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Can you explain to me how to obtain the small signal voltages of vc(vout),vb( for the upper BJT) and ve( small-signal voltage of the common emitter ? for this small-signal equivalent circuit. I have tried to do write KCL equations but I think I am not able to find the proper relationships. circuit sorry for image quality. 1 2

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  • \$\begingroup\$ I can't even understand the annotations. \$\endgroup\$ – dim Dec 12 '16 at 19:53
  • \$\begingroup\$ What does the u mean, as in r_o = 100ku? Can you show the full schematic so we can verify if your small-signal representation is correct? \$\endgroup\$ – vofa Dec 12 '16 at 19:57
  • \$\begingroup\$ sorry again for my bad hand-writing. its ohm not u. they are resistances. \$\endgroup\$ – utdlegend Dec 12 '16 at 20:00
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    \$\begingroup\$ Badly drawn circuit, not even sure that small signal model makes any sense, unreadable component value, not even showing an attempt at solving the equations. I'm moving on ! \$\endgroup\$ – Bimpelrekkie Dec 12 '16 at 20:03
  • \$\begingroup\$ Could you look at it now? I have added the main circuit, hope it will be help full. \$\endgroup\$ – utdlegend Dec 12 '16 at 20:20
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Many times it is best to use simpler equivelent circuits before writing loop or node equations. The first thing to do in this case is to find the Thevenin equivelent at the base of the npn to ground for vin and the two resistors. You know how to do this if you are to the point of analyzing circuits like this. The second thing is to notice that for the PNP i=gm*vbe is the equation of a resistor between C and E. If you substitute gm=1/r you will find that this is small enough that you can probably ignore the base emitter and collector emitter resistors. If not just do the parallel combination to replace the PNP with a single resistor.

For the PNP you have B in the wrong place. Since Vx is a short to ground for small signal, B is the ground end of the PNP r-pi.

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If I were to answer this question, I would begin by finding the equivalent input resistance looking into the base of the NPN transistor. This effectively helps reduce the system of 3 equations to a system of two equations due to the fact that, given the input resistance, call it Rb, you can find the voltage at the base of the NPN transistor. Due to this fact, you can solve for the output voltage of the circuit in terms of a test base voltage (Vb):

schematic

simulate this circuit – Schematic created using CircuitLab

This will give you $$\frac{V_o}{V_b}$$ In order to find $$\frac{V_o}{V_{in} }$$ you would need to solve for $$\frac{V_b}{V_{in}}$$ and find the product of these results. The latter can be found by using the voltage divider between the 2.2kOhm resistor and the 1890 Ohm resistor in parallel with the base resistance. Additionally, to simplify further, you can find the resistance looking down into the emitter of the PNP transistor and treat it as an AC resistance. Hopefully this helps you answer the question!

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