Variable resistors can be used in the two different ways shown in figure 2.4. Using two wires connected to the central tag and one of the ends we can make a resistor whose value can be adjusted to any value where is the pots end-to-end resistance and is takes a value from 0 to 1 depending on how we set the wiper. The volume and balance controls on Hi-Fi amplifiers are usually pots used as potential dividers. These allow us to ‘tap off’ a variable amount of an input signal.
We can use the example of a volume control to see how groups of connected resistors affect signals. Figure 2.5a illustrates a volume pot as a potential divider. The values of the two resistances in this circuit, will depend upon where the wiper is set along the curved resistive track inside the pot. Note, however, that no matter how we adjust the pot the sum of these two resistors will always equal the pot's end-to-end resistance, i.e.
for any value .
The voltages & currents in any circuit can be worked out using two basic rules
- 1) The total current flowing along the wires into & out of any node (this is what engineers call any point where things are connected together) must be zero — i.e. any current which enters by one wire must leave by another.
- 2) Components or groups of components connected in parallel (in between the same two points) must have the same voltage across them — i.e. the potential difference between any pair of nodes must be the same, no matter how we trace around the circuit between them.
The connection between and is a node. In 2.5a no other components are connected to this node. Hence any current flowing into this node via must leave through , therefore the currents in the two resistors must be the same. Giving this current the symbol, I, we can now use Ohm's Law to say that the voltage across must be and the voltage across must be . Since the two resistors are connected in series (i.e. end-to-end) the voltage across both of them must be . This is the voltage we're applying, so we can say that
The output from the volume pot is the potential difference between the wiper and one end — i.e. across one of the resistors in the potential divider, . We've already said that this voltage will be equal to , hence we can write that
We can now divide expression 3 by expression 2 to get
i.e. we can say that
where we can set any value we like in the range from 0 to 1 by adjusting the pot &rotating the wiper contact inside. This shows how we can use the volume pot to alter the signal size from a maximum (loudest) level of down to nothing.
A problem with the above analysis is that we haven't actually connected anything to the pot's output! Obviously, we have to connect an amplifier, or speaker, or something, to actually collect the output signal. Since signal information always requires the transmission of energy whatever we connect must draw at least a little current whenever the pot wants to give it an output voltage. This means that whatever we connect to the pot's output must look — so far as the pot is concerned — as if we'd connected a load resistance to its output as illustrated in figure 2.5b.
This load resistor changes the situation. There is now another way for current to flow in or out of the node (connection) between & . As a result, the currents, , in these resistances will now differ. Looking at the circuit we can see that the part of the pot and the load resistance, , are connected in parallel. The voltage across each of these will therefore be the modified output voltage, . From Ohm's Law we can therefore say that
The current flowing through into the node connecting all three resistors will be
Where is the voltage across the resistor, . Since we are applying an input, , to one end of this resistor and the voltage at the other end is we can say that
(This is just another way of saying that it has to true that since we're applying across the two resistances in series.)
combining expressions 8 & 9 we get
Since the total currents in and out of the node must balance we can also say that
combining 10 & 11 to eliminate we find that
multiplying both sides of this expression by produces the result
Expression 15 tells us the new output voltage when we connect something whose input resistance is to the output of volume control (potential divider). This ‘loads’ the potentiometer and produces a lower output, , than the voltage, , we would get if nothing were connected.
Whenever possible it makes sense to arrange that . To see why, consider what this does to expression 15. Here we find a relatively large value of means that and we can say that
Comparing this with expression 4 we can see that this is identical with the ‘unloaded’ behaviour of the volume control. Hence we can minimise the effect of the output load by making its resistance as big as possible.
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University of St. Andrews, St Andrews, Fife KY16 9SS, Scotland.