Imibuzo Eyisibonelo Exoxa Nge-Diode Ekhipha Ukukhanya (ama-LED)
I-diode ekhipha ukukhanya (i-LED) iyithuluzi le-semiconductor elikhipha ukukhanya lapho ugesi ugeleza kuwo. Isenzakalo esingemuva kwama-LED saziwa ngokuthi i-electroluminescence, lapho izinto ze-semiconductor zikhipha ukukhanya lapho zishukunyiswa ugesi. Ama-LED asethandwa kakhulu ezinhlotsheni ezahlukene zokusebenza ngenxa yokusebenza kwawo kahle kanye nokuphila isikhathi eside uma kuqhathaniswa nemithombo yokukhanya yendabuko efana nama-incandescent bulbs.
Kulesi sihloko, sizoxoxa ngezinkinga ezithile eziphathelene nama-LED futhi sichaze izixazululo zawo kanye nezingxoxo ngokuningiliziwe ukuze sisize ukuqonda umqondo ongemuva kwalesi simo.
Isibonelo Umbuzo 1: Izakhiwo Eziyisisekelo Zama-LED
Umbuzo: I-LED inamandla okukhipha ukukhanya okunobude be-wavelength obungu-650 nm (ama-nanometer). Bala amandla ama-photon akhishwa yile LED kuma-electron-volts (eV).
Ingxoxo:
Amandla e-photon angabalwa kusetshenziswa i-equation kaPlanck:
\[ E = \frac{hc}{\lambda} \]
Kuphi:
– \( E \) amandla e-photon,
– \( h \) kuyinto engaguquki kaPlanck (\(6.626 \times 10^{-34} \text{ Js}\)),
– \( c \) ijubane lokukhanya (\(3 \times 10^8 \text{ m/s}\)),
– \( \lambda \) ubude besikhathi bokukhanya (650 nm noma \(650 \times 10^{-9} \text{ m}\)).
Ukufaka la manani ku-equation kuveza:
\[
E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{650 \times 10^{-9}}
= \frac{1.9878 \times 10^{-25}}{650 \times 10^{-9}}
= 3.05 \izikhathi ezingu-10^{-19} \umbhalo{J}
\]
Okulandelayo, siguqula ama-Joules abe ama-electron-volts sisebenzisa ukuguqulwa \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\):
\[
E = \frac{3.05 \times 10^{-19}}{1.602 \times 10^{-19}}
≈ 1.90 \umbhalo{eV}
\]
Ngakho-ke, amandla e-photon akhishwa yi-LED enobude be-wavelength obungu-650 nm angaba ngu-1.90 eV.
Isibonelo 2: I-LED Forward Voltage
Umbuzo: I-LED ebomvu ine-voltage yangaphambili engu-2V futhi idinga ugesi ongu-20 mA ukuze isebenze kahle. Bala amandla asetshenziswa yi-LED.
Ingxoxo:
Ukuze sibale amandla amuncwa yi-LED, sisebenzisa i-equation eyisisekelo yamandla, okungukuthi:
\[ P = V \izikhathi I \]
Kuphi:
– \( P \) amandla ku-Watts (W),
– \( V \) yi-voltage kuma-Volts (V),
– \( I \) ingumthombo wamandla ku-Amperes (A).
Faka la manani ku-equation:
\[
P = 2 \umbhalo{V} \umbhalo{mA} izikhathi ezingu-20
= 2 \umbhalo{V} \izikhathi 0.02 \umbhalo{A}
= 0.04 \umbhalo{W}
\]
Ngakho-ke, amandla amuncwa yi-LED ebomvu esebenza nge-voltage yangaphambili engu-2V kanye nogesi ongu-20 mA angama-0.04 Watts.
Isibonelo 3: Ukusebenza Kahle Kwe-LED
Umbuzo: I-LED eluhlaza okwesibhakabhaka inamandla e-quantum angu-30%. Uma i-LED isebenzisa amandla kagesi angu-0.1 W, angakanani amandla akhishwa njengokukhanya?
Ingxoxo:
Ukusebenza kahle kwe-quantum (η) yisilinganiso samandla akhishwa ngesimo sokukhanya (P_luminous) kokufakwayo kwamandla kagesi (P_input):
\[ η = \frac{P_{\text{luminous}}}{P_{\text{input}}} \]
Ukuze uthole \( P_{\text{luminous}} \), sebenzisa i-equation elandelayo enenani lokusebenza kahle elingu-30% noma u-0.30:
\[
0.30 = \frac{P_{\text{luminous}}}{0.1 \text{W}}
\]
Ngakho-ke:
\[
P_{\text{luminous}} = 0.30 \times 0.1 \text{W}
= 0.03 \umbhalo{W}
\]
Ngakho-ke, amandla akhishwa yi-LED ngesimo sokukhanya angama-0.03 Watt.
Isibonelo Senkinga 4: Isekethe ye-LED yochungechunge
Umbuzo: Unama-LED amathathu ane-voltage yaphambili engu-2V ngalinye okudingeka axhunywe ngochungechunge. Uma umthombo we-voltage ungu-9V, nquma inani le-resistor elidingekayo ukuze unciphise ugesi ube ngu-20 mA.
Ingxoxo:
Uma ama-LED exhunywe ngokulandelana, i-voltage iyonke edingekayo iyisamba sama-voltage angaphambili e-LED ngayinye:
\[
V_{inani} = V_f1 + V_f2 + V_f3
= 2V + 2V + 2V
= 6v
\]
I-voltage esele okumele isuswe yi-resistor yile:
\[
V_{R} = V_{umthombo} – V_{isamba}
= 9V – 6V
= 3v
\]
Ngomthamo ofiselekayo ongu-20 mA, inani le-resistor libalwa kusetshenziswa uMthetho ka-Ohm:
\[
R = \frac{V_{R}}{I}
= \frac{3V}{20 \text{mA}}
= \frac{3V}{0.02A}
= 150 \Omega
\]
Ngakho-ke, inani elidingekayo lokumelana liyi-150 ohms.
Isibonelo 5: Isekethe Yomkhawulo Wamanje We-LED Ehambisanayo
Umbuzo: Kusekhethi, kunezibani ezimbili ze-LED ezihambisanayo, ngayinye ine-voltage yangaphambili engu-2V kanye ne-current engu-20 mA. Iyini i-current ephelele edingekayo emthonjeni we-voltage engu-5V uma kusetshenziswa i-resistor ye-LED ngayinye?
Ingxoxo:
Kuma-LED ahlelwe ngokulingana, i-LED ngayinye izothola i-voltage efanayo, okungukuthi i-voltage yayo yaphambili, futhi njengoba idinga i-20 mA yamanje nge-LED ngayinye, i-current iyonke yile:
\[
I_{isamba} = I_1 + I_2
= 20 \umbhalo{ mA} + 20 \umbhalo{ mA}
= 40 \umbhalo{mA}
\]
Ukuze ubale inani le-resistor le-LED ngayinye:
\[
V_{R} = V_{umthombo} – V_f
= 5V – 2V
= 3v
\]
Inani le-resistor lokunciphisa ugesi ku-LED ngayinye yileli:
\[
R = \frac{V_{R}}{I}
= \frac{3V}{20 \text{mA}}
= \frac{3V}{0.02A}
= 150 \Omega
\]
Ngakho-ke, i-LED ngayinye idinga i-resistor engu-150 ohm. Inani lamandla adingekayo kusuka kugesi ongu-5V lingu-40 mA.
Isiphetho
Ama-LED ayizingxenye ezibalulekile kuma-elekthronikhi anamuhla ngenxa yekhono lawo lokunikeza ukukhanya okusebenzayo kakhulu okuphila isikhathi eside. Ukuqonda imiqondo eyisisekelo efana ne-forward voltage, current, power, quantum efficiency, kanye nendlela yokuwahlela kwi-circuit kuzohlinzeka ngesisekelo esiqinile sezinhlelo zokusebenza ezisebenzayo. Izibonelo ezingenhla ziyisinyathelo sokuqala sokuqonda izinhlelo zokusebenza ezahlukahlukene kanye nokubala okuhlobene nokusetshenziswa kwama-LED kuma-electronic circuits.
