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.

TF EIM Lab14a.png

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

a. Try RB220Ω and IC100μA
b. IC>0.5 mA DC with no input signal
c. VCEVCC/2>2 V
d. VCC<VCE(max) to prevent burnout
e. VBE0.6V
f. ID10IB<1 mA


Let's VCC=11 V, RE=0.2 kΩ and RC=2.0 kΩ.

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 IC -vs- IEC from the previous lab 13. Record the value of hFE or β.

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


Load Line 5mA.png


My reported values of β in lab report #13 was β150. But this because we have used the approximation VE0 because small resistor value RE=100 Ω was used.

If we will do the accurate calculation of β 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 β:

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

Set a DC operating point IC 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 VBE and β 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 VB=1.18 V.

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 IR1 IB

Let's IR1=590 uA IB=14.4 uA

So

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


And R2 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 Vcc and I1 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 VB=VE+VBE, VC=VE+VEC and Vcc=VB+VR2, Vcc=VC+VRC


Also all my predicted values are in agreement with my measured DC voltage values except of the fact that my measured VBE=0.683 V instead of VBE=0.680 V as I initially assumed. That gives me the correspondent corrections to VB. 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 Av as a function of frequency and compare to the theoretical value.(10 pnts)

Table gain.png


Measure Rin and Rout at about 1 kHz and compare to the theoretical value.(10 pnts)

How do you do this? Add resistor in front of C1 which you vary to determine Rin and then do a similar thing for Rout except the variable reistor goes from C2 to ground.


Input impedance Rin measurements

To measure the input impedance of my amplifier I have set up the circulant below. Here I have attached the resistor R1 in-front of capacitor C1 and I replaced all my input internal resistor of amplifier by equivalent Rinp.


Inputresistor.png


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 my table with Rinp calculation. I did that calculation for several values of RL

Rinp table.png

We can see that the input impedance of amplifier have small dependence from RL because we change a little the base current.

Measure Av and Rin as a function of frequency with CE removed.(10 pnts)

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 npn common emitter amplifier used to amplify negative pulses?(10 pnts)
  3. What will the values of VC, VE , and IC be if the transistor burns out resulting in infinite resistance. Check with measurement.(10 pnts)
  4. What will the values of VC, VE , and IC 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 Rin if ID is increased from 10 IB to 50 IB(10 pnts)
  6. Sketch the AC equivalent circuit of the common emitter amplifier.(10 pnts)




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