Mixing Resistors

[dickjonesify]

I have this amp...



And I'm gutting the 2nd channel to put the normal channel of this in instead (with a switched bright cap)...



I plan on leaving the mixing resistors from the Gibson (R17, R18) SO: should I delete the 270K mixing resistor from the 5F6A channel? I'm sure there is math associated with this.

Side note, the Gibson's tone stack is just... awful haha so I will most likely change that out to something else as well. One problem at a time.

Thanks

[tubeswell]

 there are ‘normally’ 2 equal value mixing resistors in a passive resistor mixer. As to whether you keep the 470k in series with another resistor or not in the circuit you are planning depends on whether you like the result or not I guess.

[dickjonesify]

there are ‘normally’ 2 equal value mixing resistors in a passive resistor mixer. As to whether you keep the 470k in series with another resistor or not in the circuit you are planning depends on whether you like the result or not I guess.

Ok, cool. I guess I’m just making sure it’s more of a matter of taste and less a matter of blowing something up 

Thanks

[pdf64]

Sorry but I find your circuit description somewhat ambiguous. A schematic of what you're planning would be much better, just sketch it out on paper, take a photo and upload it.
Or photoshop those Gibson and Fender schematics together somehow.

The concern being that the mixing resistors in the Gibson are at anode V DC, whereas all likely channel output points in the Fender are at 0V DC.
Hence a blocking cap will be needed.

[dickjonesify]

Sorry but I find your circuit description somewhat ambiguous. A schematic of what you're planning would be much better, just sketch it out on paper, take a photo and upload it.
Or photoshop those Gibson and Fender schematics together somehow.

The concern being that the mixing resistors in the Gibson are at anode V DC, whereas all likely channel output points in the Fender are at 0V DC.
Hence a blocking cap will be needed.

I will make that drawing! Good idea.

To simplify with a quick question: If you were building a 5F6A but only used one channel, would you omit the 270K resistor right before the grid of V2A?

[sluckey]

If you were building a 5F6A but only used one channel, would you omit the 270K resistor right before the grid of V2A?

yes

[dickjonesify]

Sorry but I find your circuit description somewhat ambiguous. A schematic of what you're planning would be much better, just sketch it out on paper, take a photo and upload it.
Or photoshop those Gibson and Fender schematics together somehow.

The concern being that the mixing resistors in the Gibson are at anode V DC, whereas all likely channel output points in the Fender are at 0V DC.
Hence a blocking cap will be needed.

Boom! See any issues with this? Spoiler alert: I have wired it up and it does function and sound good overall but I’m sure some things could be better. I did leave that .022 coupling cap at the end of the Bassman channel before it joins the Gibson channel at the phase inverter.

Oh and I omitted the 270K where the red line is.

[sluckey]

Looks fine to me.

[mountainhick]

Sorry but I find your circuit description somewhat ambiguous. A schematic of what you're planning would be much better, just sketch it out on paper, take a photo and upload it.
Or photoshop those Gibson and Fender schematics together somehow.

The concern being that the mixing resistors in the Gibson are at anode V DC, whereas all likely channel output points in the Fender are at 0V DC.
Hence a blocking cap will be needed.

Boom! See any issues with this? Spoiler alert: I have wired it up and it does function and sound good overall but I’m sure some things could be better. I did leave that .022 coupling cap at the end of the Bassman channel before it joins the Gibson channel at the phase inverter.

Oh and I omitted the 270K where the red line is.

If you want that 5F6A channel to remain essentially the same as original, a voltage divider should be constructed at the red line of about 270K/330K. Removing the other channel removes the corresponding load. Either way it will work, but by removing that load and the 270K resistor you will have more signal voltage. That can be just fine, but will be different.

But you are mixing the resulting two channels at a different place in the chain. Where the two channels mix, the 100K resistors will not offer as much channel separation and you will lose some of that extra signal voltage. It is the same phenomenon again, each channel is a load for the other. The less impedance (100K resistors in combination with the output impedance of the other channel), the more signal attenuation. Kind of robbing Peter to pay Paul.

Personally I would increase the 100Ks to at least 220K for mixing the channels for separation, but totally fine to try the 100Ks first and swap higher after testing. YMMV.

[dickjonesify]

If you want that 5F6A channel to remain essentially the same as original, a voltage divider should be constructed at the red line of about 270K/330K. Removing the other channel removes the corresponding load. Either way it will work, but by removing that load and the 270K resistor you will have more signal voltage. That can be just fine, but will be different.

But you are mixing the resulting two channels at a different place in the chain. Where the two channels mix, the 100K resistors will not offer as much channel separation and you will lose some of that extra signal voltage. It is the same phenomenon again, each channel is a load for the other. The less impedance (100K resistors in combination with the output impedance of the other channel), the more signal attenuation. Kind of robbing Peter to pay Paul.

Personally I would increase the 100Ks to at least 220K for mixing the channels for separation, but totally fine to try the 100Ks first and swap higher after testing. YMMV.

Thank you! This is the kind of stuff I’m trying to wrap my head around. I will audition 220K mixing resistors (I did on one side but not both). 

I don’t need it to be exactly like the original but I do want it to have a little more “sauce” haha. The amp as a whole is quite clean, which is expected but I wouldn’t mind some sizzle. Do you think the voltage divider where my red line is would help that or hurt it?

I needed another triode so I stole the 7-pin socket that was an OA2 tube for the tremolo circuit. I put in a 6C4, which is only 20mu and the 5F6A used a 40mu triode. I did some Googling and found the 6AB4 tube, which has the same pinout and is 60mu. Excited to give that a try!

[mountainhick]

If you want that 5F6A channel to remain essentially the same as original, a voltage divider should be constructed at the red line of about 270K/330K. Removing the other channel removes the corresponding load. Either way it will work, but by removing that load and the 270K resistor you will have more signal voltage. That can be just fine, but will be different.

But you are mixing the resulting two channels at a different place in the chain. Where the two channels mix, the 100K resistors will not offer as much channel separation and you will lose some of that extra signal voltage. It is the same phenomenon again, each channel is a load for the other. The less impedance (100K resistors in combination with the output impedance of the other channel), the more signal attenuation. Kind of robbing Peter to pay Paul.

Personally I would increase the 100Ks to at least 220K for mixing the channels for separation, but totally fine to try the 100Ks first and swap higher after testing. YMMV.

Thank you! This is the kind of stuff I’m trying to wrap my head around. I will audition 220K mixing resistors (I did on one side but not both). 

I don’t need it to be exactly like the original but I do want it to have a little more “sauce” haha. The amp as a whole is quite clean, which is expected but I wouldn’t mind some sizzle. Do you think the voltage divider where my red line is would help that or hurt it?

I needed another triode so I stole the 7-pin socket that was an OA2 tube for the tremolo circuit. I put in a 6C4, which is only 20mu and the 5F6A used a 40mu triode. I did some Googling and found the 6AB4 tube, which has the same pinout and is 60mu. Excited to give that a try!

You will have more gain without the divider at the red. Should drive things harder.

So the 6C4 is the input triode replacing the 12AY7? Yes, I would increase that tube's gain. You can even go to a 6AV6 for 100mu, but pin outs are different. With the 6AV6, wire the diodes to ground

[HotBluePlates]

there are ‘normally’ 2 equal value mixing resistors in a passive resistor mixer. ...

If you want that 5F6A channel to remain essentially the same as original, a voltage divider should be constructed at the red line of about 270K/330K. ...

Thank you! This is the kind of stuff I’m trying to wrap my head around. I will audition 220K mixing resistors ...

If you don't play through both channels at the same time, there is no urgency to "optimize" the channel-mix resistors.

The reason Fender usually has same-values resistors is that Fender typically mixes 2 channels that are the same, or at least have the same gain-capability.

Gibson used unequal mix resistors here because turning the Trem on makes that channel sound quieter (the signal is now gone 1/2 the time).  The unequal mix resistors favor the Trem channel, making it twice the amplitude of the other channel (not "twice as loud" but it makes up for the apparent loudness-reduction caused by the trem).

The Gibson Trem channel may sound quieter because of the higher signal-loss associated with the T-filter (if you turn the Midrange down).  If that channel is quieter and you want to bring its level up, then make the resistor connected to the Trem channel smaller (or increase the value of the 5F6-A channel's mixing resistor).

Do a search for "Voltage Divider" and see that each channel looks like the "Input side" of a 2-resistor voltage divider. 

[dickjonesify]

So here’s a different angle on mixing two channels. I decided to convert this amp to a LTP phase inverter. When reading up on how they work, I learned the grid of the ‘bottom’ triode can be a 2nd input, often for another channel. I had to look around but found a few examples of that. Here’s one from Sluckey’s 18 watt Marshall…



In every case I’ve seen, there are no mixing resistors in this setup. Except that there’s always a post tone stack volume control on both channels right before the phase inverter, which counts as a resistor, unless they’re cranked all the way up I guess haha.

So, next question: if I input each of my channels separately into the two grids of an LTP phase inverter, do I need mixing resistors?

[sluckey]

So, next question: if I input each of my channels separately into the two grids of an LTP phase inverter, do I need mixing resistors?

No mixing resistors are needed because the mixing takes place inside the LTP PI tube.

[dickjonesify]

So, next question: if I input each of my channels separately into the two grids of an LTP phase inverter, do I need mixing resistors?

No mixing resistors are needed because the mixing takes place inside the LTP PI tube.

Awesome! Makes sense. I think I’ll go this route.