I'm already having trouble understanding Q-point of a transistor. I have been assigned to draw Q point as follows:

Calculate and draw a graph of Q point , VCE(cutoff) , Ic(sat) if IB = 100uA, 200uA, 300uA, 400uA, 500uA, 600uA and β = 150

Note: The values in the diagram will be replaced if mentioned in the question.

Here is the circuit

The problem is that when base current is 400(micro Amp) it gives -3.2 V VCE and negative values for other proceeding base current values. How should I plot it in the graph? Before plotting the graph I want to know if it is even possible to have negative VCE. If yes then how will current flow towards ground as it's at high potential now?

Please let me know solution asap.

Thank You.

  • 1
    \$\begingroup\$ Look at the curves of hFE versus Vce in the transistor's datasheet ... they will answer your question. Or, read about "saturation" in a transistor when Vce is close to 0. \$\endgroup\$
    – user16324
    Dec 25, 2016 at 17:46
  • 1
    \$\begingroup\$ No, in your circuit negative Vce voltage it is not possible. Try read this electronics.stackexchange.com/questions/276146/… \$\endgroup\$
    – G36
    Dec 25, 2016 at 17:49
  • \$\begingroup\$ Do you know what are the output characteristic curves of a BJT? That is the family of curves that describe the dependency of Ic (y-axis) from Vce (x-axis) with Ib as a parameter (each curve is labeled with a different value of Ib). \$\endgroup\$ Dec 25, 2016 at 22:04

1 Answer 1


As a rule of thumb, you should never have a negative \$V_{CE}\$ in your DC biasing. If you do, the emitter acts as a collector and the collector acts as an emitter. (They're not designed for that, so it won't work very well.)

In your circuit, there's no way to get a negative \$V_{CE}\$. The emitter is connected to the lowest voltage in the circuit (ground).

What actually happens when you increase \$I_B\$ is that the collector voltage drops until it reaches \$V_{CEsat}\$, which is about 0.2 volts. Then it doesn't drop anymore, regardless of how much current you pump into the base. In other words, the collector current is limited by the resistor. You have 10 volts and 220 ohms, so you'll never have a collector current higher than about 45 milliamps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.