I have this exercise of a Thevenin's equivalent that really got me confused with it's usage of a transistor and was wondering if I could get some help with understanding how to solve this particular circuit.

Circuit schematic

R1= 47 Ohms

R2= 47 Ohms

RC= 47 Ohms

RE= 47 Ohms

V1= 15 V

V1= 15 V

The exercise: Calculate the current th IB through the point B in the transistor, if the current IE through the resistor RE is 2.7 mA, and the voltage between point B and E on the transistor is 0.7 V.

Also you don't need to know about the transistor as a component to solve the exercise. It's good to know that the current IB is normally way smaller than the currents IC and IE in the transistor.

  • 2
    \$\begingroup\$ (1) Put the image in the question and not as a link on a separate page. (2) Since this is a homework question you need to show some effort at a solution and explain where you're stuck. Otherwise your question will be closed. \$\endgroup\$
    – Transistor
    Aug 13, 2016 at 22:06
  • \$\begingroup\$ What do you mean by "the current between point B and E on the transistor is 0,7 V." This is probably a typo, but is the 0,7 V the drop across the Base and Emitter? \$\endgroup\$
    – Sock314
    Aug 13, 2016 at 23:26
  • \$\begingroup\$ @Transistor I'm sorry to say but after several attempts ending in frustration, I don't seems to be getting anywhere. \$\endgroup\$
    – Codexing
    Aug 14, 2016 at 16:27
  • \$\begingroup\$ @user2125673: Work backwards. You have IE and RE so you can work out VE. From that you can work out VB. Work these details out and post the calculations in your original post. Then you're nearly there. \$\endgroup\$
    – Transistor
    Aug 14, 2016 at 16:34

1 Answer 1


The Thevenin equivalent of the circuit between common and B.

  1. Open voltage 7.5 V

  2. Equivalent resistor 23,5 ohm

This however does not solve the question. For that you need to calculate the Voltage between B and common being 2,7 mA x 47 ohm +0,7 BE = 9,8269 V. This brings the base current to 283,9 mA.

Since the emiter current is only 2.7 mA the above is not possible. Therefore the given values are wrong or incomplete


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