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linear regulator LM1086 — Parallax Forums

linear regulator LM1086

OwenOwen Posts: 100
edited 2007-02-23 01:18 in Robotics
Can anyone give me some insight on linear voltage regulators and how much heat they will generate as the output voltage and the input voltage get further apart. I designed a board for the propellar for a project of mine and used a LM1086 giving me an output of 5v but I want to input 24v is this to big of a difference? I did design extra copper on the board as a small heat sink but any idea what kind of temeratures I can expect? is the any kind of drop in solution for a to263 package?

thanks,
Owen

Comments

  • allanlane5allanlane5 Posts: 3,815
    edited 2007-02-20 18:40
    A linear regulator acts as a 'variable resistance', changing its resistance so that its output stays at a particular voltage, no matter how much current is drawn through the device.· (Within the device's limits, of course. For a LM340 TO263 package, that would be 1 Amp)

    Having said that, Ohm's law applies -- V = IR, and Power P = IV, or P = I^2 * R.

    So, if you have 24 volts in, and 5 volts out, that's a voltage drop of 19 volts.

    Now, the heat equation. Power is equivalent to Heat, in these cases -- in that P = IV equation, the P terms are 'Watts'. Each linear regulator has a certain heat dissapation it can manage. And you can increase that heat dissapation by adding a 'cooling fin' heat-sink to the device (using a suitable insulating mica wafer, and heat conductive "thermal grease"). If you don't dissapate enough heat, the temperature of the device will rise to the point where a "Thermal Crowbar" will turn off the device -- it's output will go to zero, and the current through it will go to zero, until the device cools down.

    A quick look at the LM340-5 device PDF (http://www.national.com/ds/LM/LM340.pdf) on the "Maximum average Power Dissipation" graph, shows that a TO220 form-factor device can dissipate about 2.5 Watts at zero degrees without a heat-sink. With a heat-sink, it can dissipate 5 Watts at room temperature.

    For some idea of scale, you can purchase a 5 Watt light bulb -- it's awfully dim, though.

    OK, so we have 19 volts, and 5 Watts. P = IV, so 5 = x * 19, so x == 260 mA maximum. And that's IF you put on a good heat-sink.

    Without a heat-sink, at room temperature (75 degrees or so) you have maybe 1 Watt of dissipation. 1 == x * 19, x == 52 mA.

    Also from the PDF, if you have 1 square inch of copper for the TO263 package, you have a "Theta" of 35. Power dissipation with a Theta of 35 is about 1.5 Watts -- so we're back to 79 mA or so.

    The big problem you have here is that huge, 19 volt drop, which then drives the power equation. Some people have handled this with TWO linear regulators, a 7812 and a 7805, which can give you more area for heat dissipation. I believe it's also possible to get a "switching regulator" IC, which avoids the "I'm a variable resistor" problem.

    Post Edited (allanlane5) : 2/20/2007 6:46:45 PM GMT
  • QuattroRS4QuattroRS4 Posts: 916
    edited 2007-02-20 18:49
    Owen,
    I believe the LM1086 has a fairly large Maximum allowed 'Input to Output potential difference'
    See attached doc

    but ...

    I have many stamp based designs out there with 24v Supply to LM7805CT voltage regulators..
    these have a 35v VIN max for the 5v regulator


    also see attached Doc

    Obviously the higher the VIN the more work the reg has to do - the hotter it will be - I Have used these without a heat sink - but have them attached to the pcb via tab·(TO220) to a good ground plane and some thermal paste - in this config. they get a bit warm - but not HOT. These are available in a variety of package types.


    Quattro

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    'Necessity is the mother of invention'

    Post Edited (QuattroRS4) : 2/20/2007 7:01:14 PM GMT
    pdf
    pdf
    LM1086.pdf
    822K
    pdf
    pdf
    LM7805CT.pdf
    329K
  • OwenOwen Posts: 100
    edited 2007-02-20 21:17
    thank you for the quick response!
    I wish I had asked this before sending off my pcb design to batchbcp. I wanted to be able to provide 500ma from the regulator but it doesn't seem very likely with a 24V input. Does anyone know of a to-263 3pin switching regulator I could use without a pcb re-design? I couldn't find anything on the digikey web site.

    thanks
    Owen
  • allanlane5allanlane5 Posts: 3,815
    edited 2007-02-21 02:17
    Hmm. 1.5 Watts at 500 mA is like 3 volts across the regulator. So, if you used a 7808 in a TO220 or better yet a chassis mounted TO3 form factor, the 7808 would drop 24 - 8 == 16 volts, at 500 mA that's 8 Watts. You'll need a good heat-sink, like a metal chassis. Then, the output of that regulator can feed your on-board regulator, and you're good to go.
  • FranklinFranklin Posts: 4,747
    edited 2007-02-21 03:36
    The input voltage to the propeller chip is 3.3VDC (in case you missed that)

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  • QuattroRS4QuattroRS4 Posts: 916
    edited 2007-02-21 04:19
    Not missed - the purpose is the first of the dual regulator (5v) is to take the 24v supply - which many of them can do as Vin max on a lot of the 3.3v is far less than 24v then to supply the 3.3v reg with 5V. Also I believe one of the issues is that Owen wants to use the 5v for a subcomponent in the design - well that was my take on it ..

    Quattro

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    'Necessity is the mother of invention'
  • OwenOwen Posts: 100
    edited 2007-02-21 17:13
    I am using the 5v regulator for other components then I have a 3.3v reg for the propellar eeprom ect. it looks like I will have likely have to redesign my pcb to take a more efficient regulator. I looked for aa switching regulator in the to263-3 package but it doesn't seem to exist, has anyone used a switching regulator that they could recomend to take the 24v more efficiently. Is there anything I would need to watchout for when using a switching regulator in place of a linear regulator?

    thanks,
    Owen
  • allanlane5allanlane5 Posts: 3,815
    edited 2007-02-21 17:26
    The one big functional difference is that a switching regulator can put out more high-frequency noise -- in the 40 kHz and above range, depends on the device. So you might need a pair of capacitors on the output -- a large, 22 uF one to handle current surges, and a smaller, 0.1 uF one to filter high-frequency noise.

    All the other issues cover available form-factors -- I've seen IC form-factor switchers, but not TO263. And it's hard to make a switcher as physically small as a linear regulator.

    Sadly, this highlights one of my ongoing issues -- the coolest, best designed circuit board in the world still can't be delivered until the power and enclosure issues are solved. Even though you'd have thought power and enclosure problems were trivial when you began.
  • QuattroRS4QuattroRS4 Posts: 916
    edited 2007-02-22 03:37
    Look for a post recently - Where Mike Green gave a link to an interesting article about paralleling requlators ... I will edit the post if I find it .. more for a current increasing capacity though.

    Owen,
    ······· With the examples of regulators given e.g. LM7805CT - while it may be a TO-220 package - it has a 35v max Vin - seriously consider modifying the likes of this to suit the circuit before reording boards - perhaps fit it on a seperate small pcb· - may be an option ?

    Quattro

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    'Necessity is the mother of invention'

    Post Edited (QuattroRS4) : 2/22/2007 3:47:06 AM GMT
  • Peter VerkaikPeter Verkaik Posts: 3,956
    edited 2007-02-22 07:20
    The easiest way is to put a series resistor in front of your board Vin that dissipates
    most of the power.
    To lower Vin to 7.5V @0.5A using 24V input
    Rs = (24-7.5)/0.5 = 33 ohm
    The power dissipated in Rs @0.5A
    Ps = 0.5*0.5*33 = 8.25W
    So you need a 33ohm/10W resistor
    There will be a ripple at the board Vin that depends on the current fluctuation.
    Since line regulation < 0.1% you may, or may not, require an additional
    input capacitor.

    regards peter


    Post Edited (Peter Verkaik) : 2/22/2007 9:32:11 AM GMT
  • RDL2004RDL2004 Posts: 2,554
    edited 2007-02-22 15:40
    There are integrated switching regulators available from several manufacturers that are "drop-in" replacements for the 78xx series of linear regulators. One of these may be useful in your project. They are not cheap. Here is an example:

    http://www.dimensionengineering.com/DE-SW0XX.htm







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  • QuattroRS4QuattroRS4 Posts: 916
    edited 2007-02-23 00:48
    Peter,
    That is a fair point but I think current o/p is also a consideration from what I gather ...

    ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
    'Necessity is the mother of invention'
  • allanlane5allanlane5 Posts: 3,815
    edited 2007-02-23 01:04
    That was a brilliant link! Thanks, I had no idea.
  • Peter VerkaikPeter Verkaik Posts: 3,956
    edited 2007-02-23 01:18
    Given the input voltage (24V), the required load current (0.5A), and the minimal
    input voltage to the regulator (7.5V for LM7805) the calculated values are correct.
    One does however need to check the maximum power dissipated·by the regulator.
    Pr = (24-I*Rs-5)*I = 19*I - Rs*I*I ->parabolic function -> there is a maximum
    dPr/dI = 19 - 2*Rs*I = 0 yields I = 19/(2*Rs) for which regulator dissipates maximum power
    For Rs=33ohm I=0.288A
    So the 5V regulator dissipates (24-0.288*33-5)*0.288=2.73W maximum @I=0.288A
    At I=0.4A the regulator dissipates (24-0.4*33-5)*0.4=2.32W
    So a heatsink is still required.
    If the required load current·is less, then Rs·increases, and regulator power dissipation
    decreases.

    regards peter
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