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Above circuit is taken from a book. It uses an NPN and a PNP transistor combination. In saturation, the author's answer to the Q2's collector current is 4.4mA and claims only 0.6mA of it comes from R3.

When I try to calculate this I get different answer. Apparently something wrong with my way here:

When Q2 is saturated I follow the following logic:

1-)The current comes down to Q2's qround through R3 and R2 resistors is: 15V/4.3k

2-)The current comes through emitter-base of Q3 is: (15V-0.6V)/3.3k

note: here 0.6 is the voltage in emitter-base junction in saturation

3-)The total collector current is the some of above two currents which is way above the author's answer

How can I analyse this circuit to find Ic of Q2? Whats wrong with my assumption?


1 Answer 1


When Q3 is activated you can assume that the base emitter voltage is 0.6 volts. This directly means there is 0.6 volts across R3 and this means the current through R3 is 0.6 mA.

For R2, the voltage across it is 15V - (0.6V + Vsat_of_Q2). I estimate that to be about 14.3 volts hence, the current into R2 is 14.3/3300 = 4.33 mA of which R3 supplies 0.6mA.

  • \$\begingroup\$ but why dont you add the current from R3 and the current from Q3 emitter? wont that be 4.33+0.6 = 4.93mA? \$\endgroup\$
    – user16307
    Dec 18, 2015 at 10:33
  • \$\begingroup\$ No, because there is only a single voltage at the base of Q3 - that is 0.6V below 15V - so that is the voltage across R2. The current through R2 must be the sum of the currents coming from the base of Q3 and through R3 \$\endgroup\$
    – Icy
    Dec 18, 2015 at 10:35
  • \$\begingroup\$ i dont understand it. How can R2 current include R3 current? \$\endgroup\$
    – user16307
    Dec 18, 2015 at 10:38
  • \$\begingroup\$ R2 current is 4.33 mA of which 0.6 mA is supplied via R3 leaving the remainder (3.73 mA) to come from the base of Q3. \$\endgroup\$
    – Andy aka
    Dec 18, 2015 at 10:38
  • \$\begingroup\$ can you explain me how can we write an expression to only find 3.73 mA? without using 0.6mA \$\endgroup\$
    – user16307
    Dec 18, 2015 at 11:03

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