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4. Use R1 = 975 k12 and R2 = 25 k22, determine R3 so that ILED = 20 mA. LED1 + 1 YE R2 Figure 4: Constant current source usin

\beta =100

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Solution Page 1 LEDI IC= 20 MA = I LED 0,3 12V 0:31 IE - Griven:- Ri - 975 kr R2= 25kn I LED: 20MA B = 100 Determine voltagepage 2) da De = 0,99 . IE = loa = 20m - 0.0202 A 0,99 0,99 Now . R3 = 0.3 2 0.0202 - 14.85 12 R3 = 14.852

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