Voltage splitter?

Discussion in 'General Electronics Chat' started by anne, Apr 20, 2008.

  1. anne

    Thread Starter Active Member

    Apr 20, 2008
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    I've been trying to find info on the web about this. I thought I understood well enough, but I must still be doing something wrong.

    I'm working with 12vdc and 24vdc power. At one point I need to step down the 24vdc to 12 to operate a relay (so very small amp draw) since I don't happen to have 24v relays on hand. I also need to step down 12vdc 1.0amp to 3.3 to 4 volts to power an led.

    Most recommendations for LED applications (anecdotal ones) recommend a simple series resistor. 250ohm for instance to get 12vdc down to 3 to run an led. I don't get anywhere near this. And using a resistor doesn't seem to allow any amps through at all. What am I doing wrong? (as you can see i have a very limited skill with electronics)

    Another instance, I found that a 70K series resister will bring the 24vdc down to 12vdc, but again, there is absolutely no amperage present at the 12vdc point with my meter. I hook up a fan or a light and there is nothing! How is this possible when the meter reads 12v?

    I'm really confused here because there are dozens and dozens of web pages out there describing how to step down low draw low voltage for things like LEDs using series resistors.

    I attempted a voltage splitter as well. Two or three resistors in series and the voltage varies depending on where you tap it. But again, no current seems to be available to actually run anything. (i'm really getting frustrated here).

    I suspected that it might have something to do with the fact that I'm using SMPS transformers, but I tried a conventional 12vdc wall transformer and the very same thing happened.

    Any help at all? Thank you so much!
     
  2. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    When you want to power a LED, which has voltage drop of 3V and recommended current 20ma, form 24V DC! source, you need a resistor with value U/I= (24-3)/0.02=1050 ohms.

    Do you happen to have an analog voltage meter? The have a fairly low resistance between the leads, so they act like an ideal voltage meter with a resistor in parallel. You need to count with that resistance in your measurements.
     
  3. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
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    For the relay, you can do the same type of calculation, but you need to know either the coil resistance or operating current. With operating current, say 200 mA, you need to drop 12 volts. So R = 12V/0.2A, or 60 ohms. The wattage is (0.2E2)*60 or 2.4W .

    John
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    Everybody's right - but just to make it a bit more clear:
    LEDs are usually rated for maximum voltage @ current and typical voltage @ current.
    The actual forward voltage (Vf) on an LED may vary considerably though!
    Since you have a 24v supply, you can use Ohm's Law to determine a resistance that will allow up to the maximum current to flow, in order to determine your LED's actual Vf.
    Let's say that the maximum current allowed is 20mA
    Ohm's Law:
    R = E/I (or, Resistance = Voltage/Current)
    R = 24V / 20mA
    R = 24 / 0.02
    R = 1200 Ohms
    So, if you use a 1200 Ohm resistor in series with your LED across your 24V power supply, you know that the current will be somewhat less than 20mA. But first, let's see how many Watts (power) that the 1200 Ohm resistor will be dissipating:
    P = EI (Power = Voltage * Current)
    P = 24 * 0.02
    P = 0.48 Watts
    You will need to use a single 1200 Ohm resistor that is rated for at least 1/2 Watts.
    Alternatively, you could use two 600 Ohm 1/4 Watt resistors in series.
    Once you have connected the resistor(s) in series with the LED across the power supply, you can measure the actual Vf of the LED using a multimeter.
    Then you can determine the actual value of the current limiting resistor that you need:
    Rlimit = (VoltageSupply - VfLED) / LEDCurrent

    Note that you can power several LEDs in series, and use a single current limiting resistor.
    MaxLEDsInSeries = Integer( (VoltageSupply - 1) / VfLED ) (discard the non-integer portion)
    If your voltage supply is not regulated, you should use an active current regulator instead of a simple current limiting resistor.
     
  5. anne

    Thread Starter Active Member

    Apr 20, 2008
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    I must be in some physical null zone in this shop. You are saying pretty much exactly what I have learned (not to sound unappreciative - i thank you very much). But I've gone back and tried again. I have two digital multimeters that seem to be working ok. I cannot power anything from the 24v SMPS or the twelve the moment I try to put a resistor in series with anything. The led won't light and a muffing fan will not turn. The moment I put the resistor out of the loop by jumping it with the pos lead the fan runs fine and the LED lights (just for a split second with the 12v - I know this is harsh, but I've got plenty of spares anyway).

    I've got a drawer full of resistors, so it cannot be a faulty specimen.

    Just to be quite certain, series is putting the resistor in line (either neg or pos) and parallel would be bridging the lines with a resistor before they reach the device to be powered -- correct?

    What on earth can I be doing wrong? I know It's got to be obvious and simple.
     
  6. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
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    Series is as you describe it. It is in-line with one lead.

    Is your SMPS from a computer? If so, you need to put a load on the 5V part of the supply for it to deliver any real power. Generally, you need to load the 5V to about 5 to 10 % of its rated current ability. That is also true of some other SMPS -- for example, I have one from Jameco that has a minimum current need of 10% for the 5V section, even though the 5V section isn't being used. Bad news is that it is a 50A section, so I need a 5A drain. If you do the math, that is a 1 ohm 25W resistor. I use a 50 watter, and it still keeps the bench area warm.

    So, check you power supply. Be sure its output isn't dropping as soon as you put a load on it. If so, the specs should say something about minimum load, or just assume you need 5 to 10%.

    BTW, the current I used for the example in my post may be way large for your case. I have many relays that operate on less than 50 mA. What is the current rating of your relay coil? John
     
  7. mik3

    Senior Member

    Feb 4, 2008
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    you cant put the same resistor for the led and the fan. you have to put a resistor for each separately.

    can you tell us what components you want to drive with each supply?
    it will be helpful. also it would be helpful if you tell us the rated current of the fan.
     
  8. hgmjr

    Moderator

    Jan 28, 2005
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    Many SMPS power supply designs require a minimum current draw for them to regulate properly. If the load to the supply is not above this minimum current the power supply does not output any voltage at all.

    Is it possible that this is what you are encountering? You should be able to check the datasheet for the supply and determine if this is the case with the supply you are using. Try loading the power supply with at least 10 percent of its rated output and then try your led and series resistor to see if that cures your problem.

    hgmjr
     
  9. anne

    Thread Starter Active Member

    Apr 20, 2008
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    What am I doing wrong???

    I just got up from the laptop and went to the desk. I plugged in a wall transformer (non smps) rated 12vdc 1.25A power supply. The leads measured 12.11VDC at 2.01Amp on my meter.

    I found and tested a 1.58K resistor and put it in series. This measured 12.11V !! Again, I couldn't get an amp/milliamp measurement out of it.
    It wouldn't nudge a 12v muffin fan. Won't light a bulb. I'm not trying them both at the same time.

    I tried a few other resistors. Nothing below 20K or so will affect the measured voltage, and any resistor blocks the current (seemingly).
    This is crazy. I'm doing exactly as described!!!!

    I have to go to 58K to get 7.0 volts from this 12vdc. But once again, there are no measured amps at it and it won't nudge the 12v muffing fan. 7VDC is way more than the minimum needed to spin this thing.

    I've got thousands of resistors, a couple dozen different wall transformers (std and smps) and three multimeters. I've swapped out everything for a couple of hours now.

    Is my workspace possessed? Am I a complete idiot? There must be an 'a ha!!' moment here that's just right under my nose.
     
  10. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
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    Your series resistance is way too high. 1.58K will pass only 7.6 mA from a 12V supply. Try your relay with a 200 ohm resistor and see if it works. That may be too much current (60 mA), but it probably won't burn it up right away. Do you have the relay coil resistance or current rating??? John

    Edit: I would focus on the relay or fan first. The led's come later as the calculation is a little more complex for them.

    Edit2: I am getting a little confused. Notice that many of the calculations given above are for 24V, not 12V. Rather than guessing at which of the things you want to do first. Please pick just ONE item and one voltage. We will calculate the series resistance for you. John
     
  11. anne

    Thread Starter Active Member

    Apr 20, 2008
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    Is this the a-ha moment? What I know about electronics is pretty much intuitive. I've fixed more than I've broken just by having good hunches. So I've been measuring these little breathtakingly simple circuits as they are. Intuitive?

    I went ahead and used a 185ohm resistor, even though it measured on the ohmmeter EXACTLY the same 12.1 volts that the resistorless circuit (bare leads) showed and sure enough, as expected, the LED lit and, then trying the fan, it ran as well. HOWEVER when I measured while the LED load was applied, I got the 3.4 volts as I was aiming for. No load - same voltage as without the resister -- with load -- the voltage I was aiming for.

    So why does it measure 12vcd until the load is applied? I didn't measure while the little fan was running but I suspect it was in the range as well as the speed suggested about four volts.

    My sincerest apologies to everyone for wasting this space if this is something that even a high school dropout should have foreseen. It was never mentioned in any of the dozens of pages I've read about voltage dividers and series resistance. Why not?
     
  12. jpanhalt

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    Jan 18, 2008
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    Resistors only drop the voltage when there is current flowing. That is Ohm's law: Voltage = Current*Resistance

    You don't see the voltage drop with your voltmeter alone, because it is extremely sensitive and draws essentially no current through the resistor.

    John
     
  13. SgtWookie

    Expert

    Jul 17, 2007
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    If you look closely at the face of your meter, it's likely that somewhere on it you will read:
    "10K Ohms/Volt" or perhaps "10,000 Ohms/Volt"
    Remember when you were measuring 12.11V from the wall-wart power supply? When you finally came up with a 58K resistor and got a reading of around 7V out?
    12.11V - 7V = 5.11V (dropped across the resistor)
    7V = dropped across the voltmeter
    58K / 5.11v = 11,350 Ohms per Volt.
    I suspect the zero is off a bit on your meter's needle. When using an instrument like that, I always try to take the readings with the meter lying flat on it's back, so that the needle sweeps horizontally. There is normally a small screwhead at the needle's pivot point. Turn it slightly (very carefully) in one direction or the other, until it is precisely lined up on zero. You should be looking straight down at the face of the meter, not from an angle. Good voltmeters have a mirror on the scale; adjust your viewing position until you can't see the reflection of the needle in the mirror. This ensures minimal parallax error.

    If your meter was truly 10k Ohms/volt, if you used a 60K Ohm resistor in series with your meter across 12.11V, you should measure half the voltage on the meter - or 6.055V. This is because your voltmeter is the other half of the voltage divider.

    With digital multimeters, the impedance (internal resistance) is extremely high. The advantage is that there is very little loading on source signals, so readings are much more accurate.

    If your SMPS power supply came out of a personal computer, as everyone else has said already, you will need a load. I rigged up an old ATX form factor PC power supply as a bench supply; I used a 10 Ohm 10 Watt resistor to cause a minimum load of 2A on my +5V supply. This is somewhat less than 10% of the maximum output of 25A, but it has been working fine.

    Note that in PC supplies, the -5v and -12v will have very limited current output capabilities, usually less than 1 Ampere.
     
  14. anne

    Thread Starter Active Member

    Apr 20, 2008
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    Thank you everyone for all of the assistance. I think I'm on some firmer ground now.

    One (hopefully final) question related to this matter. I've also been assuming that a device simply takes what current it needs from a source and that extra was just reserve. In other words, a 12vdc light bulb, or .25amp 12vdc fan, would perform identically whether drawing from 12vdc 1.0amp supply or a 12vdc 5.0amp supply. Is this not the case?

    If I'm energizing a miniature 12vdc relay on a 12vdc 2.0 amp supply do I have to be concerned with matching the current level? With the relay, or the LED, is it not simply a matter of supplying each with the appropriate level of specified voltage? The leds I'm using are 3.3v (typ) to 3.6v (max) with FW current of 25ma. Is it not enough to arrive, with a resistor, at 3.3v off of a 12vdc 2.0amp SMPS?

    I have a collection of power supplies in a drawer. None of them are computer power supplies - except for a couple of laptop wall transformers.
    The two laptop wall transformer SMPS's are 15vdc @3.0Amp and 15vdc @5.0amp. They are from Toshiba laptops.

    Other supplies are collected from thrift stores and garage sales and are mostly SMPS from 4.5vdc to 30vdc in a variety of amperages. The smps seem to be more robust (amperage wise) than the std wire wound transformers of similar voltage.

    I haven't used an analog meter for quite a while - still have my grandfather's little black bakelite Simpson tucked away somewhere. I have a Micronta (radio shack) 22-195 auto range desktop meter and a tiny little radio shack 22-802 digital meter. The little one has no amperage selection and I'm going to have to reread the manual on the micronta to get more familiar with it's features.

    Thank you again. I really appreciate access to a forum such as this. Someday I might be able to actually contribute.
     
  15. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
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    That is basically correct, so long as the amp capacity of the supply is greater than the amp needs of the device. I say basically, because sometimes ratings are inflated, are not for continuous duty, many devices draw more current initially than they do after being on for awhile, etc. John
     
  16. anne

    Thread Starter Active Member

    Apr 20, 2008
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    Thank you jpanhalt.

    In one instance I'm using a 12vdc 2amp smps to operate a pair of muffin fans rated at .5amps each. I expect the fans to switch on/off in an automatic temp sensing circuit approx six to ten times an hour. The relay is rated @10amps. Would it be better to keep the smps energized and switch the 12v to the fans, or better to wire fans direct and switch 120vac to the smps? Or no difference at all?

    I know the operation will remain identical. I just wonder if one or the other is recommended?
     
  17. jpanhalt

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    Jan 18, 2008
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    If you would draw your proposed circuit, it would be possible to respond with more information.

    One thing that strikes me right off is if there is no 12V line in the system, because the smps is turned off, where will the power signal come from to pull the relay in so the smps starts up? Is the thermal switch on the fans adequate to interrupt 120 VAC to turn the smps on? If so, why do you need the relay?

    Does your smps have a stand-by mode? John
     
  18. anne

    Thread Starter Active Member

    Apr 20, 2008
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    MUCH more simply put, if I were going to run a 12vdc smps out of the wall socket to power a couple of muffin fans that were going to be used intermittently, would it be wisest to put the switch between the wall and the smps, or between the smps and the fans?

    I realize it accomplishes the exact same goal. Is one method preferred?
     
  19. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
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    The answer depends on how you are going to run the switch. All else being EQUAL, why run the smps when the fans are not running.

    You have not said what the duty cycle of the fans needs to be. If they switch on and off 6 times per hour, but stay on 9 minutes each time, that is different from staying on only one minute each time. John
     
  20. anne

    Thread Starter Active Member

    Apr 20, 2008
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    Well, all things are pretty much equal. I have a situation where I have a 50/50 choice in that however they are switched does not require the addition of any more cost or complication to the project. The fan duty cycle is not calculable - only that it is not continuous.

    The fans may be running six periods for one to nine minutes each one hour, not running at all the next hour, and perhaps three periods of 19minutes the third hour... and so on, as the system is required to cool a constantly varying heat source.

    At the extreme, the fans may cycle 25 times in one hour (not likely at all but possible). If this were the case, is it actually requiring more energy, and causing more wear/tear on the smps to keep cycling it and therefore be preferrable to just leave it energized and switch its output?
    Or does it not matter at all to the smps? If it does no harm at all to cycle an smps on/off in this manner then I'll just switch the 120v to it.

    You have not said what the duty cycle of the fans needs to be. If they switch on and off 6 times per hour, but stay on 9 minutes each time, that is different from staying on only one minute each time. John

    If it were one or the other -- 9 minutes each, or one minute each --
    would that make the difference between deciding to cycle the 120v to the smps or cycling its 12v output?
     
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