Download Capacitors and Current

Capacitors and Current
Let’s start with a circuit consisting of a battery, a resistor, a capacitor and
a switch in series:
Start with no charge on the
capacitor and the switch open.
When the switch is closed at
position c, current begins to
flow, building up charge on the
capacitor plates. At first,
current flows as if there were
a wire in place of the
capacitor. As charge builds
up on the capacitor plates, the voltage difference across the capacitor
increases.
This voltage difference across the capacitor opposes the battery voltage,
decreasing the current flow. Once the voltage difference across the
capacitor reaches the battery potential, current flow stops. So the increase
in potential across the capacitor is proportional to the difference between
the battery voltage and the potential across the capacitor.
This proportionality between the rate of change of the potential and the
difference between the potential and the battery voltage leads to an
exponential curve. This curve representing the voltage across the capacitor
must begin at zero, end at the voltage of the battery and have a decrease in
the rate of change of voltage as time proceeds. Such a function is
At t = , ΔV = 10 V.
ΔV = 10 V [1-e-t/RC]
With the capacitor fully charged, the switch is now closed to position d. The
capacitor now loses charge driving current through the resistor until the
plates are back to neutral. As the charge decreases, the voltage across the
capacitor drops and the current through the resistor drops, decreasing the
rate at which the voltage decreases. This gives another exponential curve.
The charging and discharge curves for the circuit above are plotted below.
Capacitor curves
1.20E+01
1.00E+01
voltage
8.00E+00
6.00E+00
4.00E+00
2.00E+00
0.00E+00
0
20
40
60
80
100
time (s)
Notice the half time for the capacitor circuit is 5 seconds which is equal to
RC ℓn(2).
Note the units: Ohms  Farads = seconds. Let’s check that.
[Ω] = [V/A] = [V/(C/s)] = [Vs/C]
[F]= [C/V] (from V = q/C)
So, the units for RC are [Vs/C] [C/V] = [s].
So why doesn’t a break in the wire behave this way? A break in the wire
doesn’t allow any noticeable current flow, just an open circuit. Let’s try a
quick calculation using a 1mm diameter wire and a gap of 0.1mm between cut
ends with air between. Treating this as a capacitor gives the following
capacitance:
C = o A/d = A/4Kd = (5E-4)2/[4(9E9)(1E-4)] ≈ 7E-14F
Even with a 100kΩ resistor, this gives RC = 7E-9 seconds and very little
charge on the “capacitor”. This means you would never notice the charge or
discharge behavior of the open circuit, it’s over too quickly and little charge
is transferred.
RC circuits like this are often used for timing circuits. Logic circuits
operate between 0 and +5 volts. A voltage between 0 and 2 volts is counted
as a logical “zero” while a signal between 4 and 5 volts is counted as a logical
“1”.
Questions
[1] If a 2μF capacitor is charged to +5 volts and you want it to drop to a
logical 0 in 0.2 seconds. What resistor do you need to use to accomplish
this?
[2] Another use of RC circuits is in power filtering. To convert alternating
current to direct current, diodes are used to give all positive voltages. The
output from the diodes
is shown in the diagram
at right, the time
between peaks is
1/120th of a second.
The “ripple” needs to
be minimized. The resulting voltage can be used to charge a capacitor and
the discharge of the capacitor can be slowed using a resistor, shown in the
following diagram.
This signal results
from the quick
charging and slow
discharge of the
capacitor in the
diagram. Typical
capacitor values for such uses are 200 μF. If the resistor used is 330kΩ,
how much does the voltage fall from 5.0 V. in the 1/120th of a second
between peaks?