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=20 V, RE=200Ω and RC=1.8 kΩ. The load line equation becomes:

[math]I_C = \frac{V_{CC}}{R_E+R_C} + \frac{V_{EC}}{R_E+R_C} = \frac{20\ V}{2\ k\Omega} + \frac{V_{CE}}{2\ k\Omega} = (10 + 0.5 \cdot V_{CE})\ mA[/math]

This load line pretty high and give me the wide range of amplification the input signal.


Draw a load line using the IC -vs- IEC from the previous lab 13. Record the value of hFE or β.

Load Line 10mA.png


On the plot above I overlay me output lines from the previous lab report #13 and me load line I am going to use to construct amplifier. My β150 based on my previous lab report #13


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: IC=5 mA, VEC=10 mA. So I will have the wide range of amplification the input signal IB=±5 mA

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

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

So to get this operating point I need to set up VB=1.6 V. I can do it using voltage divider R1/R2. To get operating point independent of the transistor base current we want IR1 IB

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

Here are have used β=200 instead of β=150 as in my previous lab because it's my actual values of β here. I have measured this β using millivoltmeter.

Let's IR1=800 uA IB=25 uA

So

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

And R2 we can find using the voltage divider formula

[math]1.6\ V = \frac{R_1}{R_1+R_2} \cdot 20\ V [/math]

So

 [math]R_2 = \frac{20-1.6}{1.6} \cdot R_2 = 23\ k\Omega[/math]

Let's check it:

[math]20\ V = \frac{2\ k\Omega}{23+2 k\Omega}\cdot 20\ V = 1.6\ V[/math] as should be.


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]V_CE = 10\ V\lt math\gt  
 \lt math\gt I_C \approx I_E = 5\ mA[/math]

And because we have silicon transistor:

[math]V_BE = 0.6\ V\lt math\gt 

Now

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

Measure the voltage gain Av as a function of frequency and compare to the theoretical value.(10 pnts)

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.

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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