# Homework Help: BJT Biasing

1. Sep 20, 2015

### Zondrina

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

I want to plot $I_C$ versus $V_{CE}$ curves for the following:

To do so, I need to find appropriate values for $R_C, R_B$ and $R_1$ in the active region.

Note that $R_2$ is a potentiometer.

2. Relevant equations

$I_C = 1 mA$
$V_{BE} = 0.7 V$
$\beta = 150$
$R_x = R_y = 500 \Omega$
$I_{R1} = 10I_B$
$V_{CEsat} = 0.5 V$.

3. The attempt at a solution

First I need to find appropriate values for the resistors, but I've hit a snag. The analysis proceeds like so:

1) $v_{Cmax} = V_{CC} = 15V$ and $v_{Cmin} = V_{B} = V_{E} + V_{BE} = 0.7V$.

Therefore $V_{C} = \frac{1}{2} (15V + 0.7V) = 7.85 V$.

2) $R_C = \frac{V_{CC} - V_C}{I_C} = \frac{15V - 7.85V}{1 mA} = 7.15 k \Omega$.

This $R_C$ produces a $V_{CE} > 0.5 V$, so we are in active mode.

3) Now that we have $R_C$, we need to find $R_B$ and $R_1$. Let $I_B = \frac{I_C}{\beta} = \frac{1 mA}{150} = 6.68 \mu A$.

Then $I_{R1} = 10I_B = 66.8 \mu A$. This will be the current flowing across the series combination of $R_1$ and $R_x$.

Now $I_{Ry} = I_{R1} - I_{B} = 60.12 \mu A$. This will be the current flowing across $R_y$.

4) Here is where I start encountering results that don't add up. I want to find the voltage at the node between $R_1 + R_x$, $R_B$ and $R_y$ (call it $V_{BB}$), so I used:

$$\frac{V_{BB} - 0}{500 \Omega} = 60.12 \mu A \Rightarrow V_{BB} = 0.03 V$$

This result makes no sense because the voltage at the base is $0.7V$.

What went wrong here?

2. Sep 20, 2015

### Staff: Mentor

Okay. In the remaining post you fix $I_C = 1 mA$, how can you measure different values then?
You want to design your circuit with those additional constraints? Listing them as "relevant equations" makes them look like some fundamental properties of the setup. This is a crucial point because those assumptions break your analysis later.
Why?
By construction, sure.

This tells you you need more current flowing through the resistors ($I_{R1} > 10I_B$), or larger resistors.

3. Sep 20, 2015

### Zondrina

I will be varying the value of $R_C$ during the experiment, and I will be adjusting the potentiometer so it produces $I_C = 1mA$ at all times.

I'm not measuring $I_C$. When I plot $I_C$ versus $V_{CE}$ the curves will represent $I_B$.

I was told to set $R_x = R_y = 500 \Omega$. Usually we are told to make the assumption $I_{R1} = 10 I_B$.

This is due to the regions of operation of the transistor. For example, in cutoff we can observe $V_C = V_{CC} = 15 V$.

This would be much different than the usual assumption I'm told to make.

4. Sep 20, 2015

### Staff: Mentor

Well, that does not work, as you can see. A larger current in R1 does not harm. Only a smaller one could lead to problems.

Sure, but why do you choose exactly the middle of the range as working point?

5. Sep 20, 2015

### Zondrina

I'm trying to maximize the output voltage swing.

Here is the original problem statement just in case it might help:

For this section, you will generate a plot similar to Figure 1.2 by plotting IC vs VCE. The figure shows several curves for different IB values. You will produce one curve at a single constant IB value that gives IC in the active region around 1 mA.

R2 is a potentiometer (the symbol is a resistor with an arrow pointing to it). It is a 3-terminal variable voltage divider with a total resistance of 1k. This allows the base voltage to be tuned by fine increments. Thus, Rx+Ry will always be equal to 1k, but the value of Rx can be changed from 0 to 1k by turning the dial.

For this part, determine appropriate values of R1, RB and RC. You will need to justify these choices in
your report with good reasons. Obviously there will be many possible values of RC, with each representing a different VCE. Choose at least 10 values in the active region and another 5 or more values in the saturation region. Show the expected VCE for each resistance.

As a hint, for these calculations, keep in mind that VBE is around 0.7 V in the active region, VCE,sat is 0.5V, beta is around 150 and that the voltage drop across any resistor should be at least 0.25 V to be able to measure it accurately. In addition, the current flowing in the left branch should be much greater than the base current and the potentiometer should be in its middle range (i.e. Rx=Ry=500 ohms).

6. Sep 20, 2015

### Staff: Mentor

"Much greater". A factor of 10 is much greater, but a factor of several hundred is "much greater" as well. There is no reason to use 10. You can work backwards: choose Rb to satisfy the 0.25 V condition, find the voltage at the connection point, then determine the current you need in Ry to match that voltage.

Looking at the full problem statement, you'll need several different values for the resistor anyway.

7. Sep 20, 2015

### Zondrina

Yes this was not very specific at all. Would you suggest $I_{R1} = 1000 I_B$ or something along those lines? If so the voltage would be about $3.3V$ at $V_{BB}$ and the rest of the analysis would follow.

Then I would choose 10 values of $R_C$ such that $V_{CE} > 0.5 V$ and 5 values of $R_C$ such that $V_{CE} \leq 0.5 V$.

Sound good?

8. Sep 20, 2015

### Staff: Mentor

Sounds good.
If you choose the current a bit smaller than that your base current can be adjusted in finer steps.