Lab 14 RS

Go Back to All Lab Reports

The Common Emitter Amplifier

Circuit

Construct the common emitter amplifier circuit below according to your type of emitter.

Calculate all the R and C values to use in the circuit such that

a. Try [math]\displaystyle{ R_B \approx 220 \Omega }[/math] and [math]\displaystyle{ I_C \approx 100 \mu A }[/math]

b. [math]\displaystyle{ I_C > 0.5 }[/math] mA DC with no input signal

c. [math]\displaystyle{ V_{CE} \approx V_{CC}/2 > 2 }[/math] V

d. [math]\displaystyle{ V_{CC} < V_{CE}(max) }[/math] to prevent burnout

e. [math]\displaystyle{ V_{BE} \approx 0.6 V }[/math]

f. [math]\displaystyle{ I_D \approx 10 I_B < 1 }[/math] mA

Let's [math]\displaystyle{ V_{CC} = 11\ V }[/math], [math]\displaystyle{ R_E = 0.2\ k\Omega }[/math] and [math]\displaystyle{ R_C = 2.0\ k\Omega }[/math].

The load line equation becomes:

[math]\displaystyle{ I_C = \frac{V_{CC}-V_{CE}}{R_E+R_C} = \frac{(11 - V_{CE})\ V}{2.2\ k\Omega}  }[/math]

Draw a load line using the [math]\displaystyle{ I_{C} }[/math] -vs- [math]\displaystyle{ I_{EC} }[/math] from the previous lab 13. Record the value of [math]\displaystyle{ h_{FE} }[/math] or [math]\displaystyle{ \beta }[/math].

On the plot below I overlay the output transistor lines (from the previous lab report #13) and the Load Line calculated above.

My [math]\displaystyle{ \beta \approx 170 }[/math] based on my previous lab report #13

Set a DC operating point [math]\displaystyle{ I^{\prime}_C }[/math] so it will amplify the input pulse given to you. Some of you will have sinusoidal pulses others will have positive or negative only pulses.

I will set up my operating point in the middle of the load line: [math]\displaystyle{ I_C = 5\ mA }[/math], [math]\displaystyle{ V_EC = 10\ mA }[/math].

Let's calculate all bias voltage to set up this operating point.

[math]\displaystyle{ V_E = I_E \cdot R_E = 5\ mA \cdot 0.2\ k\Omega = 1\ V }[/math]
[math]\displaystyle{ V_{BE} = 0.6\ V }[/math]
[math]\displaystyle{ V_{B} = V_E + V_B = (1.0 + 0.6)\ V = 1.6\ V }[/math]

To set up the operating point above we need to set up [math]\displaystyle{ V_{B} = 1.6\ V }[/math].

We have:

[math]\displaystyle{ I_B = \frac{I_C}{\beta} = \frac{5\ mA}{200} = 25\ uA }[/math].

Here are have used [math]\displaystyle{ \beta = 200 }[/math] instead of [math]\displaystyle{ \beta = 150 }[/math] because it's my actual values here.

To get operating point independent of the transistor base current we want [math]\displaystyle{ I_{R1} \gg\ I_B }[/math]

Let's [math]\displaystyle{ I_{R1} = 800\ uA \gg\ I_B = 25\ uA }[/math]

So

[math]\displaystyle{ R_1 = \frac{V_B}{I_1} = \frac{1.6\ V}{800\ uA} = 2\ k\Omega }[/math]

And [math]\displaystyle{ R_2 }[/math] we can find from Kirchhoff voltage Low:

[math]\displaystyle{ V_{CC} = I_2 \cdot R_2 + V_B }[/math]. Here [math]\displaystyle{ I_2 \approx I_1 }[/math]

So

 [math]\displaystyle{ R_2 = \frac{V_{CC}-V_B}{I_1} = \frac{(20-1.6)\ V}{0.8\ mA} = 23\ k\Omega }[/math]

Measure all DC voltages in the circuit and compare with the predicted values.(10 pnts)

My predicted DC voltage are:

We have from operating point:

[math]\displaystyle{ V_{EC} = 10\ V }[/math] 
[math]\displaystyle{ I_C \approx I_E = 5\ mA }[/math]

And because we have silicon transistor:

[math]\displaystyle{ V_{BE} = 0.6\ V  }[/math]

Now

[math]\displaystyle{ V_E = I_E \cdot R_E = 5\ mA \cdot 200\ \Omega = 1\ V }[/math]
[math]\displaystyle{ V_C = V_E + V_{EC} = (1 + 11)\ V = 12\ V }[/math]
[math]\displaystyle{ V_B = V_E + V_{BE} = (1 + 0.6)\ V = 1.6\ V }[/math]

My measured DC voltage are:

[math]\displaystyle{ V_{EC} = (11.25 \pm 0.01)\ V }[/math]
[math]\displaystyle{ V_{BE} = (0.69 \pm 0.01)\ V  }[/math]

[math]\displaystyle{ V_E = (0.88 \pm 0.01)\ V }[/math]
[math]\displaystyle{ V_C = (12.11 \pm 0.01)\ V }[/math]
[math]\displaystyle{ V_B = (1.56 \pm 0.01)\ V }[/math]

First note that my [math]\displaystyle{ V_B = (1.56 \pm 0.01)\ V }[/math] is close to predicted values [math]\displaystyle{ V_B = 1.6\ V }[/math].

Also note that my [math]\displaystyle{ V_{EC} = (11.25 \pm 0.01)\ V }[/math] is a little higher than my initial operating point [math]\displaystyle{ V_{EC} = 10\ V }[/math]. The main reason is that my actual values of [math]\displaystyle{ V_{BE} = (0.69 \pm 0.01)\ V }[/math] instead of [math]\displaystyle{ V_{BE} = 0.6\ V }[/math] as I was assumed initially. That will reduce my [math]\displaystyle{ V_E }[/math] voltage that reduce my [math]\displaystyle{ I_C }[/math] current. So I will shift to right my operating point from [math]\displaystyle{ I_C = 5\ mA }[/math] to lower values and correspondingly will increase my [math]\displaystyle{ V_{EC} }[/math].

By direct measurements my operating point now is [math]\displaystyle{ I_C = (4.34 \pm 0.01)\ mA }[/math] and [math]\displaystyle{ V_{EC} = (11.25 \pm 0.01)\ V }[/math]

Let's check do my operating point is still on my load line. I have [math]\displaystyle{ R_E = (0.20 \pm 0.01)\ k\Omega }[/math], [math]\displaystyle{ R_C = (1.80 \pm 0.01)\ k\Omega }[/math]. So from load line equation:

[math]\displaystyle{ I_C = \frac{V_{CC}-V_{CE}}{R_E+R_C} = \frac{((20.01\pm 0.01)-(11.25\pm 0.01))\ V}{((1.80 \pm 0.01))+ (0.20 \pm 0.01))\ k\Omega} = (4.38 \pm 0.04)\ mA }[/math].

So my operating point lies in my load line within experimental error.

Measure [math]\displaystyle{ R_{in} }[/math] and [math]\displaystyle{ R_{out} }[/math] at about 1 kHz and compare to the theoretical value.(10 pnts)

How do you do this? Add resistor in front of [math]\displaystyle{ C_1 }[/math] which you vary to determine [math]\displaystyle{ R_{in} }[/math] and then do a similar thing for [math]\displaystyle{ R_{out} }[/math] except the variable reistor goes from [math]\displaystyle{ C_2 }[/math] to ground.

Questions

1. Why does a flat load line produce a high voltage gain and a steep load line a high current gain? (10 pnts)
2. What would be a good operating point an an [math]\displaystyle{ npn }[/math] common emitter amplifier used to amplify negative pulses?(10 pnts)
3. What will the values of [math]\displaystyle{ V_C }[/math], [math]\displaystyle{ V_E }[/math] , and [math]\displaystyle{ I_C }[/math] be if the transistor burns out resulting in infinite resistance. Check with measurement.(10 pnts)
4. What will the values of [math]\displaystyle{ V_C }[/math], [math]\displaystyle{ V_E }[/math] , and [math]\displaystyle{ I_C }[/math] be if the transistor burns out resulting in near ZERO resistance (ie short). Check with measurement.(10 pnts)
5. Predict the change in the value of [math]\displaystyle{ R_{in} }[/math] if [math]\displaystyle{ I_D }[/math] is increased from 10 [math]\displaystyle{ I_B }[/math] to 50 [math]\displaystyle{ I_B }[/math](10 pnts)
6. Sketch the AC equivalent circuit of the common emitter amplifier.(10 pnts)

Go Back to All Lab Reports Forest_Electronic_Instrumentation_and_Measurement