Zener Regulator Help :(

Discussion in 'General Electronics Chat' started by nvxwax, Sep 6, 2014.

  1. nvxwax

    Thread Starter New Member

    Sep 6, 2014
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    I need to build a cheap regulator to drop a voltage of about 7.5 - 13.5 to a constant around 5 volts, I will be driving two small relay coils (30 mA each) and a attinny45, I also see where some people put capacitors in there regulators, what is that for and what would i need? Any insite on my design would be appreciated

    [​IMG]
     
  2. paulktreg

    Distinguished Member

    Jun 2, 2008
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    You cannot put that much load on your circuit due to the current limiting effect of the 1k resistor.
     
  3. wmodavis

    Well-Known Member

    Oct 23, 2010
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    To properly design a zener regulator circuit you need to specify:
    1. minimum source/input voltage - your V1.
    2. maximum source/input voltage
    3. minimum load current
    4. maximum load current
    5. max ambient operating temperature

    You have sort of defined or hinted at some of those but not clearly enough to really 'design' it.

    So, were it me, I'd focus on clearly defining those actual important factors then they can be used to determine a R1 value and power rating, the appropriate power rating required of the zener and if a heatsink is required to keep it from overheating.
     
  4. MrChips

    Moderator

    Oct 2, 2009
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    Do the math. If you have 13V input and 5V output, the voltage drop across the resistor is 8V.
    The maximum current supplied to the load is I = V/R = 8V/1K-ohm = 8mA.
    When the load tries to draw more than 8mA the output voltage is going to fall below 5V.

    Now do the calculation to solve for R.
    If V = 8V and I = 100mA
    R = V/I = 8V/0.1A = 80 ohms

    Then calculate the wattage that has to be dissipated by the resistor under full load
    and the wattage to be dissipated by the zener diode under no load conditions.
     
  5. nvxwax

    Thread Starter New Member

    Sep 6, 2014
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    Thanks guys,

    MrChips, You made that easy to understand, I have seen that formula around but never explained where i could understand it. Ill breadboard this up and see what i come up with.
     
  6. jjw

    Member

    Dec 24, 2013
    173
    31
    R1 is too big.If input voltage is 7.5V then already with 2.5mA current the voltage drops to 5V.
    R1 should be selected with minimum input voltage and maximum current + some headroom.
    For example with max. 80mA current R1=(7.5-5)V/0.08A=31ohm.
    Now if the voltage is 13V and no load, zener current is
    (13-5)/31=0.26A and power consumption 1.3W.
    The resistor should be minimum 31*0.26^2=2.1W
    So quite a lot power goes to heat.
     
  7. MrChips

    Moderator

    Oct 2, 2009
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    Ohm's Law formula:

    I = V/R

    and the other two derivations

    R = V/I

    V = I x R

    are fundamental to everything electrical and electronic. Engrave these into your memory bank and never forget them.

    And include the calculations for power:

    P = I x V
    P = I^2 x R = I x I x R
    P = V^2 / R = V x V / R
     
  8. crutschow

    Expert

    Mar 14, 2008
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    You could also use the common LM317 series regulator. With that the supply current is equal to the load current. It just requires a couple resistors and a couple capacitors (connected close to the device) to go with it. Here's a page that does the resistor calculations for you (Use a value of 120 ohms for R1)
     
  9. #12

    Expert

    Nov 30, 2010
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    There is also the LM7805 chip. Still needs the capacitors, but the resistors are not necessary because the chip decides the voltage. You can get more than 500 ma out of one of these chips with 2.5 volts of head room. I prefer the TO-220 package because a TO-92 package will be struggling with the heat at 13 volts of input.
     
  10. ian field

    Distinguished Member

    Oct 27, 2012
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    I'd just go and buy one of those 13.8V to 5V car socket to USB adaptors from the £1 shop. Cheaper than the parts to go with a 78xx regulator and good for 1A output.
     
  11. crutschow

    Expert

    Mar 14, 2008
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    But will those work down to 7.5V input that the OP wants?
     
  12. MrChips

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    Oct 2, 2009
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  13. #12

    Expert

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  14. MrChips

    Moderator

    Oct 2, 2009
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    Try the LT1083.

    1V headroom at 7A.
     
  15. #12

    Expert

    Nov 30, 2010
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    A little ambitious for an attinny and 60 ma? :rolleyes: :D
     
  16. MrChips

    Moderator

    Oct 2, 2009
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    For this app maybe, but just in case you need some extra headroom.
     
    nvxwax and #12 like this.
  17. ian field

    Distinguished Member

    Oct 27, 2012
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    Probably not the ones that use the MC34063 chip, I vaguely remember rejecting that one for a project because of minimum voltage requirement.

    There are other chips used in those things, but I don't recall the number I found in the last one I took apart.

    Often the minimum voltage requirement is set by the onboard Vref generator - these are rarely more than 5.0V, the worst its going to be is that plus a dropout voltage.

    Some chips like the UC384x family have UVLO, of the 4 variants; 1 of each voltage range is 50% MSR and 1 is 100% MSR - the ones for mains need 16V to start, the automotive ones need about 9 point something to start - those are obvious examples of chips that won't do, there are numerous types that will.

    Since I never quite seem to get around to seeking out and ordering stock of chips with no UVLO, and have a junk box full of the mains variety - once or twice I've cobbled together a single transistor blocking oscillator/inverter to generate the startup voltage, it gets a bit complex though as you have to have an auxiliary winding on the inductor to generate Vcc for the chip, and if you're that bothered about it - a zener and second BO transistor to inhibit it once the chip gets going.

    Its pot luck whether *YOUR* local pound shop has adapters with a low voltage chip inside - but to the motivated experimenter, its never a total loss.
     
  18. MaxHeadRoom

    Expert

    Jul 18, 2013
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    Another solution is to use a zener supply for the low current devices, which most likely are a consistent current, and use two 12vdc relays switched off the 13v supply by a couple of 2N7000's.
    Max.
     
  19. ian field

    Distinguished Member

    Oct 27, 2012
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    Easier to get relays with decent chunky contacts too.
     
  20. nvxwax

    Thread Starter New Member

    Sep 6, 2014
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    I have some parts ordered to breadboard the original idea, of all things i did not have the correct zener lol, I need to keep this as small and cheap as possible, the items im making are only sold for about 20 bucks each and to keep that price i need the power items to be about 3 bucks max, im not making them in bulk so i cant get a good price on some of this stuff.

    Basically it will power an ATtiny45, two 5 volt relay coils and a small 3mm led, in actually thinking about it it will only have to power the ATtiny and one of the other items at the same time, I could really get away with about 40-50 mA of power.

    The other circuits i have looked at have capacitors going from just after the resister to ground with a 104 0.1uF cap, what would this be for?
     
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