Tell us about yourself!
Given that the voltage drop across the current sense resistor (R3) is 0.5v - then using Power = IV = VI = V x V/R = 0.5 x 0.5/R3 = 0.25/R3. I've built the circuit and the voltage drop across Vbe for Q1 = voltage drop for R3 - is indeed 0.5v. But how could that be predicted from the 2N5088 data sheet? What are the design steps? (it is a great tutorial BTW - just I'm frustrated by my lack of understanding - even if the light does come on!)