Voltage divider circuits may be constructed from reactive components just as easily as they may be constructed from fixed value resistors.

But just like resistors, 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 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, X_{C} ∝^{-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:

- Where:
- Xc = Capacitive Reactance in Ohms, (Ω)
- π (pi) = a numeric constant of 3.142 (or 22÷7)
- ƒ = Frequency in Hertz, (Hz)
- C = Capacitance in Farads, (F)

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:

Using the two capacitors of 10uF and 22uF in the series circuit above, calculate the rms voltage drops across each capacitor when subjected to a sinusoidal voltage of 10 volts rms at 80Hz.

Capacitive Reactance of 10uF capacitor

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

Then 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 Kirchoff’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, V_{S} 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.

Error! Please fill all fields.

I am presently working on the design stage of an amplifier. The POWER transformer is split between two unit and requires a voltage divider circuit to provide 6.3 volts at 5 amps, 5 volts at 3 amps. I have worked out how to provide the voltages but am quite stumped by the current nodes. Can you provide me with a page to study or circuit suggestions?

———————————–0

! ! !

! ! !

! ———— !

/ \ 37 mfd ? 6.3 v 5 amp / \

120 V : 24 v ———— 24 v : 120 V

! 10 amp ! ! X-frmr !

X -frmr ! \ /

\ / ———– !

! 60 mfd ? 5 v 3 amp !

! ———– !

! ! !

! ! !

! ————- !

! 150 mfd ? 12.7 v 2 amp !

! ————- !

! ! !

! ! !

!————————————0

Sorry about the crude drawing, I have not been in classes for a very, very long time and any help or direction would be greatly appreciated.

Thank you,

Steven L. Childers

It flows the same current value through both capacitors. Because they are in series. If they are connected in parallel, it will be different values of current flowing through them

check the frequency output of the dimmer if not equal to 50hz den add a good filtering capacitor around 2.2 -10uf capacitor dat should solve the problem or else get a transformer and wind it in the range of the desired voltage nd current also make sure u put tolerance into consideration. good luck

Like Thevenien resistance, V2=RL(V1) /(R1+RL), for transient responce of CVT IEC 61869-5 gives the formualae as V2=C1xV1/C1+C2.

Here C1 is coming instead of C2?

Please explain if u understand…

Best Regards,

GMF

Hello,

I have a AC to DC Trans.

input: 220V : AC : max ~15A

output: 14.2V : DC : max ~70A

(output goes to a sound system amplifier which works best with 12.5~14.0 V, DC)

well now the local AC source is 235V ( insted of 220V), so when I use the trans, the output Voltage is above even 15.5 V ( & that cause the Amplifier Protector to shutdown the Amp. )

How can I reduce the local AC Voltage around 10% ( from 230V to 200 or 205V)

by the way I used an regular Dimmer which works fine for reducing the voltage but using it caused the sound system noise! ( so please refuse advising the dimmer )

Thanks in advanced.

Considering your Demand 70 A ..their is no other way to reduce voltage efficiently without changing tarsformer. You could use smoothing capacitor with your dimmer circuit.. But if you want to go crazy just Plug a old 12 V lead acid battery (normally used in backup power ups or bus or big vhecile stareting ) parallel to transformer that will definitely work.. try it.

By their very nature cheap dimmers introduce noise, you could for example reduce the input voltage using a tapped transformer.

Helpful, Basic Electronics Tutorials. Thanks.

In My stepper motor generator voltage is 25 volts and current is 60 ma. I want 200 ma current and voltage is any.

its help me much more.

nice solve.

When I try to print the pages, the electronics tutorials title header keeps getting in the way. Is there any way to remove it from the top of the pages?.

Jim.

At the moment no. We are currently working on a printable version of the tutorials.

Ok, thanks for the reply. Do you have any idea how long that will be?.

Jim.

Does this not all just reduce down to a simple calculation?

For a Capacitance voltage divider, the voltage across C2 (for an AC voltage) is: V * ( 1-(C2/C1+C2))

“If the capacitor has a large capacitance value, then a longer RC time constant is required to charge the capacitor which means that the current will flow for a longer period of time indicating a large value of reactance, Xc.”

——You got that totally wrong bro!

He’s not totally wrong.

All that is wrong about it is that it has a lower reactance not a higher one. He was right about it taking longer to charge a larger capacitor.

You must correct it then.

Thanks Pratik.