Darlington pair circuit calculation

The formula Vout = Vin - 1.4 V assumes ideal transistors.

The calculation that was done below uses a much better model of the diode. If you look at a real diode characteristic curve you will see it does not have a absolute on voltage of 0.7 V but the is a exponential curve that governs the Vbe and Ie relationship. Larger Vbe gets larger Ie. The current through the emitters of both transistors are not the same (consider beta) and therefore Vbe is not the same for the two transistors.



Joined Aug 24, 2008
Is this a spice simulation? If you actually hooked the circuit up and tested it, did you use discrete transistors or a device such a TIP120 which has two well-matched xistors within a single case?


Joined Jul 3, 2008
Why didn't I get the supposed answer 10-1,4=8,6 V? Is this formula wrong? Why doesn't this formula work?
Your circuit simulation tool is using a transistor model that is not consistent with a typical silicon bipolar transistor. It's not even close and behaves more like a germanium transistor. Probably the saturation current is wrong in the model. Also, your assumption of Vbe equal to 0.7 is just an approximation.

With the darlinton pair arrangement the first Vbe will be about 0.2 V less than the second due to the lower current. It would be better to assume 1.1 V or 1.2 V drop across the two base-emitter junctions, but as mentioned by StayatHomeElectronics, the actual value will depend on currents.

Your basic approach to estimate the answer is good and is acceptable because the 10 V supply is much greater than the error in your assumption of Vbe. Just reduce the assumed voltage when using the Darlington, that's all.