Lab 14 RS

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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]R_B \approx 220 \Omega[/math] and [math]I_C \approx 100 \mu A[/math]

b. [math]I_C \gt 0.5[/math] mA DC with no input signal

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

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

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

f. [math]I_D \approx 10 I_B \lt 1[/math] 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 reported values of [math]\beta[/math] in lab report #13 was [math]\beta \approx 150[/math]. But this because we have used the approximation [math]V_E \approx 0[/math] because small resistor value [math]R_E = 100\ \Omega[/math] was used.

If we will do the accurate calculation of [math]\beta[/math] based on my lab report #13 measurements using exact formula:

[math]I_B = \frac{(V_{bb}-V_{BE}-V_E)}{R_B}[/math]

we will end up with value of [math]\beta[/math]:

 [math]\beta \approx 170[/math]

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 above operating point 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 [math]I_{R1} = 590\ uA \gg\ I_B = 14.4\ uA[/math]

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 end up with all my resistor values I can easily 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 correspondent 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.

Input impedance [math]R_{in}[/math] measurements

To measure the input impedance of my amplifier I have set up the circulant below. Here I have attached the resistor [math]R_L[/math] in-front of capacitor [math]C_1[/math] and I replaced all my input internal resistor of amplifier by equivalent resistor [math]R_{inp}[/math]. Here I have draw the capacitor [math]C_1[/math] only for clearness and it does not included in calculation below.

Using Ohm's Law:

[math]I = \frac{V_B-V_A}{R_L}[/math]

and

[math]R_{inp} = \frac{V_B}{I} = R_L\ \frac{V_B}{V_B-V_A}[/math]

Below is the table with my measurements and [math]R_{inp}[/math] calculation. I did that calculation for several values of [math]R_L[/math]

The minimum error for [math]R_{inp}[/math] I have for the case [math]R_L=4030\ \Omega[/math]. So my best estimate for [math]R_{inp}[/math] is:

[math]R_{inp} = (1727.1 \pm 35.7)\ \Omega[/math]

Output impedance [math]R_{out}[/math] measurements

To measure the output impedance of my amplifier I have set up the circulant below. Here I have attached the resistor [math]R_L[/math] after of capacitor [math]C_2[/math] and I replaced all my output internal resistor of amplifier by equivalent resistor [math]R_{out}[/math]. Here I have draw the capacitor [math]C_2[/math] only for clearness and it does not included in calculation below.

Here to find out the output impedance we can use the battery method from lab#1. By Kirchoff voltage law:

[math]V_B=V_A - I\cdot R_{out}[/math]

and

[math]I = \frac{V_B}{R_L}[/math]

So by graphing the current on the x-axis and the measured voltage [math]V_B[/math] on the y-axis for several values of the resistance [math]R_L[/math] we can find the output internal impedance of our amplifier as the slope of the line [math]V_B=V_A - I\cdot R_{out}[/math]

Below is the table with my measurements and current calculation:

And below is my plot of output voltage [math]V_B[/math] as function of current [math]I[/math]:

The line equation is [math]V_B[mV]=p_0[mV] + p_1\cdot I[\mu A] [/math] The slope of this line is [math]p_0=(1.971\pm 0.004)\ \frac{mV}{\mu A} = (1.971\pm 0.004)\ k\Omega[/math].

So my measured output impedance is:

[math] R_{out} = (1.971\pm 0.004)\ k\Omega[/math]

Questions

Why does a flat load line produce a high voltage gain and a steep load line a high current gain? (10 pnts)

If we have a flat load line then by changing a little the input signal (so we will change the [math]I_B[/math] and will move up or down in load line) we will change a lot [math]V_{CE}[/math]. As the result we have high voltage gain.

If we have a steep load line then by changing a little the input signal (so we will change the [math]I_B[/math] and will move up or down in load line) we will change a lot [math]I_C[/math]. As the result we have high current gain.

What would be a good operating point an an [math]npn[/math] common emitter amplifier used to amplify negative pulses?(10 pnts)

The operating point should be as higher as possible in the load line so we will have the freedom of moving down in the load line to amplify the negative pulses. But it still has to be in saturation region of transistor.

Sketch the AC equivalent circuit of the common emitter amplifier.(10 pnts)

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