Single Buffer Effects Loop

[_ej_]

I have an 18watt TMB build in working on. I really won't be using the first channel so I'm thinking about converting it to an.effects loop. I only would have a single triode to work with, are there any good single Buffer Effects loops I could incorporate into my build? I'd like a serial loop with a level control.

[cdemike]

Using a source follower, you could basically copy a Dumbleator replacing the cathode follower with an IRF820 or similar MOSFET. That would leave the triode available to function as the recovery stage.

[_ej_]

Do you really need the boost on the send? Is the return gain not enough?

[cdemike]

Well the source follower (or cathode follower in the original Dumbleator) is to reduce source impedance and does not provide any voltage gain. Either way you don’t need it — passive effects loops often work fine, though I often see them with amps with cathode followers anyway. It really depends on the application, though. A buffer (whether it be a source, cathode, or anode follower, or something else altogether) will reduce noise and maintain high end information.

[10thTx]

I've done a few amps where I let the mosfet cathode follower (& sometimes a tone stack with it) be the "send" of the FX and then use a
triode for the recovery of the psuedo-active FX loop. This has worked out really great for me & I'm quite pleased with this approach.

In the case of the V2b on the HoSo56 6BM8 " The Minimalist", it is not necessary to split the plate resistor into using two. I did that to help maintain an even cleaner tone.

IF you want series FX loop and not paralleled, then leave out the 470k going between send and recovery.

With respect, 10thtx

[Snicksound]

Do you really need the boost on the send? Is the return gain not enough?

As already mentioned, it's not about gain (in fact you usually want attenuation there), it's about output impedance. The output impedance of whatever is driving the FX send will form a low pass filter with the capacitance of the cable.

A passive loop feeding a 20' long cable for stage use will lose some details. Add a 20' return run with true bypass pedals and suddenly everything gets pretty dull. Lower the output impedance with a buffer and suddenly the sparkle is back. Also makes it easier to plug into line level fx units that have low input impedance (pedals are never an issue as they tend to have high input impedance).

When 15'+ cables get involved, it's quite a noticeable difference.

Can you compensate with EQ? Up to a point yes, but the purpose of an FX loop is often to be transparent.

[pdf64]

If the signal level whete the fx loop is to be inserted is high, then a resistive potential divider can be used to reduce the send signal level and impedance to a reasonable level. eg 470k - 10k provides a send impedance a little under 10k, even lower if a 10k pot is used for ghe send level, which may be acceptable for many applications.

[Snicksound]

If the signal level whete the fx loop is to be inserted is high, then a resistive potential divider can be used to reduce the send signal level and impedance to a reasonable level. eg 470k - 10k provides a send impedance a little under 10k, even lower if a 10k pot is used for ghe send level, which may be acceptable for many applications.

What you're suggesting would be adding a 470k in series with the signal going to the Send control correct?

This sort of didn't sound right to me because adding series resistance, AFAIK, actually increases output impedance (output impedance is not the resistance between output and ground, it's the "a measure of the source's propensity to drop in voltage when the load draws current". A resistance in series will in fact cause a drop in voltage at the output when the load draws more current.

But I figured maybe my understanding is wrong so I went and did the math.

Now, a typical location for an FX loop is between tone stack and master volume. Actual output impedance there is tricky cause it depends on how the tone stack is set. But let's say around 75k which seems about right for roughly centered controls of a cathode fed tone stack (plate fed would be higher). For this exercise we'll have to ignore the fact that this will be frequency dependent though.

We'll assume a 20' cable with a typical 25pF per foot of capacitance, so 500pF total.

Let's say we put a resistance to ground of 470k on the output (because if we go lower we completely screw up the operation of the tone stack), and it goes to a pedal with a typical 1M input impedance.

At 80Hz, almost 80% of our signal makes it through. But at 5KHz we're down 40% of our signal, aka a 5.8dB attenuation relative to 80Hz. That is a lot!

Now, if we take PDF64's suggestion, put a 10k under that 470k and tap the signal there for the Send, we now have a mere 1.8% of our signal getting to the pedal, but that's fine that's on purpose (guitar level vs line level). At 5Kz we are down to 1.5%, so still a bit of a loss but less so. Relatively it's 1.24dB down. A lot less than 5.8dB! So it works, there is still a bit of treble loss though.

Note if there are no buffered pedals on the board and you have another 20' return cable, that jumps to 2.32dB loss at 5KHz.

However, a buffered FX Send with an output Z of 10k (and it's possible to do a lot better) has a loss of .65dB at 5Khz with the 20' cable. A better designed buffer with a Z of 2k gets .23dB

So yes, the passive attenuation circuit proposed by PDF64 is much better than not having it because it reduces the impact of the cable's capacitance in the whole circuit.

[Snicksound]

Note: one dowside of the 470k/10k divider isolator circuit is the load dependant attenuation. Where you get 35dB attenuation into a 1M load, you get almost 41dB into a 10k load (e.g. line level device).

[pdf64]

output impedance is not the resistance between output and ground, it's the "a measure of the source's propensity to drop in voltage when the load draws current". A resistance in series will in fact cause a drop in voltage at the output when the load draws more current.

But I figured maybe my understanding is wrong so I went and did the math.
...

The output impedance in this scenario can't be greater than the resistance to ground

https://en.m.wikipedia.org/wiki/Norton's_theorem
I think you'd find getting your head around Thevenin's and Norton's theorems extremely beneficial