Many times, it so happens that there is only an oscilloscope to hand when an unknown capacitor's value is to be found out. Possibly, the markings on the capacitor have been removed, or the markings so faded away that what is visible is no longer reliable.
A capacitor stores charge, and like a bucket being filled with water from a small pipe, takes time to fill up. If the pipe remains unchanged, a small bucket fills in a shorter time than a larger bucket. In the same process, a smaller capacitor charges in a shorter time than a capacitor that is larger in value. The time a capacitor takes to charge through a resistance is mathematically derived as:
Here, **Vc** is the voltage on the capacitor **C**, at time **t**, **Vin** is the final voltage the capacitor will be charged to, and **R** is the resistance through which the capacitor is charging. The charging circuit looks like:
As the switch **Sw** is thrown in, the capacitor **C** starts charging through the resistor **R**. At any time t seconds after the capacitor starts charging, the voltage **Vc** on the capacitor changes as:
This graph is a mathematical plot of the equation of the capacitor charging. If the voltage **Vc** is seen on the oscilloscope, the same graph will be seen there as well. See from the graph that ultimately the capacitor reaches its full charge, represented by 9V, as **Vin**. In this example, it is assumed that:
**Vin = 9V, R = 1.0 MOhm, and C = 10 uF**
From the equation, if the value of **t** is substituted by the product of **R** and **C**, the equation reduces to:
That means, after a lapse of time **t = R*C seconds**, the voltage **Vc** on the capacitor will reach 63.2% of the final voltage, **Vin**. For 9V, this is 5.688V, which is reached by our example capacitor in 10 seconds, as is marked on the charging graph. This time is called the time constant of the Resistor-Capacitor combination.
Therefore, when trying to decipher the value of an unknown capacitor with an oscilloscope, use a known voltage for **Vin**, and a known resistor **R**. Now, determine the value of 63.2% of **Vin**, and let the capacitor charge through the resistor **R**, as in the schematic. Capture the charging waveform **Vc** on the oscilloscope and from the graph on the screen check how long it has taken for the voltage on the capacitor to reach 63.2% of **Vin**. The capacitor value is then
**C = t/R**.
If the capacitor is of a small value, the **Vin** may have to be substituted with a square wave of 50% duty cycle, which will repeatedly charge and discharge the capacitor. This has to be synchronized with the oscilloscope to obtain a steady waveform on the screen. The Switch **Sw** will no longer be necessary and **Vin** will be the peak voltage of the square wave. The oscilloscope will show both the charging as well as the discharging waveforms. Since the capacitor will be charging and discharging through the same resistor, the two waveforms will be identical but inverted. |