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Old 28th December 2004, 11:45 AM   #1
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Default Valve effects loop

Hi,

I've attached a schematic I've been given for an all valve effects loop used in guitar amplifiers.

There are a couple of things I wanted to ask your help on.

First of all, the capacitor and resistor circled in red on the right. The 22nF capacitor looks like a decoupling capacitor for the previous 'in' stage before the loop, and the 100k resistor, a grid stop resistor.

If this is the case, are these two not the wrong way around?

Secondly, the 22uF capacitor in blue. I think this is probably there to stabilise the bias resistors 220k and 33k. But this capacitor is also tied to the grid of the valve. So I would have thought it's going to try to bypass the audio signal to ground. I'm not particularly familiar with cathode followers.

The capacitors in green bothered me because they seem to be quite large values. The 47nF isn't so bad, but 4u7 seems a bit strange for a decoupling capacitor. Could this by a typo and actually be 47nF?

The two resistors in pink I think may be acting like grid stop resistors, to limit the amount of current that the effects can sink and source.

The serial / parallel, 0dB -20dB switches are supposed to be for flexibility using different effects. I'd like to also ask for your help with these. The way I see it...

In serial, the first 1M resistor after SW1 goes betweent the grid and ground. The 1M resistor to it's right goes between the grid and the anode via the 47nF cap, forming a potential divider with the attentuation resistors. When SW1 is set to 0db mode, the 100k is bypassed and the divider is composed of two 1M resistors, so they split the signal 50:50 over each other.

In 20dB mode, the 1M is bypassed and the lower leg becomes 100k, so the upper 1M resistor gets 10x more of the signal dropped across it.

What happens in parallel mode?

I'm sorry for all the questions, I just really hate the idea of making something and not knowing exactly what's happening.
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Old 28th December 2004, 03:26 PM   #2
nanana is offline nanana  Sweden
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hey eeka chu (john?),
i'll give it a try...

the stuff circled in red is not what you think... its a fixed bias arrangement for a cathode follower. the 100K is not a "grid stop", this resistor does provide a small attenuation (roughly 9/10 of input level... 100K/1M) probably because the designer understands that pedals often have too much gain. it would also limit the bandwidth much as a stooper would but its too big to have been meant for that and its on the wrong side of the cap. the voltage divider from B+ to ground puts 1/7th of whatever the B+ is on the grid of the tube through the 1M grid resistor. the 22uF cap is really big bypass... you could easily get away with .47uF. please understand that it is not directly connected to the grid... it bypasses the voltage divider.

the 4.7uF coupling cap is there to drive 10K and below input loads... many many modern gadgets for music are 10K in or less. you would lose bass and add distortion if you didn't have the means to drive this properly. it looks totally correct.

the 1K and 10K are there because of the switched jacks and they prevent pops and clicks from "hot plugging".

the serial/parallel switch mixes the effects send/return with the dry signal (para) or not (ser)....
the 47nF cap and 1M resistor is the feedback network (together with the resistors on the grid) for the mixer... the designer cleverly "reuses" the 100K feedback divider as a pad.

this circuit is not the best way of doing this job, but it is way way better than the way most amp companies do it if they do it at all!
jc
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Old 29th December 2004, 12:40 PM   #3
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Quote:
hey eeka chu (john?),
John, I hope so anyway...

Quote:
i'll give it a try...

the stuff circled in red is not what you think... its a fixed bias arrangement for a cathode follower. the 100K is not a "grid stop", this resistor does provide a small attenuation (roughly 9/10 of input level... 100K/1M) probably because the designer understands that pedals often have too much gain. it would also limit the bandwidth much as a stooper would but its too big to have been meant for that and its on the wrong side of the cap.
Interesting, I figured I was probably wrong on this one.

You mention it being too large to be for use as a grid stop. In a schematic I've seen for a guitar preamp, one grid has 470k of fixed resistance on it's grid. I've read somewhere that this is about the limit that I'll see in this position and that guitar preamplifier often have very large values here because they're usually high gain circuits, where the grid will most likely be pushed beyond 0V in normal operation.

From what I recall, they need such a large resistance to prevent blocking distortion when the grid tries to conduct. Looking at some notes on grid resistors, it seems that 470k, combined with the Miller capacitance of something like a 12AX7, would produce a filter with a roll off around 2.2kHz.

Is the 22nF cap there to function as a normal decoupling capacitor? If it is, and this circuit was going after a tone stack, would there be any point in me including it, since a tone stack capacitively decouples the stage before it from the grid connected to it's output?

How does the 100k on the grid tie in with the bias?

Quote:
the voltage divider from B+ to ground puts 1/7th of whatever the B+ is on the grid of the tube through the 1M grid resistor.
I'm not sure what you mean when you say that this voltage is put through the 1M resistor.

Say I have 300V on the B+, using just 1/7th divider rule, 43V of this would appear over the 33k resistor. At the top of the resistor, that would be 43v positive with respect to the cathode wouldn't it?

Quote:
the 22uF cap is really big bypass... you could easily get away with .47uF. please understand that it is not directly connected to the grid... it bypasses the voltage divider.
I can understand why the bypass is there, it looks like it's bypassing the grid to ground though. Why doesn't the bypass also try to filter out the modulation on the grid?


Quote:
the 4.7uF coupling cap is there to drive 10K and below input loads... many many modern gadgets for music are 10K in or less. you would lose bass and add distortion if you didn't have the means to drive this properly. it looks totally correct.
Got it!

Would this have anything to do with why the send side in almost all effects loops seem to be cathode followers? Is it to better match the input resistance of the effects? Just out of curiosity, could an interstage transformer be used to raise the input resistance of the effects and make the driving requirement less demanding? I know it'd probably end up costing more, I'm only really thinking about it to keep myself learning.

I'm not planning on using solid state parts in the loop, but isn't this something an opamp is good at doing? Driving low input impedances from a high input source?

Quote:
the 1K and 10K are there because of the switched jacks and they prevent pops and clicks from "hot plugging".
I've noticed I get quite a few clicks and pops when I change the jack my Marshall's input whilst it's turned on. The input should have a 68k or so resistor on it. Would there need to be a capacitor in series with that to remove the noise when hot plugging?

Quote:
the serial/parallel switch mixes the effects send/return with the dry signal (para) or not (ser)....
the 47nF cap and 1M resistor is the feedback network (together with the resistors on the grid) for the mixer... the designer cleverly "reuses" the 100K feedback divider as a pad.
Sorry for the dumb questions, but... a pad?

Quote:
this circuit is not the best way of doing this job, but it is way way better than the way most amp companies do it if they do it at all!
jc
Thanks jc, you've been really helpful!

Have you seen any other schematics around for valve loops, or can you think of any helpful modifications to this one? I've been looking around for as much information as I can find on loops and reverb drivers, but it seems kind of limited.

I'm not even totally sure how necessary the series parallel, attenuation switches will be. They seem to be mainly for use with old vintage pedals that don't have any level controls on them.

I think I'll build them into the circuit and if I find I don't use them much I'll take them back out.

Could the 100k attenuation resistor on the grid of V2 be replaced with a potentiometer to allow an adjustable level of attentuation? If it was a 1M pot with a 100K resistor on it's wiper, you'd be able to choose a level from 0 to -20dB of attentuation right? Or even futher, the 1M replaced with a 1M pot with a 100k resistor on it's wiper, so there is no longer a switch but it instead sweeps from 0 to -20dB.

I think the switch is there to allow players to footswitch the attentuation, or put it through a midi controller for when they change effects.

Again, it's probably not particularly useful, I'm just trying to fully understand what's happening.

Again, sorry for all the questions!
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Old 29th December 2004, 06:33 PM   #4
nanana is offline nanana  Sweden
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hey john,
i'll try to answer them one by one...

the 100K res on the input forms an "l-pad" (the letter "L", which the series and shunt res's look like...) with the 1M grid resistor, which is an attenuator that reduces output by 1/10th . it will offer some improvement in stability the way it is shown, but not much.

the miller capacitance will not be significant in this circuit because the cathode follower has no gain in the pass band and the "boot-strapped" grid resistor will look like a much higher input impedance than 1M to the source.

the 22uF is just decoupling the voltage divider centertap to ground. the 1M grid resistor is connected to the same point to put the postive voltage developed at the centertap on the grid but still provide a load for the incoming signal to work on. the black dots show connection on the schematic, please note that the 22uF cap is not directly connected to the grid. it jumps over the grid. the 1M is directly connected to the grid (you are misreading the schematic).

the 100K resistor does not "tie in" with the bias allthough you are right in guessing that it will affect overloading and clipping. but that is too much to go into here (i don't have time).

the cathode resistor is pretty big, don't you think? 10K for a 12AU7 will drop 40 to 60 something volts at the current they operate at... (4 to 6 mA) which sounds right in the ballpark if the grid was at 43 volts? and other tubes will adjust accordingly... this is a typical "fixed bias" way to configure a cathode follower.

i'll leave the opamp stuff to someone else... heehee!

the pop resistors will still pop, just not as bad... don't add a cap.

the mixer circuit/ series parallel switch is probably there for a reverb, so you can mix wet and dry.... you can remove the switch if you don't need that feature. the return pot adjusts the level of your effect... you don't need to add another pot. if you don't need the pad (attenuator), just lose the switch and the 100K res.

why don't you build it and let us know what happens?!
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Old 30th December 2004, 12:54 PM   #5
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Quote:
hey john,
i'll try to answer them one by one...

the 100K res on the input forms an "l-pad" (the letter "L", which the series and shunt res's look like...) with the 1M grid resistor, which is an attenuator that reduces output by 1/10th . it will offer some improvement in stability the way it is shown, but not much.
The other part of that L being one of the two 1M resistors at the top of the schematic?

Quote:
the miller capacitance will not be significant in this circuit because the cathode follower has no gain in the pass band and the "boot-strapped" grid resistor will look like a much higher input impedance than 1M to the source.
I'm hearing you on FM!

Quote:
the 22uF is just decoupling the voltage divider centertap to ground. the 1M grid resistor is connected to the same point to put the postive voltage developed at the centertap on the grid but still provide a load for the incoming signal to work on. the black dots show connection on the schematic, please note that the 22uF cap is not directly connected to the grid. it jumps over the grid. the 1M is directly connected to the grid (you are misreading the schematic).
I can't believe I did that!!!

I actually went through a schematic I'm using this circuit in, adding the connection dots and thinking how responsible I was being making the connections so obvious.

And here I am staring at the schematic like a retard wondering what's going on.

I always try to add the funny jump curves in the wires when I cross two over to avoid making mistakes with them.

The way I reason it is that I spend so much mind power on higher planes working on more complex problems that I don't have enough spare to keep my fingers out of door hinges and plug sockets.

That way I don't feel quite so stupid when I'm dancing around with squashed fingers... again...

Quote:
the 100K resistor does not "tie in" with the bias allthough you are right in guessing that it will affect overloading and clipping. but that is too much to go into here (i don't have time).
No problem, there's an explaination about blocking distortion and how the grid resistor, ground reference and coupling capacitor effect it on a guitar amp website I found.

Quote:
the cathode resistor is pretty big, don't you think? 10K for a 12AU7 will drop 40 to 60 something volts at the current they operate at... (4 to 6 mA) which sounds right in the ballpark if the grid was at 43 volts? and other tubes will adjust accordingly... this is a typical "fixed bias" way to configure a cathode follower.
Please check my logic...

The cathode resistor develops the output signal. It also raises the cathode by however many volts positive.

The 220k and 33k resistor also raise the grid to some positive point.

The difference between the two equals the bias voltage?

Quote:
i'll leave the opamp stuff to someone else... heehee!

the pop resistors will still pop, just not as bad... don't add a cap.
I've also noticed my amplifier whistles just as the male jack is clicking into the input jack, I think it's probably just before the live makes contact with the pin. As it makes contact, there's a snap and I just have the noise that the pickups are collecting. What might the whistling be?

It's high pitched and not particularly loud, definitly not feedback since there's no growth curve, it's just there or it isn't.

Quote:
the mixer circuit/ series parallel switch is probably there for a reverb, so you can mix wet and dry.... you can remove the switch if you don't need that feature. the return pot adjusts the level of your effect... you don't need to add another pot. if you don't need the pad (attenuator), just lose the switch and the 100K res.
I'm getting the feeling I'll end up removing a lot of these bits.

Quote:
why don't you build it and let us know what happens?!
Dammit... I think I might just do that!

[prepares Debit card for a sound whooping]
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Old 17th November 2009, 08:56 AM   #6
flood is offline flood  India
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hey all, i'm going to implement this effects loop in my souped up atomic 16 build. it looks pretty useful and versatile. if you have successfully used it, please do write back here. thanks!
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Old 16th October 2011, 02:11 PM   #7
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Hi All,
I'm thinking of buying a jet city jh20 tube amp.......there lots of mods on the net for this amp... But none to fit a effects loop.... would the schematic work on this amp? could anybody give me a list of components need for the schematic, as some values seem to be missed.

I'm a very good at soldering, but unfortunately electronics knowledge is fairly rusty.
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