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I recently heard that we use potentiometers to reduce the volume levels in audio devices. Will it waste electrical energy?

JRE
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Abin Latheef
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4 Answers4

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If you do this, then it will waste a lot of power:

schematic

simulate this circuit – Schematic created using CircuitLab

You'll also need a very large and heavy potentiometer because it will have to handle all the power that the speaker can handle.

If you do this, then almost no power will be wasted:

schematic

simulate this circuit

The signal is very low power, and you only waste a tiny bit of that tiny bit by varying the volume. The potentiometer can also be small because it only has to handle a tiny little bit of power.

JRE
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  • Technically speaking your first diagram shows a rheostat not a potentiometer. You would need to ground the top terminal. – abligh Dec 13 '20 at 08:04
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    @abligh The difference between rheostats and potentiometers is I think more nuanced than that. The big ~500W rheostats we had in sparky school were three-terminal devices. Generally, the differentiation seems to be in power rating, not whether the third terminal is available. Of course, the first example would then be a rheostat too, for any sensible speaker power. – SomeoneSomewhereSupportsMonica Dec 13 '20 at 08:51
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    @abligh: In common usage, people refer to both a rheostat and a potentiometer as a "potentiometer." A potentiometer connected directly to a speaker to control the volume would normally be wired as a rheostat (two connections) despite typically having three pins to connect to. – JRE Dec 13 '20 at 09:00
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    I originally had the potentiometer in the first diagram wired as a potentiometer but I changed it because as a child of the 1970s, I know that there were commercially available "volume controls" that you could connect between the amplifier and the speaker - and that they were wired as rheostats as well as having three terminals. – JRE Dec 13 '20 at 12:21
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    It might be worth adding that the passive-attenuation approach _can_ be perfectly sensible for very low-powered speakers: headphones! – leftaroundabout Dec 14 '20 at 00:05
  • How does the noise compare though? – detly Dec 14 '20 at 00:06
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    ...also, _burning power after the amp_ is quite commonly done by electric guitarists, because it allows driving a tube power stage at full volume (with the associated, desired distortion), without producing accordingly high acoustic volume. Avoiding power-stage noise from being unbearable compared to the small signal level, what @detly means, is also an advantage. (Still, this is of course pretty horrible from an engineering standpoint.) – leftaroundabout Dec 14 '20 at 00:18
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    Your answer is correct, but you're still falling into the same trap as the OP in saying it's "wasting" power. That dissipated power is needed for the resistance to do anything in the circuit. It's like saying that a Zippo lighter "wastes" lighter fuel, when of course if it didn't burn fuel then it wouldn't light anything. – Graham Dec 14 '20 at 11:59
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    @Graham: It **is** wasted. The amplifier must pass that power, and then you throw it away as heat. That is in most cases inefficient and stupid. It is wasteful to dump 99 watts of power through a rheostat to reduce 100 watts to 1 watt for the speaker. Why carry a 100 boards from the stack to where you are working when you only need 1 board. – JRE Dec 14 '20 at 12:35
  • @JRE I agree, power is wasted and Graham is overly pedantic. At least your answer didn't get a downvote LOL. – Andy aka Dec 14 '20 at 13:30
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\$\color{red}{\boxed{\text{Will it waste electrical energy?}}}\$

Potentiometers are variable resistors and resistors waste power: -

$$P_{dissipation} = \dfrac{V_{applied}^2}{R}$$

Or

$$P_{dissipation} = I_{applied}^2\cdot R$$

Power wasted is also energy wasted if you regard heat as a waste product.

If we use potentiometers as volume controls, don't they waste electric power?

Yes.

Andy aka
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  • No, resistors don't "waste" power. They dissipate power, but that's not "wasted". – Graham Dec 14 '20 at 11:56
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    @Graham if you read my answer fully you would see that I say "if you regard heat as a waste product". You will also see that I use the term "dissipation" in my formulas so really, a downvote seems like a little harsh/pedantic. – Andy aka Dec 14 '20 at 12:01
  • Saying "If you regard black as white" doesn't really change things. ;) – Graham Dec 14 '20 at 12:03
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    You have every right to be a pedant. – Andy aka Dec 14 '20 at 12:05
  • To be fair, EVERY electrical device wastes power. On the other hand, turning down the volume (probably) results in LESS power being wasted in the later amplification stage(s). – MikeB Dec 14 '20 at 12:47
  • I guess the point Graham was making is that the power dissipated by the resistor is necessary to perform its function. I.e. the resistor is able to limit the current because it produces heat. It is not a byproduct, but the very mechanism of the device's functioning. – Sredni Vashtar Dec 14 '20 at 13:18
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    @SredniVashtar he has every right to be a pedant. – Andy aka Dec 14 '20 at 13:23
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    @Graham you can always write your own answer you know? Instead of telling the 2 highest voted answers that they contain incorrect information. – MCG Dec 14 '20 at 13:27
  • And another downvote from possibly another pedant. [I can be a pedant too](https://electronics.stackexchange.com/questions/203557/how-to-implement-programmable-schmitt-trigger/203595#203595) – Andy aka Dec 14 '20 at 13:52
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    @Graham Then nothing is ever wasted, so why do we even need the word "wasted" in the English language? – user253751 Dec 14 '20 at 16:35
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    @user253751 If you think that, then you should familiarise yourself with what "wasted" means in the English language. ;) The dictionary definition is: "*used or expended carelessly or to no purpose.*" Since the potentiometer requires power to carry out its purpose, that power isn't wasted. – Graham Dec 14 '20 at 16:44
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    @MCG The purpose of comments is to advise the authors of existing answers on ways their answers could be improved. (In the opinion of the commenter, of course.) – Graham Dec 14 '20 at 16:50
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    @Graham but your comment hasn't advised any improvement whatsoever. Pretty much any electronics book will say phrases such as "power wasted as heat through a resistor" etc, and that is pretty much how it goes. FWIW I've upvoted this answer as it is correct. – MCG Dec 14 '20 at 16:57
  • @MCG well, most books do not get Faraday's law right either. Following the mass is not necessarily the right path, even though it will always get you the applause of the majority. Graham has made a sensible observation in that the figure of speech usually employed to describe a resistor's action is actually misleading. Other figures of speech are wrong: "the impedance *seen* from this port...", well no, ports do not have eyes; "this voltage *causes* a current" "this current *causes* a voltage", well no, there is no causation... We use them nonetheless, but we shouldn't defend them as correct. – Sredni Vashtar Dec 14 '20 at 22:22
  • @SredniVashtar now you're just being even more pedantic. I used the books as an example. Graham has not made a sensible observation whatsoever. I'll let you get back to your day of being petty and pedantic on other posts too. – MCG Dec 14 '20 at 23:02
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    @Graham what do we do with the heat produced by the potentiometer then? Oh right, it *goes to waste*. Or do you put little thermoelectric couplers on all your pots to recoup the heat "by-product" so it doesn't go to waste? Congrats on being out-pedanted ;) – Doktor J Dec 15 '20 at 15:39
  • @DoktorJ why is it so hard to understand? The heat is used to reduce the current/drop the voltage. This is how a resistor (or any metallic conductor with finite conductivity) works: loosely speaking you 'use' the lattice drag to alter the momentum and energy of the conduction electrons. I know this sounds faggy and pompous, but please don't send me to rehabilitation night! – Sredni Vashtar Dec 16 '20 at 01:48
  • Both views are correct depending on your perspective. How silly that grown men or women are arguing over this. Such is the way of self-righteousness, that it cannot see and appreciate the other point of view, but just has to be right. – KevinHJ Dec 17 '20 at 15:06
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    @SredniVashtar except the heat is not "used to" reduce current or voltage, any more so than a car's exhaust is "used to" run the engine: it is a *waste product* of the desired function (and therein, Allasso, is why "appreciating" any point of view without consideration of its basis in fact is not particularly useful in discussing fact-based subjects such as physics -- both are *not* actually correct). – Doktor J Dec 19 '20 at 07:55
  • @Allasso, I do not see anything wrong in arguing about these sort of matters. As a matter of fact, this is how science (and related fields) works: it's not democratic, it's not polite, it's not obsequious. It's frank. And no, sometimes both points of view cannot be right at the same time (even tho, in this case, it's just a matter of semantics and I wouldn't downvote an answer over such a minor quibble. Yes, surprise! I did not downvote.) I noticed, tho, that here and on YT sometimes people with high rep or many subscribers cannot accept to be wrong on fundamental matters. – Sredni Vashtar Dec 19 '20 at 13:02
  • I mean, this can be seen as an elementary application of the principles of conservation of energy and momentum: you **need to** take some out of your system in order to decelerate electrons, so the heat is not an unwanted byproduct but a necessity (kinda like heat from traditional brakes). You cannot have a resistor without power dissipation much like you cannot have an antenna without radiation. Would you call the power radiated into space 'wasted'? No, you took it out of the circuit with a purpose. Dissipating energy to make j constant is all a resistor do. Call @Graham is pedantic but right – Sredni Vashtar Dec 19 '20 at 13:10
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I recently heard that we use potentiometers to reduce the volume levels in audio devices. Will it waste electrical energy?

Anything resistive is going to waste power to a certain degree, however in audio amplifier applications a potentiometer is used to control the gain of the amplifier.

If memory serves me correctly:

You can think of an amplifier as a black box with a set of input/outputs, such described like this; High Voltage input, High Voltage outputs and pre-input stage side/control stage(LV).

The control of the volume is done by changing the resistance of the feedback circuit in the amplifier. The feedback circuit determines the level of gain/voltage allowed on the output. It is set by a series of resistors. The potentiometer is one part that can vary its resistance, and hence vary the gain of the amplifier circuit.

It all goes back to how a transistor fundamentally works, where a small current on the input can open a higher potential side of the transistor, via the changing the conduction properties of the junction layer via micro currents. Think of it as valve letting out water from a giant dam, it takes no effort to open the valve vs the energy expended by the water flowing. Do this fast enough and you can modulate a high power output via small input, in this case the amount of water flowing (voltage) over a given time.

tim
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binaryOps20
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    I am in disbelief that this is the only answer which points out that a potentiometer should be used to change the amplifiers gain, and not to be added into the signal path. This is definitively true and this is the reason why their dissipation will not waste signal power, but dissipate a constant amount of power set by the design constraints. – Horror Vacui Dec 14 '20 at 17:51
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    @HorrorVacui: Actually, many amplifiers use a fixed gain and attenuate the input signal rather than messing with the gain. – JRE Dec 14 '20 at 17:57
  • @JRE Well, then it seems to be application dependent. If signal-to-noise ratio is important than one will mess with the gain rather than attenuate the input. But probably you are right. At applications where it is not critical, it is way more simpler to put there an potentiometer. – Horror Vacui Dec 15 '20 at 10:07
  • Also the potentiometer has to be driven (requirement for the source), and it will dissipate some power. If they are OK for the use case, than fine. I've just not seen such solutions in my field. – Horror Vacui Dec 15 '20 at 10:16
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In your second circuit, you are loading the device that is supplying the signal. If that device has an decent output impedance (you have no control on that), the effective signal getting into your amplifier will be low. Yes, you can control volume, but the max volume will be low.

In your first circuit, the output impedance of the device supplying signal doesnt matter since the amplifier will offer high impedance. This might be a buffer, or mos gate or bjt base. The volume control pot now loads the output of the amplifier and we have control on the design of the output impedance of the amplifier.

Hence your first design is better. Yes, a bit of power goes as heat to warm up your room :)

OpenCircuit
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    Whose circuits are you referring to? – JRE Dec 14 '20 at 17:48
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    OP has not included any circuit diagram. Do you refer to the diagrams in @JRE's answer? – Horror Vacui Dec 14 '20 at 17:52
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    I believe you are referring to @JRE's answer. But how the max volume will be low in the second circuit if the potentiometer is set to 0 resistance (max the volume knob)? I did not get it. – Abin Latheef Dec 15 '20 at 02:56
  • @Abin - Sorry all. Yes, I was referring to JRE's answer. (Newbie to posting here). Abin, the max vol will be low because - say the signal source max (i.e. open ckt voltage) is Vs. Further say the source has an output impedance of Zs. The potentiometer resistance Zp loads this. If the pot is tapped to engage the full Zp (i.e. max volume), the signal available is Vs*Zp/(Zp+Zs). If Zs is large, the max amplitude of the signal available for amplification is small. – OpenCircuit Dec 15 '20 at 07:05