Difference between revisions of "Lab 14 RS"
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So to get this operating point I need to set up <math>V_{B} = 1.6\ V</math>. I can do it using voltage divider <math>R_1/R_2</math>.  | So to get this operating point I need to set up <math>V_{B} = 1.6\ V</math>. I can do it using voltage divider <math>R_1/R_2</math>.  | ||
| − | To get operating point independent of the transistor base current we want <math>I_{R1} \gg \I_B</math>  | + | To get operating point independent of the transistor base current we want <math>I_{R1} \gg\ I_B</math>  | 
  <math>I_B = \frac{I_C}{\beta} = \frac{5\ mA}{200} = 25\ uA</math>. Here are have used <math>\beta = 200</math> instead of <math>\beta = 150</math> as in my previous lab because it's my actual values of <math>\beta</math> here. I have measured this <math>\beta</math> using millivoltmeter.  |   <math>I_B = \frac{I_C}{\beta} = \frac{5\ mA}{200} = 25\ uA</math>. Here are have used <math>\beta = 200</math> instead of <math>\beta = 150</math> as in my previous lab because it's my actual values of <math>\beta</math> here. I have measured this <math>\beta</math> using millivoltmeter.  | ||
| + | Let's  <math>I_{R1} = 800\ uA \gg\ I_B = 25\ uA</math>  | ||
| + | So  | ||
| + |  <math>R_1 = \frac{V_B}{I_1} = \frac{1.6\ V}{800\ uA} = 2\ k\Omega</math>  | ||
| + | And <math>R_2</math> 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.  | ||
Revision as of 04:58, 20 March 2011
The Common Emitter Amplifier
Circuit
1.)Construct the common emitter amplifier circuit below according to your type of emitter.
2.)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 ,  and  . The load line equation becomes:
This load line pretty high and give me the wide range of amplification the input signal.
3.)Draw a load line using the -vs- from the previous lab 13. Record the value of or .
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  based on my previous lab report #13
4.)Set a DC operating point  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: , . So I will have the wide range of amplification the input signal 
Let's calculate all bias voltage to set up this operating point.
So to get this operating point I need to set up . I can do it using voltage divider . To get operating point independent of the transistor base current we want
. Here are have used instead of as in my previous lab because it's my actual values of here. I have measured this using millivoltmeter.
Let's
So
And we can find using the voltage divider formula
So
 
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.
5.)Measure all DC voltages in the circuit and compare with the predicted values.(10 pnts)
6.)Measure the voltage gain as a function of frequency and compare to the theoretical value.(10 pnts)
7.)Measure and at about 1 kHz and compare to the theoretical value.(10 pnts)
How do you do this? Add resistor in front of which you vary to determine and then do a similar thing for except the variable reistor goes from to ground.
8.)Measure and as a function of frequency with 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 common emitter amplifier used to amplify negative pulses?(10 pnts)
 - What will the values of , , and be if the transistor burns out resulting in infinite resistance. Check with measurement.(10 pnts)
 - What will the values of , , and 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 if is increased from 10 to 50 (10 pnts)
 - Sketch the AC equivalent circuit of the common emitter amplifier.(10 pnts)
 
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