Relevant Horowitz & Hill: Pages 131-137 Relevant Lawless: Pages 92-131 (Diode circuits start on 111)

Inductors are contained in the same cabinet as the capacitors, in the bottom row of drawers.

In this circuit, the impedance of the parallel capacitor and inductor approaches infinite as the signal's frequency approaches resonance.

The resonant frequency is given by $f_{resonance} = \dfrac{1}{2\pi \sqrt{LC}}$ (see page 114 of the Horowitz and Hill lab book for details, or All About Circuits' Tank circuit page)

Predict what the resonant frequency should be for the circuit you built.

One way we can try to identify resonance is by identifying when the circuit's ** gain** $\left(g = \dfrac{V_{out}}{V_{in}}\right)$ is maximized. While it would theoretically be equal to 1, there are second-order effects that will reduce the gain at resonance fairly significantly here.

**Unless otherwise stated, when testing the frequency behavior of a circuit you should use sine waves for your input signals.**

Experimentally determine the resonant frequency $f_{resonance}$ for this circuit by finding the frequency where $V_{out}$ is maximized

An alternate method is to find the frequency where the input and output are in phase (i.e. there is no shift in time between $V_{in}$ and $V_{out}$ signals

Experimentally determine the resonant frequency $f_{resonance}$ for this circuit by finding the frequency where the input and outputs have the same phase.

In addition to the resonant frequency, bandpass filters are also defined by a quality factor $Q$ . There are several ways to define Q, but for our purposes we'll use $Q \equiv \dfrac{f_{resonance} }{\Delta f}$, where $\Delta f$ characterizes the full-width half-power frequency for the wave. In practice, finding $\Delta f$ means finding the frequencies above/below the resonant frequency where $V_{out}$ is reduced to $V_{out,max}/\sqrt{2} \approx .7V_{out,max}$

The reason we're looking for the voltage do drop by a factor of $\sqrt2$ is because when people were creating terminology for such circuits, they were focused on **power** transfer rather than voltages. Since power is proportional to voltage squared, then when voltage drops by a factor of $\sqrt{2}$, power drops by a factor of 2.

Determine the quality factor $Q$ for your circuit

Modify your circuit by replacing $R$ with a $10k\Omega$ resistor.

Comment on the effect this has on the resonant frequency and $Q$ factor

This isn't a required part of the lab, and you won't be graded on it.

You may have noticed that there is a Fast Fourier Transform (FFT) button between the channel 1 and 2 buttons. If you press this, your scope will show you a Fourier transform of the signal, which breaks the signal down by the amplitudes of its component sine waves.

Your function generator also has a sweep function, that will let you sweep through a range of frequencies over a specified period of time.

If you set your function generator to sweep well below and well above $f_{resonant}$, and then view your signal in FFT mode, you will essentially get a plot of the gain of your circuit as a function of frequency! Note that the vertical scale is logarithmic in FFT mode.

Every periodic signal can be decomposed into a combination of sine waves of various amplitudes and frequencies. Since we've built a sensitive frequency detector, we can use it to determine some of the component frequencies of a square wave.

To do this, start with a square wave at the resonant frequency of our filter and record the amplitude. Then, slowly *decrease* the input wave's frequency until you locate another peak. The first one should be at roughly 1/3 of the resonant frequency, indicating that square waves have a sine wave component whose frequency is three times higher than the base frequency. By continuing to decrease the frequency of the input and recording relative amplitudes where you see resonance, you can reconstruct the Fourier components of a square wave.

Find the first 4 frequencies and amplitudes that make up the Fourier components of a square wave

Yes, many modern oscilloscopes can perform a Fourier transform. Similarly, a digitized signal can be analyzed to find its Fourier components.

Now we'll introduce a new component, the diode! Diodes are directional circuit elements, so the way we orient them a circuit is critical. For typical usages, there can be current from the diode's anode to cathode when the anode's voltage $V_a$ is ~$0.6 V$ higher than the cathode voltage$V_c$. In such a case, the diode's voltage (also referred to as the *“knee voltage”*) will be nearly constant at $0.6 V$ and the current will be determined by the rest of the circuit. If this condition isn't met, then there will be no current through the diode, and the voltage across the diode will be determined by the rest of the circuit.

- If $V_a - V_c \approx 0.6 V$, the diode allows (nearly) any amount of current to pass through it
- This can vary by a few hundred mV in either direction for different special purpose diodes

- Otherwise, the diode will have no current through it

Schematic symbol for a diode. The 'a' and 'c' indicate the anode and cathode, respectively. | Photograph of a 1N914
Diode, Small Signal fast switching diode, which is in your parts kit. Note that the dark band on the packaging indicates the cathode. |

Note that this is the first circuit element we've seen whose behavior depends on voltages in the circuit. This feature allows us to:

- Create digital logic
- Create circuits that can protect other components from excessive voltages
- Reshape signals

We'll explore the last two uses in the following lab exercises.

If you aren't sure which side of a diode is which, you can use your multimeter to determine this:

After getting the settings correct, connect your meter such that the red **V/$\Omega$ /Hz** terminal is connected to the anode and the **COM** terminal is connected to the cathode. On doing so, the meter will display the diode's voltage for a current of 1mA. If your diode is backwards, you'll instead see your meter read **.OL** for overload; in this instance there's no reading because there's no current through the diode.

We can use diodes to restrict the output voltage in a circuit. This is a common way to protect sensitive components. Let's make a rectifier circuit as shown in the figure below:

Using a `1kHz`

sine wave with a `14V pk-pk amplitude (7V amplitude)`

:

Predict what $V_{out}$ and $\Delta V_{diode}$ will look like as a function of $V_{in}$, including a quick picture or sketch. Note that $\Delta V_{diode}$ is the voltage difference across the two terminals of the diode, not a measurement with respect to ground.

Test your prediction, and resolve any discrepancies.

Note that to measure $\Delta V_{diode}$, you'll need to find the difference between $V_{in}$ and $V_{out}$. If you place the scope's grounding probe at $V_{out}$, you'll instead alter your circuit to behave in odd ways. Instead, you should attach the oscilloscope probe connectors to $V_{in}$ and $V_{out}$, and then use the **MATH** button to find the difference between the two signals.

Predict what would happen if you reversed the orientation

Test your prediction.

Consider the circuit below, where we use our fixed `5V`

terminals to build our circuit:

Use a `14V pk-pk (7V amplitude)`

sine wave as input again.

Predict what $V_{out}$ and $\Delta V_{diode}$ will look like as a function of $V_{in}$, including a quick picture or sketch

Test your prediction, and resolve any discrepancies.

Now suppose $V_{in}$ is a 1 kHz sinusoidal signal with a *peak-to-peak* amplitude of approximately `6 V`

(or amplitude of `3V`

)

How, if at all, do you think $V_{out}$ will change? Check your prediction, take a photo or screenshot of what you see, and resolve any discrepancies.

Modern semiconductor based devices can often be damaged by being connected to too high of voltages. You'll frequently find clamp circuits like this being used to protect the inputs to Arduino or Raspberry Pi circuits where the maximum input signal isn't known. They may use more specialized sorts of diodes, but the intent is the same.

Now suppose that the diode in the clamp circuit has been reversed, as shown in Fig. 5.

For the`same input signal`

as the previous portionpredictwhat $V_{out}$ will look like and sketch/describe your prediction.

Check your prediction, take a screenshot of what you see, and resolve any discrepancies.

In this lab, we walked you though:

- How to make a bandpass filter
- How to use the frequency selectivity of a bandpass filter to find Fourier components of a signal

- How to use a diode to rectify a signal
- How to use a diode to clamp the maximum voltage from an input
- This is extremely common (and sometimes built-in) as a means of protecting complex electronic devices from high voltages that could damage the internal semiconductors.

- How to use a diode to clam the minimum voltage from an input
- This sort of circuit is useful for running a device off of battery power only when another higher voltage power supply is disconnected.