I am struggling with the currents and their directions.

Here you can see the simple circuit part.

Since Vth(EBJ)= 0.7, Vth(CBJ)= 0.7

I could calculate the VCE by V(CBJ)-V(EBJ)= Vc-Vb - (Ve-Vb) = Vc-Ve = V(CE)= 0.2.

The thing I cannot understand is emitter current value of Q1 (Ie1) and base current of Q2 (Ib2).

We know the +10Volts, V(CE)=0.2 but I cannot still find the currents properly (especially when the base current of Q2 and collector current of Q1 meet.) Can you please help me to understand the logic?


  • \$\begingroup\$ Why you did not replace the BJT's with voltage sources? As you was given to you Vbe and Vbc values? \$\endgroup\$ – G36 May 6 at 19:27

but I cannot still find the currents properly

All currents can be found by solving a system of 4 equations.

We already know that \$\mathrm{|V_{CE}|=0.2V}\$ and \$I_E=I_B+I_C\$ for both NPN and PNP transistors.

Here are all the equations:

Equation-1 (from Vcc to PNP's emitter to base to ground):

$$ \mathrm{ 10V=10k \ I_{E1}+0.7V+10k \ I_{B1} \\ \therefore 9.3V=10k\ (2\ I_{B1}+I_{C1}) \ \ \ \ ... (1) } $$

Equation-2 (From Vcc to PNP's emitter to collector to ground):

$$ \mathrm{ 10V=10k\ I_{E1}+V_{EC-pnp}+10k\ (I_{C1}-I_{B2}) \\ \therefore 9.8V=10k\ (2\ I_{C1}+I_{B1}-I_{B2}) } $$

Equation-3 (From PNP's collector to ground and from NPN's base to emitter to ground): $$ \mathrm{ 10k\ (I_{C1}-I_{B2}) = V_{BE-npn} + 10k\ I_{E2} \\ \therefore 10k\ (I_{C1}+I_{C2}-2I_{B2})=0.7V } $$

Equation-4 (From Vcc to NPN's collector to emitter to ground): $$ \mathrm{ 10V=30k\ I_{C2}+V_{CE-npn}+10k\ I_{E2} \\ \therefore 9.8V=10k\ (4\ I_{C2}+I_{B2}) } $$

There are 4 unknowns and 4 equations. If you solve this 4-eq system via matrices (preferably) or replacements (too difficult), you'll find;





Don't forget to crosscheck.

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  • \$\begingroup\$ Thank you so much. I tried to do same thing but something went wrong and Ib2 has the negative value which is impossible if we draw currents correctly. This is clear. Thank you ! \$\endgroup\$ – Rigel May 6 at 19:45
  • \$\begingroup\$ @Rigel it's funny, at first I did get -0.026ma for Ib2. I found that I did not subtract the last two equations correctly (it is actually manageable by subtraction). I, too wrote four equations, but in Ib1, Ib2, betaF1 and beta F2 and I got the same result as Rohat. \$\endgroup\$ – Sredni Vashtar May 7 at 1:47

I cannot figure out which relationship(s) in particular you are having trouble with, so I have written fourteen unique equations to address fourteen unknowns in the circuit. You have mentioned a critical piece of info for solving this circuit: both transistors being in saturation mode allows us to know the voltage across each junction of each transistor. My equations assume the forward voltages of the EBJ's and CBJ's are 0.7v and 0.5v, respectively (as I believe you mention in the provided image). It is important to note, for the PNP transistor, VEC = 0.2v, whereas VCE = 0.2v for the NPN. I hope one of these equations are one of your missing pieces. Best of luck! enter image description here

Edit: To prove these 14 equations is sufficient to solve the entire circuit, I have written a Matlab script to solve this set of equations. There are two solutions, one of which you will see is obviously incorrect (claims Ve1 = 10v and Ie1 = 0). With this, you will see Ve1 = 3.22v. When you have the same number of (unique) equations as unknowns, you can solve the system. In this case, solving is incredibly tedious, which is what I use Matlab for :-)

 syms Ie1 Ie2 Ib1 Ib2 Ic1 Ic2 Vb1 Vb2 Vc1 Vc2 Ve1 Ve2 Bforced1 Bforced2;
 eq1 =  Ie1 - (10-Ve1)/10000  == 0;
 eq2 =  Ie1 - Ib1*(1+Bforced1)  == 0;
 eq3 = Ie1-Ic1*(1+Bforced1)/(Bforced1) == 0;
 eq4 = Vb1 - Ve1 - 0.7 ==0;
 eq5 = Ic1 - Vc1/10000 - Ib2 == 0;
 eq6 = Vc1 - Ve1 + 0.2 == 0;
 eq7 = Ic2 - (10-Vc2)/30000 == 0;
 eq8 = Ie2 - Ib2*(1+Bforced2) == 0;
 eq9 = Ic2 - Ie2*(Bforced2)/(1+Bforced2) == 0;
 eq10 = Vc1-Vb2 == 0;
 eq11 = Vb2 - Ve2 -0.7 == 0;
 eq12 = Vc2 - Ve2 -0.2 == 0;
 eq13 = Ve2 - Ie2*(10000) == 0;
 eq14 = Vb1 - Ib1*(10000) == 0;
 sol = solve(eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14);

Matlab's solutions for Ve1:

ans =



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  • \$\begingroup\$ Thank you for your time. I wrote the voltage changes on the juntions from givings. The problem is i cannot find Ve1 now. If i can find it. Every thing will be clear. \$\endgroup\$ – Rigel May 6 at 17:38
  • \$\begingroup\$ See edited answer, which should demonstrate the fourteen unique equations can be used to solve the system of fourteen unknowns. I can only imagine how bad the algebra is! \$\endgroup\$ – Mitchell Easley May 6 at 18:26
  • \$\begingroup\$ Unfortunetly, it is not kinda useful for me just now. I am a student of Electrical and Electronics Engineering. I just wonder how to analyze this things more than solve it like using KVLs or converting transistors to diode etc. I am sure, you understand me. But again thank you so much for your time and effort. \$\endgroup\$ – Rigel May 6 at 18:47
  • \$\begingroup\$ I do not understand, but I hope your issue gets resolved. \$\endgroup\$ – Mitchell Easley May 6 at 19:00

it is a trick question in a weird circuit.

2 transistors

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  • \$\begingroup\$ yeah, that is the same challenge i struggling with,if i can found the Ve1 i can find the all others. I wrote the equations from KVL. If i can finish it i will post it. Thank you ! \$\endgroup\$ – Rigel May 6 at 17:36

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