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I'm modeling a Darlington pair in QUCS to understand better how to use it in practice. The model is below:Darlington pair common emitter

The DC analysis predicts a current in the base of T1 of about 1uA and a collector current on T2 of 10.7 mA which is about what I was going for with the biasing.

AC analysis

Now when I look at the AC analysis I find an amplification of only 100 (10 mV pp in, about 800 mV pp out) instead of the 10.000 I would expect from multiplying the betas of the two transistors. Am I thinking this wrong or there is something wrong in the model?

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Voltage gain will be transconductance (GM) of Q2, times the collector resistor.

At 10mA, GM is 0.5 approximately. GM * 450 = 225.

Does the collector of Q1 load the collector of Q2?

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The current gain is about 10,000.

But the voltage gain also depends on the input and output AC equivalent resistances.

The input resistance is roughly \$r_{\pi}\$ of T2, multiplied by \$\beta\$ of T1. That's roughly

$$R_{in}=\frac{\beta^2 0.025\ {\rm V}}{10 {\rm mA}}$$

or about 25 \$\rm k\Omega\$.

The output resistance is roughly the 450 \$\Omega\$ of R5.

This calculation gives a gain of 180, which is close enough to the simulation result for a back-of-the-envelope calculation. The remaining discrepancy may be due to capacitive voltage divider effects between C2 and the input. Try tying T1's collector directly to the 5 V supply node instead of to the T2 collector and see if the gain goes up a bit.

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  • \$\begingroup\$ So that means that to increase gain I should decrease R5, but that changes the operating point of the transistor, giving other problems. How do I increase gain then? \$\endgroup\$ – Luca Jul 12 '17 at 8:54
  • \$\begingroup\$ You should increase R5 to increase the voltage gain. I'm not sure what problem you mean about changing the bias point, so I'm not sure how to address it. If you mean T2 collector voltage becomes too low, you could reduce T2's base current --- either increase R2 or add a resistor between T1's base and ground. Or AC-couple between T1 emitter and T2 base so that T2 can be biased separately from T1. Or remove T1 entirely if you don't need so much gain (adding an emitter-degeneration resistor to T2 if you need higher Rin). Or ... \$\endgroup\$ – The Photon Jul 12 '17 at 14:47
  • \$\begingroup\$ Yes, sorry, increase. I meant that if 10 mA flow through R5 than R5*10 mA will drop across the resistor, setting the collector voltage for T2. If it's too low or too high I'm out of the optimal amplification range. Oook, i get it :). Lots of options. Sorry for the seemingly idiotic questions but I can't still wrap my head around transistors... Especially because I'm a biologist by trade... \$\endgroup\$ – Luca Jul 12 '17 at 15:29
  • \$\begingroup\$ @Luca, try this \$\endgroup\$ – The Photon Jul 12 '17 at 16:04

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