Lab 9 RS

Go Back to All Lab Reports

Lab 9: Diode Circuits

Clipping Circuit

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

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

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

4.)Do another sketch for[math]\displaystyle{ 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.

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

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

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

Also because we want [math]\displaystyle{ V_{out} = 3\ V }[/math] and using [math]\displaystyle{ V_{out} = V_{in} \frac{R_2}{R_1+R_2} }[/math]. Solving this equation we get the second condition [math]\displaystyle{ R_1 }[/math] and [math]\displaystyle{ R_2 }[/math]

 [math]\displaystyle{ 2)\ \ R_2 = 1.5\ R_1 }[/math]

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

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

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

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

4) plot [math]\displaystyle{ V_{in} }[/math] and [math]\displaystyle{ 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]\displaystyle{ V_{out} }[/math] from rising above +5 V. Sketch the new circuit below.

6) plot [math]\displaystyle{ V_{in} }[/math] and [math]\displaystyle{ 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