Difference between revisions of "Lab 9 RS"

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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>
+
Also because we want <math>V_{in} = 5\ V</math> and <math>V_{out} = 3\ V</math> without any input pulse and using <math>V_{out} = V_{in}\cdot\frac{R_2}{R_1+R_2}</math>. Solving this simple equation we get the second condition for <math>R_1</math> and <math>R_2</math>
  
 
   <math>2)\ \ R_1 = 1.5\ R_2</math>
 
   <math>2)\ \ R_1 = 1.5\ R_2</math>

Revision as of 19:29, 25 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)


Below my screen save for V0 = 1.0 V

Tek00035.png


And below my screen save for V0 = 10.0 V

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 V10.3 kΩ=0.83 mA.

The resistor power is given by [math]P_R=I\cdot V_{out} = 0.83\ mA \cdot 8.6\ V = 7.14\ mW[/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 = 1.16\ mW[/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 Vin=5 V and Vout=3 V without any input pulse and using Vout=VinR2R1+R2. 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=10.15 kΩ and R2=14.90 kΩR12=R1R2(R1+R2)=6.04 kΩ. Also taking C=9.65 nF. Now I can calculate the time constant of my RC circuit as R12C=58.3 us.


By selecting the pulse width [math]\tau \approx 400.0\ ms \gg 58.3\ us[/math] 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)


Below the "screen save" of Vin and Vout as a function of time for the case Vin=1 Volts

Tek00040.png


Now I changed the input voltage to Vin=3 Volts. As we can see from the plot below the output signal does change as well.

Tek00041.png

So there is no any clipping off the output signal for the circuit above.


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

Cuircuit 2.png


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)


Below the "screen save" of Vin and Vout as a function of time for the case Vin=1 Volts

Tek00042.png


Now I changed the input voltage to Vin=3 Volts. As we can see from the plot below the output signal doesn't change at all.

Tek00043.png


Now my diode is clipping off the positive signal at about +5 V and is clipping off the negative signal at about -1 V.
And the output signal doesn't change when  we  change the amplitude of input signal.

Questions

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


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