Voltage regulator (Vin: 5-32v, Vout: 5v)

Discussion in 'General Electronics Chat' started by mikewashere05, Nov 3, 2009.

  1. mikewashere05

    Thread Starter New Member

    Oct 26, 2009
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    I have a problem that is becoming more complex than I thought it would be. I need to take input from a battery source that can consist of a 6v array, 12v or 24v and regulate it down to a usable 5v for a microcontroller. I'm having trouble finding a voltage regulator that is available and has this wide of a range. It needs to supply 500mA or so.

    Does anyone know of a regulator up to this task? I haven't been able to find one after searching Digikey and Mouser that has that range within the "Recommended" range. Most "Maximum" ratings are 35v, however I wouldn't feel comfortable using a 32v system for prolonged periods of time above the recommended rating.

    Would it be possible to have two regulators in series? One with an input range of say, 12-40v, and an output of 15v and then another with an input range of 6-25v (the typical I've found) and an output of 5v. What would happen if the source voltage is 6v, will it pass through the first regulator unchanged?

    Appreciate the help.
    Mike
     
  2. Papabravo

    Expert

    Feb 24, 2006
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    There are no linear regulators that would take a 5V input and produce a 5V output. There is a switching regulator topology called SEPIC which efficiently produces a 5V output from an input that can be above or below the output voltage. If you successfully implement one of these you will learn and worry more about magnetics than you can imagine.

    Here is a good place to start:
    http://cds.linear.com/docs/Datasheet/3757fa.pdf
     
  3. mikewashere05

    Thread Starter New Member

    Oct 26, 2009
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    Sounds too convulated for my application, appreciate the help though.

    Did you edit out a suggestion for the LM117 or am I going nuts? Although, it would be cool if I had a voice in my head that recommended circuit components. Could come in handy :D

    Both look to be possible solutions. To clarify, the range of 6-32v is more important than 4-32v, the latter is just a "it would be nice if..."

    Reading the datasheet for the LM117 I see "Vout = 1.25(1 + R2/R1) + Iadj(R2)". The MC34063 has simply "Vout = 1.25(1 + R2/R1)".

    Does this mean that with R2 of 3kΩ and an R1 of 1kΩ I can pass it any input voltage within the recommend operating range and have it output 5v? Or is it more complex?

    Thanks
     
  4. John P

    AAC Fanatic!

    Oct 14, 2008
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    Note that under your highest-voltage condition, the regulator will be dropping (32-5) or 27 volts, at half an amp. That's 13.5 watts! You'll be wanting a fine big heatsink.

    But it's either buy the heatsink, or build a switching system. I'd look through the offerings at Maxim and Linear Technology and see what's available.
     
  5. mikewashere05

    Thread Starter New Member

    Oct 26, 2009
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    What happens if I configure for 5v out and Vin drops to 6v? Will my Vout just drop proportionally or will I have Vout=0?

    Great point, I'll look into heat sinking.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    OK, here's a little bit of an explanation.

    LM117/LM317 regulators have a typical minimum 'dropout' of 1.7v.
    'Dropout' is the difference in voltage from the IN to the OUT terminal.
    There is also Vref, the reference voltage. Vref is nominally 1.25v, and is the difference between the OUT and ADJ terminals. The regulator attempts to keep Vref at a nominal 1.25v by sourcing current from the OUT terminal.

    So, the minimum output voltage the LM117/LM317 can supply from a single "rail" system is nominally 1.25v; the value of Vref.

    If the regulator is configured with R1/R2 to output 5v, the IN terminal must be supplied with at least 6.7v, or the output will fall below 5v.

    The dropout voltage will vary with current and temperature. See the graphs in the datasheet.
     
  7. mikewashere05

    Thread Starter New Member

    Oct 26, 2009
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    Cool, I appreciate the explanation. I think I understand better now.

    I'm looking at heatsinks for the TO-220 package LM317. I found this guy on Digikey, it has a power dissipation of 10w at a temp rise of 50°C. I believe this should be enough to keep the regulator at an acceptable temperature, anything that I'm unaware of that makes this a wrong assumption?

    Glad I found this site!

    EDIT: Oops, forgot link: http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=HS350-ND
     
  8. SgtWookie

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    Jul 17, 2007
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    How is 10W acceptable when you already know that dissipation may exceed 13.5 Watts?
     
  9. mikewashere05

    Thread Starter New Member

    Oct 26, 2009
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    I figured that the LM317 could handle 3.5W of heat internally? A dissipation of 3.5W would be equivalent to an input voltage of 12v with an output of 5v ((12 - 5) * .5 = 3.5). I wouldn't think that a voltage differential of that magnitude would require a heat sink, so I thought that with a heatsink that dissipates 10w with a differential of 27v is equivalent to a heatsink that dissipates 0w (no heatsink) with a differential of 7v.

    Let me know if that assumption is incorrect, this is the first time I've dealt with things of this nature.

    Thanks
     
  10. Papabravo

    Expert

    Feb 24, 2006
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    Thermal resistance, like electrical resistance, is used to compute temperature rise as a function of power dissipation. It is usually expressed in degrees C per watt. There is also a specification of two locations in space that the thermal resistance applies to.

    A subscript of J refers to the semiconductor junction, a subscript of C refers to the case, and a subscript of A refers to the ambient temperature. The Greek letter θ (theta) is used to represent thermal resistance. For example:
    A voltage regulator might have an absolute maximum junction temperature of 150 °C and a θJA of 10 °C/Watt without a heatsink. So if the ambient temperature is a comfortable 20 °C (68 °F) and the regulator is dissipating 13.5 Watts you can see that the junction temperature will be 155 °C. Clearly not a good situation. There are two things you can do in this situation, reduce the ambient temperature or add a heat sink. If you add a heat sink you need to find the θJC so you can combine the temperature rise from the junction to the case with temperature rise from the case to the ambient based on the θCA of the heatsink.

    You can dig into the datasheet for these values. They may be called something else but the units will most likely be °C/Watt
     
  11. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    The best way to solve this is to simply change the minimum spec on the 6V battery to be about 5.3V. Then, you could easily (I could easily, you could too with some learning) build what is called a linear LDO (low dropout) regulator which can regulate a 5V output with only a couple hundred mV of headroom so VIN could drop to 5.2V and still hold 5V out. At 0.5A load, you will burn about 10W with 24V input, but that can be done safely with a reasonably sized heatsink and a TO-220 size FET. You could also parallel a couple of FETs to divide the power between them.

    I hate to plug a product from the dipwad company that fired me, but here goes: I developed a product called the LP2975 for national semiconductor which is an LDO controller just for this kind of app. You select the FET you need and add the caps. I wrote an app note explaining how to use it, it's all part of the data sheet.

    http://www.datasheetcatalog.org/datasheet/nationalsemiconductor/LP2975.pdf
     
    Last edited: Nov 3, 2009
  12. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    That's a thermal resistance of about 5C/W for the heatsink alone.

    Don't forget to add about 4C/W for the rise going from the junction to heatsink for a TO-220 device. That's in series with the heatsink, so the total thermal resistance is about 9C/W. At 13.5W, that's probably too high because you will get a rise of more than 130C above ambient.
     
  13. Papabravo

    Expert

    Feb 24, 2006
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    At least up to 80 °C ambient. That'll cover most of the interesting cases.

    @bountyhunter, Sorry you were let go. I have some concept of how badly that sucks. Your initials wouldn't be RHP by any chance -- would they? If so we have several notable folks who frequent these boards. If not, still glad to have you around. I was unaware of the LP2975, I'm going to file that one away for later. Thanks.
     
  14. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    No, but now you have me squeezing my old brain to figure out who RHP is. If it was "RAP", I would know immediately as he was the most famous person at NS (drove the 50 year old Volkswagon also known as the Czar of bandgaps).

    You'll have to send me a PM and tell me who RHP is.


    I think they did me a favor. There are only two tracks at NS: the technical one and the political one, and I had come to the end of the former and I could not stomach the latter.
     
  15. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    It's a real useful part, but a bit of a fish riding a bicycle. The open loop gain on the controller is a bit too high, so you have to use pretty large output caps to roll the gain off fast enough to be stable. Problem is, the LDO control loop needs some ESR on the output cap to make the loop stable and in many cases you have to add some resistance in series with the cap. As general rule, throw a couple of thousand uF on the output with a one ohm resistor in series with the cap and it will be good to go.
     
  16. AdrianN

    Active Member

    Apr 27, 2009
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    LM3103 from National handles: Vin= 4.5V to 42V for Vout= 0.6V to 7V at 750mA. At 91% efficiency, the heatsink is the PCB.
     
  17. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    The spec says the mosfet's on resistance is 0.7 Ohms, it also can not go to 100% duty cycle. There is no way it could put out 5V unless the input voltage is about 5.8V or higher. If the input has to drop down near 5V, you need an LDO linear reg to do this design.
     
  18. mikewashere05

    Thread Starter New Member

    Oct 26, 2009
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    I think I may have found a solution: the TLE4275.

    It is a LDO linear voltage regulator with a maximum input of 45v and overhead of about 500mV. So I'm looking at 5.5-45v in, 5v out. Supplies 700mA of current. Sounds great.

    Mouser has it in stock in TO-263 packaging for an acceptable price point.

    Now onto the hypothetical scenario:
    Ambient temperature of 40°C
    Necessary dissipation of 13.5w
    Maximum operating temperature of the IC is 150°C

    This means the allowable temperature increase is Maximum - Ambient = 110°C.

    Necessary θja = 110 / 13.5 = 8°C/W

    Using the value supplied at http://www.daycounter.com/Calculators/Heat-Sink-Temperature-Calculator.phtml the TO-263 packaging has a θjc of about 23.5°C/W.

    Following the guidelines of another regulator's datasheet I've gotten:
    θha = θja - θch - θjc
    θha = 8 - 2 - 23.5
    θha = -17.5

    I think this is telling me that I need a heatsink of the rating 17.5 or greater? or lesser? The negative implies it is getting rid of heat?

    Does this sound correct to everyone that knows more than me?
     
    Last edited: Nov 3, 2009
  19. John P

    AAC Fanatic!

    Oct 14, 2008
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    Sorry, I think you'll lose big time with the TLE4275. It's a surface mount component with essentially no chance to mount to a heat sink. You might get by with a TO-220 package, but if you're really going to need the full 13.5W dissipation, you might do better with a TO-3 package.

    What's all this Bob Pease stuff, anyhow?
     
  20. Papabravo

    Expert

    Feb 24, 2006
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    Bob Pease and bountyhunter were both recently let go at National Semiconductor for reasons that are at best opaque. Regardless of the reasons it sucks to be let go.
     
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