Calculating base voltage of NPN transistor
I'm having trouble using the nodal equations to determine the base voltage Vb1 in the circuit below. I've attempted to write seven equations below, but not able to fit them in entirely.
(12Vx)/1.62k = (VxVy)/1.21k +(VxVb1)/1kequation 1
(VxVy)/1.21k = Vy/1.21k + (VyVb2)/27k equation 2
(VyVb2)/27k = Vb2/5.62k + Ib2 equation 3
(VxVb1)/1k = Vb1/1k + Ic2 + Ib1 equation 4
Ib2= (Vb20.7) /(310 * beta 2) equation 5
Ic2= (Vb20.7)/310  equation 6
Ib1=(Vb10.7)/(200* beta 1)  equation 7
Is it possible to apply the Thévenin's at nodes Vx and Vy one by one ?
While applying Thévenin's at Vx , How to deal with node Vy ?
Looking forward to suggestions / help on simplifying the circuit equations.
2 answers
For my opinion, it is not too complicated to reduce the various equations to a system of two equations with two unknowns (current Ib2 neglected), Let me try:
 Vb1Vx=I5*R5
 I5=IC4+Vb1/R3
 IC4=Ue4/R4
 Ue4=VB20.7
These 4 equations (simple insertion procedure) result in:

Vb1Vx=R5[(Vb20.7)/R4 + Vb1/R3]

Now we can Vb2 express as a function Vx (voltage division) and we have an equation with the two unknowns Vb1 and Vx.

A second equation for Vb1 and Vx can be derived from (V1Vx)/R10=I10=I5+(Vx/Rx) with I5=(Vb1Vx)/R5.

The resistor Rx is the total resistance between Vx and ground.

Last step: Solve the system with two equations for two unknowns.
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If this is a contrived homework problem where they are looking for an "exact" answer, then you have to solve a bunch of simultaneous equations. However, that would be neither useful nor anything you can rely on for a real world circuit.
Real circuits have to work with transistors with widely varying gain. The datasheet should give you a minimum you can expect in your circumstances, but the maximum gain can be an order of magnitude higher. It is usually best to assume transistor gain can be from the minimum to infinite and design the circuit to tolerate that.
The BC817 is a relatively high gain general purpose transistor. You should be able to count on a gain of 100 in this case. Let's do a firstpass analysis assuming infinite gain and a V_{BE} drop of 600 mV. That should provide some insight on its own. In addition it's a good starting point for iterative adjustment to a gain of 100.
The tricky part of this circuit is how the base voltage of Q2 depends on the current thru R5. You could write the simultaneous equations, but let's just wing it for now.
Start with no current going thru R5. Now it is easy to solve for the base voltage of Q2 since it's just a resistor network. It will be useful later to find what Vb2 is as a fraction of Vx, so let's figure that out. Vx, R9, and R8 form a Thevenin source of 1/2 Vx and 605 Ω. That is loaded with R6 + R7 = 32.62 kΩ. Due to the load, Vy is then 32,620 / 33,225 = 0.982 of Vy open circuit, or 0.491 of Vx. We apply that 0.491 Vx to the R6,R7 divider to get another attenuation by 0.172. Overall, Vb2 is therefore 0.0846 of Vx. We'll stash that away for future use.
Now let's solve for Vx with no current thru R5. That's a simple resistor network not worth going thru the details here, which comes out to Vx = 7.14 V. Using the attenuation calculated above, that means Vb2 is 604 mV. That's down in the knee area of Q2, so we can only say that it "barely" conducts. Our simplification of V_{BE} is 600 mV can't be applied here since we're right on the edge. However, even with infinite gain and 600 mV V_{BE}, there would only be 4 mV across R4, and 13 µA current thru Q2. We'll call that 0. Minor part and temperature variations make this value unpredictable, except that it will be "small" relative to other voltages in the circuit.
Now we can go back and simplify the circuit to Q2 being completely off. The voltage at Vb1 with infinite gain (no current into base) is then just a resistor network driven by 12 V. We can collapse V1, R10, R9, R8, R6, R7, R5, and R3 into a single Thevenin source. The impedance comes out to 760 Ω. I'll let you compute the voltage.
That voltage will be Vb1 for the infinite gain case. For a gain of 100, R1 will be reflected to the base as about 20 kΩ That's not going to change the voltage much against the 760 Ω source impedance.
In any case, Vb1 can't possibly be higher than 12 V, which would put about 11.4 V across R1, which results in 57 mA thru R1 and R2. That only results in 570 mV across R2. That means the collector of Q1 will always be well above its base and emitter, so we can ignore R2 completely for finding the Q1 operating point.
Anyway, this is how I would go about this. Of course it would help to know what this circuit is supposed to accomplish. That may allow more simplifications. It seems rather strange without an obvious useful function, other than maybe to keep undergrads busy with homework for a while.
is it not correct to apply the Thevenins at node Vx because that branch does not contain only resistors but non linear element like Q2 and hence theoram cant be applied ?
Thevenin equivalence is about the source, not what it's hooked up to. You can certainly find the Thevenin equivalent of Vx from V1, R10, R9, R8, R6, and R7 with R5 considered the load, especially with the approximation of infinite gain so that the base current of Q2 is 0.
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