I understand norton equivalent and I understand thevenin equivalent and how it can be used to simplify CIRCUIT A into CIRCUIT B. What I don't understand is my books explanation of simplifying CIRCUIT C into CIRCUIT D. They basically claim that Circuit B Vth is just the voltage division so that Vth=R2/(R1+R2)*(V+). I get that, and it works if you also convert (V+) and R1 into Norton equivalent, see that R1 is now parallel with R2 and combine then, and then convert the current source and Rth into Thevenin equivalent.

What I don't get is how they got Vth for Circuit D.

My Method:

  1. Norton Equivalent of V+:R1 and V-:R2

  2. Parallel simplification of Inorton(1) and Inorton(2)

  3. Parallel simplification of R1 and R2

  4. Thevenin equivalent of Inorton and Rth to get Vth and Rth

Where am I going wrong. My solution is in the picture below for Circuit C.

enter image description here

  • \$\begingroup\$ The two solutions are the same. \$\endgroup\$
    – Chu
    Commented Jul 23, 2015 at 22:11
  • \$\begingroup\$ Thank you so much, I must have entered wrong numbers when I was comparing the two solutions. But it works out when you multiply the V- by (R1+R2)/(R1+R2) in the book solution for Vth \$\endgroup\$
    – TrapLevel
    Commented Jul 24, 2015 at 0:35

1 Answer 1


Write the expression for the current through R1 + R2: (V+ - V-)/(R1+R2)

Now find the voltage drop across R2: R2*(V+ - V-)/(R1+R2)

Now find the open circuit Vth at the base connection point:

From ground: V- + (V+ - V-)*R2/(R1+R2)

(The drop across V- plus the drop across R2 is the voltage at the base connection point with respect to ground.)

  • \$\begingroup\$ What I still can't figure out is why you can make the assumption that the current through R1 is equal to the current through R2. This would mean that no current is supplied to the base. Its like your solution and the book solution assumes the two resistors are in series but they can't be since you have the base stealing some current. Any help would be VERY appreciated. \$\endgroup\$
    – TrapLevel
    Commented Jul 24, 2015 at 0:48
  • \$\begingroup\$ So we're looking for the Thevenin equivalent looking out from the base of the BJT. That means that we assume the BJT is disconnected (the "open circuit voltage") for purposes of finding Vth. \$\endgroup\$
    – John D
    Commented Jul 24, 2015 at 2:19
  • \$\begingroup\$ I get it now, that equation is basically finding the maximum current, maximum voltage and equivalent resistance to find the Thevenin equivalent circuit. Thank you so much! \$\endgroup\$
    – TrapLevel
    Commented Jul 24, 2015 at 8:48

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