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Old 29th September 2011, 06:10 PM   #1
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Default fat/boost switch

Hi, Is there any reason one could not put the bypass capacitor for the boost/fat switch in the footswitch itself? And similarily, could one put a resistor with it to change the total cathode resistor's value?

I don't want to use a relay switch to do it either.

Thanks...

Daniel
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Old 29th September 2011, 07:03 PM   #2
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Yp. It's quite common even tho it's not the best way since you risk injecting noise into the circuit. But as long as the cable is shielded it usually works fine.
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Old 30th September 2011, 10:06 PM   #3
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Default Thanks...

Just to make sure before I go through the work: Here is a schematic of what I'm planning
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File Type: jpg footswitch schematic.jpg (19.9 KB, 62 views)
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Old 30th September 2011, 10:28 PM   #4
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If I understand you right, you would have footswitch/amp cable with V1, V2 and earth leads? Could use stereo coax cable with 6.5 mm TRS plugs?

And also 3PDT switches and battery to allow indicator LEDs so you know where you are at a glance ie when the bass player trips on your rig ...

have fun
JimG
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Old 30th September 2011, 11:00 PM   #5
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I do have a 4 conductor shielded cable I was planning on using with a TRS 1/4" plug. I would use two of the wires for the cathodes to the switch and another for the ground return connecting to the jack ground.

Can the led's be hooked up w/o a battery?
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Old 30th September 2011, 11:58 PM   #6
Enzo is offline Enzo  United States
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If you are switching out the cathode parts completely like that, it is going to pop.

COnsider this approach. If you want to switch between 3k and 1.5k cathode resistors (I am making up the numbers, not using yours at the moment), don't wire a switch to select one OR the other. Wire one 3k permanently, and then just switch another 3k in and out of parallel. That way the resistance still switches from 1.5k to 3k and back, but the cathode circuit is never opened.

Same with the caps. If you want to switch between 0.68uf and 220uf, don;t select between them, just wire the 0.68uf permanetly, then switch the 220uf in and out of parallel. Leaving the 0.68uf in parallel with the 220uf won;t matter. 220uf and 220.68uf won;t sound any different. And when they are closer, say 10uf and 20uf, instead of slecting one or the other, just wire in the smaller one, then switch in another in parallel to add on what you need.

You can do this with any pair of resistances. Wire the higher choice in permanent, then calculate what parallel resistance would result in a total resistance of the desired value for the lower resistance setting.

This results in less stuff switching in and out and less stuff requiring long wires.
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Old 1st October 2011, 12:36 AM   #7
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Yes, I thought about the pop also but didn't know that trick. Could I use a larger resistor say 33K in the preamp, in parallel? That way the math doesn't affect the value much. The same w/a small cap, about 0.02?
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Old 1st October 2011, 12:37 AM   #8
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thanks for all the replies...
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Old 1st October 2011, 07:06 AM   #9
Enzo is offline Enzo  United States
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I wasn't suggesting you change the numbers you already settled on, just suggesting a different way to get there. Caps are more obvious since you find situations like 0.68uf and 220uf. You can always find a stock value resistor for any situation. If you wind up needing 1768 ohms, 1800 ohms is close enough - a standard value. And when I learned electronics, 5% was as good as resistors got other than in labs. Nowdays 1% resistors are easy to find and not expensive, so if you want some odd values, you can still get darn close.


Pops occur when unterminated points in the circuit are switched in and out - usually a cap with a free end. The trick is to keep the caps charged. SO a simple example: Let's say you want to switch a 22uf cathode bypass cap in and out as a gain boost. Just in/out, so cap or no cap. Every time you close the switch, the cap has to charge to the cathode voltage. It does this in an instant, causing a pop. If you put a high value resistor across the switch, no pop. I use 1meg since I have a bunch, but 470k or 220k would work. When the switch is open, there is the 1meg resistor in series with the cap. That basically makes the cap go away electrically. But it is sufficient to keep the cap charged. And when you close the switch, the cap was already charged, so there is no spike of charging current.

If you look at commercial guitar amps, I think you will find a resistor in that duty wherever caps are switched in and out. Same goes for places like relays switching circuit paths.
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Old 1st October 2011, 04:40 PM   #10
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I went through the numbers and came up with values that add in parallel to be close to what I'm planning. I started with the highest value resistor like you suggested and chose the rest to match with that. Same with the caps, but in series.

I have heard about the large resistor on switches, just don't fully understand it. I will check out some schematics on it though. Before I thought about this idea, I had two boost switches from two tubes. The only pop I would get would be the first time switching it in, but after that there would be none.

So, in conclusion, if I have one cathode resistor connected and choose others, in parallel, there should be no pop. But with the cap, the 0.68, there would be one initially, but then fine after that.

One last question about the cap voltage rating: Does a lower voltage rating mean less pop? Ex: 68uf@25v vs @100v...

Thanks again,
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