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Unformatted text preview: 7. Lecture 7
7.1 Capacitor charge and discharge
In the same way we can charge a capacitor through a resistor we can also discharge
it. In the circuit of ﬁg.35 we have a switch with two positions. In one position the
capacitor charges through resistor R1 and in the other it discharges through resistor R2 .
The characteristic times are given by ∆tcharge = R1 C and ∆tcharge = R2 C . Notice
that these are properties of the circuit. The capacitor itself has no characteristic time
associated with it, it depends on both the capacity C and the resistance R. A more
practical circuit substitutes the switch by an integrated circuit that does the same job.
At the end of this lecture I put a description of the circuit we used in class and how
you can build it yourself if you are interested. R2 C + −
∆V R1 Battery Figure 35: By ﬂipping the switch we can charge and discharge the capacitor. The characteristic times are given by ∆tcharge = R1 C and ∆tcharge = R2 C . 7.2 DC circuits
7.2.1 Resistors in series and parallel
More complicated circuits include many components. Let us see now what happens if
you connect several resistors in series. In that case the current going through all of – 46 – them is the same. On the other hand the potential diﬀerence across the system is the
sum of the potential diﬀerences across each resistors. For example if we look at ﬁg. 36
we see that the potential diﬀerence
∆ V = V a − V d = (V a − V b ) + ( V b − V c ) + ( V c − V d ) = ∆ V 1 + ∆ V 2 + ∆ V 3 . (7.1) Using Ohm’s law we ﬁnd
∆V1 = IR1 , ∆V2 = IR2 , ∆V3 = IR3 , (7.2) from where we obtain
∆V = IR1 + IR2 + IR3 = I (R1 + R2 + R3 ) = IR (7.3) So the total resistance is R = R1 + R2 + R3 a simple rule to remember!. Notice that
I , the current is the same for all resistors since charge is conserved therefore the same
amount of charge has to circulate per unit time in each point of the circuit. This is not
true if the circuit branches as we see in ﬁg.37 a connection known as parallel. In this
case the potential diﬀerence across all of resistors is the same as is easily seen if you
remember that all points connected by a cable are at the same potential. The total
current however is split between the three branches as:
I = I1 + I2 + I3 (7.4) Again, this is because charge is conserved so the total charge entering the circuit is
distributed among the three resistors. Again using Ohm’s law we have
∆V
∆V
∆V
1
1
1
1
I=
+
+
= ∆V
+
+
= ∆V
(7.5)
R1
R2
R3
R1 R2 R3
R
So we ﬁnd 1
1
1
1
=
+
+
(7.6)
R
R1 R2 R3
another easy rule to remember. However it requires more computations than just
adding the resistance as when they are in series. By the way, if you have the time
an interest it is easy and relatively cheap to buy a multimeter and a few resistors and
check these by yourself.
It is useful to remember the case where all resistors are equal R1 = R2 = R3 where
we get
R1
R=
(7.7)
3
This is the same if we out any number of resistors, for example with two resistors in
parallel the resistance is half etc. – 47 – R3 R2 R1
(a) (b) (c) + (d) − Figure 36: Resistors in series. The total resistance is the sum of the individual resistances. I1
I R1
I2
R2 I3
R3 + − Figure 37: Resistors in parallel. The inverse of the total resistance is the sum of the inverse
of the individual resistances. 7.2.2 Kirchhoﬀ ’s laws
More complicated circuits can be treated similarly. Many times one can split the
circuits in resistors in series and parallel. For example consider the situation in ﬁg.38
that is selfexplanatory. Another example is ﬁg.39. Notice that we get the same total – 48 – resistance. The reason is that because of the values of the resistors the points (a) and
(b) have the same potential already in ﬁg.38 so no current actually circulates through
the cable in the middle. However if one resistor changes slightly its value then a current
will circulate through the cable in the middle which can be used to detect those small
variations.
Although this can take you a long way analyzing circuits, some circuits just cannot
be analyzed in this way. Consider ﬁg.40. In this case one cannot use the idea of
resistors in parallel or series. However we can use similar rules as those used to derive
what happens for resistors in series and parallel. This can be cast in a set of rules
which are known as Kirchhoﬀ ’s laws and read as follows:
• The total potential diﬀerence between two points connected by a path is the sum
of the potential diﬀerences between the circuit components in the path. Therefore
for any close loop the sum of the potential diﬀerence of its components have to
add up to zero.
• Charge is conserved. Therefore in each node or connections where cables come
together the total current coming in has to equal the total current coming out.
These laws allow us to write as many equations as variables we need to determine.
Care should be taken to write equations which are independent namely not equivalent
to each other. If the component is a battery then the resistance across it is ﬁxed. If it
is a resistor it is given by Ohm’s law.
Going back to the example of the ﬁgure we can write the following equations:
I = I1 + I3 (7.8) I1 = I2 + I5 (7.9) I3 + I5 = I4 (7.10) I 1 R1 + I 5 R5 + ∆ V b 2 − I 3 R3 = 0 (7.11) I 1 R1 + I 2 R2 − ∆ V b 1 = 0 (7.13) I 2 R2 − I 4 R4 − ∆ V b 2 − I 5 R5 = 0 (7.12) The ﬁrst three are obtained by looking at the currents in each node. The last three by
looking at the loops shown in the ﬁgure. We imagine that we go down in potential so
we get for example for the loop colored pink, I1 R1 when we go through R1 but −I3 R3
when we go through R3 because we go against the current so the potential increases.
You can imagine you are in a hill and go up and down across each resistor or battery
and have to come back at the same height when you return to the same point. Be sure – 49 – you understand exactly how each equation is derived. These equations are not diﬃcult
to solve. One uses each equation to eliminate one variable and replace it in the other
equations. At the end we have only one equation and one variable and we can solve it.
Although not diﬃcult it is tedious, instead one can use a computer algebra program to
ﬁnd the solution (for example Maple, Mathematica, Matlab etc). For example, in this
case the current I is:
I= ((R1 + R3 )(R2 + R4 ) + (R1 + R2 + R3 + R4 )R5 )∆Vb1 + (R1 R4 − R3 R2 )∆Vb2
(7.14)
R5 (r1 + R2 )(R3 + R4 ) + R1 R2 R4 + R3 R2 R4 + R3 R2 R1 + R1 R3 R4 5k Ω 15k Ω 10k Ω 7.5k Ω
5k Ω 15k Ω 10k Ω + + − − + − Figure 38: Resistors in series and parallel. 5k Ω 10k Ω
2.5k Ω 5k Ω 5k Ω 7.5k Ω 10k Ω + − + − + − Figure 39: Resistors in series and parallel. 7.3 Example of circuit using charge and discharge of a capacitor
Here we describe how to build a simple circuit that charges and discharges a capacitor.
The time constants depend on variable resistors that we can control by hand and also
on the value of the capacity. Building the circuit is beyond the contents of the course
but I’ll brieﬂy describe how to do it because it is easy to do and you can have fun with
it if you decide to do it. It can also open the possibility to do many other small projects
you can ﬁnd in the Internet, books etc. – 50 – R R1 2
I
2 I1 R I5 5 + I3 − ∆Vb2
R4 R3 I4 I + − ∆Vb1
Figure 40: More complicated circuit. The closed loops used to write the equations are shown
in color. 7.3.1 Oscillator using the 555 chip
The circuit is diagrammed in ﬁg. 41. The 555 chip senses the voltage in one of its legs
and when it goes above 2/3 of the supply voltage (here 5v) discharges the capacitor.
When the voltage goes below 1/3 of 5 volts it charges it back again. The timing is
regulated by the value of the resistors R1 and R2 . In our case we put two variable
resistors of maximum value 10k Ω.
7.3.2 Actual construction
To build the circuit we use a breadboard. All holes in each column are connected to
each other except that the top and bottom parts are separate. The top and lower rails
area also single conductors that usually are connected to positive and negative (ground)
of the power supply. The supply voltage we need is 5V but the battery is 9V. So we
use a regulator 78L05 which takes the 9V and supplies 5V. The legs of the 555 chip are
labeled counterclockwise as in ﬁg.41. Then it is a matter of patience and putting all
components on the breadboard to have the circuit built, tested and working. One can – 51 – 5V
10k 8 4 1 8 7
555 2 3 7 555 10k 3 22µF 6 4 2
6 5 330Ω
1 LED 78L05
5V
9V Figure 41: Oscillator circuit using the 555 chip. The legs of the 555 are numbered counterclockwise. The 78L05 is used to get 5V supply from a 9V standard battery. use the multimeter to check the voltage across the capacitor and see how it increases
and decreases. This timing circuit can be used for many applications such and beepers,
timing circuit for a microcomputer or for example for a joystick. A computer can be
connected where the beeper is and by measuring the period of oscillations determine the
value of the resistor. If the variable resistor is connected to the stick of a joystick, then
the computer can sense its position. For more information on the 555 you can look at
the speciﬁcations with circuit examples http://www.national.com/ds/LM/LM555.pdf
and for the 78L05: http://www.national.com/ds/LM/LM78L05.pdf – 52 – Figure 42: Oscillator circuit using the 555 chip. The two capacitors on the right are not
part of the circuit, they are there to replace the working capacitor and check how the period
changes. Variable resistors Capacitor Beeper 78L05 555 Led Figure 43: Oscillator circuit using the 555 chip. – 53 – Figure 44: With a multimeter one can easily check that the capacitor charges and discharges. – 54 – ...
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 Fall '11
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