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\$\beta = 100\$, Transistor Q1 is PNP and Q2 is NPN. Trying to find \$V_1, V_2, V_3, V_4\$, and \$V_5\$. I assumed active mode for both.

\$\beta = 100\$
\$V_{BE} = 0.7 \mathrm{V}\$

Then $$ \alpha = \frac{\beta}{\beta + 1} = \frac{100}{101} = 0.99 $$ and $$ I_{B1} = (1 – \alpha)I_{E1} \iff I_{B1} = 0.01I_{E1} $$ Now, $$ \frac{V_1}{100000} = \frac{0.01(3 - V_2)}{9100} = 3.2967 \cdot 10^{-6} - 1.0989 \cdot 10^{-6}V_2 $$ Then $$ \frac{(V_2 - 0.7)}{100000} = 3.2967 \cdot 10^{-6} - 1.0989 \cdot 10^{-6}V_2 $$ $$ V_2 – 0.7 = 0.3296703297 - 0.1098901099V_2 \iff 1.1098901099V_2 = 1.02967033 $$ thus $$ V_2 = 0.9277227723 \mathrm{V} $$ and $$ V_1 = V_2 – 0.7 = 0.9277227723 – 0.7 = 0.2277227723 \mathrm{V} $$ Then $$ \begin{split} I_{E1} &= \frac{3 - V_2}{9100} = \frac{3 - 0.9277227723}{9100} = 2.3 \cdot 10^{-4} \mathrm{A} \\ I_{B1} &= \frac{V_1}{100000} = \frac{0.2277227723}{100000} = 2.3 \cdot 10^{-6} \mathrm{A}\\ \\ I_{C1} &= I_{E1} – I_{B1}\\ &= 2.3\cdot 10^{-4} – 2.3 \cdot 10^{-6}\\ &= 2.254455446 \cdot 10^{-4} \mathrm{A}\\ \end{split} $$ Now $$ I_{C1} = \frac{V_3 + 3}{9100} + I_{B2} $$ thus $$ \begin{split} 2.25 \cdot 10^{-4} &= \frac{V_3 + 3}{9100} + \frac{V_4 + 3}{4300}\\ 2.25 \cdot 10^{-4} &= \frac{V_3 + 3}{9100} + \frac{V_3 - 0.7 + 3}{4300}\\ 2.25 \cdot 10^{-4} &= \frac{V_3 + 3}{9100} + \frac{V_3 + 2.3}{4300}\\ 2.25 \cdot 10^{-4} &= \frac{V_3 + 3}{9100} + \frac{V_3 + 2.3}{4300}\\ 2.25 \cdot 10^{-4} &= \frac{4300V_3 + 12900}{39130000} + \frac{9100V_3 + 20930}{39130000}\\ 2.25 \cdot 10^{-4} &= \frac{134V_3 +338.3}{391300}\\ 88.21684159 &= 134V_3 + 338.3\\ -250.0831584 &= 134V_3\\ V_3 &= -1.866292227 \mathrm{V}\\ \end{split} $$ and \$V_4 = V_3 – 0.7 = -1.866292227 – 0.7 = -2.566292227 \mathrm{V}\$. Then $$ \begin{split} I_{E2} &= \frac{V_4 + 3}{4300} = \frac{-2.566292227+ 3}{4300} = 1.008622728 \cdot 10^{-4} \mathrm{A}\\ I_{C2} &= \alpha I_{E2} = 0.99(1.008622728 \cdot 10^{-4}) = 9.985365006 \cdot 10^{-5} \mathrm{A}\\ I_{B2} &= I_{E2} – I_{C2} = 1.008622728 \cdot 10^{-4} – 9.985365006 \cdot 10^{-5} = 1.008622737 \cdot 10^{-6} \mathrm{A} \end{split} $$ and $$ \begin{split} I_{C2} &= \frac{3 - V_5}{5100}\\ 9.985365006 \cdot 10^{-5} &= \frac{3 - V_5}{5100} \\ 0.5092536153 &= 3 – V_5\\ V_5 &= 2.490746385 \mathrm{V} \end{split} $$

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  • \$\begingroup\$ Try simulating it. \$\endgroup\$ – Leon Heller Feb 26 at 18:09
  • \$\begingroup\$ I tried but it's off since the BJT isn't ideal. \$\endgroup\$ – Jimmy Vailer Feb 26 at 18:41
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\frac{V_4 + 3}{4300} is the equation for I_{E2}. Therefore, it needs to be multiplied by (1 – \alpha) to obtain I_{B2}.\ Thus $$ 2.25 \cdot 10^{-4} &= \frac{V_3 + 3}{9100} + \frac{V_4 + 3}{4300}(1 – \alpha)\

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