Tusaalaha Su'aalaha Doodaha Nidaamyada Elektarooniga ah
Nidaamyada elektaroonigga ah waxay door muhiim ah ka ciyaaraan qaybaha kala duwan ee nolosha casriga ah, laga bilaabo isgaarsiinta iyo warshadaha ilaa daawada. Faham qoto dheer oo ku saabsan fikradaha aasaasiga ah iyo codsiyada nidaamyada elektaroonigga ah ayaa muhiim u ah ardayda iyo xirfadlayaasha doonaya inay horumariyaan khibrad ku saabsan arrintan. Maqaalkani wuxuu soo bandhigi doonaa dhowr tusaale oo dhibaatooyin iyo doodo ah oo ku saabsan nidaamyada elektaroonigga ah, kuwaas oo aan rajeyneyno inay bixin doonaan aragti iyo caawinta habka waxbarashada.
1. Tusaale Dhibaato: Wareegga Shaandhada ee RC-ga oo Dhamaystiran
Su'aal:
Waxaa lagu siinayaa wareegga shaandhada RC-ga ee hooseeya, halkaas oo iska caabbinta (R) ay tahay 1kΩ iyo awoodda (C) ay tahay 100nF. Xisaabi inta jeer ee la jarayo shaandhada.
Dood:
Soo noqnoqoshada goynta (f_c) ee shaandhada RC-ga ee hoose waxaa lagu xisaabin karaa qaacidada:
\[ f_c = \frac{1}{2 \pi RC} \]
Adigoo qiimaha awoodda ka beddelaya nanoFarad una beddelaya Farad:
\[ C = 100nF = 100 jeer 10^{-9} F \]
Hadda, waxaan ku beddeleynaa qiimaha R iyo C qaacidada:
\[ f_c = \frac{1}{2 \pi (1 \jeer 10^3)(100 \jeer 10^{-9})} \]
\[ f_c = \frac{1}{2 \pi \times 10^{-4}} \]
\[ f_c \approx \frac{1}{6.28 \times 10^{-4}} \]
\[ f_c \ qiyaastii 1591.55 Hz \]
Marka, inta jeer ee goynta shaandhadani waa qiyaastii 1591.55 Hz.
2. Su'aal Tusaale ah: Ka faa'iidaysiga Awood-siiyaasha Hawlgalka (Op-Amplifiers)
Su'aal:
Markaad isticmaaleyso qalab kor u qaada oo aan rogrogin oo leh qiimayaal ah R1 = 1kΩ iyo R2 = 10kΩ, xisaabi faa'iidada wareegga.
Dood:
Faa'iidada qalab kor u qaada oo aan rogin waxaa lagu xisaabiyaa qaacidada:
\[ Gaadh(A) = 1 + \frac{R2}{R1} \]
Iyada oo la raacayo qiimayaasha la bixiyay ee R1 iyo R2:
\[ A = 1 + \frac{10k\Omega}{1k\Omega} \]
\[ A = 1 + 10 \]
\[ A = 11 \]
Natiijooyinka kor ku xusan, faa'iidada qalabkan hawlgalka ee aan rogrogmi karin waa 11 jeer.
3. Su'aal Tusaale ah: Nidaamka Dijital ah oo leh Bakhtiyaa-nasiibka Calaamadaha
Su'aal:
Calaamad dijitaal ah oo shan lugood leh ayaa soo saarta qaabka koodka laba-geesoodka ah 01101. Xisaabi qiimaha jajab tobanlaha ah ee u dhigma qaabka koodka laba-geesoodka ah.
Dood:
Si aan ugu beddelno koodhka laba-geesoodka ah jajab tobanle, waxaan isticmaali karnaa isku-dhufashada awoodaha laba hab. Tiro kasta oo laba-geesood ah waxaa lagu dhufanayaa 2, iyadoo kor loogu qaadayo awoodda u dhiganta booskeeda midig ilaa bidix, iyadoo laga bilaabayo awoodda 0.
Qaabka laba-geesoodka ah ee 01101 waxaa loo xisaabin karaa sidan:
\[ 0 \jeer 2^4 + 1 \jeer 2^3 + 1 \jeer 2^2 + 0 \jeer 2^1 + 1 \jeer 2^0 \]
Noqo:
\[ 0 \times 16 + 1 \times 8 + 1 \times 4 + 0 \times 2 + 1 \times 1 \]
\[ = 0 + 8 + 4 + 0 + 1 \]
\[ = 13 \]
Markaa, qiimaha jajab tobanlaha ee qaabka laba-geesoodka ah 01101 waa 13.
4. Su'aal Tusaale ah: Wareegga Hagaajiyaha Hirka Buuxa
Su'aal:
Adigoo isticmaalaya qalab yareeya heerka hoose kaas oo yareeya danabka laga bilaabo 240V AC ilaa 24V AC, oo ku xiran hagaajiye hirar buuxa leh, xisaabi danabka DC ee ka dhasha haddii diode-ku uu yahay mid ku habboon (iyada oo aan lahayn hoos u dhac danab).
Dood:
Hagaajiye hirar buuxa leh ayaa korontada AC u beddela DC isagoo hagaajinaya wareegga AC oo dhan. Danabka DC ee uu soo saaro hagaajiye hirar buuxa leh waxaa lagu go'aamin karaa iyadoo la xisaabinayo danabka celceliska qaabka hirarka la hagaajiyay.
Si loo helo diode iyo danab RMS oo ku habboon gelinta (transformer-ka wax soo saarka), danabka DC-ga ee wax soo saarka ee hagaajiyaha hirarka buuxa ee eexda leh waa:
\[ V_{DC} \approx \frac{2V_{RMS}}{\pi} \]
Halkan, danab RMS waa 24V.
\[ V_{DC} \qiyaastii \frac{2 \times 24}{3.14} \]
\[ V_{DC} \approx \frac{48}{3.14} \]
\[ V_{DC} \qiyaastii 15.29V \]
Markaa, danab DC-ga ka dhashay waa qiyaastii 15.29V.
5. Su'aal Tusaale ah: Isku-darka Isbarbardhigga ee Wareegga Resonance-ka LC
Su'aal:
Go'aami inta jeer ee soo noqnoqoshada soo noqnoqoshada \( f_r \) ee wareegga soo noqnoqoshada LC oo ka kooban inductor L = 10mH iyo capacitor C = 10µF.
Dood:
Soo noqnoqoshada resonant (\( f_r \)) ee wareegga LC ee barbar socda waxaa lagu xisaabiyaa qaacidada:
\[ f_r = \frac{1}{2 \pi \sqrt{LC}} \]
Adigoo qiimaha L iyo C u beddelaya cutubyada Henry iyo Farad:
\[ L = 10mH = 10 jeer 10^{-3}H \]
\[ C = 10µF = 10 jeer 10^{-6}F \]
Ku beddel L iyo C qaacidada:
\[ f_r = \frac{1}{2 \pi \sqrt{(10 \jeer 10^{-3})(10 \jeer 10^{-6})}} \]
\[ f_r = \frac{1}{2 \pi \sqrt{10 \times 10^{-9}}} \]
\[ f_r = \frac{1}{2 \pi \sqrt{10^{-8}}} \]
\[ f_r = \frac{1}{2 \pi \times 10^{-4}} \]
\[ f_r = \frac{10^4}{2 \pi} \]
\[ f_r \approx \frac{10^4}{6.28} \]
\[ f_r \qiyaastii 1591.55 Hz \]
Markaa, soo noqnoqoshada soo noqnoqoshada ee wareegga LC-gan waa qiyaastii 1591.55 Hz.
Gabagabo
Laga soo bilaabo dooddii ku saabsanayd dhibaatooyinka tusaalaha ah ee kor ku xusan, waxaan aragnay sida adeegsiga mabaadi'da aasaasiga ah ee elektaroonigga ay nooga caawin karto inaan fahanno oo aan xallino dhibaatooyinka caadiga ah ee ka dhaca goobta. Fahmidda fikradaha iyo dhaqanka socda ayaa lagama maarmaan u ah barashada nidaamyada elektaroonigga ah. Waxaa la rajeynayaa in maqaalkani uu ka caawin doono akhristayaasha inay si fiican u fahmaan sida loo xisaabiyo qaybaha iyo sifooyinka aasaasiga ah ee nidaamyada elektaroonigga ah, taasoo u sahlaysa inay ku dabaqaan waxbarashadooda iyo goobta shaqada.
