How does a darlington pair work?
I am curious about how the second transistor sustains the current Ie of the first transistor.
Why is the base not damaged despite this high current?
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3\$\begingroup\$ You're assuming that the first (small) transistor decides what \$I_E\$ it will deliver to the second (large) transistor and that the second transistor just has to cope with this large current going into its base. What if it is really the second transistor determining the current? If the second transistor is in active mode, then what will its base current be? Have you read: kitronik.co.uk/blog/how-a-darlington-pair-transistor-works ? \$\endgroup\$– BimpelrekkieCommented Nov 27, 2019 at 11:46
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1 Answer
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If the first transistor is over-saturated with high base current, then \$I_{b2} = (\beta_1+1) I_{b1}\$ doesn't hold anymore, it will be: \$I_{b2} > (\beta_1+1) I_{b1}\$.
The second transistor will decide its base current \$I_{b2}\$ in darlington. Say, if the max current passing through the emitter side of the second transistor is \$I_{c2max}\$, the max base current to the second transistor will be \$I_{b2max}=I_{c2max}/\beta_2\$, as it will be in active mode.
