I'm a noob to electronics and have been reading up on transistors.
I tried to simulate Darlington (NPN) and Sziklai transistor and compare them against a single common emitter NPN configuration, shown below:
All the transistors shown have the exact same gain (ß=100) and other stats, resistors etc are all the same.
However, the collector current in the single NPN transistor is 49.3mA, which is higher than those of the Darlington and Sziklai configurations (42.88, 43.02mA resp.).
I would have expected a ß2+2ß gain for the Darlington and Sziklai pairs, so why is their collector current lower?
For the Darlington, I do understand that there is a 0.6*2 VBE voltage drop, and so the base current, Ib, is lower. (Indeed it is – – according to the simulation). Hence, the final IC might be slightly lower.
However, this is NOT the case for the Sziklai pair. I checked the base current for both NPN and Sziklai, and they are the same at 4.3mA.
So, why is the IC for the NPN transistor configuration higher? Is the simulation faulty or am I missing something?
Best Answer
Because all of those output transistors are saturated. In saturation you don’t have a linear dependency between base current and collector current. Your current is determined by the resistor and voltage drop across the saturated transistor.
With the single transistor this drop is ~0.2V. With any of your darlington configurations this drop is at least 0.7V.