Simple Autorange Capacitor Tester / Capacitance Meter With Arduino and by Hand

Hello!

For this physics-unit you need:

* a power supply with 0-12V

* one or more capacitors

* one or more charging resistors

* a stopwatch

* a multimeter for voltage measurement

* an arduino nano

* a 16x2 I²C display

* 1 / 4W resistors with 220, 10k, 4.7M and 1Gohms 1 gohms resistor

* dupont wire

Capacitors play a very important role in electronics.
They are used to store charges, as a filter, integrator, etc. But mathematically, there is a lot in capacitors. So you can practice exponential functions with capacitors and they. work out. If an initially uncharged capacitor is connected via a resistor to a voltage source, then charges flow continuously to the capacitor. With the increasing charge Q, according to the formula Q = C * U (C = capacitance of the capacitor), the voltage U across the capacitor also increases. However, the charging current is decreasing more and more as the rapidly charged capacitor is becoming more and more difficult to fill with charges. The voltage U (t) on the capacitor obeys the following formula:

U (t) = U0 * (1-exp (-k * t))

U0 is the voltage of the power supply, t is the time and k is a measure of the speed of the charging process. Which sizes does k depend on? The larger the storage capacity (that is, the capacitance C of the capacitor), the slower it fills with charges and the slower the voltage increases. The larger C, the smaller k. The resistance between capacitor and power supply also limits charge transport. A larger resistance R causes a smaller current I and therefore fewer charges per second flowing to the capacitor. The larger R, the smaller k. The correct relationship between k and R or C is:

k = 1 / (R * C).

The voltage U (t) at the capacitor thus increases according to the formula U (t) = U0 * (1-exp (-t / (R * C)))

Students should enter the voltage U at time t in a table and then draw the exponential function. If the voltage increases too fast, you'll have to increase the resistance R. On the other side if the voltage changes too slow, decrease R.

If one knows U0, the resistance R and the voltage U (t) after a certain time t, then the capacitance C of the capacitor can be calculated from this. For this one would have to logarithm the equation and after some transformations we get: C = -t / (R * ln (1 - U (t) / U0))

Example: U0 = 10V, R = 100 kohms, t = 7 seconds, U(7 sec) = 3.54V. Then C results in a value of C = 160 μF.

But there is a second, simple method to determine the capacity C. Namely, the voltage U (t) after t = R * C is exactly 63.2% of U0.

U (t) = U0 * (1-exp (-R * C / (R * C)) = U0 * (1-exp (-1)) = U0 * 0.632

What does this mean? Students must determine the time t after which the voltage U (t) is exactly 63.2% of U0. Specifically, for the example above, the time is sought after which the voltage across the capacitor is 10V * 0.632 = 6.3V. This is the case after 16 seconds. This value is now inserted into the equation t = R * C: 16 = 100000 * C. This yields the result: C = 160 μF.

At the end of the exercise, the capacity can also be determined with an Arduino. This calculates the capacity C exactly according to the method of earlier. It charges the capacitor via a known resistor R with 5V and determines the time after which the voltage at the capacitor = 5V * 0.632 = 3.16V. For the Arduino digital-to-analog converter, 5V equals 1023. Therefore, you just have to wait until the value of the analog input is 1023 * 3.16 / 5 = 647. With this time, the capacity C can be calculated. So that capacitors with very different capacitance can be measured, 3 different charging resistors are used. First, a low resistance is used to determine the charging time up to 647. If this is too short, ie if the capacitance of the capacitor is too small, the next higher charging resistance is selected. If this is also too small a 1 Gohms resistance follows at the end of the measurement. The value for C is then displayed on the display with the correct unit (µF, nF or pF).

What do students learn in this unit? You will learn about capacitors, their capacitance C, exponential functions, logarithm, percentage calculations and the Arduino. I think a lot.

This unit is suitable for students aged 16-17 years. You must have already gone through the exponential function and the logarithm in mathematics. Have fun trying it in your class and Eureka!

I would be very happy if you would vote for me in the classroom science contest. Thanks a lot for this!

If you are interested in my other physics projects, here is my youtube channel: https://www.youtube.com/user/stopperl16/videos

more physics projects: https://stoppi-homemade-physics.de/