Hi everyone
About the 4053 electronic bypass shown by R.G.
What value the capacitor should have?
From what I understand they form a HPF with the 1M resistor (but which one of the two?).
If that is true, an higher value of the capacitor should provide more bass.
But I may be totally wrong.
For guitar I wouldn't mind, but I use my stomps with analog synths and they often have very low frequencies.
Considering ONLY the 1Meg, 0.01uFd will give full bass (-1dB @ 40Hz).
More likely there is other stuff hanging on. 0.1uFd or 0.2uFd may be wise.
might be worth considering that the other end of each switch will presumably also have a 1M resistor, which will parallel with the first one (although with ~150 ohms series resistance in-between).
and what if I increase the 1M resistors to 2M, for example? This should move the cutoff down, so that you can use smaller caps, but would it still bias correctly and perform as well?
You've forgotten about source and load impedances.
It's easy tp do. There are some hidden traps in the way we learn electronics today. Nobody intentionally sets the traps, they're just part of the way we learn. Here are a few of them:
- Lines on a schematic are not wires; they're imaginary superconductors.
- Wires are not superconductors; they're very low value resistors and inductors.
- ALL wires are the leads of invisible capacitances.
In general, these arise because we have to focus our attention on one subset of all but the simplest circuits. In this case, the focus is on just the switching and de-popping. The signal source driving the input of the switching circuit has some impedance in it, and the load on the output of the switch has some impedance loading the switching. The high-pass frequency of the caps in series changes as you change the output loading.
So you have to make at least a guess at the worst-case load you're going to plug into the output of the switch before you can make any real guess at the bass rolloff. The 1M input and output resistors play into this, of course, but the output load appears in parallel with both of them when the switch is on.
So your signal source has to be able to drive the two pull-down resistors in parallel with the (unknown) eventual output load, and the capacitance it drives is that of the two capacitors in the switch in series.
I think all my stomp are low output impedance, and so are my synths.
So I'd say the worst-case load is when I plug my passive guitar direct into the switch, right? Or is it the opposite?
That's one worst case.
The other is if you connect a low-input-impedance amplifier to the output of the switch. A hifi amp with a 10K input impedance will really load it down and raise the low frequency rolloff point.
Ok so I should calculate that impedance in parallel with the impedance of the resistors on your 4053 schematic. But which one of the resistors? The one to ground or the one to +Vbias, or both? And what happens when I raise the value of them?
Also, from your article, you mention:
"actually be better to use one of the alternate true bypass setups to avoid the signal passing through two of the switches when bypassed"
Do you have any example?
the pulldown resistor also parallels with the bias resistors to some degree, check out this link for more on that: http://www.muzique.com/news/pulldown-resistor-vs-input-impedance/ (http://www.muzique.com/news/pulldown-resistor-vs-input-impedance/)
I have used 4066 switches ( the same thing, different wiring). Each switch have an ON resistance of about 300 ohms or less...
I have used
Guitar->dc blocking cap- >vref R - > cmos switch-> vref R-> opamp (buffer or gain stage)
With no issues at all. Of course i have to double the value of the Vref Rs for a desired input impedance.
Thanks
Does anyone have any example of what RG means about alternate wiring of the switch, on his article?
"actually be better to use one of the alternate true bypass setups to avoid the signal passing through two of the switches when bypassed"
http://www.geofex.com/article_folders/bypass/bypass.htm