PNP Transistor as a Current Source

  • We wanted to use a transistor as a current source.
  • Two resistors, one at the emiiter(R1) and the other at the collector(R2) were connected.
  • A voltage divider circuit with a supply of 9v was connected at the base that could supply certain voltage.

Constraints :

  • We wanted 4.5mA of current to pass through.
  • 9v supply was given to the resistor at the emitter.

Design :

  • The resistor at the collector(R2) was chosen as 1k based upon the resistance of the mic.
  • We knew that the diode would turn on at around 0.7v.
  • The voltage drop across R1 should be some resistance that gives 4.5mA of current.
  • Depending upon the availability of the resistors, 330 ohms was chosen for R1, keeping the 9v supply in mind and voltage at emiiter, it was calculated that 1.5v drop would give us 4.5mA of current.
  • Now from the voltage divider, a certain amount of voltage needed to be supplied so that the drop across R1 is 1.5v (ie, 9v – 7.5v)
  • So we wanted 6.8v from the voltage divider (ie, 6.8v + 0.7v = 7.5v)
  • Hence, the voltage divider circuit was designed with the following formula.


  • The ratio of the scaling factor was approximately 3 and depending upon the availability of the resistors, R3 was taken as 4.7k and R4 was taken as 14.5k.