Difference between revisions of "Lab 9 RS"

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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.
 
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.
  
[[File:Clipping 1.png | 600 px]]
+
[[File:Tek00034.png | 600 px]]
  
 
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)
 
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)
  
[[File:Clipping 2.png | 600 px]]
+
[[FFile:Tek00035.png | 600 px]]
  
  
[[File:Clipping 3.png | 600 px]]
+
[[File:Tek00036.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=\frac{V_{out}}{R}=\frac{6\ V}{100 \Omega} = 60\ mA</math>.
+
For the last sketch the output voltage is <math>V_{out} = 8.6\ V</math>. Let's estimate the power dissipated in resistor and diode. The current can be calculated by <math>I=\frac{V_{out}}{R}=\frac{8.6\ V}{10.3 k\Omega} = 0.83\ mA</math>.
  
  The resistor power is given by <math>P_R=I\cdot V_{out} = 60\ mA \cdot 6\ V = 0.36\ W</math>. So we are OK here.
+
  The resistor power is given by <math>P_R=I\cdot V_{out} = 0.83\ mA \cdot 8.6\ V = 0.007\ 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.
+
  The diode power is given by <math>P_R=I\cdot V_{diode} = 0.83\ mA \cdot (10 - 8.6)\ V = 0.001\ W</math>. So we are OK here as well. No any smoke out.
  
 
= Differentiating Circuit with clipping=
 
= Differentiating Circuit with clipping=

Revision as of 21:59, 24 February 2011

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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 = 10.3 \Omega[/math] 


2.) Use a sine wave generator to drive the circuit so Vin=V0cos(2πνt) where V0=0.1 V and ν = 1kHz. (20 pnts)

3.) Based on your observations using a oscilloscope, sketch the voltages Vin and Vout as a function of time.

Tek00034.png

4.) Do another sketch for V0 = 1.0 V and another for 10.0 V (DONT LET ANY SMOKE OUT!). (20 pnts)

600 px


Tek00036.png


For the last sketch the output voltage is Vout=8.6 V. Let's estimate the power dissipated in resistor and diode. The current can be calculated by I=VoutR=8.6 V10.3kΩ=0.83 mA.

The resistor power is given by [math]P_R=I\cdot V_{out} = 0.83\ mA \cdot 8.6\ V = 0.007\ W[/math]. So we are OK here.
The diode power is given by [math]P_R=I\cdot V_{diode} = 0.83\ mA \cdot (10 - 8.6)\ V = 0.001\ 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 R1 and R2 such that the current from the +5V DC source is less than 1.0 mA and the DC voltage at Vout is 3 V when there is no input pulse.


Because we want to keep the current below 1 mA and using I=VR1+R21 mA. Solving this inequality we get the first condition for R1 and R2

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


Also because we want Vout=3 V and using Vout=VinR1R1+R2. Without any input pulse Vin=5 V. Solving this simple equation we get the second condition for R1 and R2

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


I am going to use [math]R_1 = 10\ k\Omega[/math] and [math]R_2 = 15\ k\Omega[/math] which satisfy both conditions above



3) Select a capacitor (C) and a pulse width τ to form a differentiating circuit for the pulse from the signal generator. Hint: R12Cτ.


Taking (R1+R2)=15 kΩ and Cout=9.65 nF RC=0.15 ms.

And choosing the the pulse width τ1.5 ms0.15 ms I will be able to make a good differentiator circuit.


4) plot Vin and Vout as a function of time using your scope observations. (20 pnts)


5) Now add the diode circuit from part 1 to prevent Vout from rising above +5 V. Sketch the new circuit below.


6) plot Vin and Vout 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)


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