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enter image description here

I edited circuit like this:

enter image description here

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

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  • \$\begingroup\$ I am surprised textbooks still perpetuate the idea that BJT's are current-controlled devices and keep basing designs on current gain. \$\endgroup\$ – Bart Jan 21 at 7:43
  • \$\begingroup\$ @Bart Textbooks do not teach that BJTs are "current controlled", they teach that if the transistor is neither cut off nor saturated then the collector current is proportional to the base current, and the base-emitter voltage is relatively constant. As you can see from the answers below, this is a useful and appropriate model for engineering purposes. \$\endgroup\$ – Elliot Alderson Jan 21 at 18:24
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Here's a different approach. It's the way I thought about the problem:

schematic

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}\$.

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  • \$\begingroup\$ Thanks also for this solution.. !! \$\endgroup\$ – Hury H Jan 20 at 15:53
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There's an easier way to find the DC Q points and R value starting the 5V for Vce1. (or UCE1 )

enter image description here

It is simpler because we were permitted to ignore I(R) thru T2 as it is only 0.5% more (hFE=100 * 10uA vs Ie2~2mA ). 10uA/2mA*100%=0.5%

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    \$\begingroup\$ Many thanks for this! :) \$\endgroup\$ – Hury H Jan 20 at 1:26
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Well, we have the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

I used Mathematica to solve it, using the following code:

Solve[{Ix == I1 + I4, I4 == I7 + I8, I2 == I5 + I3, Ix == I3 + I6, 
  I1 == ((Vx - V2)/(R1)), I2 == ((V3 - V4)/(R2)), I3 == ((V4)/(R3)), 
  I4 == ((Vx - V1)/(R4)), I5 == ((V4 - V5)/(R5)), I6 == ((V6)/(R6)), 
  V1 - V3 == \[Alpha]1, V5 - V6 == \[Alpha]2, I1/I8 == \[Beta]1, 
  I7/I5 == \[Beta]2, I2 == I8 + I1, I6 == I5 + I7}, {Ix, I1, I2, I3, 
  I4, I5, I6, I7, I8, V1, V2, V3, V4, V5, V6}]

Your values gave and adding the fact that \$\text{V}_1-\text{V}_6=5\$:

Solve[{Ix == I1 + I4, I4 == I7 + I8, I2 == I5 + I3, Ix == I3 + I6, 
  I1 == ((25 - V2)/(7000)), I2 == ((V3 - V4)/(1000)), I3 == ((V4)/(2000)), 
  I4 == ((25 - V1)/(18000)), I5 == ((V4 - V5)/(R5)), I6 == ((V6)/(2000)), 
  V1 - V3 == 7/10, V5 - V6 == 7/10, I1/I8 == 100, 
  I7/I5 == 100, I2 == I8 + I1, I6 == I5 + I7, V1 - V6 == 5}, {Ix, I1, I2, I3, 
  I4, I5, I6, I7, I8, V1, V2, V3, V4, V5, V6, R5}]

This gives:

{{Ix -> 9399043/3051390000, I1 -> 63443/30513900, 
  I2 -> 6407743/3051390000, I3 -> 637783/305139000, 
  I4 -> 3054743/3051390000, I5 -> 9971/1017130000, 
  I6 -> 1007071/1017130000, I7 -> 9971/10171300, 
  I8 -> 63443/3051390000, V1 -> 3549896/508565, V2 -> 3187465/305139, 
  V3 -> 6387801/1017130, V4 -> 1275566/305139, V5 -> 2726133/1017130, 
  V6 -> 1007071/508565, R5 -> 352097000/2301}}

Which is:

{{Ix -> 0.00308025, I1 -> 0.00207915, I2 -> 0.00209994, 
  I3 -> 0.00209014, I4 -> 0.0010011, I5 -> 9.80307*10^-6, 
  I6 -> 0.00099011, I7 -> 0.000980307, I8 -> 0.0000207915, 
  V1 -> 6.98022, V2 -> 10.4459, V3 -> 6.28022, V4 -> 4.18028, 
  V5 -> 2.68022, V6 -> 1.98022, R5 -> 153019.}}
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