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The Darlington

If the clock in the diagram above is a wrist watch, even a 1.09mA current may overload its alarm circuitry. In that case, you may use two transistors as shown in the diagram below. This is called a darlington.

A darlington can be considered as a single transistor with the following characteristics:

VBE,darlington=VBE1+VBE2

IB,darlington=IB1. IC1= hFE1∙IB1. IC2=hFE2∙IB2=hFE2∙IE1=hFE2∙(hFE1+1)∙IB1. IC,darlington=IC1+IC2=hFE1∙IB1+hFE2∙(hFE1+1)∙IB1=(hFE1+(hFE1+1)∙hFE2)∙IB1. hFE,darlington=IC,darlington/IB,darlington=hFE1+(hFE1+1)∙hFE2. hFE1>>1 => hFE,darlington=hFE1+hFE1∙hFE2. hFE1∙hFE2>>hFE1 => hFE,darlington=hFE1∙hFE2.

Let's now recalculate a proper value for RB. Assume T1's current gain is 300 and T2's current gain is 100.

IC,darlington = 9/100=90mA. hFE,darlington=300∙100=30000. VBE,darlington=1.2V.

The voltage across RB equals 3V-1.2V=1.8V => RB < 1.8V/3µA = 600k. 560k is a safe value. IB,darlington will be 1.8/560k = 3.21µA.

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