Thats a Microelectronic Circuits Sedra/Smith's excercise , I recognized it immediately.

The way you do it is this:

The Diodes are there to counteract the \$V_{BE}\$ drop of each transistor, meaning that each diode will "cancel" each \$V_{BE}\$ drop, so the voltage at R6 is the same as the voltage of R1, and the Voltage of R4 is the same as the voltage of R7.

Look at it this way, if you do a closed loop at the base of the npn transistor from ground, going through R6, D1, the \$V_{BE}\$ drop and R1. You get something like this:

$$V_{R6}+0.7V(\text{diode drop of D1})-0.7(V_{BE} \text{ of transistor})-V_{e1}=0$$

Thus the 0.7V cancel out and you get that the Voltage of Resistor R6 is the same as \$V_{e1}\$, or in other words \$V_{R6}=V_{e1}\$. Same thing happens with the pnp transistor.

Since both voltages are equal, then the product of their currents times their resistance should also be equal, Let \$I_{d1}\$ be the current Through D1 and R6, and I1 the current through R1. Then:

$$I_{d1}\times 40k=I_1\times 2k$$

Same thing happens at \$V_{e2}\$:

Let \$I_{d2}\$ be the current through D2 and R4, and Ie2 the emitter current of Q2, then:

$$I_{d2}\times R_4=I_{e2}\times 100$$

Now, if you are very picky, you can do the node equations for the entire circuit and get an exact result which is a lot of work. Let \$I_1\$ be the current through \$R_1, I_2\$ the current through \$R_2, I_{d2}\$ the current through \$D_2\$ and so on, you get.

$$I_5=I_{d1}+I_{b1}$$

$$I_{d2}=I_{c1}-I_{b2}$$

$$I_1=I_{e1}+I_3$$

$$I_{c2}=I_3+I_2$$

You can also rely on the mesh equations. From there on its just algebra. And the exact results should be:

$$V_{b1}=3.104V$$

$$V_{c1}=6.108V$$

$$V_{e1}=2.404V$$

$$V_{b2}=6.108V$$

$$V_{c2}=2.181V$$

$$V_e2=6.808V$$

However if you make some simplifications like stating that \$I_c=I_e\$ and if you neglect the base currents of both transistors for a first approximation, then it should be a lot easier to solve the problem, and the results shouldn't vary that much from the exact method described above. In which case you could calculate the current in the voltage divider\$(I_{vd})\$ formed by \$R_5\$ and \$R_6\$ as follows:

$$I_{vd}=\frac{9-0.7}{40k+80k}\times 9= 69.17 \mu Amps$$

So the voltage at \$R_6\$ is

$$V_{R6}=69.17\times 10^{-6} \times 40k=2.766V,$$
and since \$V_{R6}\$=\$V_{e1}~,~ V_{e1}=2.766V\$, compare it to the 2.404V obtained from the "exact" method, and its not that different. However, you could in fact use this value calculated for \$V_{R6}\$ and iterate once more, but now without neglecting base current in your calculations, eventually the value you calculate will approach the exact values.

You would do that as follows:

To start, neglect base current of transistor Q2 (this current will be taken into account later on) and assume \$I_e=I_c\$, in which case the voltage in \$R_4\$ equals the voltage of the emitter resistor of Q2.

So

$$I_{e1}*2k=I_{e2}*100$$

Solving for Ie2:

$$Ie2=20*Ie1 ...... A$$

The node equation at Q2s collector node is:

$$I2=Ie2-I3 ....... B$$

Where I2 is the current through R2, I3 the current through R3

And the node equation at Q1s emitter node is:

$$I_{e1}+I_3=I_6 $$

But the voltage in \$R_6\$ is 2.766V

So

$$I_{e1}+I_3=\frac{2.766}{2k} ........C$$

The loop equation formed by R6, R3 and R2 is:

$$2.766V+I_e*2k-I2*100=0 ...... D$$

Using A on B, then on D and finally on C, you get that the current Ie1=1.383mA and the current I3=0mA

$$I_{b1}=\frac{I_{e1}}{(\beta+1)}$$

So

$$I_{b1}=13.69~\mu A$$

Now, the node equation at the base of Q1 is (notice that now im not neglecting base current of Q1):

$$I_5=I_{d1}+I_{b1} .... E$$

The voltage at the base of Q1 is:

$$V_b1=I_{d1}*40k+0.7$$

So $$I_5=\frac{(9-(I_{d1}*40k+0.7))}{80k} ....F$$

We know Ib1 is $$13.69uA$$

So using F on E yields:

$$Id1=60.04 microAmps$$

And thus the voltage at the resistor R6 and V_e1 is

$$V_e1=VR6=(60.04*10^-6)*40k$$

**$$V_e1=2.402V$$**

Which is practically the same voltage as the one found using the exact method.

You can now find V_b1

$$V_b1=2.402+.7V$$

**$$V_b1=3.102V$$**

You can now do a similar procedure to find V_e2, but now Ill consider the base current of Q2 which I neglected at the start.

Since \$V_{e1}\$ is now 2.402V, you just need to do a few modifications.

Equation "C" now becomes:

$$Ie1+I3=\frac{2.402}{2k} ...C'$$

And equation "D" becomes:

$$2.402V+I_eâ2kâI2â100=0......D'$$

Now we have to consider \$I_{b2}\$, if we consider \$I_e=I_c\$ for each transistor, the node equation at the base of Q2 is:

$$I_{d2}+I_{b2}=I_{e1}....G$$

Using equation A on G yields:

$$0.05\times I_{e2}~+~I_{b2}~=~I_{e1}$$

but $$I_{b2}=\frac{I_{e2}}{\beta+1}$$

so the equation transforms into:

$$I_{e2}*\bigg(0.05+\frac{1}{\beta+1}\bigg)=I_{e1} .......H$$

Using H on C' we get:

$$I_{e2}\bigg(0.05+\frac{1}{\beta+1}\bigg)+I_3=2.402/2k ........I$$

And using B on D' we get:

$$100\times I_{e2}-2.1K \times I_3=2.402V ......J$$

Solving the 2 by 2 system of I and J, we get:

$$I_{e2}=21.8mA$$

and

$$I_3=-0.15mA$$

Using this values on equation B, we get:

$$I_2=2.19mA$$

So \$V_{c2}\$ is:

$$V_c2=2.19mA*100$$

$$V_{c2}=2.19V$$

And \$V_{e2}\$ is:

$$V_{e2}=9-21.8mA*100$$

**$$V_{e2}=6.82V$$**

And of course if you take 0.7V from \$V_{e2}\$ you get the Base voltage of Q2 which is the same as the Collector voltage of Q1:

$$V_{c1}=V_{b2}=6.82-0.7=6.12V$$

## Best Answer

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

^{simulate this circuit â 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

mustbe taken into account and theof 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.other partWorrying 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}\$.