I created the following circuit using a npn transistor.


simulate this circuit – Schematic created using CircuitLab

After measuring all the voltages, I got: \$ V_{BE}\approx 0.6V, V_{CB}>0 \$. Are those measurements sufficient to say that the transistor is operating in the active region? How can I prove it theoretically?

Also, I'm looking for books/websites/documentation containing different transistor circuits and the explanation of their function. If someone could help me, I'd gladly appreciate it.


That's right. Vbe must be forward biased as well. The chart shows clear definition.

enter image description here

More reading from Berkeley U.

Here is a simulation of Vce=0 forced by shorted Vce with only 0 to 200mV Vbe. Note how collector current is much less than base current. This demonstrates the non-linearity of saturation but is not a practical circuit.

enter image description here

  • \$\begingroup\$ yes but Vbe could be any value forward biased, so not explicit \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 5 '17 at 17:41
  • \$\begingroup\$ +1 for nice graph showing the whole universe. I'd add to OP's list of requirements that some collector current should flow in active region. \$\endgroup\$ – glen_geek Mar 5 '17 at 17:43
  • \$\begingroup\$ In the real world, the presented circuit will never drive \$V_{CB}\$ to 0, and so it would never enter saturation according to this definition. \$\endgroup\$ – The Photon Mar 5 '17 at 17:48
  • \$\begingroup\$ For hand calculation we might use this model, but OP asked about the case where he measured the voltages on an actual circuit. \$\endgroup\$ – The Photon Mar 5 '17 at 18:04
  • \$\begingroup\$ For the circuit with Vcc = 0, how do propose to get the BJT into forward active mode to make the circuit useful as an amplifier? \$\endgroup\$ – The Photon Mar 5 '17 at 18:05

After measuring all the voltages, I got: \$ V_{BE}\approx 0.6V, V_{CB}>0 \$. Are those measurements sufficient to say that the transistor is operating in the active region?

No, it's not sufficient. You could measure the same two conditions if the BJT is in saturation.

If \$V_{CE} > ~0.2\ {\rm V}\$, then it's very likely the part is in forward active mode. If \$V_{CB} > 0\$, as you said in comments, then it's safe to say the device is in active mode.

But the definition of saturation is somewhat fuzzy, and might be defined differently by different manufacturers or for BJTs designed for different applications. Also, it's usually defined in terms of current, not voltage. A typical definition of saturation is when \$\beta < 10\$ (or 20, or some other value). So to prove the BJT is in forward-active, you'd need to work out the base and collector currents, and show their ratio is above the threshold you've chosen to define saturation.

Theoretically, there's nothing really to prove. You just have to make a clear definition of the boundary between active and saturated, and then see whether your operating point matches that definition or not.

Note: I made a mistake and gave the condition \$V_{CB} > 0.2\$ instead of \$V_{CE}>0.2\$ when I first wrote this, which might have led to much of the discussion in comments. Now corrected.

  • \$\begingroup\$ More Likely is not a clear definition \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 5 '17 at 17:42
  • \$\begingroup\$ @Tony, which is why I went on to explain "the definition of saturation is somewhat fuzzy", and the implications of that. \$\endgroup\$ – The Photon Mar 5 '17 at 17:44
  • \$\begingroup\$ @The_Photon. My definition is clear. Degree of saturation is measured many ways including a log scale of asymmetric induced sine harmonic distortion due to parabolic curve, but usually Vce(sat) @ some I. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 5 '17 at 17:50
  • \$\begingroup\$ @Tony, Clear but useless, since Vce never goes to 0 in this circuit in the real world. \$\endgroup\$ – The Photon Mar 5 '17 at 17:51
  • \$\begingroup\$ @ThePhoton .... r u sure about that.... what happens with Vcc = 0V \$\endgroup\$ – Trevor_G Mar 5 '17 at 17:54

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