How To Reduce 48VDC to 5VDC

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  #1  
Old 02-07-08, 05:41 PM
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How To Reduce 48VDC to 5VDC

I need a simple way to step down 48VDC to 5VDC. I need to use a 48VDC supply (phantom power from an XLR cable coming from a mixer). I need to tap into that to get 5VDC to run something that's designed to use 5VDC from a USB connection.

Is there a simple way to do this? Can I do it with resistors alone? I also need to use the least amount of space possible. If it's only resistors, then space won't be a concern.

I have some LEDs that came with resistors designed for stepping down 12VDC to whatever they use (1.5VDC or something... I forget). Anyway, it stands to reason to me that I can do something similar to accomplish what I need to do here unless that's too big of a reduction or whatever.

Thanks in advance!
 
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  #2  
Old 02-07-08, 05:45 PM
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A pure resistive voltage divider only works at an exact load current. If the current varies, then the voltage at the tap varies/

You can use a Zener diode to provide a more steady source of a certain voltage./
 
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Old 02-07-08, 05:46 PM
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Wouldn't a DI box work? It would take care of the phantom power plus give xlr to 1/4" adaption.
 
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Old 02-07-08, 05:54 PM
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Originally Posted by chandler View Post
Wouldn't a DI box work? It would take care of the phantom power plus give xlr to 1/4" adaption.
I'm not sure how the DI box would give me the 5VDC I'm looking for.

Problem is, even if it would, I still need a much much smaller solution than adding a component like that. If I can, I need to keep it down to the size of an XLR plug or so.
 
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Old 02-08-08, 07:43 AM
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Old 02-08-08, 07:48 AM
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Lm7805

a very simple 3 term fixed 5volt regulator, available everywhere. I forget its max input voltage, but if not rated at 48v, then you might need a input dropping resistor.
 
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Old 02-08-08, 12:59 PM
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Originally Posted by telecom guy View Post
a very simple 3 term fixed 5volt regulator, available everywhere. I forget its max input voltage, but if not rated at 48v, then you might need a input dropping resistor.
You mean like this one from Mouser?

http://www.mouser.com/Search/Refine....tt=511-L7805CV

I noticed that the max. DC input was 18V so I guess I'll need a resistor. I called Mouser and the sales person connected me to technical. The guy in technical couldn't tell me what I need so he gave me the number for the manufacturer who didn't answer. Would you happen to know which resistor I need to make this work?
 
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Old 02-08-08, 01:44 PM
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Max input V is 30

In any case, you will need to drop the 48v down before the IC sees it. What is your expected 5 volt current? You need to know or measure this before the entire circuit can be designed. There could be power dissipation concerns as well.
 
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Old 02-08-08, 01:59 PM
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Originally Posted by telecom guy View Post
In any case, you will need to drop the 48v down before the IC sees it. What is your expected 5 volt current? You need to know or measure this before the entire circuit can be designed. There could be power dissipation concerns as well.
It's a light with 3 LED's so not much. The package says something about 40ma. Does that sound right for 3 LED's?
 
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Old 02-08-08, 04:57 PM
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PST-DAA4805 cost ya $30.75 plus shipping.

http://www.powerstream.com/dc-dc.htm
 
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Old 02-08-08, 05:58 PM
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so, you have some options

if your load is fixed at .04 amps, then a simple resistor in series will work and you won't need the regulator; In your case, a 5 watt resistor, near 800ohms. I think 820 is a standard value. If your load is variable, then take Furds advice. If you want to roll your own, then, well.. It's going to take some zeners, maybe another resistor, in addition to the 7805. If you going over 40ma, then the DC/DC inverter will be much more efficient.
 

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Old 02-08-08, 06:50 PM
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Originally Posted by telecom guy View Post
if your load is fixed at .04 amps, then a simple resistor in series will work and you won't need the regulator; In your case, a 5 watt resistor, near 800ohms. I think 820 is a standard value. If your load is variable, then take Furds advice. If you want to roll your own, then, well.. It's going to take some zeners, maybe another resistor, in addition to the 7805. If you going over 40ma, then the DC/DC inverter will be much more efficient.
Ok. Yeah, I do want to keep it as simple as possible mainly to keep it as small as possible.

So you're saying that as long as it stays around .04 amps, 1 resistor will take care of it. Is that right?
 
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Old 02-09-08, 04:07 AM
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Make sure the mixer is capable of supplying enough current. Phantom power supplies are not designed for anything other than mics. Each mixer channel usually has a current limiting resistor to protect the power supply.

Also, if you change the voltage the mic won't work with 5 volts.

One more caveat: Phantom power has positive voltage on both pins 2 and 3 and zero volts on pin 1. If you connect one side of a resistor to both pins 2 and 3 you'll short the AC audio signal. You'll need one resistor connected to each pin, and those resistors must be matched to a maximum of 1% tolerance. Again, though, you'll draw down the voltage to the mic. Mics usually use only one of the two pins internally for +V, but the voltage also hits the output side of the audio transformer inside the mic body. If that voltage isn't identical on both pins the mic output will be noisy.

If you're using a mic line with no mic connected, why not disconnect the line from the mixer and use the cable with a 5-volt supply at the mixer end?

The more I think about it, the less this seems like a good idea.
 
  #14  
Old 02-09-08, 02:17 PM
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Originally Posted by Rick Johnston View Post
Make sure the mixer is capable of supplying enough current. Phantom power supplies are not designed for anything other than mics. Each mixer channel usually has a current limiting resistor to protect the power supply.

Also, if you change the voltage the mic won't work with 5 volts.

One more caveat: Phantom power has positive voltage on both pins 2 and 3 and zero volts on pin 1. If you connect one side of a resistor to both pins 2 and 3 you'll short the AC audio signal. You'll need one resistor connected to each pin, and those resistors must be matched to a maximum of 1% tolerance. Again, though, you'll draw down the voltage to the mic. Mics usually use only one of the two pins internally for +V, but the voltage also hits the output side of the audio transformer inside the mic body. If that voltage isn't identical on both pins the mic output will be noisy.

If you're using a mic line with no mic connected, why not disconnect the line from the mixer and use the cable with a 5-volt supply at the mixer end?

The more I think about it, the less this seems like a good idea.
I see. So I'll need to draw from pins 2 & 3 equally instead of just pin #2 for instance so I won't throw the current out of balance. Is that right?

Do you think 3 LEDs should draw enough current away to hinder the mic performance?
 
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Old 02-10-08, 06:29 AM
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Check out this site for samples of BA9s 48 volt LED bulbs.

They won't fit in an XLR case ...

May I ask why you're planning to do this?
 
  #16  
Old 02-11-08, 09:24 AM
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Originally Posted by Rick Johnston View Post
Check out this site for samples of BA9s 48 volt LED bulbs.

They won't fit in an XLR case ...

May I ask why you're planning to do this?
That's interesting. I was hoping to use the light I already have just because it's the right size and is already built with a flexible neck, etc. I thought about doing this because we have a small stand/lectern that's sometimes used when it's fairly dark(too dark to read what's on the stand) and it gets moved around a lot so plugging into power isn't very convenient. It has a battery powered light right now that's kind of awkward and it's prone to run out of batteries at the wrong moment. We do, however always have a mic cable running to the mic and it has phantom power so I was hoping to tap into that to solve the lighting problem.
 
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