Assume I have battery with voltage \$V = V_1\$ and a PN2222 NPN transistor. I connect the C and B to the positive side of the battery and connect the E to the negative side of the battery. Assume that at the 3 wires, I have static resistances of \$R_1\$, \$R_2\$ and \$R_3\$ as in the diagram. How can I calculate the two currents through \$R_1\$ and \$R_2\$?

(The values like 1 V, 1 Ohm etc. and 2N3904 in the diagram are just example values; I don't know how to remove them.)


simulate this circuit – Schematic created using CircuitLab

I am sorry for this dumb question, I am an absolute beginner, although I do understand math and some physics. In the data sheets of the PN2222, I could not find the necessary parameters like beta. But even if I knew beta, I would also need to know to some kind of inner resistance of the transistor, wouldn't I?

  • 4
    \$\begingroup\$ There is an absolute requirement for answering such a question: Circuit diagram. \$\endgroup\$
    – LvW
    Jul 2, 2022 at 9:16
  • \$\begingroup\$ I described it. Or which tool can I use to paint one? \$\endgroup\$
    – Kolodez
    Jul 2, 2022 at 9:19
  • 2
    \$\begingroup\$ Beta ~= hfe. You can calculate the resistance from the slope of curves in the datasheet. \$\endgroup\$
    – winny
    Jul 2, 2022 at 9:19
  • \$\begingroup\$ Kolodez, your decription is not clear. What means "I connect C and B to the positive side of the battery"? How many volts? When both are connected to a common node - short between C and B? \$\endgroup\$
    – LvW
    Jul 2, 2022 at 11:14
  • 1
    \$\begingroup\$ Edit your question and add a schematic. There is a schematic editor built into the editor. Learning starts with giving us a schematic diagram with battery voltage and resistor values. It's not that we can't handle the words, but the schematic removes ambuiguity! \$\endgroup\$ Jul 2, 2022 at 22:06

1 Answer 1


"Beta" is a Greek word that is the hFE number in a datasheet. The hFE for a transistor is a range of numbers that change with current and with temperature. You have the base and collecter shorted together and connected to battery positive then the transistor is simply the base-emitter diode in series with the emitter resistor that you do not say where it is connected to. You also do not say where RB and RC are connected.

I simulated a simple circuit that has only a single base resistor and the transistor is saturated then it is not active. I also simulated it with a voltage divider pair of resistor at the base causing it to be active but the datasheet for an old 2N2222 does not give its range of beta. Then some of them with a high beta will not work unless the emitter resistor has an increased resistance. transistor

  • \$\begingroup\$ How does the simulator calculate everything if essential values are not even provided by the datasheet? \$\endgroup\$
    – Kolodez
    Jul 3, 2022 at 7:28
  • \$\begingroup\$ @Kolodez: The simulator assumes a particular value for Beta - it can't use a range, only a single value. You have to design your ciircuit to work properly despite variaations, though. A circuit that only works for a specific single value won't be very useful - variations in real parts would cause it to fail. \$\endgroup\$
    – JRE
    Jul 3, 2022 at 7:57
  • \$\begingroup\$ More than that - the beta value plays a role that is not as important as you might think. This is because the emitter resistor provides negative feedback which makes the whole circuit less sensitive against all active parameters (like beta and Vbe uncertainties. ) In this context, it is important to realize that the base voltage divider (with 2.4kohms to ground) acts as relatively "good" (stiff) bias voltage source. \$\endgroup\$
    – LvW
    Jul 3, 2022 at 8:06
  • \$\begingroup\$ The simulator uses only a "typical" beta number and "typical" base-emitter voltage. Transistors have a wide range of beta so some with low or high beta will not work in my simple circuit unless the emitter resistor is a higher value to reduce the gain and increase negative feedback.. Simulations also allow and do not warn about severe voltage, current and heating overloads. \$\endgroup\$
    – Audioguru
    Jul 4, 2022 at 16:30

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