DOD flip flop power switching: how to always power up off?

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Poparad
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Post by Poparad »

Hi, this is my first post here, so apologies if I put this in the wrong subforum. I've recently been getting into modding pedals and I've successfully learned how to alter Boss pedals to always power up off (it's always been an annoyance of mine; if I want it on, I'll turn it on myself). I have an old DOD Grunge pedal (the original "butt/face" enclosure, the board is rev B) and looking at the schematics for it and some other DOD pedals of the same 90's era, it looks like the flip flop switching circuit is different than the Boss one. In the Boss layout, it's completely symmetrical and lowering the cap value to one of the two sides will reliably cause that side to be selected when the power is plugged in. In the DOD circuit, there are only resisters around the two transistors and it doesn't look quite as symmetrical. I'm still pretty new to understanding how circuits work, so this is beyond my knowledge at this point. Does anyone know how to modify the circuit so it always powers up in the "off" state when the power supply is plugged in? Thanks!

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modman
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Post by modman »

Schematic is exactly here: viewtopic.php?p=221070#p221070

Showing that the power is tied in to the input indeed: but I cannot really tell. Does plugging in provide a ground for the whole circuit? Short that switching bit...
DOD FX69b Grunge input
DOD FX69b Grunge input
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stolen
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Post by stolen »

First of all, welcome aboard :D!

Second of all, cool, didn't know about the 4007 - did anybody make a llama or something with one of those yet? :3

Let's make sense of the circuit. It's a lil weird, but nothing scary. Pull up the schematic and the CD4007 datasheet if you wanna follow, or feel free to skip this if you don't need the details. As you can see in the datasheet, the IC is mosfet-based, so first we gotta know a little bit about them. A very simple model is that drain and source form a switch that is controlled by the gate voltage - for a nmos, the switch is closed if the gate voltage is significantly higher (2-5V typ.) than drain/source (whichever is lower), for a pmos this is inverted. As you can see, the 4007 features a bunch of complementary pairs that are driven by the same gate voltage - one closes, the other opens. Kinda like an SPDT switch. There's also some parasitic diodes involved in mosfets that we need to be aware of in the general case, but for this simple circuit we can skip that.

The 4007 contains two different units, 2 identical, 1 with an additional internal connection. But if you look at the schematic and compare pin numbers, you'll find that this internal connection has been externally replicated on the other ones too so that all of them are functionally identical. However, there is one vital difference: Two of them are connected to the power rails, meaning they act as inverting binary buffers: With a high gate, the output is low and vice versa, and only little current has to flow into the gate regardless of what output current the buffer provides. The last one however is hooked up to a signal, and one node is floating! Referencing the datasheet reveals that we are only using the pmos side as a switch to short the input of one of the inverters to a capacitor whenever the footswitch is pressed down. From there you can put some brain cheese into how the state machine works, but we don't wanna be too verbose here.

Anyways, here's what we'd try: When pin1 of the 4007 is high you're in bypass. So at startup you want to add some inertia that keeps pin3 low at boot. We think a simple capacitor from pin3 to ground would do the trick; it needs to be large enough to affect the boot process, but small enough to not mess with C16. We'd try something like 4.7nF. For completion's sake: If this works, and someone wants an "always on" DOD for their racc system, tying the cap from pin3 to 9V should do that.

meow

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Poparad
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Post by Poparad »

Thanks! That's really helpful. I'm looking at the schematic and I think I'm reading it correctly with your explanation. In your last paragraph, you're referring to the 4007 that's connected to the signal and not either of the two connected to power, right? I'm a bit fuzzy on the pin numberings, would this be correct?
4007.png
4007.png (54.23 KiB) Viewed 174 times
Thanks!

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stolen
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Post by stolen »

Hey, uh, we're a lil confused, the pin numbers are right in the schematic - 3 is correct in your drawing (even though you put it close to pin11, but they're shorted together anyways), but you put a 2 next to pin10 and a 1 next to pin9, so there seems to be some kind of misunderstanding between the two of us?

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Post by Poparad »

Oh, I think I see what you're talking about now. I was reading the schematic as U3A, U3B, and U3C all being physically separate transistors, but now I think what you're referring to is one IC chip with many legs that just happens to visually be drawn separate, even though physically it's one unit. I didn't know what a 4007 was. I'm still pretty new to this!

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stolen
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Post by stolen »

Oh fair, don't worry! It takes some while to get used to different drawing and annotation styles :3. We didn't know what the CD4007 specifically was either, but the CD40xx series is very common, so whenever you have a 4 digit number starting with '40' chances are it's one of those.

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