Hi Jake - you have asked us to comment on a solution to a problem without telling us what the problem is!

The simplest way of wiring your solution would be to use a single-pole centre-off switch and two resistors, one of which is the lower of the resistance values and the other is the difference in required values. You wire the resistors in series and use the switch to short out the difference resistor, both resistors or none (centre position).

Give us a bit more background and we could help further.

Hey Bos, long time no see. (I was Guitarfreak here many years ago)

What I want to do is install a switch into my 1981 JCM800 2203 between the tone stack and the master volume. The Stock setting will be a pass-through with no resistor like how it is now. The Resistor 1 setting will put a resistor in the signal path to lower the output volume. Resistor 2 will do the same, but it'll be about 4x larger than the value of resistor 1 to cut volume by a lot. Right now I'm thinking values of 120k and 470k.

Attached files

Don't forget that a simple resistor on it's own often doesn't change the volume, because it's going to limit the current, that's all, so do do a volume control arrangement needs to be a bridge between the output and ground, with the point between two resistors setting the voltage. In a high impedance circuit, then two 47K resistors produces half volume at the mid point - like a volume pot. The output resistor just limits any current flowing, and with valve designs often for safety. So you could experiment with resistor value that will reduce the output, but I'd expect a small tonal difference too if you load the circuit too much. I suspect I might put two variable resistors on the back panel to set the two different gains. Quite a bit of trial and error here to make sure you retain the tone.

paulears, post: 444318, member: 47782 wrote: Don't forget that a simple resistor on it's own often doesn't change the volume, because it's going to limit the current, that's all, so do do a volume control arrangement needs to be a bridge between the output and ground, with the point between two resistors setting the voltage. In a high impedance circuit, then two 47K resistors produces half volume at the mid point - like a volume pot. The output resistor just limits any current flowing, and with valve designs often for safety. So you could experiment with resistor value that will reduce the output, but I'd expect a small tonal difference too if you load the circuit too much. I suspect I might put two variable resistors on the back panel to set the two different gains. Quite a bit of trial and error here to make sure you retain the tone.

Well the position of the resistor just prior to the MV pot would act along with the 1M value of the pot, causing a downward shift to the point at which the output comes from the voltage divider network. But you have actually given me an alternate idea. I could wire up a switch to modify the value of the 1M resistance of the MV pot using parallel resistors. A 1M resistor would make the pot into a 500k pot, and a 115k resistor would turn it into a 100k pot. Also an interesting idea, though I do feel like that may have a pulldown effect on the tone stack or may screw up the voltage seen by the phase inverter. What do you think?

I see your thinking - I wonder how much of a shift will happen - I suppose you could wire the switched resistors between the wiper and the ground end and the wiper and signal end - giving you the options, and fairly simple switching. Worth having a go, as all the mods could be tried direct onto the pot solder pads? I'd wonder if their action was going to be that silent though - might get some pops if the capacitors discharge rate changes suddenly when the load gets higher? So at worst, maybe a couple of capacitors to watch over the switch??

Hi GF - good to see you back again!

This is what I meant about needing the complete picture. You shouldn't change the value of the 1M MV pot by putting single resistors in series or parallel with it as that will affect the tone control circuit.

What you could do is have alternative L-pads that you switch into circuit with a multi-pole rotary switch. The parameters that you need in order to calculate the pad values are an input resistance of 1M (when loaded with the 1M MV pot) and your required attenuations. An L-pad is a series resistor at the top and a shunt resistor to ground.

Be aware that a consequent effect of padding at the MV pot is that you may cause signal overload in the stages prior to that point.

Jesus, this is getting too stressful lol. I'll just buy an attenuator haha. Thanks for the help though, I knew coming here I would get good advice. This is like the 4th version of this same mod I've devised, and every time I look deep enough, there are strong reasons not to mess with it. Attenuator it is. Anybody here use attenuators for tube amps? I was looking at the Weber Mass 200 because it seems the most transparent. I think it simulates a reactive load instead of being purely resistive.

Maybe we should have started here: what do you want to achieve?

Boswell, post: 444458, member: 29034 wrote: Maybe we should have started here: what do you want to achieve?

I have had the JCM for a while but it was in disrepair from the previous owner until recently when I repaired it and rebiased it and etc. It sounds great now, but it really is an absurdly loud amp, 1 on the MV is already "other-end-of-the-house" loud. Given my current living situation, I can't have the MV any higher than say 9 o clock, and that's just when the tone starts to bloom. I was looking for ways to take some edge off of the volume so that I could run the master a bit higher and explore more of the amps true character. I looked into using a Variac to drop the line voltage down a bit, also heard that this would cause the amp to sag a bit, which I like, but I was told that would mess with the bias on the stages and was not the best thing to do to a classic amp (this one is from first year of production of the 2203 series).

Also, I love switchable mods. My 5150 has a gain cut switch that I designed and installed, and the JCM is going to have a 3 way mod switch installed when the parts arrive for stock gain, more gain, and way more gain, by altering the cathode bias on one of the gain stages. There's something about increasing the versatility of a piece of audio equipment, while also retaining its completely stock character at one of the switch positions that makes me happy.

Ah - personally, I suspect that all these mods will be tone negative - until the amp starts to kick the tone will be horrible. Cutting down drive to the finals won't be a positive mod. Why don't you consider these kinds of products - much, much simpler.

https://www.thomann.de/gb/attenuators.html

paulears, post: 444675, member: 47782 wrote: Ah - personally, I suspect that all this mods will be tone negative - until the amp starts to kick the tone will be horrible. Cutting down drive to the finals won't be a positive mod. Why don't you consider these kinds of products - much, much simpler.

https://www.thomann.de/gb/attenuators.html

Yeah, that's probably the route I'm going to go. I may still try that Variac tho :p

Good to see you again, Jake! I thought you were the king of the isobox / re-amping setup, or is that not a part of your current living situation?

dvdhawk, post: 444795, member: 36047 wrote: Good to see you again, Jake! I thought you were the king of the isobox / re-amping setup, or is that not a part of your current living situation?

Good to see you too man! Yeah, those were the days! That's correct, living situation changed a few years back and I had to give up reamping... That may change in the near future hopefully.

I would strongly advise against using a Variac on the mains input. Going that way could drastically shorten the life of the output valves (tubes).

If you really want to maintain the amps's tone when driven hard yet have manageable decibel levels, then the only thing that could do this is padding at the loudspeaker terminals. This would involve some careful calculations and high-wattage resistors.

Boswell, post: 444867, member: 29034 wrote: I would strongly advise against using a Variac on the mains input. Going that way could drastically shorten the life of the output valves (tubes).

If you really want to maintain the amps's tone when driven hard yet have manageable decibel levels, then the only thing that could do this is padding at the loudspeaker terminals. This would involve some careful calculations and high-wattage resistors.

You mean like a U-Pad or an H-Pad? So like the parallel resistor is somewhere in the vicinity of 10x the cabinet load, and the series resistors are some value calculated against the speaker load which would cause a significant power drop? So if my speaker load is 8Ω and I'm going to make a U-Pad, then the parallel resistor could be 82Ω, and the two series resistors could be 12Ω each and very high wattage? I'm going to make the assumption that is not safely switchable. Ooh! I can make it into an external box though!

I also know that the 8Ω rating is more of a colloquialism than an actual hard rating, and that the actual moment to moment "resistance," while reactive, is more likely in the 18-25Ω range. So that may change the values if those are the numbers we are concerned with.

I'm just making guesses at this point, haven't done any math yet.

No issues with the switchable aspect, as mains switching voltages are unlikely to be exceeded here. In most cases this design of item presents a higher than normal impedance to the amplifier output - s0 switching extra pads in and out doesn't cause any issues.

Now I see you've been at this idea for quite a while, can I comment? You've had loads of suggestions on doing what you seem to want both internally, requiring mods and possibly extra holes in the amp chassis, plus external boxes - and you don't seem to have actually tried ANY of the - even the simplest.

You are doing a Don Quioxte quest - getting nowhere fast, but constantly increasing your options, yet failing to try any of them. Borrow an external one, or buy one from a dealer who is happy to take returns and try it out. You are going backwards. Plug one in and play! All your choices simply increase time between desire and conclusion.

Here's a page describing loudspeaker padding with a resistor calculator. I haven't checked the formulae, but at first glance, the second set of equations is what you would need. However, he doesn't give power dissipation ratios, so you would have to calculate the resistor wattages separately.

As an example, if you took his table figures for a 12dB (16th of the power) attenuation at 8 Ohms, the series resistor would be 6 Ohms (e.g. 2x 12 Ohm or 3x 18 Ohm in parallel) and the parallel resistor 2.7 Ohms. The parallel resistance comes down to roughly 2 Ohms when the speaker is connected, so the 100W power output of your amp would be split as 75W into the 6 Ohm resistor and 25W into the 2 Ohm composite, of which about 20W goes into the 2.7 Ohm resistor when the speaker is connected.

Sourcing resistors to carry this load is a slightly tricky task, and I would start by looking at the 25W THS range from TE Connectivity (Tyco). There is a value of 18 Ohm 25W, so 3 of those in parallel gives 6 Ohm at 75W. The shunt resistor is easier, so you could use the 2.7 Ohm at 25W. If there is a danger that the speaker could become disconnected from the pads, then use a higher wattage resistor for the 2.7 Ohm. Use thick wire for connecting the resistors, as it is very easy to end up unintentionally with higher resistance values if you use standard connecting wire.

paulears, post: 444983, member: 47782 wrote: No issues with the switchable aspect, as mains switching voltages are unlikely to be exceeded here. In most cases this design of item presents a higher than normal impedance to the amplifier output - s0 switching extra pads in and out doesn't cause any issues.

Now I see you've been at this idea for quite a while, can I comment? You've had loads of suggestions on doing what you seem to want both internally, requiring mods and possibly extra holes in the amp chassis, plus external boxes - and you don't seem to have actually tried ANY of the - even the simplest.

You are doing a Don Quioxte quest - getting nowhere fast, but constantly increasing your options, yet failing to try any of them. Borrow an external one, or buy one from a dealer who is happy to take returns and try it out. You are going backwards. Plug one in and play! All your choices simply increase time between desire and conclusion.

Its been like two days dude, lol. I am still researching.

Boswell, post: 445028, member: 29034 wrote: Here's a page describing loudspeaker padding with a resistor calculator. I haven't checked the formulae, but at first glance, the second set of equations is what you would need. However, he doesn't give power dissipation ratios, so you would have to calculate the resistor wattages separately.

As an example, if you took his table figures for a 12dB (16th of the power) attenuation at 8 Ohms, the series resistor would be 6 Ohms (e.g. 2x 12 Ohm or 3x 18 Ohm in parallel) and the parallel resistor 2.7 Ohms. The parallel resistance comes down to roughly 2 Ohms when the speaker is connected, so the 100W power output of your amp would be split as 75W into the 6 Ohm resistor and 25W into the 2 Ohm composite, of which about 20W goes into the 2.7 Ohm resistor when the speaker is connected.

Sourcing resistors to carry this load is a slightly tricky task, and I would start by looking at the 25W THS range from TE Connectivity (Tyco). There is a value of 18 Ohm 25W, so 3 of those in parallel gives 6 Ohm at 75W. The shunt resistor is easier, so you could use the 2.7 Ohm at 25W. If there is a danger that the speaker could become disconnected from the pads, then use a higher wattage resistor for the 2.7 Ohm. Use thick wire for connecting the resistors, as it is very easy to end up unintentionally with higher resistance values if you use standard connecting wire.

I think you forgot to post the link. The site probably has an explanation for this, but why is the shunt resistor so small? Wouldn't that pull down the impedance seen by the power tubes in much the same way that running a head set for 16 ohms into a cabinet rated at 4 ohms would do?

Hey Jake, hover over the word "Here" at the beginning of Boswell's posted. It is the link. I missed it at first too.

JakeAC5253, post: 445122, member: 46399 wrote: I think you forgot to post the link. The site probably has an explanation for this, but why is the shunt resistor so small? Wouldn't that pull down the impedance seen by the power tubes in much the same way that running a head set for 16 ohms into a cabinet rated at 4 ohms would do?

Yes, sorry, the default colour is not very visible for text that is a link, but Hawk showed you where it was.

The numbers I used are correct for an 8 Ohm loudspeaker and a 12dB loss. Those resistor values will present an 8 Ohm load to the amp. You would need to change them for other loudspeaker impedances. Note that padding in this way gives a very constant resistance as seen by the amplifier, unlike a real loudspeaker whose impedance varies quite markedly over the frequency range.

Boswell, post: 445143, member: 29034 wrote: Yes, sorry, the default colour is not very visible for text that is a link, but Hawk showed you where it was.

The numbers I used are correct for an 8 Ohm loudspeaker and a 12dB loss. Those resistor values will present an 8 Ohm load to the amp. You would need to change them for other loudspeaker impedances. Note that padding in this way gives a very constant resistance as seen by the amplifier, unlike a real loudspeaker whose impedance varies quite markedly over the frequency range.

I see what you mean about sourcing these resistors. I was able to find a 50w 2.7Ω resistor for a good price, but you were right, the 6Ω, or enough resistors to equal 6Ω with the right wattage and cheap enough, is hard to track.

It makes the purposes built ones seem pretty good value, doesn't it?

JakeAC5253, post: 445350, member: 46399 wrote: I see what you mean about sourcing these resistors. I was able to find a 50w 2.7Ω resistor for a good price, but you were right, the 6Ω, or enough resistors to equal 6Ω with the right wattage and cheap enough, is hard to track.

Any reason these high wattage aluminum housed resistors from Mouser wouldn't work? (Roundabout $3 each)

1x 25w Wirewound 1% tolerance 2.7Ω

3x 25w Wirewound 1% tolerance 18Ω

OR

1x 25w Wirewound 1% tolerance 2.7Ω

2x 50w Wirewound 1% tolerance 12w

You'll be paying extra for the high-tolerance (1%) parts - they're not necessary. Standard (5%) are fine and a lot cheaper.

You'll need to mount the resistors on a large heatsink.

dvdhawk, post: 445479, member: 36047 wrote: Any reason these high wattage aluminum housed resistors from Mouser wouldn't work? (Roundabout $3 each)

1x 25w Wirewound 1% tolerance 2.7Ω

3x 25w Wirewound 1% tolerance 18Ω

OR

1x 25w Wirewound 1% tolerance 2.7Ω

2x 50w Wirewound 1% tolerance 12w

Ahh, I didn't see the 50w 12Ω ones, those would be good enough. I didn't like the 18Ω ones because Boswell mentioned that they may have to dissipate heat in the range of about 75w, and with those resistors combined, they would be able to dissipate exactly 75w. I know its a good practice to rate your parts to run at no higher than 80% of their capability, so I multiply what I need by 1.33 and try and source that value.

I've actually been very busy lately. Working 60-70 hour weeks, and in my free time trying to refurbish my Marshall, fix my car, and also preparing to build a new PC. I will probably just buy a pre-made attenuator, I don't yet trust my execution skills to risk my 35yr old amp if I do something wrong. I mean there should be ventilation and a fan or two ideally, it should be designed with some sort of a soft-failure mode etc... Maybe keep the parts on a wish list for a later project if I get more time freed up.