# Bipolar Junction Transistor problem. Where did I go wrong?

$$\\beta = 100\$$, Transistor Q1 is PNP and Q2 is NPN. Trying to find $$\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, $$\frac{V_1}{100000} = \frac{0.01(3 - V_2)}{9100} = 91(3 – V_2)$$ Then $$\frac{(V_2 - 0.7)}{100000} = 273 – 91V_2$$ $$V_2 – 0.7 = 27300000 – 9100000V_2 \iff 9100001V_2 = 27300000.7$$ thus $$V_2 = 2.999999747 \mathrm{V}$$ and $$V_1 = V_2 – 0.7 = 3 – 0.7 = 2.299999747 \mathrm{V}$$ Then $$\begin{split} I_{E1} &= \frac{3 - V_2}{9100} = \frac{3 - 2.999999747}{9100} = 2.3 \cdot 10^{-3} \mathrm{A} \\ I_{B1} &= \frac{V_1}{100000} = \frac{2.3}{100000} = 2.3 \cdot 10^{-5} \mathrm{A}\\ \\ I_{C1} &= I_{E1} – I_{B1}\\ &= 2.3\cdot 10^{-3} – 2.3 \cdot 10^{-5}\\ &= 2.277 \cdot 10^{-3} \mathrm{A}\\ \end{split}$$ Now $$I_{C1} = \frac{V_3 + 3}{9100} + I_{B2}$$ thus $$\begin{split} 2.277 \cdot 10^{-3} &= \frac{V_3 + 3}{9100} + \frac{V_4 + 3}{4300}\\ 2.277 \cdot 10^{-3} &= \frac{V_3 + 3}{9100} + \frac{V_3 - 0.7 + 3}{4300}\\ 2.277 \cdot 10^{-3} &= \frac{V_3 + 3}{9100} + \frac{V_3 + 2.3}{4300}\\ 2.277 \cdot 10^{-3} &= \frac{V_3 + 3}{9100} + \frac{V_3 + 2.3}{4300}\\ 2.277 \cdot 10^{-3} &= \frac{4300V_3 + 12900}{39130000} + \frac{9100V_3 + 20930}{39130000}\\ 2.277 \cdot 10^{-3} &= \frac{134V_3 +338.3}{391300}\\ 890.9901 &= 134V_3 + 338.3\\ 552.6901 &= 134V_3\\ V_3 &= 4.124552985 \mathrm{V}\\ \end{split}$$ and $$\V_4 = V_3 – 0.7 = 4.124552985 – 0.7 = 3.424552985 \mathrm{V}\$$. Then $$\begin{split} I_{E2} &= \frac{V_4 + 3}{4300} = \frac{3.424552985+ 3}{4300} = 1.49408209 \cdot 10^{-3} \mathrm{A}\\ I_{C2} &= \alpha I_{E2} = 0.99(1.49408209 \cdot 10^{-3}) = 1.479141269 \cdot 10^{-3} \mathrm{A}\\ I_{B2} &= I_{E2} – I_{C2} = 1.49408209 \cdot 10^{-3} – 1.479141269 \cdot 10^{-3} = 1.494082134 \cdot 10^{-5} \mathrm{A} \end{split}$$ and $$\begin{split} I_{C2} &= \frac{3 - V_5}{5100}\\ 1.479141269 \cdot 10^{-3} &= \frac{3 - V_5}{5100} \\ 7.543620472 &= 3 – V_5\\ V_5 &= -4.543620472 \mathrm{V} \end{split}$$