Difference between revisions of "Lab 9 RS"

From New IAC Wiki
Jump to navigation Jump to search
Line 8: Line 8:
  
 
[[File:TF_EIM_Lab9.png | 400 px]]
 
[[File:TF_EIM_Lab9.png | 400 px]]
 +
 +
 +
I am going to use 1) Zener diode 4.7 V 1N5230B-T 2) the resistor <math>R = 100.5 \Omega</math>
 +
 +
  
 
2.) Use a  sine wave generator to drive the circuit so <math>V_{in} = V_0 \cos(2 \pi \nu t)</math> where <math>V_0 = 0.1</math> V and <math> \nu</math>  = 1kHz. (20 pnts)
 
2.) Use a  sine wave generator to drive the circuit so <math>V_{in} = V_0 \cos(2 \pi \nu t)</math> where <math>V_0 = 0.1</math> V and <math> \nu</math>  = 1kHz. (20 pnts)
Line 21: Line 26:
  
 
[[File:Clipping 3.png | 600 px]]
 
[[File:Clipping 3.png | 600 px]]
 +
 +
 +
For the last sketch the output voltage is <math>V_{out} = 6\ V</math>. Let's estimate the power dissipated in resistor and diode. The current can be calculated by <math>I=\farc{V_{out}}{R}=\frac{6\ V}{100\ Omega} = 60\ mA</math>.
 +
 +
The resistor power is given by <math>P_R=I\cdot V_{out} = 60\ mA \cdot 6\ V = 0.63\ W</math>. So we are OK here.
 +
 +
The diode power is given by <math>P_R=I\cdot V_{diode} = 60\ mA \cdot (8 - 6)\ V = 0.12\ W</math>. So we are OK here as well. No any smoke out.
  
 
= Differentiating Circuit with clipping=
 
= Differentiating Circuit with clipping=

Revision as of 04:59, 24 February 2011

Go Back to All Lab Reports


Lab 9: Diode Circuits

Clipping Circuit

1.) Construct the circuit shown below using a silicon diode.

TF EIM Lab9.png


I am going to use 1) Zener diode 4.7 V 1N5230B-T 2) the resistor [math]R = 100.5 \Omega[/math]


2.) Use a sine wave generator to drive the circuit so [math]V_{in} = V_0 \cos(2 \pi \nu t)[/math] where [math]V_0 = 0.1[/math] V and [math] \nu[/math] = 1kHz. (20 pnts)

3.) Based on your observations using a oscilloscope, sketch the voltages [math]V_{in}[/math] and [math]V_{out}[/math] as a function of time.

Clipping 1.png

4.) Do another sketch for [math]V_0 [/math] = 1.0 V and another for 10.0 V (DONT LET ANY SMOKE OUT!). (20 pnts)

Clipping 2.png


Clipping 3.png


For the last sketch the output voltage is [math]V_{out} = 6\ V[/math]. Let's estimate the power dissipated in resistor and diode. The current can be calculated by [math]I=\farc{V_{out}}{R}=\frac{6\ V}{100\ Omega} = 60\ mA[/math].

The resistor power is given by [math]P_R=I\cdot V_{out} = 60\ mA \cdot 6\ V = 0.63\ W[/math]. So we are OK here.

The diode power is given by [math]P_R=I\cdot V_{diode} = 60\ mA \cdot (8 - 6)\ V = 0.12\ W[/math]. So we are OK here as well. No any smoke out.

Differentiating Circuit with clipping

1) Construct the circuit below.

TF EIM Lab9a.png


2) Select [math]R_1[/math] and [math]R_2[/math] such that the current from the +5V DC source is less than 1.0 mA and the DC voltage at [math]V_{out}[/math] is 3 V when there is no input pulse.


Because we want to keep the current below [math]1\ mA[/math] and using [math]I = \frac{V}{R_1+R_2} \leq 1\ mA[/math]. Solving this inequality we get the first condition for [math]R_1[/math] and [math]R_2[/math]

 [math]1)\ \ R_1+R_2 \geq \frac{5\ V}{1\ mA} = 5\ k\Omega[/math]


Also because we want [math]V_{out} = 3\ V[/math] and using [math]V_{out} = V_{in}\cdot\frac{R_2}{R_1+R_2}[/math]. Without any input pulse [math]V_{in} = 5\ V[/math]. Solving this simple equation we get the second condition for [math]R_1[/math] and [math]R_2[/math]

 [math]2)\ \ R_2 = 1.5\ R_1[/math]


I am going to use [math]R_1 = 2\ k\Omega[/math] and [math]R_2 = 3\ k\Omega[/math] which satisfy both conditions above



3) Select a capacitor [math](C)[/math] and a pulse width [math]\tau[/math] to form a differentiating circuit for the pulse from the signal generator. Hint: [math]R_{12}C \ll \tau[/math].


Taking [math](R_1 + R_2) = 5\ k\Omega[/math] and [math]C_{out} = 1.255\ \mu F\ \Rightarrow RC = 6.28\ ms[/math].

And choosing the the pulse width [math]\tau \approx 80\ ms \gg 6.28\ ms[/math] I will be able to make a good differentiator circuit.


4) plot [math]V_{in}[/math] and [math]V_{out}[/math] as a function of time using your scope observations. (20 pnts)


5) Now add the diode circuit from part 1 to prevent [math]V_{out}[/math] from rising above +5 V. Sketch the new circuit below.


6) plot [math]V_{in}[/math] and [math]V_{out}[/math] as a function of time with the diode circuit you added using your scope observations. (the diode should clip off positive spikes) (20 pnts)

Questions

  1. Explain your results in parts 1 & 2 in terms of the diode turn-on voltage. (20 pnts)


Forest_Electronic_Instrumentation_and_Measurement Go Back to All Lab Reports