Tambayoyi Misali Game da Diodes Masu Fitar da Haske (LEDs)
Diode mai fitar da haske (LED) na'urar semiconductor ce da ke fitar da haske lokacin da wutar lantarki ke ratsa ta cikinta. Abin da ke faruwa a bayan LEDs ana kiransa da electroluminescence, inda kayan semiconductor ke fitar da haske idan wutar lantarki ta motsa su. LEDs sun shahara sosai a aikace-aikace daban-daban saboda ingancinsu da tsawon rayuwarsu idan aka kwatanta da tushen haske na gargajiya kamar kwararan fitilar incandescent.
A cikin wannan labarin, za mu tattauna wasu misalan matsalolin da suka shafi LEDs kuma mu yi bayani dalla-dalla game da mafita da tattaunawarsu don taimakawa fahimtar manufar da ke bayan wannan lamari.
Misali Tambaya ta 1: Muhimman Halayen LEDs
Tambaya: LED yana da damar fitar da haske mai tsawon nisan nm 650 (nanometers). Lissafa kuzarin photons da wannan LED ke fitarwa a cikin volts na lantarki (eV).
Tattaunawa:
Ana iya ƙididdige kuzarin photon ta amfani da lissafin Planck:
\[ E = \frac{hc}{\lambda} \]
Ina:
– \( E \) shine makamashin photon,
– \(h \) shine daidaitaccen Planck (\(6.626 \times 10^{-34} \text{ Js}\)),
– \( c \) shine saurin haske (\(3 \sau 10^8 \text{ m/s}\)),
– \( \lambda \) shine tsawon hasken (650 nm ko \(650 \times 10^{-9} \text{ m}\)).
Sauya waɗannan dabi'u cikin lissafin yana haifar da:
\[
E = \frac{6.626 \sau 10^{-34} \sau 3 \sau 10^8}{650 \sau 10^{-9}}
= \frac{1.9878 \sau 10^{-25}}{650 \sau 10^{-9}}
= 3.05 \sau 10^{-19} \text{ J}
\]
Na gaba, muna canza Joules zuwa volts na lantarki ta amfani da juyawar \(1 \text{eV} = 1.602 \times 10^{-19} \text{ J}\):
\[
E = \frac{3.05 \sau 10^{-19}}{1.602 \sau 10^{-19}}
≈ 1.90 \rubutu{ eV}
\]
Don haka, ƙarfin photon da LED mai tsawon zango na 650 nm ke fitarwa yana kusan 1.90 eV.
Misali na 2: Wutar Lantarki ta Gaba ta LED
Tambaya: Jajayen LED suna da ƙarfin lantarki na gaba na 2V kuma suna buƙatar wutar lantarki ta 20 mA don yin aiki yadda ya kamata. Lissafa ƙarfin da LED ɗin ke amfani da shi.
Tattaunawa:
Don ƙididdige ƙarfin da LED ke sha, muna amfani da daidaitattun ma'auni don iko, wato:
\[ P = V \sau I \]
Ina:
– \( P \) shine ƙarfin da ke cikin Watts (W),
– \(V \) shine ƙarfin lantarki a cikin Volts (V),
– \( I \) shine wutar lantarki a cikin Amperes (A).
Sauya waɗannan dabi'u cikin lissafin:
\[
P = 2 \rubutu{ V} \sau 20 \rubutu{ mA}
= 2 \rubutu{ V} \times 0.02 \rubutu{ A}
= 0.04 \rubutu{ W}
\]
Don haka, ƙarfin da jajayen LED ke sha wanda ke aiki a ƙarfin lantarki na gaba na 2V da kuma ƙarfin lantarki na 20 mA shine Watts 0.04.
Misali na 3: Ingantaccen Hasken LED
Tambaya: LED mai launin shuɗi yana da ingancin kwatancen 30%. Idan LED ɗin yana amfani da wutar lantarki mai ƙarfin 0.1 W, nawa ne wutar lantarki da ake fitarwa a matsayin haske?
Tattaunawa:
Ingancin Kwatancen (η) shine rabon wutar da aka saki a cikin nau'in haske (P_luminous) zuwa shigarwar wutar lantarki (P_input):
\[ η = \frac{P_{\text{luminous}}}{P_{\text{input}}} \]
Domin nemo \(P_{\text{luminous}} \), yi amfani da lissafin da ke ƙasa tare da ƙimar inganci na 30% ko 0.30:
\[
0.30 = \frac{P_{\text{luminous}}}{0.1 \text{ W}}
\]
Saboda haka:
\[
P_{\text{luminous}} = 0.30 \times 0.1 \text{ W}
= 0.03 \rubutu{ W}
\]
Don haka, ƙarfin da LED ke fitarwa a cikin nau'in haske shine Watt 0.03.
Misali Matsala ta 4: Da'irar LED ta Jeri
Tambaya: Kuna da LED guda uku masu ƙarfin lantarki na gaba na 2V kowannensu wanda ke buƙatar a haɗa shi a jere. Idan tushen ƙarfin lantarki shine 9V, a tantance ƙimar juriyar da ake buƙata don iyakance wutar zuwa 20 mA.
Tattaunawa:
Idan aka haɗa LEDs a jere, jimlar ƙarfin lantarki da ake buƙata shine jimlar ƙarfin lantarki na gaba na kowane LED:
\[
V_{total} = V_f1 + V_f2 + V_f3
= 2V + 2V + 2V
= 6V
\]
Ƙarfin wutar lantarki da resistor zai wargaza shine:
\[
V_{R} = V_{tushe} – V_{jimla}
= 9V – 6V
= 3V
\]
Da zarar an samu ƙarfin lantarki mai ƙarfi na 20 mA, ana ƙididdige ƙimar juriya ta amfani da Dokar Ohm:
\[
R = \frac{V_{R}}{I}
= \frac{3V}{20 \rubutu{ mA}}
= \frac{3V}{0.02A}
= 150 \Omega
\]
Saboda haka, ƙimar juriya da ake buƙata shine 150 ohms.
Misali na 5: Da'irar Iyaka Mai Layi Mai Layi Mai Layi Mai Layi
Tambaya: A cikin da'ira, akwai LED guda biyu a layi ɗaya, kowannensu yana da ƙarfin lantarki na gaba na 2V da kuma ƙarfin lantarki na 20 mA. Menene jimillar ƙarfin lantarki da ake buƙata daga tushen ƙarfin lantarki na 5V idan ana amfani da resistor ga kowane LED?
Tattaunawa:
Ga LEDs ɗin da aka shirya a layi ɗaya, kowace LED za ta fuskanci irin wannan ƙarfin lantarki, wato ƙarfin gaba nasu, kuma tunda suna buƙatar 20 mA na wutar lantarki a kowace LED, jimillar wutar lantarki shine:
\[
I_{jimla} = I_1 + I_2
= 20 \rubutu{mA} + 20 \rubutu{mA}
= 40 \rubutu{ mA}
\]
Don ƙididdige ƙimar juriya ga kowane LED:
\[
V_{R} = V_{tushe} – V_f
= 5V – 2V
= 3V
\]
Ƙimar juriya don iyakance wutar lantarki a kowace LED ita ce:
\[
R = \frac{V_{R}}{I}
= \frac{3V}{20 \rubutu{ mA}}
= \frac{3V}{0.02A}
= 150 \Omega
\]
Saboda haka, kowace LED tana buƙatar resistor mai ƙarfin ohm 150. Jimillar wutar lantarki da ake buƙata daga wutar lantarki mai ƙarfin 5V shine 40 mA.
Kammalawa
LEDs muhimmin abu ne a cikin na'urorin lantarki na zamani saboda ikonsu na samar da haske mai inganci tare da tsawon rai. Fahimtar ra'ayoyi na asali kamar ƙarfin lantarki na gaba, wutar lantarki, ƙarfin lantarki, ingantaccen kwantum, da kuma yadda ake saita su a cikin da'ira zai samar da tushe mai ƙarfi don aikace-aikacen aiki. Misalan da ke sama mataki ne na farko don ƙwarewa a fannoni daban-daban na aikace-aikace da lissafi da suka shafi amfani da LEDs a cikin da'irori na lantarki.
