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ee2:task_y7dozgdsljqvnqge_with_calculation [2024/07/05 01:16] (aktuell) mexleadmin angelegt |
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+ | {{tag> | ||
+ | |||
+ | # | ||
+ | <fs medium> | ||
+ | Given is the multilayer capacitor shown below, with the following dimensions: | ||
+ | * Length of layer overlap: | ||
+ | * Distance between single layers: $d=1.0 ~\rm \mu m$ | ||
+ | * Depth of component: $w=0.7 ~\rm mm$ | ||
+ | * Number of layers (as shown in the picture): 3 left-side and 3 right-side layers. | ||
+ | |||
+ | {{drawio> | ||
+ | |||
+ | The material shall have a dielectric permittivity of $\varepsilon_r=3$. \\ | ||
+ | The following calculations shall ignore boundary effects on the end of the layers. \\ \\ | ||
+ | $\varepsilon_0 = 8.854 \cdot 10^{-12} ~\rm As/Vm$ | ||
+ | |||
+ | 1. What is the field strength in the dielectric material between the layer, when a voltage of $U=6.3 ~\rm V$ is applied? | ||
+ | |||
+ | # | ||
+ | The electric field strength $E$ is given by: | ||
+ | \begin{align*} | ||
+ | E &= {{U}\over{d}} \\ | ||
+ | &= {{6.3 ~\rm V}\over{1 \cdot 10^{-6} ~\rm m}} \\ | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | $E = 6.3 {{\rm MV}\over{\rm m}}$ | ||
+ | # | ||
+ | |||
+ | 2. Calculate the capacity $C$. | ||
+ | |||
+ | # | ||
+ | |||
+ | The capacity can be derived from the geometry by: | ||
+ | \begin{align*} | ||
+ | C = \varepsilon_0 \varepsilon_r {{A}\over{d}} | ||
+ | \end{align*} | ||
+ | |||
+ | For the area $A$ we have multiple plates with the area $A_0= l \cdot w$ facing each other. | ||
+ | |||
+ | {{drawio> | ||
+ | |||
+ | How many " | ||
+ | For this, we have to count facing areas $A_0$. There are $N=5$. | ||
+ | |||
+ | {{drawio> | ||
+ | |||
+ | Therefore, the formula is | ||
+ | \begin{align*} | ||
+ | C &= \varepsilon_0 \varepsilon_r {{N \cdot l \cdot w}\over{d}} \\ | ||
+ | &= 8.854 \cdot 10^{-12} ~\rm As/Vm \cdot 3 \cdot {{5 \cdot 1.5 \cdot 10^{-3} {~\rm m} \cdot 0.7 \cdot 10^{-3} {~\rm m} }\over{1 \cdot 10^{-6} {~\rm m} }} | ||
+ | \end{align*} | ||
+ | |||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | $C = 0.139 ~\rm nF$ | ||
+ | # | ||
+ | |||
+ | 3. Due to a production problem the left-side layers are covered with $d_{\rm c}=0.1 ~\rm \mu m$ of air ($\varepsilon_{r, | ||
+ | What is the new capacity $C_\rm c$? | ||
+ | |||
+ | # | ||
+ | The air builds another capacitor in series to the dielectric material. | ||
+ | Therefore, the capacity can be calculated as | ||
+ | \begin{align*} | ||
+ | C_{\rm c} &= {{C \cdot C_{\rm Air}}\over{C + C_{\rm Air}}} | ||
+ | \end{align*} | ||
+ | |||
+ | The capacity of air is | ||
+ | \begin{align*} | ||
+ | C_{\rm Air} &= \varepsilon_0 \varepsilon_{r, | ||
+ | &= 8.854 \cdot 10^{-12} ~\rm As/Vm \cdot 1 \cdot {{5 \cdot 1.5 \cdot 10^{-3} {~\rm m} \cdot 0.7 \cdot 10^{-3} {~\rm m} }\over{0.1 \cdot 10^{-6} {~\rm m} }} \\ | ||
+ | &= 0.465... ~\rm nF | ||
+ | \end{align*} | ||
+ | |||
+ | By this the overall capacity is | ||
+ | \begin{align*} | ||
+ | C_{\rm c} &= {{0.139... ~\rm nF \cdot 0.465... ~\rm nF}\over{0.139... ~\rm nF + 0.465... ~\rm nF}} | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | $C_{\rm c} = 0.107~\rm nF$ | ||
+ | # | ||
+ | |||
+ | |||
+ | # |