TRS Phantom Power box to power onboard buffer?

[Big Fat Walrus]

Have you the contract to be signed prepared..??

Ha ha, that's up to you my friend. I believe you have earned some compensation for all your help here.  Perhaps we should name the circuit after you?

Now I just need to sludge through the Eagle part of this. Once I have something that seems good I'll ask for someone on the board's help to double check it all, and then have a few printed. And as promised, I will post all the files so anyone can produce it if needed. Share and share alike, no?

Thanks!

[Big Fat Walrus]

So I have attached my schematic below.

The only question I have left is how to calculate what the voltage out will be. I know the R values can change that, but (as a novice) I am having a hard time with this because it is connected to another circuit. 12V goes in, and hopefully 10 or so comes out, but I can't seem to actually calculate this.

And also- C1 is obviously an Elec cap. But what type should the other two be. Both ceramic? One ceramic and one film?

[antonis]

R1 + LED should be connected to +12V before R2..

For Vout (actually Vin - R2 voltage drop) implement Ohm's Law..
V = I x R, where V is the voltage drop across R and I is the current flowing through R..

If I recall it right, your circuit has 3mA (0.003A) current need..
So, voltage drop across 470 Ohms simply is 1.41V..
12V -1.41V = 10.59V = Vout..

P.S.
Make C3 1nF (in case of μ isn't a typo..)
Both C2 & C3 can be either film or ceramic..
(the later suggested..)

[Big Fat Walrus]

Yes, that is a typo. It is in fact 1nf. Should C2 and C3 both be ceramic? Or only C2?

And- what if I left R1 after the switch? Still works, right? I wanted to be able to use a SPDT to switch it all

[antonis]

Make both C2 and C3 ceramic and walk in peace..

If you place R1 + LED after R1, you'll have to add LED's current on 3mA to calculate total voltage drop across R1..
('cause both circuit and LED current pass through R1)

[Big Fat Walrus]

OK great; thank you.

So, that means the voltage drop across R2 should be adjusted to 23mA (for a red LED)? And keeping the 4k7 value for R1, that means that Vout= 1.19v?

Boo.

[Big Fat Walrus]

Wait a minute-

I admit that this might be a stupid question, but I come here to learn. So here we go-

I don't understand why it matters where R1 and LED are placed. They are still connected to the same 12v IN source, and will still drop the V OUT voltage the same no matter where they are placed. Right?

The switch is used to turn the power on or off to the jack (OUT+ solder pad). The LED is there to indicate when power is going to the + V OUT pad, NOT to indicate when the circuit has 12v IN connected. So isn't it the same regardless of whether it is placed before or after R2?

All this circuit needs to do is provide more than 9v to the V OUT pad (but ideally more than that) after filtering the ripple, hum, etc.

Am I missing something?

Thanks!

[anotherjim]

Well, the "on" LED seems to be the right side to me (has to be after the "on" switch). However, if you use a high efficiency LED, you might use as high as 10k instead of 4k7 & still get good light from it and less loading on the output voltage.

I see you put a reverse polarity protection diode on the input - so do you expect reverse power as a possibility? If so, a power supply capable of over 1A can either blow the diode open circuit or cause it to short circuit. Open circuit won't do as the reverse power can continue to the rest of circuit. If you place the diode after the 470R filter resistor, the resistor will either limit the reverse supply current or get hot/smoke and die. Not necessarily a bad outcome if it protects against further damage and a damaged resistor usually just cuts the feed like a fuse.

[antonis]

Case closed..
(I hope so..)



P.S.
> If you place the diode after the 470R filter resistor, the resistor will either limit the reverse supply current or get hot/smoke and die.<

Jim, in the former case any circuit semiconductor will suffer from reverse voltage where in the later case nobody can guarantee ultra-fast resistor death..

[Gus]

links
https://web.archive.org/web/20100824183740if_/http://www.ebs.bass.se/2009/manuals/NeoGorm.pdf

links in the thread below don't work but it should help
https://www.diystompboxes.com/smfforum/index.php?topic=73492.msg596553#msg596553

IIRC SD or another(s) pickup company had wiring diagram for active circuits powered via a circuit like in the EBS link

[antonis]

Let's make it more graspable..

[Big Fat Walrus]

Thanks, guys

I see you put a reverse polarity protection diode on the input - so do you expect reverse power as a possibility?

Not really, I just thought it would be safer than not having anything at all there

Case closed..
(I hope so..)

Hmm. This is a dilemma

Let's make it more graspable..

Yes, thank you. I need crayons at this point ha ha

[Big Fat Walrus]

You know what, I think I am wasting everyone's time here. So I will buy a book on electronics and try to learn some of these principles on my own. Building pedal kits is basically 'paint by numbers,' and gives you (me) the idea that you understand more than you really do.

Anyway, as promised I have a G*rb*r file for this board that I will send to anyone who wants it. Just send a PM. A small thank you for all the help I have received here. As antonis points out the R values might need to be tweaked to get the desired output, but as Jim points out different LEDs might help with that. As anyone who has read through this thread knows, that is above my level of understanding. But I digress...




Here is an image of the board. It is 1.44" (36.58mm) wide by 2.17" (55.12mm) tall. It is arranged so that it can hang from a solder lug 3PDT switch inside an enclosure, but a SPDT toggle switch in the center column will work as well. The status LED is located just above the switch. The input and outputs are all solder pads, so you can use the jacks that you prefer. There is also an extra GND solder pad on the output side of the board, so a mute switch could be added between the in and out jack if desired. The board drilling will accept 1w resistors and a 50v Electrolytic cap (C1).

So thanks to everyone who has helped along the way, especially antonis. My apologies for bugging everyone!

[antonis]

I shouldn't design boards before electronics book purchase (and studing.. ) or there should be some space/cost waste..
e.g.
Neither resistors need to be 1W nor C1 needs to be rated at 50V..
Caps should be rated at working voltage across their legs (ok.. a bit higher) and resistors at their value times current squared..

[Big Fat Walrus]

I shouldn't design boards before electronics book purchase (and studing.. ) or there should be some space/cost waste..

Yes, that is a fair point. But I have a need for something like this now, and did not like the other options currently available to me.

Neither resistors need to be 1W nor C1 needs to be rated at 50V..

And yes, I know that. But it won't hurt to use larger values, and it is safe. There is plenty of empty space inside my enclosure, so why not? And, in the event that someone else has a similar need but at a higher voltage, this will work for them too right 'out of the box.' Smaller value components can still be used if desired.

Please don't take my previous post as 'I'm taking my ball and leaving.' I did not mean it that way at all. I am very appreciative for all your help and patience, and it led me to the realization that I really should build a better foundation of knowledge before I stray from just building others' circuits. This forum has been very kind and welcoming to me, which is why I wanted to offer the board to anyone who wanted on at no cost. While this circuit might still need a little fine tuning, it is probably still a better and cheaper option for anyone who can solder and drill an enclosure than what is commercially available (say, this: https://www.emgpickups.com/parts/es-918.html)...

But again, sincere thanks to anyone who participated here and helped me learn more about a hobby I really like