Lab 9 RS

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Lab 9: Diode Circuits

Clipping Circuit

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

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 [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)

600 px

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} = 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.

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_1}{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_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 [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) = 15\ k\Omega[/math] and [math]C_{out} = 9.65\ nF\ \Rightarrow RC = 0.15\ ms[/math].

And choosing the the pulse width [math]\tau \approx 1.5\ ms \gg 0.15\ 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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