Capacitors in Series
Capacitors are said to be connected together "in series" when they are effectively "daisy chained"
together in a single line. The charging current ( iC ) flowing through the
capacitors is THE SAME for all capacitors as it only has one path to follow and
iT = i1 = i2 = i3 etc.
Then, Capacitors in Series all have the same current so each capacitor stores the same amount of charge
regardless of its capacitance. This is because the charge stored by a plate of any one capacitor must have come
from the plate of its adjacent capacitor. Therefore, capacitors connected together in series must have the same charge
Q.
QT = Q1 = Q2 = Q3 ....etc
Consider the following circuit in which the three capacitors, C1,
C2 and C3 are all connected
together in a series branch across a supply voltage between points A and B.
Capacitors in a Series Connection
In the previous parallel circuit we saw that the total capacitance, CT of the circuit was equal to the sum of all the individual capacitors added together. In a series connected circuit however, the total or equivalent capacitance CT is calculated differently. In the series circuit above the right hand plate of the first capacitor, C1 is connected to the left hand plate of the second capacitor, C2 whose right hand plate is connected to the left hand plate of the third capitor, C3. Then this series connection means that in a DC connected circuit, capacitor C2 is effectively isolated from the circuit.
The result of this is that the effective plate area has decreased to the smallest individual capacitance
connected in the series chain. Therefore the voltage drop across each capacitor will be different depending upon the values
of the individual capacitances. Then by applying
Kirchoff's Voltage Law,
( KVL ) to the above circuit, we get:
Since Q = CV or V = Q/C, substituting
Q/C for each capacitor voltage VC in the above
KVL equation gives us
dividing each term through by Q gives
Series Capacitors Equation
When adding together Capacitors in Series, the reciprocal ( 1/C ) of the individual capacitors are all added together ( just like resistors in parallel ) instead of the capacitances themselves. Then the total value for capacitors in series equals the reciprocal of the sum of the reciprocals of the individual capacitances.
Example No1
Taking the three capacitor values from the above example, we can calculate the total
circuit capacitance for the three capacitors in series as:
One important point to remember about capacitors that are connected together in a series configuration,
is that the total circuit capacitance ( CT ) of any number of capacitors connected
together in series will always be LESS than the value of the smallest capacitor in the series and in our example
above C = 0.055uF where as the value of the smallest capacitor in the series
chain is only 0.1uF.
This reciprocal method of calculation can be used for calculating any number of capacitors connected
together in a single series network. If however, there are only two capacitors in series, then a much simpler and
quicker formula can be used and is given as:
With series connected resistors, the sum of all the voltage drops across the series circuit will be equal to the applied voltage VS ( Kirchoff's Voltage Law ) and this is also true about capacitors in series.
With series connected capacitors, the capacitive reactance of the capacitor acts as an impedance due to
the frequency of the supply. This capacitive reactance produces a voltage drop across each capacitor, therefore the series
conncted capacitors act as a capacitive voltage divider
network. The result is that the voltage divider formula applied to resistors can also be used to find the individual voltages
for capacitors in series. Then:
Where: CX is the capacitance of the capacitor in question,
VS is the AC supply voltage across the whole series chain and VCX
is the voltage drop across the target capacitor(s).
Example No2
Find the overall capacitance and the individual rms voltage drops across the following sets of two
capacitors in series when connected to a 12V a.c. supply.
- a) two capacitors each with a capacitance of 47nF
- b) one capacitor of 470nF connected in series to a capacitor of 1uF
a) Total Capacitance,
Voltage drop across the two identical 47nF capacitors,
b) Total Capacitance,
Voltage drop across the two non-identical Capacitors: C1 = 470nF
and C2 = 1uF.
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, VS. Then 8.16 + 3.84 = 12V.
Note also that if the capacitor values are the same, 47nF in our example,
the supply voltage is divided equally across each capacitor as shown, because each capacitor shares an equal amount of charge
( Q = C x V = 0.564uC ) and therefore has half ( or
percentage fraction for more than two capacitors ) of the applied voltage, VS.
When the capacitor values are different, the larger value capacitor will charge itself to a lower voltage
and the smaller value capacitor to a higher voltage, and in our example above this was 3.84 and 8.16 volts respectively, to
maintain the same amount of charge on the plates of both capacitors as shown.
Note that the ratios of the voltage drops across the two capacitors connected in series will always remain
the same regardless of the supply frequency. Then the two voltage drops of 8.16 volts and 3.84 volts above in our simple example
will remain the same even if the supply frequency is increased from 100Hz to 100kHz.
Although the voltage drops across each capacitor will be different for different values of capacitance,
the coulomb charge across the plates will be equal because the same amount of current flow exists throughout a series circuit
as all the capacitors are being supplied with the same number or quantity of electrons.
In other words, if the charge across each capacitors plates is the same, as Q
is constant, then as its capacitance decreases the voltage drop across the capacitor increases, because the charge is large
with respect to the capacitance. Likewise, a larger capacitance will result in a smaller voltage because the charge is small
with respect to the capacitance.
Summary
Then to summerise, the total or equivalent capacitance, CT of a
circuit containing Capacitors in Series is the reciprical of the sum of the reciprocals of all of the
individual capacitances added together.
Also for capacitors connected in series, all the series connected capacitors will have
the same charging current flowing through them as iT = i1 = i2 = i3
etc. Two or more cpacitors in series will always have equal amounts of coulomb charge across their plates. As the charge,
( Q ) is equal and constant, the voltage drop across the capacitor is determined by
the value of the capacitor only as V = Q ÷ C. A small capacitance value
will result in a larger voltage while a large value of capacitance will result in a smaller voltage drop.
Great!
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