0
\$\begingroup\$

I am trying to power a circuit with two op-amps that require multiple voltages. The voltage requirements are -14v, +14v, ground/24+. I am trying to power this circuit from a 12vdc power supply.

I am using a 14vdc to 24dc boost module to obtain 28v and a rail splitter circuit to create a virtual ground for the -14v and +14v and its working. I am using 24v from the dc to dc boost to supply the board for the 24vdc requirment.

The problem I am having is the input/output negative(common) from the power supply through the DC TO DC board connects to the virtual ground circuit input and output 14v-. This rail cannot connect to the negative ground common, this is my problem.

So my logic is to isolate the dc to dc input from the output and this will in turn eliminate the -14v from being connected to the 12v negative ground common?

I think this will work but how can I achieve this? I already have the dc to dc board and the virtual ground board. Is there a simple electrical circuit I can make to add isolation or a whole other idea that doesnt require me to have to use a dc to dc board that has isolated negative input/output?

Any suggestion would be greatly appreciated, thank you.

Newbie

enter image description heredual rail splitter

\$\endgroup\$
9
  • \$\begingroup\$ Me thinks you are asking way too much of those components. You can do one (splitter) or the other (booster) but not in conjunction at the load end. Does it really need to be 14 or can u get away with +/-12V \$\endgroup\$
    – Trevor_G
    Mar 7, 2017 at 20:20
  • \$\begingroup\$ i could get a way with +/-12 at slightly diminished performance. Can you explain splitter? At this point im all ears.... \$\endgroup\$ Mar 7, 2017 at 20:28
  • \$\begingroup\$ See answer with my comment. \$\endgroup\$
    – Trevor_G
    Mar 7, 2017 at 20:33
  • \$\begingroup\$ See single-supply op-amp application-note on Google. It is your friend. \$\endgroup\$
    – skvery
    Mar 7, 2017 at 20:33
  • \$\begingroup\$ What is the 24V used for if I might ask....? \$\endgroup\$
    – Trevor_G
    Mar 7, 2017 at 20:45

2 Answers 2

2
\$\begingroup\$

Use two DC-DCs instead.. much simpler and saves all that messing around with the ground.

enter image description here

\$\endgroup\$
1
  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ Mar 7, 2017 at 23:03
0
\$\begingroup\$

It would be MUCH simpler to use DC-DC voltage inverter to get -12V from your +12V and use that to power your device.

If it works with +14V it will probably work with +12V.

You can use:

  • a charge pump
  • an inverting DC-DC (called inverting buck boost) which can be designed to output +14V if you want
  • an isolated DC-DC

Choice of topology will depend on current draw from device.

\$\endgroup\$
10
  • \$\begingroup\$ Yes @peufeu that was where I was headed. Use 2 DC-DCs. One inverting to -12V the other Boosting to +24. \$\endgroup\$
    – Trevor_G
    Mar 7, 2017 at 20:32
  • \$\begingroup\$ Sorry, saw your message after writing. Also, requirements are not clear: if OP uses a virtual ground, then we dont know if the +24 shall be referenced to virtual GND or to the other GND... not to mention to what GND the analog IO shall be referenced! \$\endgroup\$
    – bobflux
    Mar 7, 2017 at 20:34
  • \$\begingroup\$ the -/+ 14 is 300-500ma and the 24v is 5a....I was looking at the charge pump and islolated dc-dc before the post...I already have the dc-dc modules so I am trying to utilize them. They work on seperate 12v supplies but im trying to get it all to work on one supply. \$\endgroup\$ Mar 7, 2017 at 20:35
  • \$\begingroup\$ OK; in the schematic you posted, please note that the orange "+28V", seen from your board's point of view, is actually +14V, since your board's GND is a virtual GND which is 14V above the power supply's GND... (ie, problem!) \$\endgroup\$
    – bobflux
    Mar 7, 2017 at 20:36
  • \$\begingroup\$ A whooping 5A on 24V???? Why not use a 24V power supply instead of an expensive high current boost DCDC? \$\endgroup\$
    – bobflux
    Mar 7, 2017 at 20:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.