Difference between revisions of "Lab 9 RS"

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=Clipping Circuit=
 
=Clipping Circuit=
1.) Construct the circuit shown below using a silicon diode.
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1.) \large Construct the circuit shown below using a silicon diode.
  
 
[[File:TF_EIM_Lab9.png | 400 px]]
 
[[File:TF_EIM_Lab9.png | 400 px]]

Revision as of 04:38, 23 February 2011

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Lab 9: Diode Circuits

Clipping Circuit

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

TF EIM Lab9.png

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.

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)

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)


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