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)

Below my screen save for [math]V_0 [/math] = 1.0 V

And below my screen save for [math]V_0 [/math] = 10.0 V

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 = 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.

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_{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]

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=10.15\ k\Omega[/math] and [math]R_2=14.90\ k\Omega \Rightarrow R_{12}=\frac{R_1 R_2}{(R_1+R_2)} = 6.04\ k\Omega[/math]. Also taking [math]C = 9.65\ nF[/math]. Now I can calculate the time constant of my [math]RC[/math] circuit as [math]R_{12}C = 58.3\ us[/math].

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 [math]V_{in}[/math] and [math]V_{out}[/math] as a function of time using your scope observations. (20 pnts)

Below the "screen save" of [math]V_{in}[/math] and [math]V_{out}[/math] as a function of time for the case [math]V_{in} = 1\ Volts[/math]

Now I changed the input voltage to [math]V_{in} = 3\ Volts[/math]. As we can see from the plot below the output signal does change as well.

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 [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)

Below the "screen save" of [math]V_{in}[/math] and [math]V_{out}[/math] as a function of time for the case [math]V_{in} = 1\ Volts[/math]

Now I changed the input voltage to [math]V_{in} = 3\ Volts[/math]. As we can see from the plot below the output signal doesn't change at all.

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)

All explanation is based on current voltage diagram for diode used in this laboratory work. I used the ZENER silicon diode type # 1N5230B-T. The diode current vs. diode voltage plot I measured in previous lab (see my pictures below):

The critical parameters here is reverse turn on voltage and forward turn on voltage. Reverse turn on voltage for this type of diode equal about 4.7 Volts and forward turn on voltage about 0.9 Volts. When the voltage drop on diode is beyond this two points the diode becomes good conductor with internal resistance about 4-6 [math]\Omega[/math]. On the other side when voltage below 0.9 Volts in forward direction or below 4.7 Volts in reverse direction the diode becomes good isolator with current less than 1 mA through the diode. All explanation of the clipping current can be done by using the voltage current diagram above and by using reverse and forward turn on voltage points.

1) Let see what happens for the first clipping circuit:

When the input voltage is about 0.1 Volts the diode is below the reverse and forward turn on voltage points. The diode is working as a good isolator and there are no any signal at output because all voltage drop is on the diode.

When the input voltage is about 1 Volts we are above the forward turn on voltage. So the diode becomes conduct in forward direction when the voltage drop on diode exceed 0.9 Volts. On the other side we are still bellow 4.7 Volts on the negative side of input signal. So the diode cut off the all negative signal.

When the input voltage is about 10 Volts we are above the forward turn on voltage and we are above the reverse turn on voltage as well. So diode becomes to conduct as soon as the voltage exceed 0.9 Volts in forward direction and as soon as the voltage exceed the 4.7 Volts in reverse direction

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