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

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

• The potentiometer is implemented before the output stage of the amplifier so almost no energy is dissipated. Commented Dec 12, 2020 at 15:39
• Yes, but you should calculate how much power is wasted. Commented Dec 12, 2020 at 16:05
• Short of some sort of digital doohickey, how would you adjust volume without a potentiometer? Commented Dec 13, 2020 at 22:47

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

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:

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.

• @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
Commented Dec 13, 2020 at 9:00
• 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
Commented Dec 13, 2020 at 12:21
• ...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.) Commented Dec 14, 2020 at 0:18
• 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. Commented Dec 14, 2020 at 11:59
• @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
Commented Dec 14, 2020 at 12:35

$$\\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.

• @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. Commented Dec 14, 2020 at 12:01
• You have every right to be a pedant. Commented Dec 14, 2020 at 12:05
• @Graham you can always write your own answer you know? Instead of telling the 2 highest voted answers that they contain incorrect information.
– MCG
Commented Dec 14, 2020 at 13:27
• @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
Commented Dec 14, 2020 at 16:57
• @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 ;) Commented Dec 15, 2020 at 15:39

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.

• 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. Commented Dec 14, 2020 at 17:51
• @HorrorVacui: Actually, many amplifiers use a fixed gain and attenuate the input signal rather than messing with the gain.
– JRE
Commented Dec 14, 2020 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. Commented Dec 15, 2020 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. Commented Dec 15, 2020 at 10:16

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 :)

• Whose circuits are you referring to?
– JRE
Commented Dec 14, 2020 at 17:48
• OP has not included any circuit diagram. Do you refer to the diagrams in @JRE's answer? Commented Dec 14, 2020 at 17:52
• 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. Commented Dec 15, 2020 at 2:56
• @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. Commented Dec 15, 2020 at 7:05