A few circuits that are interesting and that might help with your project:
Using some components you haven’t used before:
The DC power in this experiment comes from your DC power supply. You do not need an Arduino for this lab, so make sure none is connected to your circuit. We don’t want >5V accidentally going into (and frying) a Arduino pin by accident.
Set a current limit (e.g. 50 mA) on your DC supplies, to reduce the chance of frying miswired components.
Simple passive RC or LR circuits can act as circuits that passively differentiate their input, i.e. \(V_{out} \propto \frac{dV_{in}}{dt}\), but op amp differentiators allow more control (with careful thought about the details of the circuit).
An ideal op amp has infinite input impedance and draws no current, so in the ideal simple circuit below, the current through the resistor and capacitor is:
\[I_C = C \frac{d\left(V^- -V_{in}\right)}{dt} = I_R = \frac{V_{out}-V^-}{R}\]
Assuming an ideal op amp, then for this circuit \(V^- = V^+ = 0\), so
\[C \frac{dV_{in}}{dt} = \frac{-V_{out}}{R}\]
\[\therefore V_{out} = - RC\frac{dV_{in}}{dt}\]
Build the above circuit on your breadboard, and play with various inputs and observe the outputs.
R-1) If you input a 50 Hz, 1 Vpp amplitude, 0 V Offset Ramp input, is the output what you expect?
R-2) How does the output get better or worse if you put a 100nF capacitor in parallel with the 1k resistor? Can you qualitatively explain any differences?
An ideal op amp has infinite input impedance and draws no current, so in the ideal simple circuit below, the current through the resistor and capacitor is:
\[I_R = \frac{V^- -V_{in}}{R} = I_C = C \frac{d\left(V_{out}-V^-\right)}{dt}\]
Assuming an ideal op amp, then for this circuit \(V^- = V^+ = 0\), so
\[\frac{-V_{in}}{R} = C \frac{dV_{out}}{dt}\]
\[\therefore \frac{dV_{out}}{dt} = -\frac{1}{RC}{V_{in}}\]
\[\therefore V_{out} (t) = -\frac{1}{RC}\int_0^t V_{in}(t')\,dt'\]
Build the above circuit on your breadboard with Wavegen input, and play with various inputs and observe the outputs.
R-3) If you input a 1 kHz, 1 Vpp amplitude, 0 V Offset Square input, is the output what you expect? If not, can you explain any unexpected characteristics?
The above simple integrator is not a practical circuit because it is sensitive to even tiny DC offsets on the input or non-zero currents flowing through the op amp. To improve the circuit we can try:
R-4) Which of above solutions most improves the output of your circuit?
The LM555CN is one of the family of 555 timer chips.
Build the circuit in Figure 6-5 of the NE555P Spec Sheet, with nominal \(R_A=330k\), \(R_B=100k\), and \(C=47nF\).
R-5) Include a screen capture of the output.
R-6) Try a few different resistors and capacitors to confirm that the frequency and duty factor change as expected according to the relations following Figures 6-5 and 6-6 on page 12 of the NE555P Specifications.
Let’s expand the above circuit into a tunable, battery powered, push-button buzzer. Build this circuit.
Notes:
The piezo connector wires may be too soft to insert into the breadboard sockets, so you may want wrap them around some small jumper wires inserted in the sockets. (Soldering the wrapped joins would be best, but is probably not neccessary for this temporary circuit.)
Connect the piezo to the output and confirm that it buzzes when the button is pressed.
Replace one of the resistors with a 100k trim-pot, and confirm that the buzzer frequency changes as the trim-pot resistance is changed.
R-7) Explain why you chose that resistor to change and not the other.
R-8) Make a short (<30s) video with audio, showing your push button buzzer circuit working.
This is the last regular lab. Next week you are working on your project.
Once again this week’s exercise could again be the basis for a final project.
For example, for the buzzer, could you:
See you next week!