**October 22, 2008 Educational Radio Net, PSRG 22nd session, Lee Bond N7KC**

The impedance series is now history. During the course of 13 weeks we looked at several of the most fundamental ideas in the physics of electrical phenomenon and, hopefully, gained some practical knowledge of how these ideas link together to form a basis for our understanding of all things electrical. Let's exercise some of this earlier impedance series material and see how it can be applied to solve practical problems which are routinely encountered on the bench. My choice for the first study is the potentiometer or "pot" in the vernacular.

There is one wee problem with the word potentiometer which we must clear up before proceeding... there are two devices which share the same name but perform different duties in the electrical world. In early laboratories one could find a very elegant device, with many knobs, generally in a nicely crafted wooden box, and which was used to measure electrical potential differences with great accuracy. Today this function is performed by sophisticated digital voltmeters and one rarely sees the older instrument except in museums. The potentiometer that we will study is the familiar device commonly found on the front panels of our radios, which can be rotated to produce some desired action.

These devices are everywhere. Virtually all "level" controls such as audio volume, AGC, squelch, power supply voltage output, and many more are based on the lowly pot so a good grasp of the underlying operational details is a must for your bag of tricks. We all know what a pot looks like physically. It is a resistive device which has 3 contact points. Basically each end of the resistive "element" is attached to one of the points. The remaining contact point, generally known as the "wiper", connects to a sliding assembly which is controlled by some knob or motor, and which makes a mechanically movable contact which is adjustable from one end of the resistive element to the other.

The resistive element proper was carbon in the early days of this device but modern materials have largely replaced carbon. More common today is the very robust cermet element and the even more robust wire wound element. The carbon element tended to abrade as the wiper slid along its surface and they would become "scratchy" and very annoying. Cermet has much less tendency to abrade and also offers what is called infinite resolution. In contrast is the wire wound pot which may or may not offer infinite resolution depending on construction. If the resistance element is just a length of resistance wire formed in a circle then the resolution would be deemed infinite since the slider can find any point on the wire. If the wire resistance element is a helically wound structure which is then formed in a circle then the wiper can only contact discrete points along the main wire and the pot cannot be set infinitely fine.

The most common pot is structured with a linear taper meaning that doubling the angle of rotation will double the resistance from the wiper with respect to a designated end of the element. Additionally, there log taper pots where the resistance changes logarithmically with rotation angle. The log class includes the audio taper pot which produces a uniform change of loudness to your ear with uniform shaft rotation if used as a volume control. Variations here are log clockwise or counter clockwise.

There is a class of very high precision multiturn wire wound potentiometer devices called Helipots by Beckman. Bourns and others offer similar devices. These offer 5 turn, 10 turn, and 20 turn rotations so the total resistance can be controlled over as much as 7200 degrees of rotation. The linearity of these devices as deviations from a straight line are specified and they all offer extraordinary precision. Generally used with turns counting dials.

One last point concerning the use of wire wound pots is in order. In addition to the desired resistance mechanism there is an added component of inductance present. If the element is helically wound then the inductive component is much larger than that encountered with the simple wire element. Inductance/reactance effects limit the use of these sort of pots in AC circuits hence they are more commonly found in DC circuits.

Ok, let's build a circuit. We will need a power supply of some sort so how about using a 10 volt battery. 10 volts will be convenient for our discussion even though a 10 volt battery would be an oddity for sure. Then we need a pot to work with. Lets choose a simple 1000 ohm carbon unit rated at 2 watts and which is structured as a linear device. That's it for our circuit parts... just a battery and a pot. Lets connect the battery and pot in series in the following manner. Pot contacts are normally labeled 1, 2, and 3. Contact 1 is commonly the low potential reference so it will go to the battery negative terminal. Contact 2 is always the wiper and common convention states that the wiper, contact 2, moves toward the contact 3 end with clockwise rotation of the knob. So, to complete our circuit we connect contact 3 to the battery positive terminal.

Now we need a measuring device so let's choose a VOM as in Volt-Ohm-Milliamp meter. In fact we will need two of these meters so let's use the common Simpson 260 VOM. We want to measure the series current flowing in our circuit so disconnect the pot contact 3 from the battery positive and insert, in series, one of the VOM's with positive lead going to the battery positive and the negative (common) lead going to contact 3 on the pot. From Ohm's Law we expect the series current to be 10 volts divided by 1000 ohms and, sure enough, the series meter shows the current to be 0.01 amperes or 10 milliamperes. (Note: let me assert that we are using a "perfect" meter here... one that does not influence the circuit being measured. In real life no such device exists and all measuring instruments change the circuit to some degree. The effect is commonly described as "loading".)

From earlier discussions we know that 1 ampere is defined as 1 coulomb of charge per second past a given point so the 0.01 ampere represents 0.01 coulombs per second flowing in our circuit. Also from earlier discussions we know that you cannot impress any voltage on a resistor without the resistor becoming warmer than it's surrounding environment. The job of a resistor is to convert the energy of moving electrical charge to heat energy. Remember Joule's Law? The power calculation for our little circuit is voltage squared divided by resistance or 0.1 watt. From earlier discussions we know that 1 watt is one joule per second so we conclude that 100 milli-joules of electrical energy per second is being converted to heat in our pot resistive element and a sensitive thermometer would show some upscale movement.

Now, adjust the pot shaft fully clockwise, and let's add the second meter as a voltmeter and place the meter negative lead on the battery negative and the meter positive lead on contact 2 of the pot. Fully clockwise moves contact 2 to contact 3 and we see 10 volts on the meter as you would expect since both are in contact with the battery positive terminal. Now rotate the shaft to mid rotation, half way between rotational extremes, and notice that the meter indicates 5 volts or 1/2 of the previous initial reading. Moving the shaft again such that the wiper moves toward contact 1 shows that voltage goes toward zero whereas moving from midpoint toward contact 3 shows the voltage going toward maximum.

Now consider the shaft at mid position where we measured 5 volts on the meter. Since the pot is linear the mid position resistance should be 1/2 of the 1000 ohms or 500 ohms. At mid point we would expect 1/2 of the total power to be dissipated above the wiper position and 1/2 dissipated below. At mid point we measure 5 volts and we know that the resistance is 500 ohms. So, 5 squared divided by 500 from Joule's Law gives 0.05 watts which is 1/2 of the total 0.1 watts dissipation.

The idea of "voltage drop" follows directly from the lesser amount of energy dissipated as the wiper approaches the reference terminal or contact 1 on the pot. By extension, one could partition the resistor element into 10 equal sections and then argue that the total must be the sum of the parts so each part would then dissipate 0.01 watts. Each partition would then "drop" 1 volt over 100 ohms which computes to 0.01 watts.

My point here is to show that, yes, one can talk glibly about voltage drops around a resistive circuit, but the underlying principle is directly related to energy conversion to heat. Resistors always throw something away but the wiper on a pot allows you to choose at what level you want to save. Basically a pot is an attenuator. The output signal will never be larger than the input because of the energy conversion into heat phenomenon.

The pot is a simple voltage divider and the output voltage can be easily calculated. The fraction of the tapped off resistance divided by the total resistance times the input will yield the output voltage. For example, using our 1000 ohm pot, if the tapped resistance is 133 ohms and the input voltage were 8.5 volts then the output voltage is 133 ohms divided by 1000 ohms times 8.5 volts or 1.1305 volts. Conversely, if one knows the output voltage and the tap ratio then computing the input voltage is a piece of cake. In like fashion, if you know the input voltage and desired output voltage then calculating the tap point is one more piece of cake.

One last point... sometimes you will see a pot symbol wired with terminals 1 and 2 or 2 and 3 connected together. In this case the pot is wired as a rheostat and is nothing more than a variable resistor. It is not possible to voltage divide with a single rheostat. You must have at least two units to achieve voltage division.

In summary, all resistors dissipate energy and will be measurably warmer than their environment. Voltage drops are a direct consequence of energy dissipation in a resistive element. Kirchoff's voltage law which states that the algebraic sum of the voltages in a closed loop is zero is simply a restatement of conservation of energy where total energy converted to heat equals total input energy. Given that work and energy are identical it follows that work in equals work out hence the net work is zero.

This concludes the set up for the discussion of potentiometers or pots. Are there any questions or comments?

Something to ponder: Two atoms are leaving a bar when one says to the other "I left my electrons in the bar". The other says to the first "are you sure?" The first replies "I am positive".

This is N7KC for the Wednesday night Educational Radio Net.

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