + [ 298 ⋅ l n ( R ( T ) 1000 ) ⋅ 1 3530 ] {\displaystyle \mathbf {T_{(K)}={\frac {T_{0}}{1+\left[T_{0}\cdot ln\left({\frac {R(T)}{R(298)}}\right)\cdot {\frac {1}{B}}\right]}}\rightarrow \ T_{(K)}={\frac {298}{1+\left[298\cdot ln\left({\frac {R(T)}{1000}}\right)\cdot {\frac {1}{3530}}\right]}}} }
R ( T ) = R ( 298 K ) e x p B ⋅ ( T 0 − T T ⋅ T 0 ) {\displaystyle \mathbf {R(T)=R(298K)exp^{B\cdot \left({\frac {T_{0}-T}{T\cdot T_{0}}}\right)}} }
R t e r m i s t o r e N T C = ( V ⋅ 1000 ) ( 5 − V ) {\displaystyle \mathbf {R_{termistoreNTC}={\frac {(V\cdot \ 1000)}{(5-V)}}} }
R f o t o r e s i s t e n z a ( K O h m ) = ( V ⋅ 10 ) ( 5 − V ) {\displaystyle \mathbf {R_{fotoresistenza(KOhm)}={\frac {(V\cdot \ 10)}{(5-V)}}} }
R f o t o r e s i s t e n z a ∝ 1 I l l l u x {\displaystyle \mathbf {R_{fotoresistenza}\propto \ {\frac {1}{Ill_{lux}}}} }
( y 2 − y 1 ) = m ( x 2 − x 1 ) → m = y 2 − y 1 x 2 − x 1 {\displaystyle \mathbf {(y_{2}-y_{1})=m(x_{2}-x_{1})\ \rightarrow \ m={\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}} }
R g a s R a i r ⋅ R a i r = R g a s {\displaystyle \mathbf {{\frac {R_{gas}}{R_{air}}}\cdot \ R_{air}=R_{gas}} }
C = 10 ( − 0 , 085 − 2 , 146 ⋅ L o g R g a s ) {\displaystyle \mathbf {C=10^{(-0,085-2,146\cdot \ LogR_{gas)}}} }
L o g C = L o g 0 , 3 + ( L o g R g a s − L o g 1 , 6 ) − 0 , 466 → L o g C = − 0 , 085 − 2 , 146 ⋅ L o g R g a s {\displaystyle \mathbf {LogC=Log0,3+{\frac {(LogR_{gas}-Log1,6)}{-0,466}}\ \rightarrow \ LogC=-0,085-2,146\cdot \ LogR_{gas}} }
m = ( L o g 0 , 18 − L o g 1 , 6 ) ( L o g 30 − L o g 0 , 3 ) = − 0 , 466 {\displaystyle \mathbf {m} ={\frac {(Log0,18-Log1,6)}{(Log30-Log0,3)}}=-0,466}
R g a s ( k O h m ) = ( 5 − V ) ⋅ 10 V {\displaystyle \mathbf {R_{gas}(kOhm)={\frac {(5-V)\cdot \ 10}{V}}} }
V = 0 , 0317 ⋅ R h ( % ) + 0 , 827 → R h ( % ) = ( V + 0 , 827 ) 0 , 0317 {\displaystyle \mathbf {V=0,0317\cdot \ Rh(\%)\ +0,827\ \rightarrow \ Rh(\%)={\frac {(V+0,827)}{0,0317}}} }
R h ( t e r r e n o ) ∝ 1 R {\displaystyle \mathbf {Rh(terreno)\propto \ {\frac {1}{R}}} }
R R h ( t e r r e n o ) = ( V ⋅ 100 ) ( 5 − V ) {\displaystyle \mathbf {R_{Rh(terreno)}={\frac {(V\cdot \ 100)}{(5-V)}}} }
I B > I B ( m i n i m a ) = I C h F E {\displaystyle \mathbf {I_{B}>I_{B(minima)}={\frac {I_{C}}{h_{FE}}}} }
V B E ≤ 0 {\displaystyle \mathbf {V_{BE}\leq 0} }
B = μ 0 N l i {\displaystyle \mathbf {{B}=\mu _{0}\ {\frac {N}{l}}\ i} }
i d i f f = i d e r i v a {\displaystyle \mathbf {i_{diff}=i_{deriva}} }
e n = N D {\displaystyle \mathbf {e_{n}=N_{D}} }
h p = N A {\displaystyle \mathbf {h_{p}=N_{A}} }
n i = 1 , 45 {\displaystyle \mathbf {n_{i}=1,45} } 10 10 {\displaystyle \mathbf {10^{10}} } { c m − 3 {\displaystyle \mathbf {cm^{-3}} }
V 0 = K B T e l n ( N D N A n i 2 ) {\displaystyle \mathbf {V_{0}={\frac {K_{B}T}{e}}ln\left({\frac {N_{D}N_{A}}{n_{i}^{2}}}\right)} }
i = d Q d t {\displaystyle \mathbf {i={\frac {dQ}{dt}}} }
V B = V 0 − V {\displaystyle \mathbf {V_{B}=V_{0}-V} }
I = I 0 [ exp ( e ⋅ V K B T ) − 1 ] {\displaystyle \mathbf {I=I_{0}\left[\exp \left({\frac {e\cdot V}{K_{B}T}}\right)-1\right]} }
e ⋅ V K B T ≃ − 4 {\displaystyle \mathbf {{\frac {e\cdot V}{K_{B}T}}\simeq -4} }
I ≃ I 0 ( exp − 4 − 1 ) → e x p 4 ≃ 0 , 014 − 1 ≃ 1 → I ≃ − I 0 {\displaystyle \mathbf {I\simeq I_{0}(\exp {-4}-1)\rightarrow \ exp{4}\simeq 0,014-1\simeq 1\rightarrow \ I\simeq -I_{0}} }
e x p 4 ≃ 0 , 014 << 1 {\displaystyle \mathbf {exp{4}\simeq 0,014<<1} }
I ≃ − I 0 {\displaystyle \mathbf {I\simeq -I_{0}} }
I C + I B = I E {\displaystyle \mathbf {I_{C}+I_{B}=I_{E}} }
V C E = V C B + V B E {\displaystyle \mathbf {V_{CE}=V_{CB}+V_{BE}} }
I C = α I E + I 0 ≃ α I E {\displaystyle \mathbf {I_{C}=\alpha \ I_{E}+I_{0}{\displaystyle }\simeq \alpha \ I_{E}} }
h F E = I C I B {\displaystyle \mathbf {h_{FE}={\frac {I_{C}}{I_{B}}}} }
V C E < V B E {\displaystyle \mathbf {V_{CE}<V_{BE}} }
A B C D E a b c d e 1234 A B C D E a b c d e 1234 {\displaystyle \mathrm {ABCDEabcde1234} {\mathit {ABCDEabcde1234}}}
V B E ( s a t ) ≃ 0 , 7 V {\displaystyle \mathbf {V_{BE(sat)}\simeq 0,7V} }
V C E ( s a t ) ≃ 0 , 2 V {\displaystyle \mathbf {V_{CE(sat)}\simeq 0,2V} }
n i = 2 , 1 ⋅ 10 10 c m − 3 {\displaystyle \mathbf {n_{i}=2,1\cdot \ 10^{10}\ cm^{-3}} }
n e n h = n i 2 = 1 , 45 ⋅ 10 20 c m − 6 {\displaystyle \mathbf {n_{e}n_{h}={n_{i}}^{2}=1,45\cdot \ 10^{20}\ cm^{-6}} }
# Il soldato tedesco ha stupito il mondo, il bersagliere italiano ha stupito il soldato tedesco.
Immagine:AKrommel.jpg
--Lord web 19:40, 9 mag 2006 (CEST) \simeq
![{\displaystyle \mathbf {T_{(K)}={\frac {T_{0}}{1+\left[T_{0}\cdot ln\left({\frac {R(T)}{R(298)}}\right)\cdot {\frac {1}{B}}\right]}}\rightarrow \ T_{(K)}={\frac {298}{1+\left[298\cdot ln\left({\frac {R(T)}{1000}}\right)\cdot {\frac {1}{3530}}\right]}}} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e7d5104d6d680e9f3b91dd2d868b7d87f6acca95)
{
![{\displaystyle \mathbf {I=I_{0}\left[\exp \left({\frac {e\cdot V}{K_{B}T}}\right)-1\right]} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/adda42050a5a4c42b28ae1d55b04acb7f5b103ce)
