Su'aalo Tusaale ah oo Ka Hadlaya Diodes-yada Iftiinka Soo Saara (LEDs)
Diode iftiin soo saaraya (LED) waa qalab semiconductor ah oo iftiin soo saara marka koronto koronto ku socoto. Dhacdada ka dambeysa LED-yada waxaa loo yaqaan electroluminescence, halkaas oo walxaha semiconductor-ku ay iftiin soo saaraan marka ay ku farxaan koronto koronto. LED-yadu waxay noqdeen kuwo aad caan ugu ah codsiyo kala duwan sababtoo ah hufnaantooda sare iyo cimrigooda dheer marka la barbar dhigo ilaha iftiinka dhaqameed sida nalalka incandescent.
Maqaalkan, waxaan kaga hadli doonnaa tusaalooyin dhibaatooyin ah oo la xiriira LED-yada waxaanan si faahfaahsan u sharxi doonnaa xalalkooda iyo doodahooda si aan u fahanno fikradda ka dambeysa dhacdadan.
Su'aal Tusaale 1: Sifooyinka Aasaasiga ah ee LED-yada
Su'aal: Nalalka LED-ku waxay leeyihiin awood ay ku soo saaraan iftiin leh hirar dhererkoodu yahay 650 nm (nanometers). Xisaabi tamarta fotonada ay soo saaraan nalalka LED-kani ee volts-ka elektarooniga ah (eV).
Dood:
Tamarta fotonka waxaa lagu xisaabin karaa iyadoo la isticmaalayo isle'egta Planck:
\[ E = \frac{hc}{\lambda} \]
Halkee:
– \( E \) waa tamarta sawir-qaadista,
– \( h \) waa joogtada Planck (\(6.626 \times 10^{-34} \text{ Js}\)),
– \( c \) waa xawaaraha iftiinka (\(3 \times 10^8 \text{ m/s}\)),
– \( \lambda \) waa hirarka iftiinka (650 nm ama \(650 \times 10^{-9} \text{ m}\)).
Qiimahan ku beddelashada isleegta ayaa soo saara:
\[
E = \frac{6.626 \jeer 10^{-34} \jeer 3 \jeer 10^8}{650 \jeer 10^{-9}}
= \frac{1.9878 \jeer 10^{-25}}{650 \jeer 10^{-9}}
= 3.05 \jeer 10^{-19} \qoraal{ J}
\]
Marka xigta, waxaan Joules u beddelnaa volts-ka elektarooniga ah annagoo adeegsanayna beddelka \(1 \text{eV} = 1.602 \times 10^{-19} \text{ J}\):
\[
E = \frac{3.05 \jeer 10^{-19}}{1.602 \jeer 10^{-19}}
≈ 1.90 \qoraal{ eV}
\]
Markaa, tamarta photon-ka ee ay soo saarto LED-ka leh hirarka 650 nm waa qiyaastii 1.90 eV.
Tusaale 2: Danabka Hore ee LED-ka
Su'aal: Nalalka cas waxay leeyihiin danab hore oo ah 2V waxayna u baahan yihiin koronto 20 mA ah si ay si sax ah u shaqeeyaan. Xisaabi awoodda uu isticmaalo nalalka LED-ka.
Dood:
Si loo xisaabiyo awoodda ay nuugto LED-ka, waxaan isticmaalnaa isla'egta aasaasiga ah ee awoodda, kuwaas oo kala ah:
\[ P = V \jeer I \]
Halkee:
– \( P \) waa awoodda Watts (W),
– \(V \) waa danabka ku jira Volts (V),
– \( I \) waa hadda ku jirta Amperes (A).
Qiimahan ku beddel isla'egta:
\[
P = 2 \qoraal{ V} \jeer 20 \qoraal{ mA}
= 2 \qoraal{ V} \times 0.02 \qoraal{ A}
= 0.04 \qoraal{ W}
\]
Markaa, awoodda ay nuugto LED cas oo ku shaqeeya danab hore oo ah 2V iyo hadda ah 20 mA waa 0.04 Watts.
Tusaale 3: Waxtarka LED-ka
Su'aal: LED buluug ah wuxuu leeyahay hufnaan ku dhawaad 30%. Haddii LED-ku uu isticmaalo 0.1 W oo koronto ah, immisa awood ayaa iftiin ahaan loo soo saaraa?
Dood:
Waxtarka Quantum (η) waa saamiga awoodda lagu sii daayo qaabka iftiinka (P_luminous) iyo gelinta korontada (P_input):
\[ η = \frac{P_{\text{luminous}}}{P_{\text{input}}} \]
Si aad u hesho \( P_{\text{luminous}} \), isticmaal isla'egta soo socota oo leh qiimaha waxtarka 30% ama 0.30:
\[
0.30 = \frac{P_{\text{luminous}}}{0.1 \text{ W}}
\]
Sidaas darteed:
\[
P_{\text{luminous}} = 0.30 \times 0.1 \text{ W}
= 0.03 \qoraal{ W}
\]
Markaa, awoodda ay sii deyso LED-ka qaabka iftiinka waa 0.03 Watt.
Tusaale ahaan Dhibaatada 4aad: Wareegga LED-ka Taxanaha ah
Su'aal: Waxaad haysataa saddex LED oo leh danab hore oo 2V ah midkiiba oo u baahan in si taxane ah loogu xidho. Haddii isha danabku tahay 9V, go'aami qiimaha iska caabbinta loo baahan yahay si loo xaddido qulqulka ilaa 20 mA.
Dood:
Marka LED-yada si taxane ah loogu xidho, wadarta guud ee danabka loo baahan yahay waa wadarta danabka hore ee LED kasta:
\[
V_{wadarta} = V_f1 + V_f2 + V_f3
= 2V + 2V + 2V
= 6V
\]
Danabka haraaga ah ee uu iska caabintu kala diri karo waa:
\[
V_{R} = V_{ilaha} – V_{wadarta}
= 9V – 6V
= 3V
\]
Marka la helo koronto la doonayo oo ah 20 mA, qiimaha iska caabbinta waxaa lagu xisaabiyaa iyadoo la adeegsanayo Sharciga Ohm:
\[
R = \frac{V_{R}}{I}
= \frac{3V}{20 \qoraal{ mA}}
= \frac{3V}{0.02A}
= 150 \Omega
\]
Sidaas darteed, qiimaha resistor loo baahan yahay waa 150 ohms.
Tusaale 5: Wareegga Xaddidaadda Hadda ee LED-ka ee Isbarbar socda
Su'aal: Wareegga, waxaa jira laba LED oo is barbar socda, mid walbana leh danab hore oo 2V ah iyo hadda oo 20 mA ah. Waa maxay wadarta hadda ee looga baahan yahay isha danab ee 5V haddii iska caabin loo isticmaalo LED kasta?
Dood:
Nalalka LED-ka ee isku midka ah ee loo habeeyey si is barbar socda, nalalka LED-ka kasta waxay la kulmi doonaan danab isku mid ah, gaar ahaan danabkooda hore, maadaama ay u baahan yihiin 20 mA oo hadda ah nalalka LED-ka kasta wadarta hadda waa:
\[
I_{wadarta} = I_1 + I_2
= 20 \text{ mA} + 20 \text{ mA}
= 40 \qoraal{ mA}
\]
Si loo xisaabiyo qiimaha iska caabbinta ee LED kasta:
\[
V_{R} = V_{ilaha} – V_f
= 5V – 2V
= 3V
\]
Qiimaha iska caabbinta si loo xaddido hadda ku jira LED kasta waa:
\[
R = \frac{V_{R}}{I}
= \frac{3V}{20 \qoraal{ mA}}
= \frac{3V}{0.02A}
= 150 \Omega
\]
Sidaas darteed, LED kasta wuxuu u baahan yahay iska caabin 150 ohm ah. Wadarta guud ee hadda looga baahan yahay korontada 5V waa 40 mA.
Gabagabo
Nalalka LED-ku waa qaybo muhiim u ah qalabka elektaroonigga casriga ah sababtoo ah awooddooda ay ku bixiyaan iftiin aad u hufan oo leh cimri dheer. Fahmidda fikradaha aasaasiga ah sida danabka hore, hadda, awoodda, hufnaanta quantum, iyo sida loogu habeeyo wareegga waxay siin doontaa aasaas adag codsiyada wax ku oolka ah. Tusaalooyinka kor ku xusan waa tallaabada ugu horreysa ee lagu barto codsiyada iyo xisaabinta kala duwan ee la xiriira isticmaalka nalalka LED-yada wareegyada elektaroonigga ah.
