SuperminarenandClaude Sonnet 5.5
f7fa8cfbc1
- Sections and slides rewritten in Swedish; German law and institutions replaced by Swedish/CEPT/ITU equivalents (PTS, SSA, PTSFS 2025:1, SSMFS 2008:18), unverified statements marked "% TODO VERIFY" - New toc: INSTEG, HAREC, UPPGRADERING; section idents are Swedish (see ident-map.tsv) - Exam-question tags, BNetzA question pool, solutions and the German SWL course removed - Static pages, snippets, applets, READMEs and image alt-texts translated - TikZ labels translated; SVG drawings regenerated (drawing 689 still fails to build upstream: it references a missing photo) - Changed from the original: not endorsed by DARC e.V., SSA or PTS Co-Authored-By: Claude Sonnet 5.5 <[email protected]>
43 lines
1.8 KiB
Markdown
43 lines
1.8 KiB
Markdown
<left>
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[photo:299:StepUpDownWandler: Nedväxlande (buck) och uppväxlande (boost) omvandlare]
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</left>
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<right>
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* Omvandlar likspänningar $\rightarrow$ DC/DC-omvandlare
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* t.ex. från $\qty{13,8}{\volt}$ till $\qty{5}{\volt}\rightarrow$ step-DOWN (nedväxlare)
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* t.ex. från $\qty{12}{\volt}$ till $\qty{19}{\volt}\rightarrow$ step-UP (uppväxlare)
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</right>
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<note>
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Buck-boost-omvandlaren på bilden kan ställas in från 0,5 V till 25 V på utgången. Maximal effekt är 25 W.
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</note>
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---
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### Verkningsgrad
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* Förluster uppstår i kopplingens komponenter
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* Verkningsgrad $\eta$, oftast angiven i $\%$
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<fragment>
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$\eta = \frac{P_{\mathrm{out}}}{P_{\mathrm{in}}}$
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</fragment>
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---
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#### Lösningsgång
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* givet: $U_{\textrm{in}} = \qty{12}{\volt}$
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* givet: $U_{\textrm{out}} = \qty{5}{\volt}$
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* givet: $I_{\textrm{in}} = \qty{2}{\ampere}$
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* givet: $I_{\textrm{out}} = \qty{3}{\ampere}$
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* sökt: $\eta$
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<fragment>
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$\begin{split} \eta &= \frac{P_{\mathrm{out}}}{P_{\mathrm{in}}} = \frac{U_{\mathrm{out}} \cdot I_{\mathrm{out}}}{U_{\mathrm{in}} \cdot I_{\mathrm{in}}}\\ &= \frac{\qty{5}{\volt} \cdot \qty{3}{\ampere}}{\qty{12}{\volt} \cdot \qty{2}{\ampere}} = \frac{\qty{15}{\watt}}{\qty{24}{\watt}} = \num{0,625} = \qty{62,5}{\percent} \end{split}$
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</fragment>
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---
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* givet: $U_{\mathrm{in}} = \qty{5}{\volt}$
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* givet: $U_{\mathrm{out}} = \qty{12}{\volt}$
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* givet: $I_{\mathrm{in}} = \qty{3}{\ampere}$
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* givet: $I_{\mathrm{out}} = \qty{1}{\ampere}$
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* sökt: $\eta$
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<fragment>
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$\begin{split} \eta &= \frac{P_{\mathrm{out}}}{P_{\mathrm{in}}} = \frac{U_{\mathrm{out}} \cdot I_{\mathrm{out}}}{U_{\mathrm{in}} \cdot I_{\mathrm{in}}}\\ &= \frac{\qty{12}{\volt} \cdot \qty{1}{\ampere}}{\qty{5}{\volt} \cdot \qty{3}{\ampere}} = \frac{\qty{12}{\watt}}{\qty{15}{\watt}} = \num{0,8} = \qty{80}{\percent} \end{split}$
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</fragment>
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