I know that for an NPN transistor to be in the active region, the base emitter should be forward biased and the base collector should be reverse biased while for the NPN transistor to be in the saturation region, the base emitter and the base collector should both be forward biased. This appears clearly in the view of the NPN transistor as 2 diodes put back to back as in this sketch:

enter image description here

I can't understand the case where it is in a circuit as in this sketch: (had to remove the link due to lack of reputation but it's just the basic BJT circuit)

Assuming there is some voltage on the base. Why do we have to assume it's active and then see if$$V_{be} > 0.7$$ and $$ I_{b} , I_{e} > 0 $$ and check if: $$V_{ce} > 0.2$$ If not then we assume it's saturated and check the same conditions except that: $$V_{ce} < 0.2$$ Shouldn't the condition for saturation or activism(if that's the correct word to use) be the biasing of the base collector and base emitter. What made me think about this is this PDF here page 2, the circuit drawn:

enter image description here


They're using it as it will always be in saturation mode although I see the base emitter is forward biased while the base collector is reverse biased so that suggests an active mode not saturation one.

  • \$\begingroup\$ In your first sentence, did you mean to compare an NPN with a PNP instead of saying two contradictory things about NPN? \$\endgroup\$
    – The Photon
    Oct 26, 2013 at 16:50
  • \$\begingroup\$ OP is probably long gone, but future readers with a similar question might want to see this question also: electronics.stackexchange.com/questions/51405/… \$\endgroup\$
    – The Photon
    Oct 26, 2013 at 16:54

1 Answer 1


In general, it is the states of the PN junctions inside the transistor which will determine what operation region it is in. However, after gathering some experience, one can deduce the states of the above junctions by inspecting the circuit itself without actually measuring the voltages at the terminals.

An example:

Lets analyze the circuit you've referenced.

enter image description here

Once the switch is closed a current of approximately \$1mA\$ will flow into the base, which will cause:

$$V_{BE} \approx 2V$$

Since this is higher than the minimum of \$0.6V-0.7V\$ for being out of cut-off - the transistor is in one of its operational modes. In reality, the Base-to-Emitter voltage will not rise much beyond \$0.6V-0.7V\$ (due to presence of protection resistor R1), which means that the Base current will be a bit higher than \$1mA\$.

Knowing that the motor is \$12V, 100mA\$, and that the transistor is capable of handling \$100mA\$ Collector-to-Emitter current, we can deduce that:

$$I_C = I_{Motor} \approx 100mA$$

Given that we know (from motor's specs) that the motor will consume \$100 mA\$ at \$12V\$, the voltage on the motor:

$$V_{Motor} \approx 12V$$

Which leads to:

$$V_C \approx 0V$$

But this means that Collector-to-Base junction is forward biased which implies that the transistor in saturation.

The above analysis is quite general for this configuration (full voltage rated motor switched by matching BJT), therefore, in majority of circuits like this one, the transistor will be in saturation.

Experienced engineers perform the analysis above at a glance, knowing that the transistor in saturation a second after they see the schematics.

  • \$\begingroup\$ Will \$V_{BE}\$ really be about 2V? \$\endgroup\$
    – Andy aka
    Sep 26, 2013 at 8:51
  • \$\begingroup\$ @Andyaka, no way - the Base-Emitter junction will melt long before it is forward biased with 2 volts. I see what you mean - I should've explained this point in a bit more details. Thx \$\endgroup\$
    – Vasiliy
    Sep 26, 2013 at 9:20

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