Multiple BJT Analysis: Homework Statement

[STEMucator]

Homework Statement

Hi everyone, this is not homework, but I can't seem to see something.

I have a multiple BJT circuit I've drawn, and I want to analyze it:

Homework Equations

The Attempt at a Solution

Assume the transistor $\beta$s are equal for both transistors and are known. Assume all resistances and capacitances are known.

DC Analysis

The DC analysis would be straightforward. The capacitor would eliminate the left branch of the circuit, and the rest would be analyzed as follows:

1. Use Thevenin's theorem to obtain $R_{th}$ and $V_{th}$ at the base. Write a loop equation around the BE-loop and use the relationship $I_{B1} = \frac{I_{E1}}{\beta + 1}$ to find $I_{E1}$:

$$I_{E1} = \frac{V_{th} - V_{BE1}}{R_{E1} + \frac{R_{B1}}{\beta + 1}}$$

2. Then we can find $I_{C1} = \alpha I_{E1}$ and $I_{B1} = \frac{I_{C1}}{\beta}$ assuming the transistor is active.

3. To verify $Q_1$ is active, we must show $V_{CE1} = V_{C1} - V_{E1} = I_{C1}R_{C1} - I_{E1}R_{E1} > 0.7 V$, assuming $V_{BE1} = 0.7V$ as usual.

So the first transistor is done.

4. Now we can write a KCL equation that says:

$$I_{B2} = I_{B1} + I_{C1} - I_{E1}$$

To find $I_{B2}$.

5. Then we would simply use $I_{C2} = \beta I_{B2}$ and $I_{E2} = I_{C2} + I_{B2}$ to find the remaining DC currents.

6. To make sure $Q_2$ is active, we must show $V_{CE2} = V_{C2} - V_{E2} = I_{C2}R_{C2} - I_{E2}R_{E2} > 0.7 V$, assuming $V_{BE2} = 0.7V$ as usual.

All the DC analysis would be done at this point I believe.

AC Analysis

I have drawn the simplified hybrid-pi model for both transistors to make the analysis easier.

As you can see the in the image, finding $v_{\pi 1}$ is easy by using a voltage divider. Then finding $i_{c1}$ would be easy using the bias conditions.

Next I would find $i_{b1}$ using Ohm's law. Now I have $i_{b1}$ and $i_{c1}$ in hand. These two currents form the $I_x$ current I have labelled in the image.

My question is, how do I analyze the bottom portion of the circuit?

All I have is the two current relations I wrote down in the image. I can't seem to see what's next. If I had the voltage at the node I would be able to continue.

 

[STEMucator]

Wait, I had a brain malfunction earlier. I think I studied for too long without a break.

Would the voltage at the node where $I_x$ is simply be $v_{C1} = i_{C1} R_{C1} = g_m v_{\pi 1} R_{C1}$?

If so, most of the remaining problem can be solved using Ohm's law.

I have learned something very interesting about straightening the topology of a circuit.

 

[STEMucator]

Sorry for this post, but something dawned on me. Doesn't the dependent current source also have a voltage drop? I have never been asked to consider the voltage drop across one of these sources.

This would affect the node voltage I'm looking for slightly. Replacing the dependent current source by an impedance $Z_{\pi 1} = \frac{1}{g_{m 1}}$ the voltage at the node would be:

$$V_x = v_{C1} - v_x = i_{C1}R_{C1} - i_{C1} Z_{\pi 1} = i_{C1} \left[ R_{C1} - \frac{1}{g_{m 1}} \right]$$

With that being said, I think the problem is solved. I will take a break now.