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The R3 is connected to the output of the PIC, I could understand the NPN transistor but PNP was a bit confusing. When the transistor starts to conduct, some current will flow towards the base. My question:

  1. What is the use of R3 in PNP transistor?
  2. There is current coming from the PIC as well as base, now the two currents will be in opposite direction. So what happens?
  • \$\begingroup\$ R7 is a pull-up, R3 is a current limiting resistor. It protects the chip output from over-current. \$\endgroup\$
    – DerStrom8
    May 16, 2016 at 14:18
  • \$\begingroup\$ @derstrom8 ok, if possible could you also look at question 2? That one is a bit confusing to me. will the current limiting resistor R3 limit the current going in to the B-E junction? \$\endgroup\$ May 16, 2016 at 14:20

1 Answer 1


1) R3 limits the current flowing through the base of the transistor and helps protect the microcontroller pin. You can find more information on base resistors here:


2) When the PIC is sourcing current, the PNP transistor will be switched off, and no current (in theory) will be flowing through it (or through the base). This is based on the PNP theory of operation. The PNP will be switched off when the PIC pulls its base HIGH, and switched on when the PIC sinks the current from the base.

  • \$\begingroup\$ Ah ok, for my clearance "R3 limits the current flowing through the base of the transistor (towards the pin of the PIC)" right? \$\endgroup\$ May 16, 2016 at 14:27
  • \$\begingroup\$ Well, technically the current flows from ground through the load to the positive side of the supply, but I believe you understand the concept. \$\endgroup\$
    – DerStrom8
    May 16, 2016 at 14:28
  • \$\begingroup\$ If this answers your question sufficiently, consider marking it as "accepted". Thanks! \$\endgroup\$
    – DerStrom8
    May 16, 2016 at 15:02
  • 2
    \$\begingroup\$ This answer is only applicable if the emitter and the top or R7 are connected to the same voltage as the PIC supply. \$\endgroup\$ May 16, 2016 at 15:04
  • \$\begingroup\$ Excellent point, thanks Olin! I assumed both were connected to the same supply but I suppose I should not have made that assumption. \$\endgroup\$
    – DerStrom8
    May 16, 2016 at 15:09

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