# Electronic – How to solve this BJT

bjtnpntransistors

I edited circuit like this:

I tried representing currents like:
\begin{aligned} &I_{R1}=100_{B1}+I_{B2}\\ &I_{R2}=101 I_{B2}-I_{B1}\\ &I_{i}=101 I_{B2}+100 I_{B1} \end{aligned}

Also I get:
$$U_{BC1}=U_{BE1}-U_{CE1}=4,3(V)$$

Then setting up equations from second Kirchoff law:

$$I: U_{CB1}-U_{BE2}+I_{E 2} \cdot 1 k+I_{B 1} \cdot R=0$$
$$II:-I_{R 1} \cdot 18 k+I_{C 2} \cdot 7 k-U_{C B 2}=0$$
$$III:-25-I_{R 2} \cdot 2 k-I_{E 2} \cdot 1 k+U_{C E 2}-I_{C 2} \cdot 7 k=0$$
$$IV:-I_{E_{1}} \cdot 2 k+U_{CE1}-U_{BE2}+I_{E2} \cdot 1 k+I_{R 2} \cdot 2 k=0$$

After solving system of equations I get wrong numbers so where is problem in this my method?

I'm safer with Symbolab so these are wrong results which I get: $$x=I_{B1}, y=I_{B2}, z=U_{CE}, q=R$$

$$x=I_{B1}=-\left(\frac{13.61731 \ldots}{2040000}\right), y=I_{B2}=-0.0000186856, z=U_{CE}=6.271711, q=R=1.503483 \cdot 10^{8}$$

https://www.symbolab.com/solver/system-of-equations-calculator/3.6%2B101%5Ccdot%20y%2B1000%2Bx%5Ccdot%20q%3D0%2C-18000%5Ccdot%5Cleft(100x%2By%5Cright)%2B7000%5Ccdot100%5Ccdot%20y-%5Cleft(-07%2Bz%5Cright)%3D0%2C-25-%5Cleft(101y-x%5Cright)%5Ccdot2000-101y%5Ccdot1000%2Bz-100y%5Ccdot7000%3D0%2C-101x%5Ccdot2000%2B4.3%2B101y%5Ccdot1000%2B%5Cleft(101y-x%5Cright)%5Ccdot2000%3D0

Here's a different approach. It's the way I thought about the problem:

simulate this circuit – Schematic created using CircuitLab

In the problem text, they suggest that $$\I_{B_1}\$$ and $$\I_{B_2}\$$ can be ignored ($$\I_{B_1}\ll I_{C_1}\$$ and $$\I_{B_2}\ll I_{C_2}\$$) for the purposes at hand. But that's obviously wrong. If it really were true, then there would be no current in $$\R\$$ and therefore no voltage drop and therefore no way to specify its value. So they lied, there. The base currents must be taken into account and the other part of the problem statement mentioning $$\\beta_1=\beta_2=100\$$ must be used. You can see that I took this last part into account in the above schematic.

Worrying about $$\R_3\$$ is pointless, as it is at $$\Q_2\$$'s collector. So all KVL analysis will start with the positive rail voltage, work it's way through $$\R_1\$$, and then from there along three separate paths. Let's assign the collector voltage for $$\Q_1\$$ the name, $$\V_{C_1}=25\:\text{V} - 18\:\text{k}\Omega\cdot\left(100\cdot I_{B_1}+I_{B_2}\right)\$$. Then, it follows that the three equations along the three different paths are:

\begin{align*} V_{C_1}-5\:\text{V}-2\:\text{k}\Omega\cdot 101\cdot I_{B_1}&=0\:\text{V}\\\\ V_{C_1}-700\:\text{mV}-1\:\text{k}\Omega\cdot 101\cdot I_{B_2}-2\:\text{k}\Omega\cdot \left(101\cdot I_{B_2}-I_{B_1}\right)&=0\:\text{V}\\\\ V_{C_1}-700\:\text{mV}-1\:\text{k}\Omega\cdot 101\cdot I_{B_2}-R\cdot I_{B_1}-700\:\text{mV}-2\:\text{k}\Omega\cdot 101\cdot I_{B_1}&=0\:\text{V} \end{align*}

This solves out as: $$\I_{B_1}\approx 9.8031\:\mu\text{A}\$$, $$\I_{B_2}\approx 20.79151\:\mu\text{A}\$$, and $$\R\approx 153.019\:\text{k}\Omega\$$. From those, you find that $$\V_{E_2}\approx 6.28\:\text{V}\$$ and $$\V_{C_2}\approx 10.446\:\text{V}\$$ and therefore that $$\V_{\text{CE}_2}\approx 4.166\:\text{V}\$$.