What is the difference between a 15A pot and a 30A?

[Bordonbert]

Hi. Can anyone enlighten me as to how a 250k 15A pot and a 250k 30A pot are different?

As an example we are looking at the schematic number 050384 Rev B for the Hot Rod Deville here. The tonestack show the treble control as the 30A and the Bass control as the 15A. I just can't imagine any significant difference in those pots unless it's to do with "15% halfway point" or "30% halfway point" tapers. I had never come across this before but when I found it and looked online there are people selling the two as different parts.

[Offshore Angler]

Hi. Can anyone enlighten me as to how a 250k 15A pot and a 250k 30A pot are different?

As an example we are looking at the schematic number 050384 Rev B for the Hot Rod Deville here. The tonestack show the treble control as the 30A and the Bass control as the 15A. I just can't imagine any significant difference in those pots unless it's to do with "15% halfway point" or "30% halfway point" tapers. I had never come across this before but when I found it and looked online there are people selling the two as different parts.

It has to do with how much current the potentiometer is rated for. Generally, you can replace with higher Amperage rating but not the other way around. The other important thing you need to know is what the taper is. Audio vs linear taper makes a big difference - especially on volume controls.

[Bill Moore]

I can't imagine any reason for either 15 or 30 amp rated pots in a circuit with a 3A line fuse!
Most of the "regular" CTS pots we use in builds are about 1/3A, and the smaller snap-in pots I suspect are rated much less.

[Bordonbert]

It has to do with how much current the potentiometer is rated for. Generally, you can replace with higher Amperage rating but not the other way around. The other important thing you need to know is what the taper is. Audio vs linear taper makes a big difference - especially on volume controls.

Come come now, think about it, that's absolutely impossible. A 15amp pot and a 30amp pot? Maybe on the Planet Express in Futurama but not here on Earth. Let's do a bit of calculation to look into that. Say we force 5A through a 250k pot. That would take a signal of 1.25MegV! A 25V signal across it would mean a current of only 100uA, (that's 1uA=1millionth of 1amp. At just 1A a 250k resistance is dissipating 250W. [EDIT: That's a dozey late night mistake to make. I missed out the 'k' didn't I. ] I suppose it might do it for a few milliseconds. It can't be anything to do with current as there isn't any current worth even considering in the tonestack or any of the controls in the HRD.

I still think it could be that the taper is adjusted so the breakpoint is at either 15% or 30% of its rotation. Most audio/log pots are made as simple approximations based on linear sections which are easier to produce. There are 2 or 3 areas of different steepness which switch from one to the other at a specified point along the track. Maybe the 15A or 30A stands for an Audio taper with its breakpoint at 15% or 30% of travel. It could mean that it breaks at 50% travel where the resistance is set to climb to 15% or 30% of its total value. Measuring one would give us the answer of course but I don't have one out of circuit.

[Offshore Angler]

Interesting, let's examine that. At 250K yep, I'd need the Flux Capacitor at 1.2 gigiWatts. (What was I thinking Marty?) But - it's not a resistor it's a potentiometer, and it goes down to zero Ohms when you turn it the other way. So Since I=V/R (ok, it's really impedance because we're in AC but you get the gist ) when R goes to zero I goes to infinity, right? Ok, we have capacitors, pickup resistance and inductance too so there's a time constant involved so it's more than just Ohms law to consider.

[phantomman]

Ok, we have capacitors, pickup resistance and inductance too so there's a time constant involved so it's more than just Ohms law to consider.

Correct.

It's a variable RCL circuit plus the dynamics of the pickups themselves.

[Bordonbert]

Interesting, let's examine that. At 250K yep, I'd need the Flux Capacitor at 1.2 gigiWatts. (What was I thinking Marty?) But - it's not a resistor it's a potentiometer, and it goes down to zero Ohms when you turn it the other way. So Since I=V/R (ok, it's really impedance because we're in AC but you get the gist ) when R goes to zero I goes to infinity, right? Ok, we have capacitors, pickup resistance and inductance too so there's a time constant involved so it's more than just Ohms law to consider.

No, sorry but that thinking is not correct. The potentiometer does not go down to 0ohms, unless it is specifically configured as a variable resistor by connecting the wiper to one end. A potentiometer is a method of tapping off variable levels of a fixed resistor so what I said holds true. As to the idea of bringing in impedance and time constants with the other capacitance and inductance in the circuit, they are irrelevant as that makes absolutely no difference to the internal voltage and current of the potentiometer. Sure they affect what is going on in the overall circuit block externally but to the potentiometer itself they can make no difference at all. Within the pot V=IR holds absolutely, as does P=V^2/R and P=I^2*R. That's the Physics!

I only quoted the calculations as a rough and ready way to show the ridiculousness of a 250k pot, (effectively a resistor as we can ignore the wiper position the overall resistance of the track staying absolutely constant), having even 1A through it let alone 15A or 30A. It is inevitable that if you drive 1A through a 250k potentiometer's outer lugs across the full track it would be dissipating 250kW - period! As I said Power = I^2*R = 1A*1A*250k = 250kW and nothing outside of the pot affects that if it has that 1A passing through it.

[Bordonbert]

Correct.

It's a variable RCL circuit plus the dynamics of the pickups themselves.

We are only considering the potentiometer. It, as an individual component, is specified as 15A or 30A whatever that means. You seem to be considering the situation where the pot is used within a guitar as a volume or tone control. How much current are you imagining the pickup is capable of generating? The currents inside the guitar are vanishingly small, microamps would be good! There is absolutely no situation where a pot inside the guitar could pass even 1mA of current.

This pot is used inside the amp. Again, every one is fed from a high impedance source. That's from the anode of the preceding valve. That's an output impedance in the ballpark of 100k. That alone limits the current that could even possibly pass through any of the pots. How much current could that supply to the pot with even a huge signal passing through in amplifier terms? Here is a link to the datasheet for a common or garden Bourns pot just like the ones we use. https://docs.rs-online.com/03f3/0900766b8157241c.pdf. Note that the spec sheet sets maximum voltage at 250V and maximum power dissipation at 0.25W. (Left hand side under Environmental Characteristics.) 0.25W for a 250k pot means a maximum permissible current of 1mA. Think about that. 1A?

[Offshore Angler]

A, correct, we are considering a single component, as that is indeed the subject of the conversation.

As for the rest of your editorial, show me the math. Pretty sure I can follow it. If you want to really impress me do phasor diagram for your circuit too and show me the phase shift. And don't forget the capacitors.

And 30A does mean 30%, not 30 Amps on the pot. I checked.

[Bordonbert]

A, correct, we are considering a single component, as that is indeed the subject of the conversation.

As for the rest of your editorial, show me the math. Pretty sure I can follow it. If you want to really impress me do phasor diagram for your circuit too and show me the phase shift. And don't forget the capacitors.

And 30A does mean 30%, not 30 Amps on the pot. I checked.

Thanks for acknowledging that the 30A is nothing to do with amps, we are certainly in agreement on that. But the little bits of analysis in which I DID show you the maths, (the very simple maths and all that is relevant), should really have been enough for it to be clear. That should really be engineering common sense if you realise that any current at all in a 250k resistor, (which is what the pot is), means a huge dissipation, far more than the pot could ever stand in real life. Remember, to get it to draw just 2mA you would need to put 500V across it and that would dissipate 1W which is certainly well over its maximum dissipation as the datasheet I linked to showed.

Phasor diagrams? Why do you need to test my knowledge of phasor diagrams which are irrelevant to the discussion? I'm not interested in chipping out what you actually know, I can assess that from your current responses, neither am I interested in showing off my mathematical background for no reason. I don't understand why you would be testing someone who only asked a simple question about a feature of specification in that way? My education, work background and experience gives me a very good working knowledge of phasor diagrams thank you. They are absolutely irrelevant here as, I say once again, we are only considering the pot for the benefit of its specifications to be isolated from any other circuitry. The phasor diagram under those conditions would show a steadily rotating vector footed on the origin representing the signal voltage, assuming it is a sine wave, with a second vector representing the current through the pot absolutely in line and on top of the first. As I said, it's a case so basic it is irrelevant.

You still seem to be missing the point by a mile. All external components, capacitors and inductors, certainly would produce phase shifts in the signal within the external circuitry. That is again obvious but, that is absolutely irrelevant here. The parameter that is under discussion is nothing to do with the pot being used in a real circuit. It is purely an indication of the shape of the track taper. That is based purely on the component's construction out of any circuit. No capacitance! No Inductance! At least at audio frequencies. Phasor diagrams are irrelevant for the voltage and current of a simple pot where we are only discussing its own internal actions! It is a purely resistive device and what happens internally to it is simple if that is completely divorced from what may be happening outside or even inside it when it is applied in a complex circuit. There are no useful phasor diagrams to be drawn for a purely resistive device under conditions of isolation. At last look Ohms Law still rules in our part of the Universe. The current through the pot track as far as the pot is concerned is entirely dependent on one single factor. The voltage it sees across it. It is scaled according to its resistance and it is in phase with the voltage. In isolation, the relationship between its voltage across and current through is purely and absolutely linear - always. They have no phase shift involvement at all. Now put it in a real circuit with capacitance and inductance in the mix and, of course, you will see phase shifts, I am of course very aware of that. I design utilising that principle regularly. BUT! The point is, once again, that that is not what we are discussing here.

The "15/30" designation is nothing to do with current, we have established that. I knew that without even considering it due to years of experience in circuit design as I am sure you did. The 'A' is of course the taper designator. It is part of the isolated specification of the pot. How that affects and is affected by any circuitry it is put into is a different story. An irrelevant one for this discussion.

Do you not agree?



EDIT: In the spirit of clearing this up, how about this? Tell you what Offshore Angler. Would it be easier if YOU SHOW ME where you think I am wrong in this. Show me the phasor diagram which you think is relevant to the discussion and which displays the phase shift you seem to think is inherent in action of the pot. Is that an easier option? I may perhaps be missing something so obvious it is below my eyeline and out of sight. I am never too old to learn even after many years working in electronics design and will happily acknowledge openly any mistake I make when it is clearly pointed out to me.

Here is a schematic of the tonestack with approximate source and load values inserted. I can't seem to get it into the actual text of the post but you can view it on Dropbox. How do you visualise the phasor analysis of the 25k Bass pot as it currently stands in this schematic?

https://www.dropbox.com/s/fcy2lygtbe...lysis.jpg?dl=0

[Bordonbert]

Incidentally, my apologies, in my haste I made a silly mistake in that post #4. 1A through 250k is NOT 250W. I typoed out something very relevant. The 'k'. It makes a bit of a difference I'm sure you will agree.

[Bordonbert]

Ok, we have the answer so here it is for anyone else curious about this.

I asked Fender support about this:

I have a number of 250kΩ pots in the amp. On the schematic the component types are designated 250kΩ15A and 250kΩ30A. I have also now found 250kΩ20A pots in other schematics. I am fully aware of what the 250kΩ and 'A' designators mean but the 15/20/30 figure is new to me.

I am guessing it would be a specification of the taper of the pot, perhaps the rotation %age of the breakpoint between two linear tapers to approximate the log/audio curve, or even the height at 50% track rotation, but that is guessing. No one I can find seems to understand what his implies. Can you please shed some light on this for me and clear up my confusion?

This is their reply:

I have spoken to our technicians regarding your enquiry and they have advised that you are indeed correct about the taper on the pots. They also recommended visiting this article here which explains this using some diagrams - https://www.alphapotentiometers.net/...er_curves.html

There it is confirmed from the horses's mouth. That link should make it clear for everyone, it's pretty interesting for us tech types. If you check out the first set of curves for Taper 'A' types you will see that at 50% travel the resistance has increased by the percentage of the pot rating, i.e. a '20' pot has increased by 20% at 50% travel. It doesn't seem to set the breakpoint in any way but at least we know what it actually does define now.