BJT Amplifier Analysis: Questions and Solutions

[STEMucator]

Homework Statement

Homework Equations

The Attempt at a Solution

I was not too sure about the analysis and thought I would ask some questions.

Do I short out the AC source $v_i$ when doing the DC analysis?

The input capacitor with $C = \infty$ is designed to couple the AC signal and behaves as a short circuit.

Then I thought to simplify the base of the transistor. The Thevenin voltage and resistance are given by:

$$V_{th} = \left[\frac{100k}{200k} \right] (5 V) = 2.5V$$
$$R_{th} = 100k || 100k = 50k$$

Now there would be a DC voltage of $2.5V$ and a $50k$ resistor in the base (assuming $v_i$ is shorted).

Does this sound reasonable so far?

Now If the transistor is active, does that imply $V_{BE} = 0.7V$?

If so I'm guessing the next step would be:

$$I_B = \frac{2.5V - 0.7V}{50k} = 0.036 mA$$

I am not entirely sure how to proceed from here and I'm wondering if this is even correct.

 

[milesyoung]

Do I short out the AC source $v_i$ when doing the DC analysis?

There isn't any AC signal present in the DC equivalent circuit, so yes.

Due to the property of superposition, you're allowed to split up your analysis for signals at different frequencies and combine the results.

Then I thought to simplify the base of the transistor. The Thevenin voltage and resistance are given by:

$$V_{th} = \left[\frac{100k}{200k} \right] (5 V) = 2.5V$$
$$R_{th} = 100k || 100k = 50k$$

Now there would be a DC voltage of $2.5V$ and a $50k$ resistor in the base (assuming $v_i$ is shorted).

Hmm, I'm not sure what it is you want to accomplish here. You've found the Thévenin equivalent circuit looking into the voltage divider from the base of the BJT.

Now If the transistor is active, does that imply $V_{BE} = 0.7V$?

Approximately, and it's fine to assume that since you have some resistance from the emitter to ground.

A better model is the BJT as a transconductance amplifier where $V_{BE}$ controls $I_C$.

If so I'm guessing the next step would be:

$$I_B = \frac{2.5V - 0.7V}{50k} = 0.036 mA$$

Edit: In your equivalent circuit, you've only replaced the part looking "out" from the base of the BJT. Nothing about the part involving the BJT has changed, so this calculation doesn't make a lot of sense. You're also rarely interested in the base current, since it depends on parameters that vary wildly. It's usually enough just to know that it's "small enough".

You know the voltage at the base, and you know $V_{BE}$, so what is the emitter voltage?

If you know the emitter voltage, what is the emitter current? Since $I_C \approx I_E$..

 

[STEMucator]

Due to the property of superposition, you're allowed to split up your analysis for signals at different frequencies and combine the results.

This.

You know the voltage at the base, and you know ##V_{BE}##, so what is the emitter voltage?

I calculated $V_B$ using the voltage divider scheme to be $2.5V$. The voltage $V_B$ would be located at the node in between the two 100k resistors.

So I know $V_E = V_B - V_{BE} = 2.5V - 0.7V = 1.8V$.

Then $I_E = \frac{V_E - 0}{3.6k} = \frac{1.8V}{3.6k} = 0.5 mA$.

Now we know $I_C = \alpha I_E$, but with the transistor $\beta$ presumed to be large, we make the approximation $\alpha ≈ 1$. So $I_C = \alpha I_E ≈ I_E = 0.5 mA$.

Then replacing the BJT with its voltage controlled current source T-model would be the next step. Here are the parameters for future reference:

$$g_m = \frac{I_C}{V_T} = \frac{0.5 mA}{25 mV} = 20 mS$$
$$r_e = \frac{V_T}{I_E} = \frac{25 mV}{0.5 mA} = 50 \Omega$$

So the question asks to find $\frac{v_{o1}}{v_i}$. Since the DC source is shorted, $V_B = v_i$, and so this can be done with a voltage divider:

$$v_{o1} = \left[ \frac{R_E}{R_E + r_e} \right] v_i \Rightarrow \frac{v_{o1}}{v_i} = \left[ \frac{R_E}{R_E + r_e} \right]$$

Now for $\frac{v_{o2}}{v_i}$:

$$v_{o2} = -g_m v_{\pi} R_C$$

Where we know:

$$v_{\pi} = \left[\frac{r_e}{r_e + R_E} \right] v_i$$

And so:

$$\frac{v_{o2}}{v_i} = -g_m \frac{r_e}{r_e + R_E} R_C$$

We also know $g_m = \frac{\alpha}{r_e}$, so the expression reduces to:

$$\frac{v_{o2}}{v_i} = - \frac{\alpha R_C}{r_e + R_E}$$

Finding the values of the gains is easy.

 

[milesyoung]

Great.

I know what you mean with $v_\pi$, but I'd personally use $v_{be}$ instead for the base-emitter voltage.

With regards to what sources are in what state for whatever part of the analysis:
If you have a linear network with two voltage signals at frequencies $\omega_1$ and $\omega_2$, then you construct two equivalent circuits, one for each frequency.

The equivalent circuit at $\omega_1$ has no component in it at $\omega_2$, so there's always 0 V across the terminals of those sources, which is equivalent to a connection with an ideal wire. If it was current signals instead, there would always be 0 A through those sources, which is equivalent to an open circuit.

 

[STEMucator]

Great.

I know what you mean with $v_\pi$, but I'd personally use $v_{be}$ instead for the base-emitter voltage.

With regards to what sources are in what state for whatever part of the analysis:
If you have a linear network with two voltage signals at frequencies $\omega_1$ and $\omega_2$, then you construct two equivalent circuits, one for each frequency.

The equivalent circuit at $\omega_1$ has no component in it at $\omega_2$, so there's always 0 V across the terminals of those sources, which is equivalent to a connection with an ideal wire. If it was current signals instead, there would always be 0 A through those sources, which is equivalent to an open circuit.

I'm using the Sedra convention for $v_{\pi}$, usually I would probably say $v_e$ or something along those lines.

Now that that's correct, connect $v_{o1}$ to ground. The resistor$R_E$ is now bypassed with no voltage across it due to zero current.

The question then asks what is $\frac{v_{o2}}{v_i}$ at this point. Once again, we know:

$$v_{o2} = - g_m v_{\pi} R_C$$

This is obtained using either the pi model or the T model. The pi model is a little bit nicer in this case since $R_E$ is bypassed. Now:

$$v_{\pi} = v_i$$

because we would have to drop $v_i$ volts across $r_{\pi}$ to reach ground. Hence:

$$\frac{v_{o2}}{v_i} = - g_m R_C$$

This is the usual gain of a BJT CE amplifier with $R_E = 0$.