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 and

b. mA DC with no input signal

c. V

d. to prevent burnout

e.

f. mA

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

The load line equation becomes:

[math]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]I_{C}[/math] -vs- [math]I_{EC}[/math] from the previous lab 13. Record the value of [math]h_{FE}[/math] or [math]\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]\beta \approx 170[/math] based on my previous lab report #13

Set a DC operating point [math]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]I_C = 2.5\ mA[/math], [math]V_{EC} = 5.5\ V[/math].

Let's calculate all bias voltage needed to set up this operating point. Because the knowing of [math]V_{BE}[/math] and [math]\beta[/math] is very important for this calculation I did the preliminary set up to measure this quantities. They are the only parameters which depends from transistor. I was able to find:

[math]V_{BE} = 0.68\ V[/math]

[math]\beta = 173[/math]

Now

[math]V_E = I_E \cdot R_E = 2.5\ mA \cdot 0.2\ k\Omega = 0.5\ V[/math]
[math]V_B = V_E + V_B = (0.50 + 0.68)\ V = 1.18\ V[/math]

To set up the operating point above we need to set up [math]V_{B} = 1.18\ V[/math].

We have:

[math]I_B = \frac{I_C}{\beta} = \frac{2.5\ mA}{173} = 14.4\ uA[/math].

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

Let's

So

[math]R_1 = \frac{V_B}{I_1} = \frac{1.18\ V}{590\ uA} = 2\ k\Omega[/math]

And [math]R_2[/math] we can find from Kirchhoff Voltage Low:

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

and Kirchhoff Current Low:

[math]I_2 = I_1 + I_B[/math]

So

 [math]R_2 = \frac{V_{CC}-V_B}{I_1+I_B} = \frac{(11-1.18)\ V}{(590 + 14.4)\ mA} = 16.25\ k\Omega[/math]

I tried to adjust my calculation by varying the fee parameters [math]V_{cc}[/math] and [math]I_1[/math] to get good values for resistors I can easily to set up.

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

My predicted DC voltages are: (from the calculation above):

[math]V_{EC} = 5.50\ V[/math] 
[math]V_{BE} = 0.68\ V [/math]

[math]V_E = 0.50\ V[/math]
[math]V_B = 1.18\ V[/math]
[math]V_C = V_E + V_{EC} = (0.50 + 5.50)\ V = 6.00\ V[/math]

My measured DC voltages are:

Here is very important to set up all resistor values as close as possible to my assumed values.

After many tries and errors I was able to end up with the following values of my resistors:

[math]R_E = (200.0 \pm 0.1)\ \Omega[/math]
[math]R_C = (2.002 \pm 0.001)\ k\Omega[/math]
[math]R_1 = (2.004 \pm 0.001)\ k\Omega[/math]
[math]R_2 = (16.26 \pm 0.01)\ k\Omega[/math]

And my measurements of DC voltages looks like:

[math]V_{cc} = (11.00 \pm 0.01)\ V[/math]

[math]V_E = (0.500 \pm 0.001)\ V[/math]
[math]V_B = (1.183 \pm 0.001)\ V [/math]
[math]V_C = (6.03 \pm 0.01)\ V [/math]

[math]V_{BE} = (0.683 \pm 0.001)\ V[/math]
[math]V_{EC} = (5.53 \pm 0.01)\ V[/math]

[math]V_{R_2} = (9.82 \pm 0.01)\ V[/math]
[math]V_{R_C} = (4.97 \pm 0.01)\ V[/math]

All my measurements are in agreement with each other within experimental errors.

I mean here that [math]V_B = V_E+V_{BE}[/math], [math]V_C = V_E+V_{EC}[/math] and [math]V_{cc} = V_B+V_{R_2}[/math], [math]V_{cc} = V_C+V_{R_C}[/math]

Also all my predicted values are in agreement with my measured DC voltage values except of the fact that my measured [math]V_{BE} = 0.683\ V[/math] instead of [math]V_{BE} = 0.680\ V[/math] as I initially assumed. That gives me the correspond corrections to [math]V_B[/math]. But if I will consider only the one significant sign all my predicted and measured DC voltage values are in total agreement.

Below are my current measurements which I did using millivoltmeter and which are also in agreement with each other and with all my previous calculation:

[math]I_E = (2.48 \pm 0.01)\ mA[/math] 
[math]I_C = (2.49 \pm 0.01)\ mA[/math] 
[math]I_B = (14.4 \pm 0.1)\ uA[/math]

[math]I_1 = (588 \pm 1)\ uA[/math] 
[math]I_1 = (603 \pm 1)\ uA[/math]

Measure the voltage gain [math]A_v[/math] as a function of frequency and compare to the theoretical value.(10 pnts)

Measure [math]R_{in}[/math] and [math]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]C_1[/math] which you vary to determine [math]R_{in}[/math] and then do a similar thing for [math]R_{out}[/math] except the variable reistor goes from [math]C_2[/math] to ground.

Measure [math]A_v[/math] and [math]R_{in}[/math] as a function of frequency with [math]C_E[/math] removed.(10 pnts)

Questions

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

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Lab 14 RS

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