When applying voltages across resistors, remember that \(P=V^2/R\).
Design and build a circuit with a resistor, capacitor, and inductor in series with the wave generator, such that the Output taken across the resistor has
Study the attenuation and phase vs frequency characteristics of the circuit.
Include in your report:
R-1) A Falstad simulator circuit simulation screenshot or other circuit diagram. Be sure the component values are all clearly visible.
R-2) A photo of the actual circuit.
R-3) A Bode plot of the attenuation/Magnitude and phase of the Output signal, over a frequency range from 10Hz to 20MHz, with the output taken across the resistor.
For the Bode plot,use the scope’s Frequency Reponse Analysis (FRA) function, either directly or remotely via USB control.
R-4) Is the spectrum consistent with expectations from the simulation? If not, modify your Falstad simulation so that it roughly agrees with what you see. Include a screen shot of the modified circuit simulation.
R-5) Include in your report a Bode plot showing the attenuation/Magnitude and phase of the Output signal, over a frequency range from 10Hz to 20MHz, with the output taken across the inductor+capacitor.
Diodes are not linear devices. The current (I) through them is not simply proportional to voltage (V), and they preferentially allow current to flow in the forward direction, but not in the backward direction.
The effective reverse resistance is not infinite, but we will not try to measure it since this is difficult with the breadboard set-up.
The simplest model for the forward current of a real p-n junction diode is that it grows roughly exponentially
\[i = i_s\left(e^\frac{eV}{k_BT} - 1\right)\]
where \(e\) is the electron charge, \(V\) is the voltage across the diode, \(k_B\) is the Boltzmann constant, \(T\) is the diode temperature, \(i_s\) is the reverse bias saturation current.
R-6) Is the I-V curve of the blue LED in the forward direction roughly consistent the above exponential model?
A quick but less precise why to indirectly see the I-V characteristics of a diode is put the diode in series with a resistor, apply a ramp voltage, and use your oscilloscope in X-Y mode.
Channel 2 shows the voltage across the resistor is proportional to the current flowing through the circuit, since \(I=V/R\). The voltage across the diode can be calculated from \(V_{LED}=V_{in}-V_R\), so the display tells us about the I-V characteristics of the diode. You should see something like this:
\(V_R\) vs \(V_{in}\) for blue LED in series with a 1k resistor. When current is flowing in the forward direction (on the right), the LED has almost zero effective resistance so all the voltage drop is across the resistor and is proportional to the applied voltage. When the current flows in the diodes backward direction, the diode has extremely high effective resistance so almost all the voltage drops across the diode and almost none across the resistor - hence the flat line on the left.
Replace the blue LED with a nominal 3.9V zener diode. At high enough reverse voltage, all diodes eventually break down and conduct, but zener diodes have relatively low and well defined breakdown voltages that can be used to limit and control voltages in a circuit.
R-7) Include a photo of the display for the zener diode and qualitatively explain what you see.
A basic, albeit not very efficient, AC to DC converter can be created by simply inserting a diode to rectifyat the input of an RC Low-Pass filter. The diode rectifies the voltage, allowing the capacitor to charge when current is flowing though the diode in the forward direction, but preventing the capacitor from discharging when the voltage is reversed. As long as the RC time constant is much longer than the AC input frequency, the capacitor will charge to close to the peak AC voltage. A zener diode can be put across the output for a fixed output voltage lower than the peak AC voltage.
R-8) Using a 1N4001 rectifier diode, a 3.9V zener, a capacitor, and whatever resistor(s) you need, design and build an circuit that takes a 20Vpp 60Hz square-wave input and outputs about 3.9V DC across a 1k output resistor, with less than about 5% ripple.
Include in your report a Falstad circuit simulation of your final design, and include a photo of your scope showing the AC input and DC output signals.
See you next week!