# Plotting ##I_C## vs. ##V_{CE}## for Resistors in Active Region

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In summary, the student is trying to find the appropriate values for the resistors and potentiometer in order to enable the transistor to operate in active mode. They are having some trouble understanding what is going on and need more information.
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## Homework Statement

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.

## Homework 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##.

## 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?

Zondrina said:
I want to plot ##I_C## versus ##V_{CE}## curves for the following:
Okay. In the remaining post you fix ##I_C = 1 mA##, how can you measure different values then?

## Homework Equations

[...]
##R_x = R_y = 500 \Omega##
##I_{R1} = 10I_B##
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.
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##.
Why?
This ##R_C## produces a ##V_{CE} > 0.5 V##, so we are in active mode.
By construction, sure.

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

mfb said:
Okay. In the remaining post you fix IC=1mAI_C = 1 mA, how can you measure different values then?

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##.

mfb said:
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.

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

mfb said:
Why?

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

mfb said:
This tells you you need more current flowing through the resistors (IR1>10IBI_{R1} > 10I_B), or larger resistors.

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

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

This is due to the regions of operation of the transistor. For example, in cutoff we can observe ##V_C = V_{CC} = 15 V##.
Sure, but why do you choose exactly the middle of the range as working point?

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

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).

"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.

Zondrina said:
I'm trying to maximize the output voltage swing.
Looking at the full problem statement, you'll need several different values for the resistor anyway.

mfb said:
"Much greater".

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?

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

STEMucator

## What is the purpose of plotting ##I_C## vs. ##V_{CE}## for resistors in active region?

The purpose of this plot is to analyze the behavior of a resistor in active region, specifically how the collector current (##I_C##) changes in response to the collector-emitter voltage (##V_{CE}##).

## What is the active region for a resistor?

The active region for a resistor is the range of operation where the transistor is amplifying the input signal. In this region, the collector current is directly proportional to the collector-emitter voltage, and the transistor behaves like a linear device.

## How does the plot of ##I_C## vs. ##V_{CE}## change for different resistors?

The slope of the ##I_C## vs. ##V_{CE}## plot will vary for different resistors due to their different resistance values. However, the overall shape and behavior of the plot should remain similar for all resistors in active region.

## What can we learn from the slope of the ##I_C## vs. ##V_{CE}## plot?

The slope of the plot represents the resistance value of the transistor in active region. A steeper slope indicates a higher resistance, while a flatter slope indicates a lower resistance. This information can be useful for selecting resistors for specific applications.

## What other factors can affect the ##I_C## vs. ##V_{CE}## plot for resistors in active region?

Other factors that can affect the plot include temperature, biasing conditions, and the properties of the transistor itself. It's important to consider these factors when interpreting the plot and making design decisions.