# Analyzing a PNP BJT

1. Mar 15, 2015

### Zondrina

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

I have several thoughts about this circuit. My first thought was to decide whether or not the transistor is in active or saturation mode.

Assuming active mode operation and neglecting the base current, I see $V_B ≈ 0V$. So the emitter voltage will be $V_E = V_{EB} ≈ 0.7V$.

Hence the emitter current $I_E = \frac{5V - 0.7V}{2.2k} ≈ 2mA$.

Since $R_C = R_E$, we know $I_C ≈ I_E$. So:

$$2 mA = \frac{V_C + 5V}{2.2k} \Rightarrow V_C = (2 mA)(2.2k) - 5V = - 0.6V$$

Hence $V_{EC} = V_E - V_C = 0.7V + 0.6V = 1.3V$

So $V_{EC} > 0.3V$ and I would think the transistor is in active mode.

Does this seem reasonable? Am I approaching this correctly?

2. Mar 15, 2015

### LvW

It`s an approximate solution - and I cannot detect any fundamental error. However, as you certainly know - the correct solution requires a finite base current (other than 0).
By the way: We can set IC≈IE (if beta is large enough) - indpendent on the condition RC=RE.

3. Mar 15, 2015

### milesyoung

I think the idea with this assignment is to not assume VB = 0. You can set up a system of equations to give you the device currents as a first step.

4. Mar 15, 2015

### Zondrina

Okay so assuming the transistor is active, we have:

$$I_B = \frac{V_B - 0V}{R_B} = \frac{V_B}{20k}$$
$$V_E = V_B + V_{EB} = V_B + 0.7V$$
$$I_E = \frac{5V - V_E}{R_E} = \frac{5V - (V_B + 0.7V)}{2.2 k} = \frac{4.3V - V_B}{2.2 k}$$

From here I am not entirely sure what the next step should be. Should I use:

$$V_C = V_E - V_{EC} = V_B + 0.7V - V_{EC}$$

Then I figure:

$$I_C = \frac{V_C + 5V}{R_C} = \frac{(V_B + 0.7V - V_{EC}) + 5V}{2.2k}$$

Then $I_E = I_C + I_B$ yields:

$$\frac{4.3V - V_B}{2.2 k} = \frac{(V_B + 0.7V - V_{EC}) + 5V}{2.2k} + \frac{V_B}{20k}$$

Usually this would spit out $V_B$, which I could solve all the equations with. The problem is I don't know $V_{EC}$ for this transistor.

5. Mar 15, 2015

### milesyoung

Consider where $\beta$ is in all this. You need to involve it somehow to solve your assignment.

$I_C = \beta I_B,I_C = I_E - I_B$ is two equations in three unknowns. You should be able to find a third equation using KVL to solve for the device currents.

6. Mar 15, 2015

### Zondrina

So I somehow need a KVL equation involving $I_E$? Would that mean I apply KVL to the top loop?

Also is my prior method incorrect?

7. Mar 15, 2015

### milesyoung

Consider the loop starting at the 5 V source and moving down through the base-emitter junction to ground.

It is if you need to involve $\beta$. If you assume $V_B = 0$, then $\beta$ has no effect on your circuit.

8. Mar 15, 2015

### Zondrina

I thought the circuit depended on $\beta$ if $V_B = 0$. If $V_B = 0$, the whole base is a ground. Then $V_{EB} = 0.7V$ for active mode operation would imply $V_E = 0.7V$, and you can find $I_E$ from there. Then since $I_C = \alpha i_E = \frac{\beta}{1 + \beta} i_E$, the collector current does depend on $\beta$.

I get the feeling $\beta = 50$ is somewhat large, so I think that's why you're saying it has no effect.

As for the loop:

$$\sum V = 0 \Rightarrow 5V + I_E R_E + V_{EB} + I_B R_B = 0$$
$$I_E R_E + I_B R_B = - 4.3V$$

So the three equations would be:

$$I_E R_E + I_B R_B = - 4.3V$$
$$I_C = \beta I_B$$
$$I_E = I_C + I_B$$

9. Mar 15, 2015

### milesyoung

Sure, but I was thinking more in terms of your assignment. To put it another way: If you just short out RB, what actually changes in your circuit when you set RB = 100 kΩ?

There's a sign issue with the source, but otherwise it looks fine.

10. Mar 15, 2015

### Zondrina

Oh derp on the sign there sorry. I forgot the source is oriented so the negative side is traversed. So it would be $+4.3 V$.

After solving the equations and obtaining the currents, simple application of KVL will be enough to find $V_E, V_B, V_C$.

$$I_E R_E + I_B R_B = 4.3V$$
$$I_C = \beta I_B$$
$$I_E = I_C + I_B$$

Plugging the second into the third:

$$I_E = \beta I_B + I_B = (\beta + 1)I_B$$

Now plugging the above into the first:

$$(\beta + 1)I_B R_E + I_B R_B = 4.3V$$
$$I_B \left[(\beta + 1) R_E + R_B \right] = 4.3V$$
$$I_B = \frac{4.3V}{\left[(\beta + 1) R_E + R_B \right]} = 32.53 \mu A$$

Plugging back:

$$I_E = (\beta + 1)I_B = 1.659 mA$$

One more plug:

$$I_C = I_E - I_B = 1.627 mA$$

Now simple application of KVL will find the voltages.

As for the change in $R_B$, nothing would happen because $I_B = 0$ if $V_B = 0$.

11. Mar 15, 2015

### milesyoung

If $V_B \equiv 0$, then you've effectively just replaced $R_B$ with an ideal wire, so there can still be base current.

The point of this assignment is to show you the effect of bad biasing when $R_B$ becomes large compared with the input impedance $\beta R_E$ of the BJT (it's a voltage divider).