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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:

Sziklai, Darlington, NPN transistors 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?

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  • \$\begingroup\$ Every BE voltage is approx constant at 0.6V (-0.6V for the PNP in the Sziklai). I'm using a simulator to do the "calculations", so maybe it's the simulator's fault? \$\endgroup\$
    – Samleo
    Commented Feb 19, 2019 at 13:59
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    \$\begingroup\$ You talk about common collector yet I only see common emmitter circuits! \$\endgroup\$
    – Bimpelrekkie
    Commented Feb 19, 2019 at 14:15
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    \$\begingroup\$ so maybe it's the simulator's fault? The simulator is nearly always correct, that does not mean it never lies (it does). But when you're a beginner, you're most likely misinterpreting what is happening. You want to see \$\beta\$ which is the current gain in active mode. Your circuits force these transistors into saturation mode so then you will not see that active mode's \$\beta\$. \$\endgroup\$
    – Bimpelrekkie
    Commented Feb 19, 2019 at 14:19
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    \$\begingroup\$ I fix the simulation for you tinyurl.com/y2xoa24n \$\endgroup\$
    – G36
    Commented Feb 19, 2019 at 14:24
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    \$\begingroup\$ In active mode the transistor sets \$I_c\$ where \$I_c = \beta * I_b\$. In saturation mode that is no longer valid. In sat. mode we make \$I_b\$ large, so large that \$I_c\$ becomes the maximum current which the load allows to flow. It is no longer the transistor which sets \$I_c\$ but the load! Then we can still say \$\beta = I_c / I_b\$ but that beta will be much smaller and it will not be a property of the transistor but a parameter we determine outside the transistor because we define both \$I_b\$ and \$I_c\$ externally. \$\endgroup\$
    – Bimpelrekkie
    Commented Feb 20, 2019 at 12:38

1 Answer 1

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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.

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