I have not worked with transistor much, and therefore need some guidance in deriving the base resistor and pull-up resistor values.

I have the following circuit and I want to calculate base resistor R1, and the R2 resistor, which I am using to pull up the PNP base to 3.3V to prevent it from false turn ON when the GPIO pin is in the high impedance state.

As per the BC857C datasheet, the minimum current gain is 420, Vbe(sat) is -700mV, and Vce(sat-max) is -300mV.

With the following circuit, I intend to enable/disable the 1.5K resistor as pull-up to the load. So basically I want to use the transistor as a switch with 1 (3.3V) from GPIO putting it in the cut-off and 0 (0V) from GPIO putting it in saturation.

Could anyone please help me with the calculations?


simulate this circuit – Schematic created using CircuitLab


1 Answer 1


If your load current is 1mA (guesswork on my part) then this will produce 1.5 volts across R3. The voltage on the base to produce this must be about 2.2 volts below Vcc (1.1 volts above 0V when Vcc is 3.3 volts).

If you assume R2 is 10 kohms then R1 will have 1.1 volts across it and R2 will have 2.2 volts across it - this means a current of 220uA. This means R1 is 5 kohms.

  • \$\begingroup\$ Thanks for the answer, Andy. But I am not clear with how you decided that the base voltage should be 2.2V below Vcc. Could you please elaborate it with relevant calculation, if there is any? \$\endgroup\$
    – LoveEnigma
    Commented Jul 14, 2014 at 11:11
  • 1
    \$\begingroup\$ If your load is 1mA, then that current is also flowing through the emitter resistor (R3) - this drops the emitter down from 3.3 volts to 1.5 volts lower due to the emitter resistor. If the transistor is being activated then the base is about 0.7 volts lower than the emitter hence the base is about 2.2 volts away from 3.3 volts. \$\endgroup\$
    – Andy aka
    Commented Jul 14, 2014 at 11:40
  • \$\begingroup\$ I think you mean 1.8V instead of 1.5V, because 1mA will cause a drop of 1.5V across R3 thereby reducing emitter voltage to 3.3V-1.5V=1.8V. Now, since the emitter must be at least 0.7V higher than base, the base should be <=1.8V-0.7V, i.e. <=1.1V. Is my understanding correct? \$\endgroup\$
    – LoveEnigma
    Commented Jul 14, 2014 at 12:08
  • \$\begingroup\$ In absolute terms the emitter will be 1.8 volts above ground (0V) - I said "1.5 volts lower" but, you have got the idea now. \$\endgroup\$
    – Andy aka
    Commented Jul 14, 2014 at 12:11
  • \$\begingroup\$ What does R2 being 10k have to do with the voltage across R1? If the base if 1.1V then wouldn't both resistors have 2.2V across them? EDIT: Nevermind, PNP transitors have base current leaving the terminal. Therefore 0V (when transistor is on) means 1.1V across R1 and still 2.2V across R2. Both have 1.1V across them when the transistor is off, right? But if the transistor is off how can we assume the base is still at 1.1V? \$\endgroup\$
    – sherrellbc
    Commented Jul 14, 2014 at 17:44

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.