# Electronics 12v DC to 9v DC 600ma

Discussion in 'Modding' started by nry, 19 Nov 2011.

1. ### nryMember

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I have a KVM switch I want to power from 12v but the switch requires 9v 600ma

How easily can this be done? Assuming I could build a fairly simple circuit up to do this?

Thanks

2. ### confusisKiwi-modder

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3. ### LennyRhysOink!

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You don't need to regulate the amperage, just the voltage - an appliance will draw what it needs, as long as the power is available (600mA is nothing).

As for 9v, you don't even require a circuit - you could do a "dirty" mod and use +12v for supply voltage and +3.3v (orange) for ground, giving you effective 9v (or 8.7v which is near enough). Just don't short the 12v and 3.3v or you will fry everything.

Safer alternative would be to buy a fixed 10v regulator - after voltage dropout, it will be about 9v.

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5. ### Wicked_SludgeMy eyes! The goggles do nothing!

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for such a low current, couldnt he just use resistor to drop the voltage? it wouldnt necessarily be the most energy efficient way, and the resistor might require a small heat sink since it would be dissipating something like 5-6 watts....but surely itd be cheapest.

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6. ### LennyRhysOink!

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The problem with using a resistor would be that the output is not regulated - if the input fluctuates, so will the output.

@OP there are a number of circuits that you can put together but the best ones require quite a bit of work and a few specific components.

Assuming that your switch would normally run from a mains transformer, can you not just buy a 9vdc 600mA mains transformer? They aren't expensive.

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7. ### mvagustaDid a skid that went for two weeks.

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I don't think a resistor would work fine, because I expect that current draw would fluctuate proportionately to data flow.

One of these would work fine, just make sure you use a small heatsink to keep temps down to a safe/reliable level. If you used a relatively large heatsink, such as on old/unused chipset/oem cpu heatsink, then the regulator should run very cool.

A 9v regulator will output 9v, as the voltage dropout is taken into account in the design. This is why regulator chips need an input at least 1v above the output voltage.

Another option is to use four high current diodes in series, which would be a similar or possibly cheaper option than a regulator.
Each diode will drop ~.7v, dropping your 12v to ~9.2v, and provided the diodes are rated at say ~4a, there will be no need for heatsinking. You can just trim the positive lead appropriately, solder them in line, heatshrink over both the diodes and the negative lead, and you've got a very neat solution.

Last edited: 20 Nov 2011
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8. ### LennyRhysOink!

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The diode solution is a nice idea.

A 9v regulation module will output 9v, but a 9v regulator (just the component) will output less than 9v, hence my suggestion for 10v regulator.

9. ### nryMember

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I would go down the route of a 9v transformer, but I have limited number of sockets available in the equipment rack, and have quite a beefy 12v power supply which supplys a few things, so wanted to use that.

10. ### TealcNew Member

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I'd probably go with the diodes idea too if I didnt already have a load of LM317T variable voltage regulators to hand.

Based on 600mA you could probably get away with 1A diodes but it's usually a good idea to use larger ones just to be safe.

What about some 1N5401, they are cheap enough 3A diodes.

You could do the job for less than a £/\$/€

11. ### nryMember

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Showing my complete and utter noobness in electronics here. I can only put a circuit diagram into a working thing, wouldnt know the first thing to do with these different components you mention!

12. ### TealcNew Member

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For the diodes you'd just break the positive wire going from the 12v power supply, solder 3 of the diodes in a line so that the grey bands all sit towards the KVM switch. Then attach the diode string between the break you made. Insulate with insulation tape and job done.

For the 9 or 10v regulator you'd need to again break the positive connection, placing one wire on the transistors input and one for output ( the KVM switch). The regulator will also need a ground connection I believe.

13. ### mvagustaDid a skid that went for two weeks.

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I've never come across that, I've found that regulator chips just need an input which is at least 1v above the output, connect the ground, and enjoy regulated voltage - if the input voltage is too low, then voltage dropout will cause the output to be below spec.
All the other stuff like heatsinking, filtering, current boosting, etc, is needed to suit the current load, but there's no need to help with the voltage regulation - the circuit within the regulator chip takes care of that.

There's detailed explanations of all this in regulator datasheets.

14. ### arakisNew Member

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my advice use a regulator chip, 7809 add two 0.1uF (16+V) caps, one from Vin to gnd, and the other from Vout to gnd, keep them as close to the IC leads as possible.

My second advice is not to use the diode as a voltage drop, the reason why is that diodes change their voltage drop with current(0.5-0.7) first you'd need at least 4 of them to drop 2.8V, and that is when high current is drawn, at low current you'd only get a 2V drop...that's,
BTW the 7809s are dirt cheap and realty the best solution for you.

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15. ### mvagustaDid a skid that went for two weeks.

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That's true arakis, which is why I suggested using 4 diodes. 4 of the 1N5401 would be good, I wouldn't go any lower in the rating, unless you want diodes that would run warm, and would get hot when wrapped in heatshrink.

The 7809 is a 1a regulator... so @600ma, a small to-220 style heatsink would run quite warm.
Forget running it without a heatsink, as it would soon run hot and eventually die.
Sure using an old chipset heatsink or something about the size would be an easy solution, and adding a couple of caps is a good thing.... but all this basically needs to go in a case!

This is another reason why I suggested simply using a few diodes, heatshrinked together with the power lead, as it's so neat and compact.

I wouldn't worry if the diodes only dropped 2v, since 10v is close enough, but @600ma, I doubt things would be that far off.
The switch would have a 5v regulator inside, and it's possible that you could just give the switch 12v without a problem!
I wouldn't recommend this without having seen the schematic of the switch of course, my point is that getting the voltage down to anywhere near 9v would be fine, so even if the diodes only did drop the supply down to 10v, it wouldn't be a problem.

But hey, I guess you might as well order 5 diodes, test it all out before you heatshrink them, and you'll find out how many diodes you'll need - I'd rather have a tad more voltage than not enough btw.

Last edited: 21 Nov 2011
16. ### arakisNew Member

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My concern had more to do with the noise that will result for the power supply voltage swinging in tune with the current used, even with a secondary voltage regulator it could introduce noise which could hamper the performance of the device..

17. ### mvagustaDid a skid that went for two weeks.

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Yeah, there would be some noise, but not any worse than the noise that comes out of any typical plugpack, or even a high quality pc psu. A HQ bench psu is pretty clean, but not much else running from AC is, that's what all the noise filtering circuits on electronic devices are for.

18. ### LennyRhysOink!

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Yes, and that's precisely why some regulators bear the "low dropout" designation - they output what they claim to output without any dropout, unlike crap regulators such as the LM317T which have ridiculous dropout.

My suggestion for a 10v regulator assumes that it is a cheap one with high dropout

19. ### arakisNew Member

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the dropout for 78xx IC is 2V, as long as the input is above 11V the regulator will keep its output at 9V.

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