How to remove popping when switching cathode resistors?

[psychepool]

I am making a multi-channel amp that can switch channels with a foot switch.

For configuration reasons, the cathode resistor must be switched when switching channels.
Another resistor is connected in parallel through a switch to the originally connected resistor.

There is some popping noise at this point, is there a way to eliminate or reduce it a lot?

I know how to reduce pop noise when switching the cathode bypass cap by connecting a resistor of about 100K between the cathode and ground of the cap.
I don't think this can be used in the case of switching resistors.

Is there a special way to avoid popping when switching by adding a resistor in parallel to the cathode resistor?

[R.G.]

If I understand you correctly, you're changing the actual cathode resistor's value by switching in another one.

If that's the case, you're always going to get a shift in DC bias on the tube. That's always going to make a signal change at the plate. Well, at cathode too, but you knew that. I can think of three ways to get around this. One is to do it very, very slowly, so that the change in bias point happens so slowly that the plate change is sub-audio. The circuitry to do this would be complicated. Another is to rig the new resistor to run through a BFC so that it does not change the DC bias point, only the gain. The third is to rig a timed mute so that the pop can still happen, but the mute keeps it from getting out into the audio path. This makes for a small fraction of a second of silence, which is usually not noticeable. Still some circuitry, but manageable. A 555 and a JFET could probably handle the mute; maybe you could do something with diodes, resistors and caps to do a short mute.

[roberto]

IIRC peavey was using a triac to mute the circuit while changing channel. On the 6505 maybe?
Not sure, I saw it a long time ago!

[psychepool]

If I understand you correctly, you're changing the actual cathode resistor's value by switching in another one.

If that's the case, you're always going to get a shift in DC bias on the tube. That's always going to make a signal change at the plate. Well, at cathode too, but you knew that. I can think of three ways to get around this. One is to do it very, very slowly, so that the change in bias point happens so slowly that the plate change is sub-audio. The circuitry to do this would be complicated. Another is to rig the new resistor to run through a BFC so that it does not change the DC bias point, only the gain. The third is to rig a timed mute so that the pop can still happen, but the mute keeps it from getting out into the audio path. This makes for a small fraction of a second of silence, which is usually not noticeable. Still some circuitry, but manageable. A 555 and a JFET could probably handle the mute; maybe you could do something with diodes, resistors and caps to do a short mute.

Below is the circuit diagram of the amplifier I am making. (It's a preamp circuit, but I am making a complete amp including a power amp.)
The part marked in color is the part that is switched when switching channels.
The parts marked in blue are related to the question content.

pop_question.jpg

There is no marked blue point in the original circuit, but when I tested it with the original circuit, the amount of gain was too insufficient.
So I added a resistor and a cap only when switching the crunch channel to increase gain and thickness of the tone.
I was satisfied the result, but I am dissatisfied with the popping noise.
Popping noise does not occur only in this position, but when switching only this point separately during the test, the popping noise at this point was severe.
I asked if there is a way to switch this point without noise.



I have asked here about the mute circuit before.
https://ampgarage.com/forum/viewtopic.php?f=6&t=34133

I asked a mute circuit that only works instantaneously when switching channels, and the answers are shown as contents for permanent muting.
I tried to ask again by making sure what I wanted to ask, but it was beyond my foreign language skills.

I think that applying the mute circuit is the most ideal way to evade the pop noise, so I would like to take this opportunity to ask a question again.
Below is the schematic of Mesa Lone Star I want to study.
(If uploading this image is a problem, please request a deletion.)

This is the switching circuit diagram,

boogie_lonestar-6.jpg

This is the circuit at the point where the channel mute is applied (just before the FX Loop, i.e. near the rear end of the preamp and the reverb pot applied independently for each channel).

boogie_lonestar-5.jpg

This may be a difficult answer, but can I ask you to explain how this mute circuit works?
If you go beyond what can be explained through this page, you don't have to.
If it's not difficult, please answer me.

IIRC peavey was using a triac to mute the circuit while changing channel. On the 6505 maybe?
Not sure, I saw it a long time ago!

Thanks for the answer.
I found and looked at the Peavey 6505 circuit.
However, this circuit is so complex that it feels too difficult for me to understand.
As knowledge increases, I will try to understand it by looking at it from time to time.

[R.G.]

Below is the circuit diagram of the amplifier I am making. (It's a preamp circuit, but I am making a complete amp including a power amp.)
The part marked in color is the part that is switched when switching channels.
The parts marked in blue are related to the question content.
[...]
Popping noise does not occur only in this position, but when switching only this point separately during the test, the popping noise at this point was severe.
I asked if there is a way to switch this point without noise.

Well, the first step would be to put a 100k to 1M resistor across that switch. That would at least keep the DC voltage on the cap at closer to it's final value when the switch is closed. The change in bias point will still happen, but you might get it to be considerably less.

The first rule of suppressing switching transients in switching signals is to not have a sudden change in DC conditions in the circuit when the switch changes!

I tried to ask again by making sure what I wanted to ask, but it was beyond my foreign language skills.

I'm impressed that you can converse on technical matters in a language that is not your native one. My foreign language skills are not sufficient for that. Congratulations to you for your ability to do that!

I think that applying the mute circuit is the most ideal way to evade the pop noise, so I would like to take this opportunity to ask a question again.
Below is the schematic of Mesa Lone Star I want to study.
[...]
This may be a difficult answer, but can I ask you to explain how this mute circuit works?
If you go beyond what can be explained through this page, you don't have to.
If it's not difficult, please answer me.

Sure.
Everything except the section titled “Mute Circuit” in the lower left corner of the diagram is just the switching circuits that change the relays. The two transistors and diodes in the middle of the diagram make a short pulse appear on the “Mute Pulse Bus”. This pulse turns on the transistor in the “Mute Circuit” for a very short time, pulling its collector low. The 0.1uF cap across that transistor to ground, and the 2.2uF capacitor at the left side are pulled to ground ( or, for the 2.2uF, nearly to ground through the diode) and so the points labeled J175M and J175RM are pulled to ground, then rise slowly as the capacitors re-charge up towards 12V.

There are two J175 JFETs n the second schematic. The J175 is a P-channel device, so if you have its source tied to ground and its drain tied to a signal with an average level of 0Vdc, you can leave the JFET turned off (not conducting) by holding the gate 6 or more volts above ground. When this is true, the JFET is effectively an open circuit, and does not mute the signal. When the JFET's gate is pulled down to ground, it changes to it's on stage, and it looks like a resistor of Rdson across the JFET. For the J175, this resistance is on the order of 100 ohms. This would mute the signal that the JFET is attached to.

So all of the circuit there is to sense changes to relays and footswitches and generate a pulse on the Mute Pulse Bus when something happens, then to trigger the Mute Circuit to make a sudden pull to ground for both JFET gates, with a slow rise as the capacitors recharge on the gates. This has the effect that the JFETs must the signals quickly. Then as the voltage on the gates rise again, the JFETs let the signal fade back in.

[psychepool]

Well, the first step would be to put a 100k to 1M resistor across that switch. That would at least keep the DC voltage on the cap at closer to it's final value when the switch is closed. The change in bias point will still happen, but you might get it to be considerably less.
The first rule of suppressing switching transients in switching signals is to not have a sudden change in DC conditions in the circuit when the switch changes!

I know that the method of connecting a resistance of 100K~1M between the ground is a method that is commonly used when switching the cathode bypass cap.
(In a dual rectifier, the 1uF cap is separated from the ground through a 47K resistor.)
But is it okay to switch only the cathode resistor without a capacitor in this way?
For example, the current default cathode resistance is 39K, but if put 15K+100K resistors in parallel, doesn't it make a difference?
Currently, my circuit is applied with a capacitor, but the results are expected to be quite different.
When the value of this point is changed in the lead channel (when the resistance value is lowered), severe oscillation occurs, so I think it should be separated without error, so I ask a question.

I'm impressed that you can converse on technical matters in a language that is not your native one. My foreign language skills are not sufficient for that. Congratulations to you for your ability to do that!

If Google Translator does most of the work and adds a little bit of my will to it, there isn't much difficulty with text conversations. What a beautiful world!

Sure.
Everything except the section titled “Mute Circuit” in the lower left corner of the diagram is just the switching circuits that change the relays.
[...]
So all of the circuit there is to sense changes to relays and footswitches and generate a pulse on the Mute Pulse Bus when something happens, then to trigger the Mute Circuit to make a sudden pull to ground for both JFET gates, with a slow rise as the capacitors recharge on the gates. This has the effect that the JFETs must the signals quickly. Then as the voltage on the gates rise again, the JFETs let the signal fade back in.

Honestly, I didn't expect an answer to this question, but thank you for your kind answer!

I lack language skills, but above all, I am not an electrician, so despite your kind answer, there are still areas that I do not understand.
I would like to ask a little more about this.

1.
I know the NPN TR works as a switch, but I don't know the exact theory.
In the circuit diagram, RY1 and RY2M/R are switching through TR, but I don't know in what state the collector and emitter are conducting.
When the select switch is open, +12V is supplied to the base, and when the select switch is shorted, the base is connected to the ground.
Does the relay turn on when the select switch is shorted? Or does it turn on when open?

2.
I am curious about the theory that Mute Pulse only occurs temporarily when the relay is turned on and off.
As I understand it, it seems that the signal will continue to be sent when the relay is turned on, and the signal will stop when it is turned off. Why is the pulse being sent and cut off only momentarily?

3.
First of all, I apologize because it may be a question in a way that is too conscienceless.
There are parts in the schematic that I am curious about their role, so I numbered them with red numbers.
Can you explain the role of each part?
Sorry for the hassle.

boogie_lonestar-6_mute.jpg

Thanks to the writing you written above, I learned a lot of knowledge.
If you explain a little more about the requested part, I think I will understand more about the circuit.
Depending on the situation in the future, it seems that it may reach a stage where it can be applied.

Please answer. Thank you.

[R.G.]

But is it okay to switch only the cathode resistor without a capacitor in this way?
For example, the current default cathode resistance is 39K, but if put 15K+100K resistors in parallel, doesn't it make a difference?
Currently, my circuit is applied with a capacitor, but the results are expected to be quite different.
When the value of this point is changed in the lead channel (when the resistance value is lowered), severe oscillation occurs, so I think it should be separated without error, so I ask a question.

In thinking about your circuit's operation, it is necessary to think about it both the local circuit (that is, just the circuit around that one tube section) as well as the whole amplifier.
In the local circuit, just that one tube section, changing the value of the cathode resistor does two things. Making it smaller changes both (1) the AC gain of the section and (2) the DC conditions in that one section.
My whole point about using a resistor in series with the capacitor was to avoid the change in DC conditions (item (2) above). Any sudden change in the DC conditions will make a sudden DC change on the tube plate, and this will be heard as a pop. To avoid the pop, you must do one of the three options I mentioned: (a) make the change so slowly that the movement of DC is too slow to hear, (b) not actually change the DC conditions, or (c) mute the pop that happens, so it is not heard.
So it is OK to change the cathode resistor without a capacitor, but if you do, it >WILL< have a sudden DC condition change, and that >WILL< cause a pop that you will have to either suppress by muting, or live with. “OK” is different from “not making any popping sound”.

The issue with severe oscillations when the cathode resistor is lowered is I think a problem with the complete amplifier and perhaps its wiring. Electrical engineers have a saying that everything will oscillate if the gain is high enough. This happens because some oscillation feedback paths can happen through the air around the parts. By lowering the cathode resistor, you increase the AC gain of the local stage, and the combined gain through the whole amplifier is enough to make the whole amplifier oscillate.

1.
I know the NPN TR works as a switch, but I don't know the exact theory.

NPN transistors only conduct when there their base is at least 0.5 to 0.7V more positive than their emitter, and current flows in the base. The base current allows a current in the collector to flow. The collector current will be the current gain (hfe) of that transistor times the base current, as long as the stuff outside the transistor will let that much collector current flow.

When the external circuit in the collector limits how much current can flow into the collector, the collector-emitter voltage can get very low, a fraction of a volt. This is how most NPNs used as switches work. And it is how all three of the NPNs in the schematic work. They have different amounts of base current, and different external limits on their collector currents, but they all share that the base current is big enough when the base is raised from ground by 0.5 to 0.7V, it makes the voltage across the collector-emitter drop to nearly zero.

In the circuit diagram, RY1 and RY2M/R are switching through TR, but I don't know in what state the collector and emitter are conducting.

The collector-emitter conducts when the base is pulled 0.5 to 0.7V above the emitter. Otherwise, the collector-emitter acts like an open circuit.

When the select switch is open, +12V is supplied to the base, and when the select switch is shorted, the base is connected to the ground.
Does the relay turn on when the select switch is shorted? Or does it turn on when open?

When the select switch is shorted to ground, the base is connected to ground. The voltage on the base of TR is then LESS than one silicon diode drop of 0.5V to 0.7V, so the transistor's collector-emitter is NOT conducting, and no current flows through the transistor. Relay 1 has a small coil resistance, much smaller than the 220K to the base of the second TR, so relay 1 does not have enough current to be pulled into the on condition. However, the 220K is fed current through relay 1's coil (although not enough to turn Relay 1 on), and that current is enough to raise the base of the second TR above ground, turning the second transistor on.
This makes the voltage on the second TR's collector-emitter fall to nearly zero, turning on the coils to both RY2M and RY2H.

So when the select switch is closed to ground, Relay 1 is off, and RY2M and RY2H are on.
When the select switch opens, Relay 1 is turned on, and RY2M and RY2H are turned off.

Notice that any time the select switch changes,

2.
I am curious about the theory that Mute Pulse only occurs temporarily when the relay is turned on and off.
As I understand it, it seems that the signal will continue to be sent when the relay is turned on, and the signal will stop when it is turned off. Why is the pulse being sent and cut off only momentarily?

You are correct that in steady state, one or the other of the two diodes feeding the Mute Pulse Bus is on, and making the Mute Pulse Bus high. This holds the base of the Mute Circuit transistor high and the collector-emitter at nearly ground. However, when the select switch changes from open to closed OR from closed to open, which diode is conducting changes. There is a short time while the changing from one diode to the other where neither one is driving the TR enough to keep it turned on (that is, collector-emitter voltage nearly zero) and during that short moment while the voltages on the Trs feeding the relays are changing, the Mute Circuit TR conducts. This only happens for the short time when the voltages are changing, not at either steady state. It's a change only in the middle while the states are changing.

3.
First of all, I apologize because it may be a question in a way that is too conscienceless.
There are parts in the schematic that I am curious about their role, so I numbered them with red numbers.
Can you explain the role of each part?

Resistor 1 and capacitor 1 form the time delay for muting controlled by J175RM. When the Mute Circuit Transistor turns on from the pulse from the Mute Circuit Bus, the capacitor is pulled nearly to ground. This turns the J175RM on, and the slow rise of voltage on the 2.2uF capacitor releases the mute slowly.
Diode 3 lets the Mute Circuit Transistor pull down capacitor 2 quickly, but prevents the transistor from letting capacitor 2 charge back up when the transistor turns off.
4 is a bidirectional zener which suppresses spikes from the sudden switching and prevent the voltage to the gates of the two J175s from going too high.
Capacitor 5 and resistors 6 and 7 form the time delay for J175M turning back off after it's been turned off by conduction of the Mute Circuit Transistors, much like the operation of J175RM. The use of diode 3 lets these have two different mute timings.
Resistor 8 pulls the base of the Mute Circuit Transistor to ground when there is not enough current coming through resistor 10. Capacitor 9 is a “speed up” capacitor so that changes across resistor 10 make the transistor switch quickly.
Diodes 11 and 12 allow the steady state high voltage on their respective sides of the circuit to hold up the voltage to the Mute Circuit Transistor's base when switching is not happening. A pulse on the Mute Circuit Bus is a sudden drop from a high voltage to a much lower voltage.
Resistor 13 limits the amount of base current into TR to a safe level, and is picked so that the current from resistor 14 and the relay coil for RY1 do not damage the transistor, yet enough current flows to fully switch the TR on and off.
Resistor 15 provides enough current into the base of the first TR to make it switch reliably. This current goes either into the TR base, or through the select switch if the switch is closed.

If this does not make perfect sense to you, don't get discouraged. I had a full semester course in switching transistor circuits, and then years of working with them in practice. Keep trying to understand the basics of circuits - where the currents flow, what the voltages do across components, and how transistors work. It may take some time, but little "simple" circuits like this are not really all that simple. Study! You'll learn.

[psychepool]

NPN transistors only conduct when there their base is at least 0.5 to 0.7V more positive than their emitter, and current flows in the base. The base current allows a current in the collector to flow. The collector current will be the current gain (hfe) of that transistor times the base current, as long as the stuff outside the transistor will let that much collector current flow.
[...]
If this does not make perfect sense to you, don't get discouraged. I had a full semester course in switching transistor circuits, and then years of working with them in practice. Keep trying to understand the basics of circuits - where the currents flow, what the voltages do across components, and how transistors work. It may take some time, but little "simple" circuits like this are not really all that simple. Study! You'll learn.

Thank you very much for another detailed reply!

To conform what I understand is correct, let me summarize your explanation:
1. TR is 'on' as power (0.5~0.7V higher voltage compared to emitter) is supplied to the base. (Collector and emitter are shorted.)
2. Each 6426-1 and 2 are turned on alternately by the channel select switch.
3. The current flowing through the relay coil is by supplying constant voltage to the 6426M through the 11/12 diode.
Accordingly, the J175M/RM is pulled down to ground to release the mute connected to the preamplifier.
4. However, as the channel is switched, a short empty moment occurs in the Mute Pulse supplied to the 6426M, and at that moment, the 6426M turns'off' and the power is supplied to the J175M/RM to momentarily mute the preamp.
5. Momentary Mute Pulse After the supply gap ends, the J175M/RM is pulled back to ground, but delays the unmuted time as much as the pop disappears through various parts of the mute circuit.

I think I can summarize it like this. Is it correct?

For an additional question,
How are the values ​​of various parts to set the numbered TR operating point and mute delay time?
What I want to ask is, is there a possibility that the amplifier that will be made in the future does not operate properly depending on the amount of current supplied from the power supply even if it is configured with the same value as that circuit?
Or, if only the same voltage is supplied, does it have the same operating point (mutation delay time, etc.) as in Lonestar even if other variables such as difference in current amount occur?

[R.G.]

[To conform what I understand is correct, let me summarize your explanation:
1. TR is 'on' as power (0.5~0.7V higher voltage compared to emitter) is supplied to the base. (Collector and emitter are shorted.)

That is an oversimplification, but yes, it's a useful way to look at the situation. The more-complete version is hidden inside how bipolar transistors work. Bipolars (NPN or PNP) don't conduct from collector to emitter until the base-to-emitter voltage is large enough to let base current flow. The base-emitter "looks like" just a semiconductor diode from the base lead's point of view. So to get base current to flow, you need to forward bias the base-emitter diode junction. This requires that 0.5 to 0.7V to get base current to flow.

Once base current flows, this allows current to flow from collector to emitter. The amount of current that flows is the base current times the current gain of the transistor.

The transistor can be destroyed by making too much current flow in either the base lead or the collector-to-emitter path. So just connecting a voltage source to the base lead can make too much current flow in the base lead. This is why there are resistors attached to the base leads in the circuit - the resistors not only provide the possibility of the 0.5 to 0.7V on the base-emitter, they also limit the base current to safe levels for the base to conduct without being destroyed.

Strictly speaking, the collector and emitter are not shorted, only conducting a lot of current, with a voltage across collector and emitter that drops to a fraction of a volt. Exactly how much current flows is determined by the resistors and resistance of the relay coils in series with the collectors. But it is a useful simplification to think of the collector and emitter as being approximately shorted.

Again, this is all a simplification. I highly recommend that you search for and study some educational tutorials on how bipolar transistors work.

2. Each 6426-1 and 2 are turned on alternately by the channel select switch.

Yes.

3. The current flowing through the relay coil is by supplying constant voltage to the 6426M through the 11/12 diode.
Accordingly, the J175M/RM is pulled down to ground to release the mute connected to the preamplifier.

Almost correct. The J175M/RM gates are pulled to ground to cause the muting, then they release the mute as the voltage on the gates rise more than a few volts.

The whole Mute Pulse Bus and Mute Circuit Transistor circuit is a little bit of a sneaky, tricky circuit. First, the Mute Pulse Bus relies on the slowness of the 6426 transistors switching alternately to make a small pulse drop to a low voltage and then rise back to the near-12V level for normal operation. Then, the Mute Circuit Transistor is biased off normally.

The Mute Circuit Transistor has its base fed from the Mute Pulse Bus through a 220K resistor and has a 10K to ground. In normal operation, the Mute Pulse Bus has a voltage of nearly 12V on it. So the base of the Mute Circuit Transistor is fed by a voltage of something less than 12V divided down by the 220K and 10K resistors. This means that the transistor's base is nearly, but not quite at the minimum voltage to conduct, and is not pulling its collector down towards its emitter. When the Mute Pulse Bus drops down, then back up, the capacitor across the 220K resistor first pushes the base voltage down lower than ground, then pulls it up as the Must Pulse Bus rises again toward 12V. It is this sudden rise in voltage coupled through the capacitor across the 220K that turns on the Mute Circuit Transistor for just a very short time. That is enough to pull the gates of the J175s low, causing a mute, which then releases slowly as the gate voltages rise slowly.

4. However, as the channel is switched, a short empty moment occurs in the Mute Pulse supplied to the 6426M, and at that moment, the 6426M turns'off' and the power is supplied to the J175M/RM to momentarily mute the preamp.

Yes - see the explanation in 3 above.

5. Momentary Mute Pulse After the supply gap ends, the J175M/RM is pulled back to ground, but delays the unmuted time as much as the pop disappears through various parts of the mute circuit.

That is almost correct. The J175 gates normally sit high, which has the J175s turned off and not muted. The action of the momentary mute pulse is to cause the J175 gates to be pulled to ground and mute the signal, then slowly rise up and release the mute.

Again, this is a little bit of a tricky circuit. The action is not immediately obvious. I had to sit and think about it for a while to figure out how it works, and I'm used to doing this kind of thing.

For an additional question,
How are the values ​​of various parts to set the numbered TR operating point and mute delay time?
What I want to ask is, is there a possibility that the amplifier that will be made in the future does not operate properly depending on the amount of current supplied from the power supply even if it is configured with the same value as that circuit?
Or, if only the same voltage is supplied, does it have the same operating point (mutation delay time, etc.) as in Lonestar even if other variables such as difference in current amount occur?

The values are set up with the assumption that the DC power supply to the whole mute circuit will be approximately 12V. If that changes very much, then the resistor values would need to be re-calculated. I think that the biggest effect would be on the 220K/10K on the base of the Mute Circuit Transistor, but there could possibly be other changes needed. One of the big changes would be that the relays would have to be changed to relays with different coil voltages. If the 12V power supply voltage gets too low, perhaps lower than 6-8V, then the J175 JFETs might not turn fully off and UN-mute.

I would suggest that if you want to use this circuit, you provide 12Vdc to supply the whole circuit, with enough current to supply the relays. That is - don't change it. It is possible to change the power supply and recalculate all the values, but it would take some detailed thinking about every part, checking to see if the part values needed changed.

[psychepool]

[To conform what I understand is correct, let me summarize your explanation:
1. TR is 'on' as power (0.5~0.7V higher voltage compared to emitter) is supplied to the base. (Collector and emitter are shorted.)

That is an oversimplification, but yes, it's a useful way to look at the situation. The more-complete version is hidden inside how bipolar transistors work. Bipolars (NPN or PNP) don't conduct from collector to emitter until the base-to-emitter voltage is large enough to let base current flow. The base-emitter "looks like" just a semiconductor diode from the base lead's point of view. So to get base current to flow, you need to forward bias the base-emitter diode junction. This requires that 0.5 to 0.7V to get base current to flow.
[...]
I would suggest that if you want to use this circuit, you provide 12Vdc to supply the whole circuit, with enough current to supply the relays. That is - don't change it. It is possible to change the power supply and recalculate all the values, but it would take some detailed thinking about every part, checking to see if the part values needed changed.

Thank you.
Anyway, even when making a tube amp or a stompbox pedal, I don't fully understand the electrical theory and make it, so just the explanation you gave me was helpful enough.
The detailed knowledge you gave me seems to be a lot of hints when I search for the part I'm curious about in the future.
Even webpages in my native language are full of data on electric knowledge.
However, it was difficult to understand the content until I read your article, but now I have the confidence to understand it more easily.

This is just a hobby for me, so I don't think I will gain much knowledge in a short time.
However, the next time I make a two-channel amplifier, I think that I can apply this circuit as it is with the following thoughts.
"The values ​​of the various components of Lone Star's switching circuit work well with a 12VDC supply standard."


The amplifier made this time seems to be difficult to apply the mute circuit.
It has a small sized chassis compared to the size of the circuit, so space is insufficient.
And it is not a two-channel amplifier.
It is not a method of selecting directly by pressing the button of the corresponding channel by the latch circuit,
There are clean channel and dirty channel, and the dirty channel has two channels of crunch/lead.
It seems difficult to me right now to make a suitable circuit by applying it because it is a configuration that can switch clean/dirty with a foot switch and switch between crunch/lead.


So I wanted to talk about the basic popping noise countermeasures again.
Below is the circuit diagram of the Orange Rockerverb.

rv50.jpg

The relay part for channel switching is marked in color.
Particularly, the part marked in red was impressive for me.
As resistors constantly connected in series from each channel output switch, the connection is not disconnected and is connected to ground and then dropped.
In the clean channel, 470K is connected to the ground through 220K in series at the end of the treble pin 2.
In dirty channel, 220K is connected to the ground through 470K in series at the end of pin 2 of the master volume.
It feels like a configuration that keep to the basic theory of avoiding popping noise you told me (don't change the DC condition).

So, I thought, I would like to configure such a type of switching to reduce popping noise.

srt_sw.jpg

The part marked with the color box is the newly constructed part, and the one drawn in the square box next to it is in its original state.
This is roughly the same result as adding one more B1M pot set at 12 o'clock after the end of each channel.(In dirty channel, two B1M pots are placed in a row.)

I am wondering if this will be effective in dramatically reducing popping noise,
Also, I am curious if there will be a lot of sound change (change in tone shape, large change in output level, etc.)

[R.G.]

The part marked with the color box is the newly constructed part, and the one drawn in the square box next to it is in its original state.
This is roughly the same result as adding one more B1M pot set at 12 o'clock after the end of each channel.(In dirty channel, two B1M pots are placed in a row.)

I am wondering if this will be effective in dramatically reducing popping noise,
Also, I am curious if there will be a lot of sound change (change in tone shape, large change in output level, etc.)

It appears to have DC blocked from the switches by capacitors, and the capacitors leading to the switches pulled to ground by resistors, so yes, I think it will stop any capacitor-opened popping. Clicks and pops, like hum, are sneaky and have ways to get inside audio signals. Relays can cause small clicks because there is capacitance from the coil to the signal contacts. There are even relays with shields inside to help reduce this. But if you're using switches, that won't happen. There is another sneaky way to get clicks. metal switch contacts "make" and "break" instantly. They also bounce open and closed when being "made". If there is a signal voltage across the switch when it "makes" or "breaks", the signal after the switch will change to follow the signal instantly, making a little instant jump to the new level. This is heard as a little "click". In electric organs, this is a valuable thing, as it helps approximate the "chiff" sound from air-organs as the keys open the air path to the pipes. In guitar music, this isn't such a good thing.

Try out your new scheme - it should help. I don't think there will be much change to tone.

[psychepool]

Try out your new scheme - it should help. I don't think there will be much change to tone.

It seems not difficult to change from current state.
When the results aren't good, it's not difficult to reverse it, so I'll try it once after finishing the build.
Thanks for your help so far!

[SacredGroove]

I am making a multi-channel amp that can switch channels with a foot switch.

For configuration reasons, the cathode resistor must be switched when switching channels.
Another resistor is connected in parallel through a switch to the originally connected resistor.

There is some popping noise at this point, is there a way to eliminate or reduce it a lot?

I know how to reduce pop noise when switching the cathode bypass cap by connecting a resistor of about 100K between the cathode and ground of the cap.
I don't think this can be used in the case of switching resistors.

Is there a special way to avoid popping when switching by adding a resistor in parallel to the cathode resistor?

I have a similar situation going on. But, I'm not sure about your question by how it's worded. Are you switching from one resistor to the other resistor? Or are you adding one resistor to the other so, there is always a resistor connected to ground?

I have been switching between 10k and 2.7k//.68uF at the cathode and there is a pop sound. The sound wasn't really bad until recently though. So, I am wondering if I left the 10k resistor fixed to the cathode and then, via toggle switch, add a 3.3k//.68uF combo that the pop will decrease or even disappear?

[psychepool]

I have a similar situation going on. But, I'm not sure about your question by how it's worded. Are you switching from one resistor to the other resistor? Or are you adding one resistor to the other so, there is always a resistor connected to ground?

I have been switching between 10k and 2.7k//.68uF at the cathode and there is a pop sound. The sound wasn't really bad until recently though. So, I am wondering if I left the 10k resistor fixed to the cathode and then, via toggle switch, add a 3.3k//.68uF combo that the pop will decrease or even disappear?

Breaking the connected resistor and switching to another resistor has not been attempted this time.
Probably, the popping noise will be much larger than the parallel adding method, and most of all, may be the desired effect is likely not to appear in real time.
So this method is not considered at all.