Resistor or Voltage Regulator to drop voltage and/or current?

Discussion in 'General Electronics Chat' started by jellytot, Jun 8, 2016.

  1. jellytot

    Thread Starter Member

    May 20, 2014
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    I want to limit voltage (and current) to my load. Normally I use a voltage regulator (with 2 capacitors). But then I realized I can just use a resistor to drop the voltage (and subsequently, current). But then I thought, actually whats the difference between the two choices? Both drop voltage (and my load will draw less current because of it), and both solutions generate waste heat. Is there a difference? I'm still an electronics noob, so answers with a bit more detail would be greatly appreciated (because there's a greater chance I'll understand the answer that way).

    Thanks!
     
  2. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    If your load is constant and your power supply is constant (and regulated), then not much.

    If your power supply is swinging all over the place or your load draws various amount of current, then yes, there is a difference. A resistor follows ohms law and will drop voltage equal to I x R. A voltage regulator outputs constant voltage.
     
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  3. dl324

    Distinguished Member

    Mar 30, 2015
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    If the input voltage and load current aren't constant, you need voltage regulator with current limiting. Some regulator datasheets show how to implement current limit.
     
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  4. jellytot

    Thread Starter Member

    May 20, 2014
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    Thanks for the responses! They are great answers. :)

    Any other pros/cons of either solution? For example, would the resistor solution generate more heat, compared to the voltage regulator solution? How could I calculate the heat generated by either solution?
     
  5. AlbertHall

    Well-Known Member

    Jun 4, 2014
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    In the same conditions they generate the same amount of heat.
    The amount of heat, aka power, in the resistor/regulator is the current through it multiplied by the voltage across it (i.e. the difference between the power supply voltage and the load voltage).
     
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  6. RamaD

    Active Member

    Dec 4, 2009
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    A switching regulator can reduce the amount of heat, though it will be more expensive.
     
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  7. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    Both the linear regulator and resistor generate about the same power loss equal to the load current times the difference in supply and load voltages. The resistor needs a very slightly lower power but the regulator well regulates the load voltage.

    As stated a switching regulator would be the most efficient.
     
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  8. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
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    Please state/explain what your load is and its current/voltage requirements and what your power source is too..
    While you got good information none of that "may" be needed and they are just going with your assumptions that something may be needed..

    For example many "newbies" think that because their load requires 12V @ 500mA and because they have a power supply thats rated at 12V @ 2 Amps that they think they need to limit the current..
    Thats typically NOT the case as the load will only draw the current it needs from the power supply..
     
  9. jellytot

    Thread Starter Member

    May 20, 2014
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    Thanks for the answers, everyone!

    I made that mistake before, not so long ago. But nowadays I need to limit current due to too much current draw (either my voltage regulator is overheating, or the total current draw is above the max spec of the current source).

    I've never tried using a switching regulator (yet). They seem to require more components, and I worry about the potential noise they generate causing issues with my MCU or the attached sensors. I may need to go down that route though, as my voltage regulators are generating a lot of heat. But is my concern valid? Could a switching regulator cause issues with my MCU or the sensors?
     
  10. crutschow

    Expert

    Mar 14, 2008
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    Yes, switching regulators do generate more noise which can affect analog signals, but generally it's not enough to bother digital signals or MCU digital operation.
     
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  11. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    Not when used properly with sufficient filtering on the output.
     
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  12. BobTPH

    Active Member

    Jun 5, 2013
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    If you want any better help, tell us the characteristics of the load.

    Bob
     
  13. jellytot

    Thread Starter Member

    May 20, 2014
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    Thanks. This is completely new territory for me. By "filtering", do you mean use a choke? If so, how do I know what kind of choke to use? And the choke is put on the output, in series I presume?
     
  14. ian field

    Distinguished Member

    Oct 27, 2012
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    The heat generated depends on volts x amps - all things being constant, it will be the same for resistor as regulator.

    If reducing generated heat is most important; you'll need a switching regulator.
     
  15. Tonyr1084

    Active Member

    Sep 24, 2015
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    Are you saying your available current is below the requirements of your circuit? If so then that's probably the reason for excessive heat.

    All components are designed to handle a "Max" current (or some other maximum in their rating). That "Max" is the specified survival limit for that component. If a device is rated at 200 mA then you can reasonably draw 200 mA through it at all times. However, if you draw 201 mA - that doesn't mean you'll have instantaneous failure. The device can actually sustain periods above their rated values but doing so comes at a cost - the component will eventually break down. The more you exceed the rating the quicker it will break down.

    When I was a young teen I came across a car radio. I wanted to use it so I found a 12 volt transformer and an old diode - one with the fins - and figured it would work. It did. But horribly noisy. I tried to filter it by using a capacitor. Someone told me to use a cap to filter the hum. So I found a capacitor - happened to be an electrolytic (probably rated for 16 volts) and I put it across the output of the transformer. Did nothing. So I put it across the input. - - - Long story short, the capacitor failed spectacularly. All that remained was the empty shell of what once housed a capacitor.

    The moral? All things have ratings that are safe and reliable. Depending on how much beyond those ratings makes the difference between lasting a day, an hour, a minute or sudden and spectacular failure.

    If your supply is not capable of handling the task then you need a bigger supply.
     
  16. RamaD

    Active Member

    Dec 4, 2009
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    As MCGYVR said, Please state your source voltage and load voltage and current requirements. There are even drop in switching regulators to replace the 3 pin 78xx regulators.
     
  17. jellytot

    Thread Starter Member

    May 20, 2014
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    No, requirements of the circuit are below the available current. the I think the source of the heat is that I'm drawing ~1 amp from my 78XX.

    Source voltage is a 12V, 2 amp wall adapter. Load requires 5V, 1 amp. Load is an MCU, small motors, IR sensor.

    That would be awesome! To just swap out the 78XX would be such a simple solution. I had found a 3 pin switching regulator, but it's maximum output was 500ma (I need at least 1 amp). Do they make these higher than 500ma?

    Can anyone help me with this question?
     
  18. AlbertHall

    Well-Known Member

    Jun 4, 2014
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    If this is powering an MCU then presumably it needs a pretty constant voltage supply. If you use a resistor to drop the voltage and the motor current varies then the voltage will vary resulting in an unhappy MCU.
    So, going back to the original question, in this application you NEED a regulator - a resistor is not an option.
     
  19. EM Fields

    Member

    Jun 8, 2016
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  20. BobTPH

    Active Member

    Jun 5, 2013
    783
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    12V in 5V out at 1A means a loss of 7W. Not unmanageable, but you will need a moderate sized heatsink.

    Bob
     
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