LDO currents

Discussion in 'General Electronics Chat' started by Col, Dec 18, 2012.

  1. Col

    Thread Starter Member

    Apr 19, 2012
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    If I have a high input voltage and low output voltage, is it so that I will be able to provide proportionally higher output currents? Another way to phrase it, is the power equal at the input and output sides of the LDO.

    Lets say my IP is 25V and 1mA, if my output is 2.5V can I provide 10mA to my load?? Assuming 100% efficiency of course which will never be the case

    Intuitively I would think so. Or at least Pout = Pin-Ploss, but I notice LDOs get pretty warm, and the bigger the voltage difference the warmer they get. Also there is no mention of losses, efficiency or I/O current ratios in any LDO datasheets
     
  2. MrChips

    Moderator

    Oct 2, 2009
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    To convert with the highest efficiency from one voltage to another voltage, whether increasing or decreasing voltage, one should use a switching power supply.

    A low drop out (LDO) regulator only mean that your minimum input voltage only needs to be a slightly higher than the desired output voltage. Any voltage drop between the input voltage and the output voltage must be given up as wasted heat, determined as (Vin - Vout) x current.
     
  3. JDT

    Well-Known Member

    Feb 12, 2009
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    A carefully designed switching circuit should be able to reach 90% efficiency or even better. I can see a couple of problems with your input and output voltages and currents though.

    • The switching control circuit will take some operating current which could easily be in the region of 1mA. That would drop the maximum efficiency to 50%.
    • With an output voltage of 2.5V synchronous rectification (or low-side switching) will have to be used. The Schottky diode normally used in a buck switcher will drop at least 0.4V considerably reducing the efficiency still further.
    Have a look at:- http://www.linear.com/product/LTC3632 for an example.
     
  4. Col

    Thread Starter Member

    Apr 19, 2012
    44
    2
    Maybe I should describe my actual application. I want to harvest energy from a bus with 18-30V, I can only take 100uA from the bus. I will be charging a 10F 'super' cap over a long period (roughly 12 hrs) and then discharging to provide energy for an RF transmission burst.

    I looked briefly at switching regulators, but as you mention the quiescent currents are in the mA range. I would not have power to supply the regulator itself never mind charge my cap.

    So I guess this means the current is constant, and power is burned off in the LDO.

    I had found this LDO, (http://www.farnell.com/datasheets/610151.pdf) as I can tolerate its quiescent current of 12uA. If an LDO is too inefficient, and a switching regulator is too power hungry is there a better way?
     
  5. Col

    Thread Starter Member

    Apr 19, 2012
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    Ah.. this guy looks like a great start!! The bus I take my energy from is a 4mA-20mA current loop. However this guy consumes 125uA in active mode. So not charge left for my cap :(

    Am I trying an impossible task?
     
    Last edited: Dec 18, 2012
  6. MrChips

    Moderator

    Oct 2, 2009
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    A current loop supplies current not voltage.
    You have minimum of 4mA available, not 100μA.
    Depending on the other loads that are already in the loop, you can add an additional 100Ω load in series and this would drop 0.4 - 2V which should not be a problem then use something like LTC3105 or LTC3108 to step up the voltage.

    Just an idea.
     
  7. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    Let's first do some underlying physics and assume you get every bit of that 100uA into your 10F cap for 12 hours. How much voltage will your circuit get?

    since: I = C dv/dt

    dv = I dt / C = (100 uA) (12 * 60 * 60) / (10F) = .432 volts


    That is it, the simple physics. If you think you can get more voltage given this cap, this time, and this current, you are looking for a perpetual motion machine and not a circuit.

    energy = 1/2 C V^2 = 1/2 (10F) (.432)^2 = .933 joules

    Now a joule is 1 amp for 1 second, meaning if you can figure out a way to extract every bit of energy from your cap (and you can't) you can run your circuit with 1 amp of current for 1 second.

    Note: it may be counter intuitive but a smaller cap actually accumulates energy faster, and the voltage is higher too. A 1F cap yeilds 4.3V and 9.3 joules after the 12 hours.

    Rather then trying to convert the source voltage down a simple constant current source would be simpler, though I don't know of any offhand that give 100uA. A 300K resistor alone would limit the current to 100uA for 30V, but drops to 60uA at 18V.
     
    Last edited: Dec 18, 2012
  8. bountyhunter

    Well-Known Member

    Sep 7, 2009
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    The thread title is "LDO" which is a linear regulator, not a switcher. The input and output current are the same, raising VIN does nothing to change that.
     
  9. Col

    Thread Starter Member

    Apr 19, 2012
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    2
    True, but the main load in the loop requires a loop current of 3.9mA, so I have been given 100uA to play with. My initial understanding was that I would be powering in parallel and drawing this current at the loop voltage. (I admit this is my first experience with current loops). Now I get it that we will be in series, with an allowed voltage drop of up to 2.5V. I understand the current is the signal, and 4mA is the lower limit. Surely if I start sending this current into my cap then I'm going to mess with the signal

    Thanks Ernie, lesson learned... start with the basic physics before jumping right into the problem with misguided assumptions

    Absolutely right, we digress from the title of the thread. Maybe I should change the title.

    So now im looking at a constant current source to 'steal' away 100uA from the current loop and use it to charge my capacitor

    Thanks guys your comments are very helpful
     
    Last edited: Dec 19, 2012
  10. MrChips

    Moderator

    Oct 2, 2009
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    The current delivered by a 4-20mA current loop is determined by the transmitter.
    Hence, to a limited degree, you cannot mess up the current.

    You really need to sit down and figure out the circuit topology in order to draw power from the 4-20mA current loop. You will need to place your load in series with the loop in which case you will see the same 4-20mA.

    I still have not figured out for myself what kind of circuit will be required in order to harvest any power from this loop.

    Edit: If you look at the data sheet for LTC3105 you will find a circuit for a 4-20mA power tap.
     
    Last edited: Dec 19, 2012
  11. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    It would be easy enough, you could power any circuit like a 5v device (PIC etc) using a 5.1v zener diode in the loop. Or a 12v zener gives you 12v power.

    The catch of course is that you can only draw <4mA to power your device at all times, although you can draw more current at times when the current loop has more than 4mA flowing.
     
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