Volume control - Page 1
Forum section: electronics / beginners

author: fabiodl
author_rank: Frequent Contributor
timestamp: July 17, 2023, 11:47:36 am
content: How would you implement an analogue volume control? I am placing the volume control between the line out of a device and the input of another. I am using a 50K potentiometer, but that seems higher than the input impedance of the device downstream.Clearly the output impedance of the volume stage is too high. What is the usual way to do this? Would you place an opamp or a simple emitter follower after the potentiometer?It is in no way an audiophile setup, but I'd like avoiding degrading the signal too much if possible.Nearby I have only a +5V power source, so if I use an opamp I would have to ac couple it in both input and output.p.s. in the current configuration I have a dual log potentiometer with the brush connected to the output, one terminal to the input and the other to gnd.

author: themadhippy
author_rank: Super Contributor
timestamp: July 17, 2023, 12:04:03 pm
content: Try a 10K pot

author: Vovk_Z
author_rank: Super Contributor
timestamp: July 17, 2023, 12:23:32 pm
content: Or 4.7-5.0k pot.

author: fabiodl
author_rank: Frequent Contributor
timestamp: July 17, 2023, 02:42:33 pm
content: The next stage has 10K input impedance.

author: Vovk_Z
author_rank: Super Contributor
timestamp: July 17, 2023, 03:12:39 pm
content: So you may try 2...10k pot.

author: fabiodl
author_rank: Frequent Contributor
timestamp: July 17, 2023, 03:14:33 pm
content: thank you, I'd like to do it the proper way though

author: Vovk_Z
author_rank: Super Contributor
timestamp: July 17, 2023, 03:24:27 pm
content: There can be different approaches, so there can be several 'proper ways'. It depends on a line output possibities and how far you are from objective mesured characteristics. Possibly you prefer audiophile way of designing things by 'how it sounds'. So, in reallity, the pot may have resistance from 50 R to 10 k. I've seen it all.Typical engineering is your pot resistanse has to be 10-20 times smaller than next device input impedance (and at the same time large than you line output impedance) so log characteristics of your pot still was log. But if it want be log it will work too just a bit different from optimal log.

author: TimFox
author_rank: Super Contributor
timestamp: July 17, 2023, 03:28:20 pm
content: An interesting cheap way to make a "log-ish" volume control is to use a load resistor after a linear pot.Linear pots are easier to obtain, and are probably more reliable than the variable track width on a true audio (log) pot.For example, if your 10 k input impedance is connected to a 20 k pot (ratio 2:1), it's a not-bad control in dB/rotation from 20% to 100% of rotation, with about 17 dB at 20%.For a good treatment of this, see Merlin BlencoweDesigning High-Fidelity Tube Preamps(2016), pp 316 ff.You can easily compute the ratioVIN/VOUTfor different values of the ratio between total pot resistance and load resistance using Excel and graph it in dB to see what you get.In your case, with a log pot driving a low-ish input resistance, the result is a non-ideal dB/rotation, but no degradation of the audio signal.Exception: the pot with low load resistance presents a variable load to the driver.If the output capacitor at the driver is too low for the minimum value of that variable driver load (pot in parallel with load) is too low, there will be loss of low frequencies due to the high-pass filter of the capacitor into the total driver load.However, that driver load for your case is only a bit lower than the 10 k input resistance of the next stage.In audio mixers (recording studios), it is often more important that the various gain controls match each other than match an ideal dB/rotation, so this trick is used with linear pots that are better matches to each other than log pots.

author: antonyant
author_rank: Newbie
timestamp: July 18, 2023, 02:18:58 pm
content: There are a few ways to implement an analogue volume control. One way is to use a potentiometer, as you are doing. However, as you mentioned, the output impedance of a 50K potentiometer is too high for some devices. In this case, you could use an opamp or a simple emitter follower to buffer the output of the potentiometer.

author: donlisms
author_rank: Frequent Contributor
timestamp: July 19, 2023, 06:34:09 am
content: I notice some complex answers to this one; I was just going to say that in real-world audio circuits, the output doesn't come off a pot (except n electric guitars, which have their own rules!). So there's an amplifier of one kind of another after the pot. It can be a simple buffer, or it might do some more interesting work. The goal is to get the output impedance down to a pretty small value, theoretically as close to zero as you can manage. Practically, it might be maybe 100 ohms or some such, to limit the output current to what the amplifier is more comfortable with.It's all about a voltage divider, made from the output and input impedances stacked up.