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I have to interface a specific motor, it has a weird driving board, I was in contact with the technical team from the manufacturer, and they told me that I have to send a -24V on a pin to allow it to turn. I have a +24V power supply, several step down modules, on 5V and 3.3V for my microcontrollers and other things.So here is my question: What is the simple way to produce a -24V voltage source from a 24V or lower one? I have searched a lot, but found nothing... My first idea was an inverting opamp, but if I were able to power it down to -24v then I would'nt need it at all :)

Any Ideas are welcome.

My ideas are now oriented to step down ICs able to output negative voltage from a positive one... I guess something around -18V should be enough to be understood by the driver.

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  • \$\begingroup\$ how much current do you need on the -24V? \$\endgroup\$ – Jasen Jul 26 '16 at 9:36
  • \$\begingroup\$ google polarity inverting buck boost e.g. ti.com.cn/cn/lit/an/snva022e/snva022e.pdf \$\endgroup\$ – JIm Dearden Jul 26 '16 at 9:57
  • \$\begingroup\$ Are you sure that this is merely a signal and not a power supply rail? \$\endgroup\$ – Oleksandr R. Jul 26 '16 at 10:03
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Linear tech produce a number of inverting switchers like this: -

enter image description here

Check the data sheet to see if it can produce -24V. If not then use the search engine on this page here

There's also this one: -

enter image description here

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  • \$\begingroup\$ No, it can't. Its specification is 45V maximum between Vin and ground. \$\endgroup\$ – Oleksandr R. Jul 26 '16 at 10:05
  • \$\begingroup\$ @OleksandrR. I did say in my answer to check the data sheet to see if it can produce -24V. Anyway I've added another option. \$\endgroup\$ – Andy aka Jul 26 '16 at 10:06
  • \$\begingroup\$ Yes, and I checked it and found it's not suitable. The second option is better and I upvoted your answer. I wonder, if not the datasheets, where are you getting these images from? \$\endgroup\$ – Oleksandr R. Jul 26 '16 at 10:07
  • \$\begingroup\$ @OleksandrR....... enter your search phrase into google then select images from the top line! Is anyone of these pictures you: google.co.uk/… \$\endgroup\$ – Andy aka Jul 26 '16 at 10:09
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Assuming the -24v is a control input, so doesn't need much current, and only needs approximately -24v, then a cheap, quick and dirty way is to use a charge pump. Like this.

schematic

simulate this circuit – Schematic created using CircuitLab

What's drawn as an opamp is any circuit capable of making a rail to rail square wave.

The output is unregulated, and drops with increasing current.

If you do need exactly -24v, then you can use a 2 stage charge pump and regulator.

If you need more current than a few mA, then it's worth going to a magnetic solution, an inverting boost, rather than a charge pump.

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  • \$\begingroup\$ It takes an oscillator to drive "OA2", and diodes D1 and D2 are reversed. As shown, this would generate positive 24V. \$\endgroup\$ – Whit3rd Jul 26 '16 at 19:42
  • \$\begingroup\$ OA2 could be the oscillator, that's not a full circuit of course. Diodes the wrong way round ... I'd like to say 'congratulations, you're the first to spot my deliberate mistake', but I must actually say 'doh! facepalm'. \$\endgroup\$ – Neil_UK Jul 26 '16 at 20:12
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An easy way is to use a DC-DC converter with an isolated output. For example, an PDS1-S24-S24-S. This particular one, while cheap, has a minimum output load requirement of 4mA so add an LED indicator or whatever if your load is insufficient (the consequence of insufficient loading is that the magnitude of the output voltage rises above the expected value).

enter image description here

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What is the simple way to produce a -24V voltage source from a 24V or lower one?

You don't have to produce a new source. Because +24 and GND/0V is just a name for a potential we can switch the postitive and negative potential to get a negative voltage. Connect the "0V/GND" for your -24 Voltage to +24 and your -24V connector to 0V/GND.

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  • \$\begingroup\$ It's probably not that simple. If the motor needs positive voltages, too, this trick won't work. But generally spoken you are right. \$\endgroup\$ – Ariser Jul 26 '16 at 12:15

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