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I need to find the operating point of this circuit (Ic and Vce). There is an NPN transistor and Ec=15V, Vbe=0.65V, Rb1=120Ω, Rb2=60Ω, Rc=6Ω, Re=6Ω an β=150. As far as I know, Ic=β*Ib and Vce should be Ec-Ic*Rc-Ie*Re, in this case. My problem is that I don't know how to determine Ib.

What are the equations for determining Ib and is
Vce = Ec - Ic * Rc - Ie * Re?

  • \$\begingroup\$ To solve this in closed form you will have two simultaneous equations to determine Ib. You may find the Thevenin equivalent of the base divider useful. \$\endgroup\$ Jan 4 at 9:09

2 Answers 2



  1. Convert the circuit at the base of T1 (Ec, Rb1 & Rb2) to its Thevinan equivalent (voltage with series resistance). This now sets the voltage at the T1 base for Ib = 0.
  2. Calculate V at T1 emitter (Ve) for Ib = 0.
  3. Calculate Ie: Ie = Ve / Re.
  4. Calculate Ib: Ib = Ie / (β + 1).
  5. Re-calculate the new value for V at T1 emitter using the value of Ib obtained at step 4. This will change due to the new value of base current causing a voltage drop in the series resistance of the Thevinan equivalent of step 1.
  6. Re-calculate new value for Ie (same formula as step 3).
  7. Re-calculate new value for Ib (same formula as step 4).
  8. Repeat steps 5, 6, 7 for as many times as required for the accuracy needed. For your example here I suggest doing this 2 or 3 times at most.
  9. Once you are happy with accuracy, calculate Ic: Ic = Ie - Ib.
  10. Calculate V at T1 collector: Vc = Ec - (Ic * Rc).

I propose to use the superposition principle for finding - as the first step - the base potential Vb=Vb1+Vb2. The two voltage sources contributing to Vb are the supply voltage (Vb1) and the base-emitter voltage (Vb2), which will be treated as a constant DC voltage source of Vbe=0.65...0.7 volts.

  • This method requires nothing else than to use twice the voltage divider principle.

1.) For Vb1 (simple voltage divider) we need the parallel connection of Rb2 and the reflected emitter resistor Re" (which must be transferred to the base side: Re"=Re(1+βRe).

2.) For Vb2 we set the supply voltage to zero and calculate the voltage at the base node - driven by Vbe (again, a simple voltage divider).

3.) Now we have Vb=Vb1+Vb2 and Ve=IeRe because of Ve=Vb-Vbe.

4.) As the last step we can find Ie (via Ve) and Ic.

5.) Comment: As you can see, it was NOT necessary to find the base current Ib. This is not surprising because the current Ie is controlled by the voltage Vbe - and Ib is nothing else than an (unwanted) by-product. Of course, if you wish you can calculate Ib as the last step: Ib=Ic/β.


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