Diodes to Improve Regulator Efficiency?

Discussion in 'General Electronics Chat' started by jwilk13, Sep 9, 2011.

  1. jwilk13

    Thread Starter Member

    Jun 15, 2011
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    Hello all,

    I've got what seems to be a fairly simple question regarding linear voltage regulators, I'm just not sure if my logic is correct. According to the image shown, the efficiency of the regulator would be approximately 42%, which is pretty bad. The regulator shown (MIC2954) is capable of producing 5V with anything up to about 30V, but the efficiency is obviously even worse.

    [​IMG]

    My questions are these:

    1. Is it possible to use diodes to drop the input voltage enough to improve the efficiency of the regulators, and is that even practical?
    2. What are the things to consider if this approach is taken?

    I know it's probably better to use a switching regulator, but they're costly and require quite a few more external components (I know, so does stringing together diodes :p). Any thoughts or alternate approaches?
     
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    1) You can use a number of diodes, or even a resistor, to drop voltage - however, you're just shifting the power loss from the regulator to the diodes/resistor.

    With any linear regulation scheme, your efficiency will be just a tad worse than Vout/Vin; where Vin is wherever you start dropping voltage. I say worse, because there is a small amount of current that is used in the regulation itself, whether through a resistor divider to set the voltage, or through an internal divider to ground for fixed regulators.
     
  3. paulktreg

    Distinguished Member

    Jun 2, 2008
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    You could reduce the voltage to the regulator this way and hence improve its efficiency but you'd just be moving the losses to the diodes.

    Snap!
     
  4. jwilk13

    Thread Starter Member

    Jun 15, 2011
    228
    12
    That doesn't seem like a horrible tradeoff though, right? Consider the following situation:

    12V input, 5V output

    This means that 41.67% of the power will be expended in the load and 58.33% in the regulator. If the source current is 150 mA, this means ~7V * 150 mA = 1.05W dissipated in the regulator. That seems like an awful lot to ask of this particular regulator.

    So it seems like adding in a few diodes with a total Vf of about 5V would help quite a bit, actually bringing down power dissipation from 1.05W in the regulator to about 300 mW.

    Is there anything tricky about selecting diodes for this purpose?
     
  5. jwilk13

    Thread Starter Member

    Jun 15, 2011
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    Or maybe you're trying to tell me that the diodes can't handle that amount of power dissipation :p
     
  6. bertus

    Administrator

    Apr 5, 2008
    15,647
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    Hello,

    The efficiency will not be better when you use extra diodes.
    The power dissipation will be the same, but be split into diode power and regulator power.

    If you want higher efficiency, you should take a look at a switching regulator, like a buck regulator.

    Bertus
     
  7. jwilk13

    Thread Starter Member

    Jun 15, 2011
    228
    12
    Thanks Bertus,

    You're right, better efficiency would be achieved with a switching regulator. The wording in my first post wasn't the best. I probably should have worded it along the lines of reducing the power that the regulator has to dissipate. I'm just concerned that >1W is going to make the performance of the regulator really poor, or just destroy it altogether (depending on heat sink).
     
  8. tom66

    Senior Member

    May 9, 2009
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    Well, *technically* diodes do improve the efficiency of the regulator (lower Vdrop), but not the entire circuit; the efficiency stays the same.

    Now, you need to calculate if the temperature is too high with 1.05W dissipation:-

    MIC2954 has a junction (the internal chip) to ambient (air) thermal resistance of around 62.5°C/W in the TO-220 package. This means for each watt, the regulator rises 62.5°C above ambient. This figure is common to most TO-220 devices. If we assume a room temperature lab, at 25°C, the maximum power dissipation is given by (125°C [max temp] - 25°C [ambient temp]) / 62.5°C/W = 1.6W. So you're in the clear at 1.05W.

    Most people worry when electronics get hot. 125°C is enough to burn your finger, but the device has been rated to handle that. Power electronics get hot; they are often small and dissipate a lot of power in that small area - but it's not a problem that they do because they are designed to handle that. And worst case, there is a thermal protection circuit, which will turn off the regulator if the device exceeds a set limit.
     
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  9. jwilk13

    Thread Starter Member

    Jun 15, 2011
    228
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    So if the PCB were designed with an on-board heat sink (basically a top-side ground plane connected to the tab of the MIC2954) that was connected with vias to a bottom layer ground plane, the max power dissipation could be increased (or the max ambient temperature could increase)?
     
  10. tom66

    Senior Member

    May 9, 2009
    2,613
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    Yes but you would need to take into account the thermal resistance of the PCB to get accurate results.
     
  11. Adjuster

    Well-Known Member

    Dec 26, 2010
    2,147
    300
    To drop (say) 5V, you would have to use quite a few diodes, and note that as they heated up their voltage drop would decrease. You might be better off with a suitably rated resistor.
     
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