In below circuit I sweep Vb from zero to 5V Vcc. And I obtain the following plots for the currents:

enter image description here

I have a bit problem with understanding two things in saturation region:

1-) Why/how does the Ic decreases with Vb after the saturation point?

2-) How can I quantify/formulate Ib at saturation region?

  • \$\begingroup\$ Once Vce aproaches 0V Vc rises with Ve and Vb thus V(Rc) drops with rising Vb and Ic=V(Rc)/Rc so you expect Ic to drop with such a large Re. It is neither a switch or a linear amp during this entire sweep, but both. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 24 '17 at 17:09

The collector current in your circuit decreases because, when the BJT is in saturation, by definition, the junction B-C is forward biased.

Therefore you'll have that the collector voltage will be just \$V_C=V_B-V_{BC}\$.

As a result, the current in the resistor Rc will be:

$$I_C=\frac {V_{CC}-V_C}{R_C} = \frac{V_{CC}-(V_B-V_{BC})}{R_C}$$

Therefore the higher \$V_B\$, the smaller \$I_C\$.

To answer the second question, simply use KCL at the BJT. You know that \$I_C+I_B-I_E=0\$ (assuming base and collector currents entering the BJT. Emitter current exits the BJT). Hence \$I_B=I_E-I_C\$. In fact in your plot, \$I_B\$ is the difference of the other two currents.

As a first order approximation, you can assume constant \$V_{BC}\$ and \$V_{BE}\$. This allows you to calculate \$I_C\$ and \$I_E\$ easily.


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