Question: Series resistor into Vin of buck regulator.

Discussion in 'General Electronics Chat' started by Wormius, Mar 18, 2014.

  1. Wormius

    Thread Starter New Member

    Jul 25, 2007
    4
    0
    Hi,

    This is my very first post, so go easy on me!

    I am designing a product for use in a hazardous location environment, and that requires current limiting of the power source (in this instance a 12V SLA battery). The lowest value of resistance that I am able to use for current limiting in the event of a short-circuit is about 4.2 ohms. The issue is that I am also using a buck regulator (TPS62130) to generate a 4.7V rail at up to 2A.

    What I am trying to wrap my head around is how to predict the input current and thus the input IR drop into the regulator across the 4.2 ohm resistor during different load conditions.

    Conversely - is there also a method for predicting this in a boost regulator, or does it create a runaway situation with 100% duty cycle or something?
     
  2. Dodgydave

    Distinguished Member

    Jun 22, 2012
    4,986
    745
    well if your design spec is 2Amp across a 4.2ohm, that's 8.4Volts max drop. Have you got a diagram?
     
  3. joeyd999

    AAC Fanatic!

    Jun 6, 2011
    2,678
    2,737
    Your average input current can be determined by computing the input power based upon the output power and the switcher's efficiency.

    The peak input current, on the other hand, is going to depend on the implementation.

    You may run into stability issues unless you tack a large, low esr cap onto the input after the resistor, but this is going to be a problem when bucking up against the energy storage limits as defined by the hazloc standard you are designing for. This is compounded by the fact that you also need an inductor and output cap, adding to the problem of total stored energy exceeding the standard.

    For these reasons, I've avoided inductive switchers for hazloc designs in the past.

    Not saying you can't do it -- but the agency that'll be testing for you will be going over your solution with a very fine tooth comb. Expect a headache...
     
  4. Papabravo

    Expert

    Feb 24, 2006
    10,142
    1,790
    You predict the input current by starting with the power output and an assumption about the efficiency of the switching regulator. For example

    4.7 Volts * 2 A = 9.4 Watts, Power Out

    Start with an assumption that the regulator will be 80% Efficient

    9.4 / 0.80 = 11.75 Watts, Power In

    The voltage from the SLA battery will vary according to it's particular discharge curve. At 12.8 Volts the 11.75 Watts will imply a current of ≈ 918 mA.

    When the battery is discharging to say 11.5 volts the current will rise to ≈ 1022 mA.

    The series resistor will of course negate the efficiency of the switching regulator by dissipating 4 Watts all by itself, further increasing the current demand from the battery.

    IMHO you are on the wrong track here.
    I would be thinking of an LDO with a pre-regulator to share some of the heat that will be generated. You're also going to want a substantial heat sink to keep the temperature rise in check regardless.

    OOPS..joeyd999 beat me to it. What he said!

    Ask yourself what current will be drawn as Vin approaches Vout. That's right -- it gets big pretty quick and you don't want that in a hazardous location. Trust me on this.

    Also ask yourself what the current in the inductor might be. It can be two times the input current.
     
    Last edited: Mar 18, 2014
  5. joeyd999

    AAC Fanatic!

    Jun 6, 2011
    2,678
    2,737
    I agree with you that he is on the wrong track...but a linear regulator won't cut the mustard either -- he'll require at least a 2A input for a 2A output...this leaves a headroom of only 3.6V on the far side of the input resistor.
     
  6. Wormius

    Thread Starter New Member

    Jul 25, 2007
    4
    0
    That's 2A at the output of the buck. I am not sure if it will work out proportionally to the input based on the resistor being there.

    My assumption is that if the output current is 2A, then the input current would be (4.7V/12V)*2A = 783mA. However, with the series resistor that isn't necessarily 12V at the input now, it is 12V-(I*4.2ohms). This is what is kind of throwing me.

    [​IMG]
     
  7. Papabravo

    Expert

    Feb 24, 2006
    10,142
    1,790
    I was thinking he could replace the resistor with an active preregulator so there would be another silicon device to share the load.
     
  8. Wormius

    Thread Starter New Member

    Jul 25, 2007
    4
    0
    The 2A max is not *likely* but possible, as we're dealing with powering a cell modem and other circuitry. It would also be very short bursts, but still needs to be accounted for because I don't want to have a potential brown-out.
     
  9. joeyd999

    AAC Fanatic!

    Jun 6, 2011
    2,678
    2,737
    Not allowed! The current limiting device *must* be passive and reliable against all possible modes of failure (i.e. fail open, not short).
     
  10. Wormius

    Thread Starter New Member

    Jul 25, 2007
    4
    0
    I wish!

    The resistor itself can't be replaced. The current limiting to provide a safe short-circuit current has to be infallible, and a resistor is pretty much it. Active devices are assumed to be short-circuits when the circuit is analyzed.

    In this case, at 12V (13.2V) is 3.33A. There are PCB directives not shown to deal with the heat-sinking, clearances, etc.
     
  11. Papabravo

    Expert

    Feb 24, 2006
    10,142
    1,790
    You might be fighting a battle you can't win which is not uncommon in these applications. Now I remember why we walked away from developing a DeviceNet node for hazloc. We couldn't price it to make any money.
     
Loading...