A simple Relay Bypass

i'm guessing there isn't an equivalent of a CD40106 with less inverters in it in a smaller package? say, a 3x in a DIP-8 or even a dual in a DIP-6?
 
also, i'm sure this has been covered extensively elsewhere (so please feel free to just drop a link), but what is the advantage of using a relay over, say, a CD4053?
 
CD4053 is not a perfect switch. One has to consider on-resistance, off isolation, linearity, crosstalk and charge injection. They can and have been used successfully when applied correctly. They are more reliable than relays, although a relay does not significantly impact the reliability of a guitar pedal. Analog switches require more parts than a relay solution because they do not support true bypass. What Boss & Ibanez do with JFET switches could be done with a CD4053 or another CMOS switch.
 
Chuck D. Bones said:
In theory, yes. And it might work in some cases. Do you ever wonder why Boss & Ibanez put buffers before and after their FET switches?
Click to expand...
i remember reading about that on electric druid's page about designing a tubescreamer-style overdrive pedal, but re-reading it now all it really says is that the FET switches need isolating from the outside world. doesn't go into details.

i think the little SPDT diagrams on the CD4053 circuit i posted above also made me forget that they're FET switches rather than physical ones.
 
okay, here is my full circuit for my input-output and switching board. it has input, output, DC jack, power conditioning, LED and switching on one board, gives the option of true or buffered bypass, grounds the pedal input in bypass mode and bypass buffer in true bypass mode, and leaves the other half of the TL072 available as either an input or VREF buffer:

dedicated bypass buffer.png.png
 
Last edited:
R101 is too large for the amount of relay current you'll have. I recommend providing a separate filtered power rail for the relay so that changes in relay coil current don't bounce the pedal's Vcc.
 
Chuck D. Bones said:
R101 is too large for the amount of relay current you'll have. I recommend providing a separate filtered power rail for the relay so that changes in relay coil current don't bounce the pedal's Vcc.
Click to expand...
should i have any series resistor for the relay power, or just skip it entirely? and are 100u+100n the right filtering caps for the relay?
 
I'd use 10Ω for the relay power. 100μF + 100nF is good for relay filtering. If you have a high quality power supply, like a Voodoo Labs, then you can forgo the 10Ω resistor. I like to have a little bit of series resistance in case I plug into a lesser power source.
 
i have a couple of basic questions to help me understand the circuit and make sure i've wired it up correctly and not backwards (i.e. LED turns on in bypass mode) -- the datasheets show the positions of the switches in "energized" and "non-energized" states for the coil which is connected to pins 1 (+) and 10 (-). am i correct in assuming that "energized" means there is current flowing through the coil, which in the case of this circuit means that pin 10 is connected to ground, and "non-energized" means there is no current flow because the path to ground is disconnected? or is it the other way around?

secondly, i understand the role of the cd40106 as an electronic latch thanks to r.g. keen's article on CD4053 switching, but that article kind of makes it sound to me like the state of the latch at startup is random. you state on your schematic that your circuit powers up with the relay de-energized, so how is that achieved?
 
thunderaxe said:
i have a couple of basic questions to help me understand the circuit and make sure i've wired it up correctly and not backwards (i.e. LED turns on in bypass mode) -- the datasheets show the positions of the switches in "energized" and "non-energized" states for the coil which is connected to pins 1 (+) and 10 (-). am i correct in assuming that "energized" means there is current flowing through the coil, which in the case of this circuit means that pin 10 is connected to ground, and "non-energized" means there is no current flow because the path to ground is disconnected?
Click to expand...
correct.

thunderaxe said:
i understand the role of the cd40106 as an electronic latch thanks to r.g. keen's article on CD4053 switching, but that article kind of makes it sound to me like the state of the latch at startup is random. you state on your schematic that your circuit powers up with the relay de-energized, so how is that achieved?
Click to expand...
With the power off, C5 & C6 are discharged, i.e. the voltage across each of them is zero. When the power comes on, C5 hold U2-9 is at zero volts. This forces pins 8 & 11 high and therefore pin 10 low. U2-10 low keeps C5 discharged. C6 charges thru R5 and after 50ms or so is ready to toggle the flip-flip to the other state when the stomp switch is pressed. That is how C5 ensures that the circuit always powers-up with the relay in the de-energized state. One could probably put the flip-flop in a random state by rapidly switching the power on and off, but that's not a normal situation.
 
thunderaxe said:
will any N-channel MOSFET work for the ground switch, or are there specific specs i should be looking for?
Click to expand...

I use 2n7000 for this application because it's current-production, cheap, available from many sources, and is still readily available in a TO-92 through-hole package (2n7000), as well as a SOT-23 SMD package (2n7002). In my experience, it tends to be cheaper than BS170, though I would expect that to work just fine in this application.

I get the impression 2n700x is the go-to N-MOS for applications like this.
 
You must log in or register to reply here.
Back
Top