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Darlingtons

Neem aan dat hFE3 = hFE4 = 100. De totale hFE van elke darlington is 2000. IRL,max = 1.4A => IE1,max = 1.4A. IB3,max = IE1,max/(hFE+1) = 1.4A/2001 = 0.7mA. We verwaarlozen weer de stroom door D1 en D3, dus IR1 = 0.7mA. UB3,max = 11.3 + 2∙0.7 = 12.7V => UR1 = V1 - UB3 = 15 - 12.7 = 2.3V. R1 = UR1/IR1 = 2.3V/0.7mA = 3.3kΩ.

Laten we nu de maximale vermogensdissipatie van T1 en T2 berekenen. PT1 = UCE1∙IE1 = (V1 - URL)∙(URL/RL) = (V1∙URL - U2RL)/RL. PT1 bereikt z'n maximum als dPT1/dURL = V1 - 2∙URL = 0 => URL = V1/2 = 7.5V. PT1,max = (15∙7.5 - 7.52)/8 = 7W. Aangezien T1 alleen gedurende de positieve helft van het ingangssignaal werkt, is de maximumdissipatie van T1 (en T2) 3.5W.

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