Opinion desired, linear regulator voltage size (automotive application)

Discussion in 'The Projects Forum' started by ke5nnt, Oct 31, 2009.

  1. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    Just a simple question of opinion here. Would you say that using a LM7810 10V linear regulator that gets its input voltage from the battery in a vehicle is cutting it kind of close, given that generally input voltage needs to be maintained at 2V over output voltage. I mean, realistically, even when a car is off, as long as the battery isn't on its last leg, they usually have 12.3V or better, and my circuit isn't designed to operate with the vehicle off. My truck seems to maintain a steady 14.6V while running.

    I can get away with 9V regulator (lm7809) I would prefer 10V as it would make things easier for me.

    What do you guys think?


    Edit: As an after thought, let me add another question. Using the TO-220 package with a heat sink (regulator gets pretty hot but stays within limits), the package takes up a lot of space on the board (board is low-profile so the TO + heatsink lay flat on the PCB). The SMT version of the regulator takes up less space, but I'm concerned about heat distribution using nothing but a copper area under the device on a single sided 1oz board. Thoughts on this as well would also be much appreciated. I'd prefer the DPAK if I can get away with it.

    To answer any questions about airflow, it will be in open air (enclosure has several air vents) but no fan or anything, mounted inside the passenger cabin.
     
    Last edited: Oct 31, 2009
  2. spacewrench

    Member

    Oct 5, 2009
    58
    1
    You might want to add info about the current drawn by your circuit. If the regulator is getting hot with a drop of 2.5-4.5V, I'd guess you're pulling quite a bit. Maybe you can use a buck supply instead. I've been using the 33063 controller for various projects recently, and it's served me well.
     
  3. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    Although I certainly appreciate anyone who takes the time to post a response, yours doesn't answer either of my questions. Despite the warmth of the lm7809 I'm using currently, my amp draw is just over half of the 1A continuous output rated for the device, .580A or so. The LM7809 specifies (2.2A typ) so I'm not concerned with that.
    I said I was concerned with space on my design, and though I'm on my phone atm, where it's difficult to look up parts, I seem to remember that buck supplies take up significantly more space than a TO-220 packaged regulator, so that would be a step backwards for me.
    I'm simply asking 2 questions; opinion on 10V regulator from automotive supply, and heat dissipation by a copper area on the PCB.
    Please don't take my response as me bashing you for trying to help, as that is certainly not the case, and I appreciate your response sincerely, it's just the wrong kind of help at the moment.

    Thanks.
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    Depends; if you'll never use it with the engine off, it might be OK. Power dissipation will be around 2.7W if your input is 14.6v.
    1oz copper won't be much help in dissipating the power your regulator will be wasting. A small heatsink might do it for you.

    There ARE switching regulators available in a 3-lead TO-220-like package that are much more efficient than a standard 7810 regulator, but they're not cheap.
     
  5. Wendy

    Moderator

    Mar 24, 2008
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    It also depends on how stable your voltage has to be. Will a ½V drop be catastrophic to the design, or will it shrug it off?

    Cell phone chargers, which are increadibly cheap, tend to use switching regulators, and they fit inside a cigarette lighter jack. Switching doesn't generate much heat, linear does.

    How much amperage are we talking here?

    Using the PCB as heat sink is basically worthless if the heat has no place to go. This also applies to a heat sink inside a box. There must be a mechanism to dump the heat to atmosphere somewhere, if it is in a box its an oven. If the box is metal, and the component is touching the wall, you might have something.

    Can't sleep either heh?
     
  6. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    Thanks for the response Sarge. I gather that regulators like the lm78xx series are able to output a steady voltage because they dissipate the difference in input voltage as heat, which is why they get warm?

    Sometimes I feel my questions are elementary at best, but I'm trying to understand.
     
  7. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    I work midnights Bill. 10p to 6a.

    To answer a couple of your questions, total amp draw for the 9 or 10V source is in the area of about .580A. 24LEDs on each side (48 total) 2 per series so 12 sets, Vf = .024A so 12 x .024 = .288 x 2 sides = .576A.

    Now a second related question I have is this; 10V source goes into each series through a resistor then to the LEDs then out to a NPN which does the switching from a uC. As far as Power goes, its the LEDs and the resistors that eat up the wattage right? Since each side has its own NPN, .288A x 9V supply is 2.6W, the NPN's have a power dissipation of half that, 1.3W. I'm just making sure that the NPNs don't dissipate all that power, I don't think they do.

    Regarding the enclosure, it's plastic with a lot of vent holes. Think jail bar type enclosure.

    I've looked around for one of these "switching regulators" and I haven't really found anything. Want to point me in the right direction?

    Ryan
     
  8. Wendy

    Moderator

    Mar 24, 2008
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    I've been trying to come up with some switching regulators with mixed success rates, nothing I'd brag about yet. I'm still working on it, at the moment I'm thinking of a 2 8 pin chip design with a MOSFET and largish coil, not what you're looking for if I read your post correctly.

    You need to have the power disappating components have some access to air, otherwise everthing is going to get hot. This may mean your project box is going to have an aluminum side with some small fins.

    I could be wrong, but the µC doesn't need the regulator very badly (correct me if I am), but you need to dump the waste heat from current limiting the LEDs. Even with a regulator, if the LEDs were outside the regulator the regulator itself wouldn't get very warm, since the µC is pretty low power. Does that sound right?

    You can control LEDs on/off and current separately with something like this.

    [​IMG]

    The LEDs don't require the voltage regulator, the transistors Q1 and Q2 regulate current separately, and will the components getting hot.

    We work the same hours. My Mon-Fri is Sunday night to Thursday night.
     
  9. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    V+ 12V automotive
    C1 & C2 - 22pF ceramic disk
    C3 - 220μF 25V electrolytic
    C4 - 47μF 10V electrolytic
    C5 - 0.1μF tantalum

    R1 - 10K
    R2 & 4 - 270Ω
    R3, 5 & 6 - 10K

    4MHz crystal
    Q1 & 2 - LL MOSFETs
    IC1 - 12F683

    -------------------------------------------------

    As has been said many times by various people on this forum, the DC system from an automobile is extremely harsh. I chose 5V into uC for obvious reasons. 9 or 10V (still deciding) to provide positive voltage into the LEDs.

    LED Vf is 3.2V so I can only put 2 LEDs in series with a 9V supply (2.6V drop by resistor). As stated before, there are 24 LEDs per side, 2 per series so 12 series drawing .024A each, so .288A per side and .576A total from the 9V regulator.

    Each series needs 2.6V dropped by resistor so 2.6 / .024 = 108.3 so I think next standard resistor size is 120. P = .024 * 2.6V drop = 62.4mW dissipated by resistor and 76.8mW by each LED. Each side of 24 dissipates 1.67W.

    If 14.6V goes into 7809 (typical for my truck) and 9V comes out that's 5.6V being dropped and .576A being drawn so 3.23W being dissipated by the linear regulator.
    I've only ever tested this circuit with 1.82W being dissipated by the 7809. Gets pretty warm with that, but I can hold my finger on it (with heat sink attached) without getting burned. They also have thermal shutdown protection so I guess I'd know if it got too hot. They're good to 125C.
     
  10. MaxSmoke

    Active Member

    Oct 29, 2009
    35
    0
    Good to see your circuit Ryan, as you say an automotive electrical system is a harsh environment. When you switch the headlamps from on to off, there will be a load dump from the alternator and this can produce a large spike, 80V is not uncommon from what I have read. The other component missing from your diagram is a fuse on the supply input, V+. But, I guess you are using an inline fuse or similar.

    Also, I see there is no output capacitor on the 7809, I think you will find you need a capacitor on the output to keep the regulator stable under all conditions, like temperature, etc. Take another look at the datasheet.

    The voltage rating of C3 looks too low to me, for an automotive application.
    Have you considered some transient protection on the input power supply after the fuse? A suitable MOV or avalanche diode?

    Hope you don't mind my comments, just trying to be helpful.

    Mike.
     
  11. ELECTRONERD

    Senior Member

    May 26, 2009
    1,146
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    Hey Bill,

    In your circuit, which I think you designed:

    [​IMG]

    What is the purpose of the 36Ω resistors at the emitter and the 1k resistors? In addition, what is the purpose of the diodes?

    Thanks,

    Austin
     
  12. Wendy

    Moderator

    Mar 24, 2008
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    If you want to pursue this we probably ought to take it off this thread, but for now...

    Transistors make excellent constant current devices. The diodes make a fairly stable voltage referance. This creates a constant voltage across the 36Ω diodes, around .6V. This in turn translates into a constant current on the collector.

    By taking the diodes voltage drop to zero, the transistors emitter goes to zero volts, which will cut off the current.

    The transistors can get pretty hot, but they are pretty rugged devices overall.

    Like I said, if you want to go into this I have several circuits similar to this in my albums, but this is a subject we need to take off this thread.
     
    Last edited: Nov 1, 2009
  13. SgtWookie

    Expert

    Jul 17, 2007
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    Ryan,
    Have you been following this thread at all? Link: http://forum.allaboutcircuits.com/showthread.php?t=29506

    The MC34063A is a pretty interesting switching regulator that can be configured as a step-up converter aka voltage booster. Instead of regulating down to 9v or 10v, you could boost up to 28v or so @ 175mA, using the parts shown.

    This would enable you to run more LEDs in series; instead of 12 strings, you'd have 4 strings - and very little power wasted. Also, your input voltage could be anywhere from 8v to 16v, and the output would vary only a small amount (perhaps 30mV, around 1%). It would be far more efficient than using the old 7810 regulator.
     
  14. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    @Mike, definately don't mind your comment. I very much appreciate all input I get from members. Regarding the fuse, I do intend on including one, whether in-line or on the board I haven't decided yet. I'm quite familiar with the datasheet for the 78xx regulators, I'm really not sure now why I decided not to include a cap on the output of the 7809, i'll re-evaluate doing that. As to the diode or mov after the fuse, I didn't know to do that, still new to electronics as a whole. Can you tell me how either of those options would help my circuit? what is a mov?

    @Sarge, thanks for the suggestion. I'll look at that thread and see what it's all about (circuits.com) ha
     
  15. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    Ok I'm looking at that device's data sheet, page 5 fig 9 like you mentioned on Quick2Scramble's thread. On my board layout (image attached) I'm running out of space. Obviously going with the MC34063 would replace the V9 component, in this case a TO-220 package LM7809, and the heat sink attached to it. Notice the mount hole H3 so that the IC and heat sink lay flat on the board.

    Looking at the figure on the data sheet, I can replace 9V, H1, and heat sink with the 8pin dip, 4 additional resistors, 3 additional capacitors (although one of the caps would clearly replace C1 on my current board so actually only 2 extra caps), 1 inductor and 1 Schottky diode.

    Obviously the benefit of using the step-up converter is great. As you can see on the board, I have 2 arrays of LEDs, 24 in each array. Left side = 3.2Vf @ 24mA, right side = 2.0Vf @ 24mA. With 28V from the step-up, for 3.2V LEDs I'm looking at 8 diodes per string, 3 strings for total current draw of 72mA compared to 12 strings @ 2 each with the 7809 totaling a whopping 288mA, quite a difference.

    The 2.0Vf diodes only require 2 strings at 14 diodes each, for 48mA instead of 6 strings with 4 diodes each totaling 144mA.

    So, with the MC34063 I'm drawing 120mA, with the LM7809 I'm drawing 432mA. Clearly both are within Io limits, but like I mentioned before, with the 7809 Vi being anywhere from 12 to 15V it's dissipating power as heat for a voltage drop ranging between 3 to 6V @ almost half an amp.



    I should also mention that I'm not done with the board yet, so there's no need to point out to me that there's no traces, or resistors for the LEDs.
     
  16. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    With regards to the MC34063, with the component values listed in figure 9 on the data sheet, it shows Vinput as 12V. We all know that the 12V power source from an automobile is never actually 12V, will these values still hold a 28V output with a V input varying by a few volts?

    Also, what kind of inductor would the 170uH inductor be? I'm not familiar with inductors, and I'm seeing Common Mode, Power, and RF Inductors. A search on mouser for 170uH Power inductors gets 2 results, both obsolete.
     
  17. SgtWookie

    Expert

    Jul 17, 2007
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    Did you read onwards in the datasheet? I guess not...
    Vin can range from 8v to 16v with a variance of 30mV in the output.

    Here's one from Mouser: http://www.mouser.com/ProductDetail...EpiMZZMsg%2by3WlYCkU%2bCbm/yM2yLsnWwZBWWlp7k=

    Here's one from Digikey: http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=553-1402-ND

    It doesn't have to be exactly 170uH; +/- 10% or so (even more) would be OK.
     
  18. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    At 14.5V in, you will be burning about 0.6 x 4.5 = 2.7W. This is easily doable with a TO-220 device, but well beyond the range of what most surface mount devices can handle. Use the TO-220.

    Will work. The 7805 needs 2VBEs + SAT to operate. At normal temps, that's about 1.8V. At 12V you have it made, if it drops below the output of the 7810 will fall with it.


    PCB copper is remarkably poor for heat sinking because it is so thin. With a T0-220 soldered down to a very large copper plane, the lowest thermal value you can expect to get is about 24C/W. That means for 3W of power disspation, the TO-220 gets about 72C hotter than the ambient temp. That is acceptable.

    For smaller surface mount devices, thermal values are more typically in the 50 - 100C/W range. Check the data sheet to get the specific value.
     
  19. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    Although I've become a lot more fluent in reading datasheets, sometimes things still skip by me. I wasn't entirely sure that "Line Regulation" was what I needed. Still, thanks for pointing that out. Every day I sit in front of the computer working on this project and others, I learn new things.

    As always, thanks for the knowledge and the recommendations. I think I found something that will work. I'll be sure to post again down the road after I have everything set on the board and schematic to see if folks want to pick at any errors that might be present.


    To bountyhunter:

    Thank you for the reply on those things. It helps to reaffirm the conclusions I already came to, but hearing it from someone else makes me feel a lot better, so again, thanks for the response.

    New Question:

    If you look at my schematic in this thread, regarding the +5V IC in it (it's a LM7805 TO-220), I'm really wondering just how much current it needs to be able to support. I'm not familiar with how to determine how much current the stuff powered off 5V will need (2 LL MOSFETS, the microcontroller, a crystal, and passive components, plus +5V to pin 5 of the uC whenever the momentary button is pressed) without building the circuit and using an amp meter. I'm hoping that all of those things will be under 100mA, because I'd love to get rid of that TO-220 package and replace it with a TO-92 .1A regulator if I can. Do you guys think it would be less than .1A?

    Thanks for all your help and support, you all make this hobby so much more fun than it would be without all of your help. My wife thinks I'm crazy, but I tell her "electronics is far more fun than any video game, and far more rewarding". Seeings as she is a "gamer", I think she's confused by my logic :rolleyes:

    EDIT: Added Schematic attachment
     
    Last edited: Nov 4, 2009
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