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Designing a power supply with +10V/-10V/-12V

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Wounded Paw

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This is for a guitar effects mixing/switching box.
Currently using a Meanwell DCW03A-12 encapsulated power supply which supplies -12V/12V, 125mA on each rail, from 9 to 18V DC input which is common for guitar effects pedals adaptors.
Then using 2 x 1N4001 diodes in series to drop each rail to under 11V. The reason for this is that I need a negative voltage more than 1 volt higher than the rails for analog JFET switching using J112's.
This set up currently works well as the Meanwell power supply has very constant voltage but I realize using the diodes is not really the way to go.
Should I try instead using LM317/LM337 adjustable regulators or 7810/7910 fixed regulators to get my rails to a lower voltage than the 12V supplied? The dropout voltage of the regulators may be a problem.
Is there a better way to go while keeping the current consumption to a minimum?
 
Meh, your diode method isn't bad and I wouldn't change it to be honest. Yeah you're going to be dropping some power over them as heat, but its minimal.
Aside from the dropout requirement being low, any linear regulator is going to have exactly the same power wastage so there's no point, and for guitar effects I wouldn't go switchmode to try and drop that negligible power loss either, coz you just risk adding any noise and they're never 100% efficient anyway!
 
That's kind of what I was thinking about the power wasted. I was more worried about the fact that the voltage drop across the diodes would vary with current used in the rest of the circuit. The power consumption of the circuit consists of 9 dual op-amp ICs which are always powered from the +/- rails and the -12V for either 4 J112 JFETs or LEDs depending on 4 switch positions.
 
More a quick diagram than a schematic but there's no part for the DCW03A-12 encapsulated power supply**broken link removed**
 
Unless you go to a tiny switching circuit that is way too complex for this application, any series element used to drop the output voltage will dissipate the same amount of power, delta-V times I. Diodes, resistors, low dropout regulator chips, same, same, same.

ak
 
That's kind of what I was thinking about the power wasted. I was more worried about the fact that the voltage drop across the diodes would vary with current used in the rest of the circuit. The power consumption of the circuit consists of 9 dual op-amp ICs which are always powered from the +/- rails and the -12V for either 4 J112 JFETs or LEDs depending on 4 switch positions.
You'll drop anything from 1V (from <10mA) to 2V (from 2A) combined depending on the current you're pulling if you look at the Vf/If curves. There will be a minimum current draw you need, but its minimal and pretty sure your opamps will do it - if not a quick way is to add in an LED, also gives you a visual power indication.
 
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