How might I produce two 'isolated' power supplies from one source?

[kimelopidaer]

Hello!
Right now i'm playing with a few different phase shift oscillators, seeing how they can introduce different tremolo type sounds
when i interface an oscillator with a simple booster on my breadboard.
I encountered that ticking problem when i feed both the booster and oscillator with the same power supply.
So, I connected the oscillator to its own completely separate power source ( another 9v battery) and the ticking went away.

Okay I am interested now in making it work with just one power supply again.
How does one create two isolated power supplies from one source? I would appreciate even a few keywords to search for.
There's no problem disabling the oscillator when I don't need it - footswitch or toggle accomplish that task and then the tick is gone; I just hear it through my playing when the effect is engaged.

thankyou All,
K

[R.G.]

It's worth thinking a bit.

I would (and often do  ) wonder how they couple together in the first place. It's not just that they both connect to the same power supply. There is some way that the operations of one get through to the other.

There exist only three known ways that signal can get from one place to another in electronics. Those are conduction, electrical field coupling and magnetic field coupling. By showing that a second power supply stops the signal transfer, you've removed electrical and magnetic (or their combination, electromagnetic/radio). So it's got to be conduction. But in what?

The two commonest suspects are power supply impedance - a non-zero "resistance" inside of the power supply which lets signal currents make the power supply voltage wiggle around, and shared ground currents/resistances, which does the same thing on the ground side.

In this context, "isolating" may take the flavor of reducing the internal impedance of the power supply at some frequencies with inserted isolation resistance/inductance and shunt capacitances to lower the impedance again outside of the resistance/inductance. Or it may take the path of separating ground return currents into separate wires so the can't cause a signal transfer because any ground-shift voltages are in different wires.

If any of these work, a second/isolated power supply is an expensive way to solve the problem.

But to answer your question, you can make an isolated supply from another supply. You chop/amplify it into AC, run that through a transformer for isolation, the rectifiy/filter back to DC. It's done all the time.

[kimelopidaer]

Thankyou,

i really appreciate the detailed response.
I'll try separating the grounding current paths first.


Sometimes I get lost in my little piles of pedal parts. The more I know the less I know. Electronics is humbling but fascinating.
regards,
K

[darron]

as R.G. is hinting, hopefully there's a MUCH easier solution than making an isolated power source.

breadboards can be great for carrying noisy signals with capacitance. maybe try some filtering caps close to the critical areas too?

ANYWAY, to your question. Like R.G. said, you'd have to oscillate it - turn it into AC, then put it through a transformer, rectify it, filter it, probably regulate it.... sounds like a LOT of work to run a boost with a phaser, the circuitry just tripled. but I've been playing with the isolated DC-DC convertors (or sometimes DC-DC regulators) which do all of this in tiny scale in a little four pin package. I just made a power supply with 10 isolated outputs that can handle 100mA @9VDC from them. They are a bit costly though.

But I'd look at a fixing the problem rather than a semi-workaround (:

[seedlings]

[speculation]
This might work:


Adjust the resistors depending on current needs so you don't lose too much voltage.
[/speculation]

CHAD

[R.G.]

[speculation]
This might work:

Adjust the resistors depending on current needs so you don't lose too much voltage.
[/speculation]

Correct. The series 10K's isolate with series impedance, the 100uF's lower the apparent impedance of the supply.

Also correct in that the series resistors drop voltage. Ohm's law says R = V/I. In fact, this is the very definition of resistance. So 1 ohm resistor can be thought of as one volt per ampere. A 1K can be thought of as one volt per milliampere.  A 10K resistor is ten volts per milliampere. That's a lot. If you shorted the outputs, you'd only see currents of 9V/10k = 900uA. That's not enough to run a pedal.

Once you know how much current the loads want, you can then make a decision about how much resistance you can stand. I'm guessing that the right answer will be between 10 ohms  and 100 ohms.

The frequencies this is effective for also change. So a 10K and a 100uF will be most effective above F = 1/(2*pi*10K*100uF) = 0.159Hz, which  is massively below audio. But a 100 ohm with the same 100uF will only be effective in shunting above 15.9Hz. Still, not too shabby. Actually, it's better than that, as you have two rolloffs, one from each output to the power supply filter 100uF, and another to the other output. So 100R (that is 100mV/milliampere) might be a good place to start.

Isolating the return ground wires from each output so they only meet back at the 9V supply will also help.

[seedlings]

So get some big 10W 100 ohm resistors!  If you're drawing more than 200mA (4W).

CHAD

[PRR]

> get some big 10W 100 ohm resistors!  If you're drawing more than 200mA (4W).

Ohm's Law is your friend.

200mA is 0.2A. And 0.2A times 100 ohms is 20 Volts.

You can't have a 20V drop in a 9V world.

What's the worst-case, resistor takes ALL the 9V leaving nothing for the pedal? 9V/100 is 0.090A. And Power: 9V times 0.090A makes 0.81 Watts. A 1 Watt (preferably 2W) part is fine for 100 ohms in a worst-case 9V situation.

But we don't want the resistor to take ALL of our 9V. Say we want 8V left to work with. So the resistor drop is 1V. Assume we get 1V drop with 100 ohms. Then the current must be 1V/100 or 0.010A; the heat is 1V*0.010A or 0.01 Watts. A 1/8th-watt part is way more than we need (but is the cheapest resistor so we use 1/8W).

[seedlings]

> get some big 10W 100 ohm resistors!  If you're drawing more than 200mA (4W).

Ohm's Law is your friend.

200mA is 0.2A. And 0.2A times 100 ohms is 20 Volts.

You can't have a 20V drop in a 9V world.

What's the worst-case, resistor takes ALL the 9V leaving nothing for the pedal? 9V/100 is 0.090A. And Power: 9V times 0.090A makes 0.81 Watts. A 1 Watt (preferably 2W) part is fine for 100 ohms in a worst-case 9V situation.

But we don't want the resistor to take ALL of our 9V. Say we want 8V left to work with. So the resistor drop is 1V. Assume we get 1V drop with 100 ohms. Then the current must be 1V/100 or 0.010A; the heat is 1V*0.010A or 0.01 Watts. A 1/8th-watt part is way more than we need (but is the cheapest resistor so we use 1/8W).

P=I^2 * R = .2*.2 * 100 = 4W was what I plugged in.  Yes, limited to 9V - forgot about that....

So... if his circuit draws more than 10mA he'll have to lower the resistor?

CHAD

[bonaventura]

to get max 10mA, he should use 900 ohm resistors, if im not mistaken.

any thought against applying this trick to branch off an output from a power supply?

weber quoted 70 dollars to ship their transformer to this part of the world 

[PRR]

> to get max 10mA, he should use 900 ohm resistors,

That leaves _zero_ voltage in the pedal-guts.

Yes, if the pedal normally draws much less than 10mA, and the power supply burns at 10mA, 900 (or 1K) is the answer.

[bonaventura]

mmm...burns at 20mA,perhaps?

[Mick Bailey]

I've had success eliminating LFO ticking some time ago in feeding the DC into separate LDO regulators and treating the outputs as separate supplies. Doesn't work with all types of regulator, though.