Voltage divider circuits may be constructed from reactive components just as easily as they may be constructed from fixed value resistors
But just like resistive circuits, a capacitive voltage divider network is not affected by changes in the supply frequency even though they use capacitors, which are reactive elements, as each capacitor in the series chain is affected equally by changes in supply frequency.
But before we can look at a capacitive voltage divider circuit in more detail, we need to understand a little more about capacitive reactance and how it affects capacitors at different frequencies.
In our first tutorial about Capacitors, we saw that a capacitor consists of two parallel conductive plates separated by an insulator, and has a positive (+) charge on one plate, and an opposite negative (–) charge on the other.
We also saw that when connected to a DC (direct current) supply, once the capacitor is fully charged, the insulator (called the dielectric) blocks the flow of current through it.
Typical Capacitor
A capacitor opposes current flow just like a resistor, but unlike a resistor which dissipates its unwanted energy in the form of heat, a capacitor stores energy on its plates when it charges and releases or gives back the energy into the connected circuit when it discharges.
This ability of a capacitor to oppose or “react” against current flow by storing charge on its plates is called “reactance”, and as this reactance relates to a capacitor it is therefore called Capacitive Reactance ( XC ), and like resistance, reactance is also measured in Ohm’s.
When a fully discharged capacitor is connected across a DC supply such as a battery or power supply, the reactance of the capacitor is initially extremely low and maximum circuit current flows through the capacitor for a very short period time as the capacitors plates charge up exponentially.
After a period of time equal to about “5RC” or 5 time constants, the plates of the capacitor are fully charged equalling the supply voltage and no further current flows. At this point the reactance of the capacitor to DC current flow is at its maximum in the mega-ohms region, almost an open-circuit, and this is why capacitors block DC.
Now if we connect the capacitor to an AC (alternating current) supply which is continually reversing polarity, the effect on the capacitor is that its plates are continuously charging and discharging in relationship to the applied alternating supply voltage.
This means that a charging and discharging current is always flowing in and out of the capacitors plates, and if we have a current flow we must also have a value of reactance to oppose it. But what value would it be and what factors determine the value of capacitive reactance.
In the tutorial about Capacitance and Charge, we saw that the amount of charge, ( Q ) present on a capacitors plates is proportional to the applied voltage and capacitance value of the capacitor. As the applied alternating supply voltage, ( Vs ) is constantly changing in value the charge on the plates must also be changing in value.
If the capacitor has a larger capacitance value, then for a given resistance, R it takes longer to charge the capacitor as τ = RC, which means that the charging current is flowing for a longer period of time. A higher capacitance results in a small value of reactance, XC for a given frequency.
Likewise, if the capacitor has a small capacitance value, then a shorter RC time constant is required to charge the capacitor which means that the current will flow for a shorter period of time. A smaller capacitance results in a higher value of reactance, XC.
Then we can see that larger currents mean smaller reactance, and smaller currents mean larger reactance. Therefore, capacitive reactance is inversely proportional to the capacitance value of the capacitor, XC ∝-1 C.
Capacitance, however is not the only factor that determines capacitive reactance. If the applied alternating current is at a low frequency, the reactance has more time to build-up for a given RC time constant and oppose the current indicating a large value of reactance.
Likewise, if the applied frequency is high, there is little time between the charging and discharging cycles for the reactance to build-up and oppose the current resulting in a larger current flow, indicating a smaller reactance.
Then we can see that a capacitor is an impedance and the magnitude of this impedance is frequency dependent. So larger frequencies mean smaller reactance, and smaller frequencies mean larger reactance. Therefore, Capacitive Reactance, XC (its complex impedance) is inversely proportional to both capacitance and frequency and the standard equation for capacitive reactance is given as:
Now that we have seen how the opposition to the charging and discharging currents of a capacitor are determined not only by its capacitance value but also by the frequency of the supply, lets look at how this affects two capacitors connected in series forming a capacitive voltage divider circuit.
Consider the two capacitors, C1 and C2 connected in series across an alternating supply of 10 volts. As the two capacitors are in series, the charge Q on them is the same, but the voltage across them will be different and related to their capacitance values, as V = Q/C.
Voltage divider circuits may be constructed from reactive components just as easily as they may be constructed from resistors as they both follow the voltage divider rule. Take this capacitive voltage divider circuit, for instance.
The voltage across each capacitor can be calculated in a number of ways. One such way is to find the capacitive reactance value of each capacitor, the total circuit impedance, the circuit current and then use them to calculate the voltage drop, for example:
1. Capacitive Reactance of 10uF capacitor
2. Capacitive Reactance of 22uF capacitor
Total capacitive reactance of series circuit – Note that reactance’s in series are added together just like resistors in series.
or:
Circuit current
Thus the voltage drop across each capacitor in series capacitive voltage divider will be:
When the capacitor values are different, the smaller value capacitor will charge itself to a higher voltage than the larger value capacitor, and in our example above this was 6.9 and 3.1 volts respectively.
Since Kirchhoff’s voltage law applies to this and every series connected circuit, the total sum of the individual voltage drops will be equal in value to the supply voltage, VS and 6.9 + 3.1 does indeed equal 10 volts.
Note that the ratios of the voltage drops across the two capacitors connected in a series capacitive voltage divider circuit will always remain the same regardless of the supply frequency. Then the two voltage drops of 6.9 volts and 3.1 volts above in our simple example will remain the same even if the supply frequency is increased from 80Hz to 8000Hz as shown.
Using the same two capacitors, calculate the capacitive voltage drop at 8,000Hz (8kHz).
While the voltage ratios across the two capacitors may stay the same, as the supply frequency increases, the combined capacitive reactance decreases, and therefore so too does the total circuit impedance. This reduction in impedance causes more current to flow.
For example, at 80Hz we calculated the circuit current above to be about 34.5mA, but at 8kHz, the supply current increased to 3.45A, 100 times more. Therefore, the current flowing through a capacitive voltage divider is proportional to frequency or I ∝ ƒ.
We have seen here that a capacitor divider is a network of series connected capacitors, each having a AC voltage drop across it. As capacitive voltage dividers use the capacitive reactance value of a capacitor to determine the actual voltage drop, they can only be used on frequency driven supplies and as such do not work as DC voltage dividers. This is mainly due to the fact that capacitors block DC and therefore no current flows.
Capacitive voltage divider circuits are used in a variety of electronics applications ranging from Colpitts Oscillators, to capacitive touch sensitive screens that change their output voltage when touched by a persons finger, to being used as a cheap substitute for mains transformers in dropping high voltages such as in mains connected circuits that use low voltage electronics or IC’s etc.
Because as we now know, the reactance of both capacitors changes with frequency (at the same rate), so the voltage division across a capacitive voltage divider circuit will always remain the same keeping a steady voltage divider.
Where it says:
“Then we can see that larger currents mean smaller reactance, and smaller currents mean larger reactance. Therefore, capacitive reactance is inversely proportional to the capacitance value of the capacitor, XC ∝-1 C.”,
Shouldn’t it say “Then we can see that larger CAPACITANCE means smaller reactance, and smaller CAPACITANCE means larger ewctance” ?
I’m a complete novice, but it seems this is either a simple typo, or I’m mightily confused!
For a pure capacitor, its reactance value, XC is inversely proportional to both its capacitance and applied frequency. For a given voltage, IC = V/XC (Ohms Law). Thus a smaller reactance value at a higher frequency will allow a greater current to flow through it, while a larger reactance value at a lower frequency will result in a smaller current flow as stated. Therefore, the tutorial is correct as given.
Can I divide in half, with capacitors, variable AC up to 250 V, from a wind turbine; then rectify it for a max 150 VDC input regulator to charge a lithium battery of 36 V/25 A. If so:
1. what type of material is better for the caps, electrolyte, ceramic, film, etc.;
2. What mf /volts?
3. Diodes capacity in amps/volts.
Thanks in advance.
Hello good time In half-bridge circuits, two capacitors of the same shape are used to divide the voltage But in some of them, their capacity is a few nanofarads And in some others it is a few microfarads What difference does increasing the capacity of these two axes make?
I understand the theory behind voltage divider using two capacitors in series, supply voltage say 10 volt 60 Hz sinewave. I am using two equal capacitors and obviously expecting 5 volt across each cap. The circuit works fine as long as capacitor values are something like .1 uF or so. But as I use low value caps, say 1 or 2 pF, the circuit does not work as expected. May I know what is importance of cap. value here ?
Obviously the reactance of the 1pF capacitors is much higher compared to the 0.1uF as the 0.1uF capacitors are 100,000 times greater than the 1pF, therefore the reactance is 100,000 times greater. Current flow through the 1pF capacitors will be in the nano-amperes, thus I*Xc will be different. Also another point, as the reactance of the 1pF capacitor is extremely high, about 2650 MΩ at 60Hz, the impeadance of your parallel connected multimeter will affect the readings
Yes, Wayne..I fully understand your explanation. However, I tried to see the waveforms on Scope, which hopefully has a higher impedance, but obviously it did not work there as well, as current thru circuit is too low ( in nA range).
I am now stuck in my project, where I am interested in running a yarn thru’ a small capacitor (3 x 4 mm plates separated by 2 mm gap., a textile machine application). I want to measure the yarn thickness variation, using 4 caps in a Wheatstone bridge, then an Instrument Amplifier, a voltage amplifier and finally the an Arduino or raspberry pi.
Can you then suggest me any alternate solution, Many thanks for your help…Arun
Arun,
I think the dielectric medium will affect the capacitance.
You’re saying that you will measure the thickness of yarn in 2mm gap of the plates. Here yarn not only the medium but the contained air and its moisture will also affect the capacitance.
Your article is wonderful, and explains the capacitor theory so nicely.
However I am trying to make an AC square wave voltage divider ckt, using two home made identical caps, 3mm x 4mm area with 3 mm distance.( C = .035 pF…very small value)
The ckt works fine if I use standard 2 x 0.1 uF capacitors, but does not work when I use my home made capacitors.
The Vss supply voltage is 5 volts peak to peak, at 2.6 MHz, while I get voltage across each capacitor (as shown on O’scope) is only 240 mV peak to peak. Though not so important but I tried lower frequencies and different duty cycles as well.
Can you suggest where is the problem please ? Thank you…Arun Karkare
Yeah, 0.035pF (35 femtofarad) is ridiculously small. Make something in the picofarad (pF) or nanofarad (nF) ranges
Hi Wayne…I tried the value of 1 or 2 pF, but did not work as well, any idea where could be the problem ? Thanks for ur support….
Super
The best and simplest explanation I have gotten so far about capacitors. Thanks.
We want to develop voltage measurement unit which can measure voltage from 400 to 2500Vrms , frequency range fron 15KHz to 300KHz
We tried it with capacitive divider but not succeed yet
Crystal clear explanation 👍
Yes. Thank you. I am interested in this subject because I have a ceiling fan, the speed of which is changed by a wall controller consisting of 3 capacitors. Problem is, the controller is not the original which malfunctioned some time ago.
Thank you very much for a very helpful and very well-written article which the common man can understand without having a physics degree.
Je cherche utiliser diviseur capacitif pour 800v ac en 250v et 350v
Well explained, thank you.
You are simply the best, please publish a book.
What would be result for capacitor devider when damping oscilator is applied?
Please advice me ,
How to check the capacitor.?
Thank You.
many ways to check it:
– using multimeter capacitor check
-multimeter diode and peep check, when capacitor doesn’t result beep it’s okay
– multimeter ohm check, it must gives you infinity.
Very good tutorial
It was a great article man
Best explanation I’ve ever since or reactance. Why did they not tech like this in university?
Thank you. Needed brush up to understand ceiling fan circuitry. Studied electronics in Navy, long time ago.