Capacitive voltage dividers are circuits, which employ capacitors in series with an alternating current (AC) power supply to produce a voltage drop across each capacitor.
The most common use for these circuits is, to safely decrease extremely high voltages for measurements or protection. High frequency capacitive voltage dividers are increasingly being used in touchscreens and display devices in smartphones and tablets in present-day applications.
Capacitive voltage dividers only work with an AC source that changes continuously and waveform-shaped, generally referred to as a sine wave. In contrast to this, resistive voltage dividers may operate with both AC and DC power supplies.
This is due to the fact that capacitive reactance, denoted by the symbol XC, and measured in ohms, determines how voltage divides between capacitors connected in series. The following equation describes the inverse relationship between capacitive reactance and the individual capacitors capacitance and AC supply frequency:
XC = 1 / 2πfC
In the above formula:
- XC represents the Capacitive Reactance in Ohms, (Ω)
- π (pi) is a well known numeric constant of 3.142
- ƒ indicates the Frequency in Hertz, (Hz)
- C refers to the Capacitance in Farads, (F)
Basically we can calculate the reactance of the capacitors, or how much they resist the current, by knowing the voltage and frequency of the AC supply. The voltage drop across each capacitor may be found by entering these reactance values into the above formula, which functions similarly to the voltage divider rule but with capacitors in place of resistors.
Solving a Capacitor Divider Problem
Calculate the rms voltage drops across each capacitor in terms of their reactance when it is connected to a 100 volt, 50Hz rms supply by using the two capacitors in the series circuit above, which have capacity values of 10 μF and 22 μF.
Solving the problem:
- XC1 = 1 / 2πfC1 = 1 / 1 / (2π * 50 * 10 * 10-6) = 318.3 Ω
- XC2 = 1 / 2πfC2 = 1 / 1 / (2π * 50 * 22 * 10-6) = 144.7 Ω
- XCT = XC1 + XC2 = 318.3 Ω + 144.7 Ω == 463 Ω
- VC1 = VS (XC1 / XCT) = 100(318.3 / 463) = 69 Volts
- VC2 = VS (XC2 / XCT) = 100(144.7 / 463) = 31 Volts
The principle applies equally to series resistances and pure capacitors: the sum of all individual voltage drops across the series elements always equals the source voltage. In this context, the reactance of each capacitor ultimately dictates how much voltage drop occurs across it, whereas the capacitance value itself determines the relative magnitude of that drop.
Consequently the smaller 10μF capacitor, with its higher reactance of 318.3Ω, experiences a greater voltage drop of 69 volts compared to the larger 22μF capacitor, which has a 31-volt drop and a reactance of 144.7Ω. Importantly due to their series connection, both C1 and C2 will always share the same current (IC) in the circuit, which in this case is 216 mA.
Finally it's important to consider the phase relationship in capacitive voltage dividers. When the capacitor voltages are in phase, the 69 and 31 volt drops across C1 and C2 will indeed sum directly to the 100 volt supply voltage, provided theres no series resistance present. However if the capacitor voltages become out of phase due to external factors, a phasor addition of the two waveforms becomes necessary. In such cases, simple voltage addition following Kirchhoffs voltage law is no longer sufficient.
References: Capacitive Voltage divider