I'm making a simple booster, I want to keep component count to a minimum and also avoid large-sized resistors (even if shunted to ground, they might contribute noise). It's academic but still... (also, I'd like to keep the input impedance around 1Mohm). So, what do I do? I either have shunt resistors on either side of the input cap and double their size, or what? I would like the booster to be true bypass (otherwise, the solution is just to go with a non-true bypass design, use a "Millennium bypass" which I don't really like, or...?).

Shunt resistors on either side of input cap is the way to go. Don't worry about large resistors going to ground adding noise. Whatever noise they create is shunted to ground through the much lower source impedance (pickups). Just avoid large resistors in series with the source. Even a relatively low 10k in series with the pickups will add noticeable noise - and I see this situation a lot in various designs.

IN & OUT grounded when effect is OFF..



Personally, I use the below wiring..
(you can ommit LED anti-pop configuration..)

I don't get that LED anti-popping circuit. Yes, the current inrush can cause a "DC offset" of sorts, but why not just wire the LED and resistor independent of the main circuit (which will have its own current limiting resistor + capacitor)? This way you should essentially be "isolating" the LED from the circuit. Unless the power supply has high internal resistance, but even a battery has an internal resistance measured in milliohms.

Thanks for sharing Antonis, just tried it (without the led anti pop) on today's build and I really like it. No pop at all and it's a very high gain circuit so I was expecting at least a little bit.

Think I'll make it my standard from now on

Quote from: fryingpan on May 31, 2024, 08:21:14 AMI don't get that LED anti-popping circuit. Yes, the current inrush can cause a "DC offset" of sorts, but why not just wire the LED and resistor independent of the main circuit (which will have its own current limiting resistor + capacitor)? This way you should essentially be "isolating" the LED from the circuit.

PCB has (or should have) its own RC LP filtering (after V+), which serves for circuit's voltage fluctuations related to circuit's instant current needs..
(additionally, circuit's current draw is usually less than LED's one and series resistor in LPF is sized accordingly, for affordable voltage drop..)

But both LED and PCB are fed from the very same source (Power jack)..
So, a sudden drop on power jack can't be effectively faced by circuit's LPF 'cause there is no time for filter's cap to charge..


The above configuration effectivelly acts as a LED current amount delay circuit..

Quote from: stallik on May 31, 2024, 08:43:26 AMThanks for sharing Antonis,

Yor're welcome but I didn't share anything but an aggregate draw of various techniques already advised (and satisfactorily implementend) by various experienced persons..

Quote from: antonis on May 31, 2024, 09:18:00 AM

Quote from: fryingpan on May 31, 2024, 08:21:14 AMI don't get that LED anti-popping circuit. Yes, the current inrush can cause a "DC offset" of sorts, but why not just wire the LED and resistor independent of the main circuit (which will have its own current limiting resistor + capacitor)? This way you should essentially be "isolating" the LED from the circuit.

PCB has (or should have) its own RC LP filtering (after V+), which serves for circuit's voltage fluctuations related to circuit's instant current needs..
(additionally, circuit's current draw is usually less than LED's one and series resistor in LPF is sized accordingly, for affordable voltage drop..)

But both LED and PCB are fed from the very same source (Power jack)..
So, a sudden drop on power jack can't be effectively faced by circuit's LPF 'cause there is no time for filter's cap to charge..


The above configuration effectivelly acts as a LED current amount delay circuit..

Yes, but if you have the LED powered "before" the circuit's RC power filter, there shouldn't be a sudden drop, unless the power supply is "slow" (high internal resistance), and anyway the significantly sized cap on the power supply should readily compensate. I mean, we're talking about 5-10mA at worst (most LEDs are made to run at much lower currents, according to what I see on schematics). A power supply rated for 500mA (it's a small one), even a linear one, shouldn't bat an eyelid.

Quote from: fryingpan on May 31, 2024, 09:36:04 AMthe significantly sized cap on the power supply should readily compensate

It depends on both the size of the capacitor and the value of series resistor (if any)..
Without that resistor, power supply capacitor looks like a short for start-up power..

Anyway, LED anti-pop configuration is only needed for high gain circuits fed from power supply without adequate filtering..
(there is no reason for such implementation if you trust your power supply..)

Quote from: fryingpan on May 31, 2024, 08:21:14 AMI don't get that LED anti-popping circuit. why not just wire the LED and resistor independent of the main circuit ? This way you should essentially be "isolating" the LED from the circuit. Unless the power supply has high internal resistance, but even a battery has an internal resistance measured in milliohms.

You're absolutely right, but the diagram already shows the LED suplied from the raw power input, before any filter resistor. Anyway a wall wart has long wires that may not be such low impedance as a local battery.