zener regulator calculations

Discussion in 'The Projects Forum' started by ebeowulf17, Mar 12, 2015.

  1. ebeowulf17

    Thread Starter Active Member

    Aug 12, 2014
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    I'm working on a simple 555 circuit that needs to be able to run on a range of input voltages, ideally anywhere from 3-12VDC. My plan is to run the whole circuit on 3V and regulate the incoming voltage down to 3V. Even LDO regulators seem to require more dropout voltage than I'd like when starting with such low voltages, so I was thinking zener diode.

    My entire knowledge of zener regulators so far is from this website (article here) and I get the basic idea, but I'm not sure how to calculate the zener regulator's resistor (R3) value when the load resistance is unknown. Can I estimate the current draw of the post-regulator circuit (fairly easy to add up 555 current, LED current, relay coil current, etc.) and calculate what resistance at 3V would draw that amount of current, and then treat the circuit as a dummy resistance for the sake of calculating my zener regulator's resistor value?

    I know on the one hand that higher value resistors will waste less energy, but also that they can potentially choke off power to my circuit when running on only 3V in. Maybe after doing all the math on this I'll realize that I can't really outperform the LDO regulators anyway, but I thought it was worth checking out.

    Thanks in advance for any insights!

    Here's a rough version of what I'm working on:
    555_relay-out_3-12VDC_01.jpg
     
  2. MrChips

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    Why not scrap the resistor and Zener altogether and operate 3-12V directly?
     
  3. Dodgydave

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    Jun 22, 2012
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    Why do you need to regulate the 555 ,it will run on 3V, if you use the CMOS version, just use a 3V relay coil.
     
  4. ebeowulf17

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    Aug 12, 2014
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    I was planning on CMOS 555 and 3V relay coil. I thought regulation would be necessary because, if I design the whole circuit for 3V (as I planned to do,) the two LEDs and the relay coil will draw far too much current on 12V, right? Is it possible to make a circuit with LEDs and a relay coil that are totally fine with anything from 3-12VDC? I know the 555 would be fine; it's all the other stuff I was worried about.
     
  5. MrChips

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    Your (not so simple) solution (if only solution) would be to use a buck-boost switching regulator such as LTC3114-1
     
  6. Dodgydave

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    You can only make a voltage regulator that drops the "wasted " voltage but will still dissipate heat as the current or supply increases, whether its a simple Zener, Regulator, or op-amp and transistor,if your supply is varying from 3V to 12V, and you only need 3V, use a buck regulator.
     
  7. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    A zener is a poor choice for "large" currents since the basic circuit is essentially a current wasting unit. The resistor is chosen to drop the voltage for the sum of max load current plus min zener current at the lowest input voltage. Raise the input and the excess current goes down the zener. Turn the load off and the current goes (you guessed it) down the zener.

    On the other hand a LDO regulator only needs a small current, and delivers whatever the load needs so it is far more efficient in terms of not wasting more current than is necessary for supporting the load.

    Now about that darlington driving the relay: darlingtons have a rather high saturation voltage of about .8V. At 3V you only have about 2.2V across the relay to make it click. Does your relay spec support that? A single transistor may be able by itself to drive the load, or a 2 device amp where it is not in the darlington config so you get a low saturation voltage C to E.

    Or... a nice low voltage MOSFET would be great for the relay driver.
     
  8. ebeowulf17

    Thread Starter Active Member

    Aug 12, 2014
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    Ernie, thanks for pointing out the Darlington issues. Went with that just cause I have some lying around, but didn't think about the voltage drop. Will definitely look at other options there.

    Still reading up on buck converters. Starting to wonder if the 3V spec isn't asking just a bit more than it's worth for this project. Maybe wouldn't be the end of the world if it only worked down to 3.3 or even 5.
     
  9. Dodgydave

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  10. MCU88

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    Mar 12, 2015
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    Guys you haven't given him the maths that he is asking for. He just wants to know the get ready to drop resistor value.

    RZ = (VDC - ZV) / (IZ + I)

    RZ =
    Zener Resistor
    VDC = Supply Voltage
    ZV = Zener Voltage i.e. 3V3
    IZ = Current to Keep the Zener in Regulation (typically 5mA)
    I = The Desired Supply Current to Power the Circuit

    Any questions? This is basic ohms law and algebra transposition for this.
     
    DickCappels likes this.
  11. MrChips

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    Just use a 3V LDO linear regular and be done. Make sure the regulator is mounted on a suitably sized heat sink.
     
  12. BobTPH

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    Jun 5, 2013
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    All of which is totally useless when the input voltage can range from 3V to 12V.

    Bob
     
  13. Dodgydave

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    I think we've come full circle on this, back to posts #2,3...
     
  14. SgtWookie

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    Jul 17, 2007
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    Just a small problem; the LM555's operating voltage range is 4.5v to 16v:
    http://www.ti.com/lit/ds/symlink/lm555.pdf
    See item #9 on page 15.
    3v simply won't work for a BJT (transistorized) 555 timer.

    There are CMOS versions that might operate that low; however that's pushing it.
    Also, you show a Darlington configuration driving the relay; unfortunately it's saturation voltage will likely exceed 1.3v even with a light load. If your supply voltage is 3v, then the relay will have to operate with 3v - ~1.3v ~= 1.7v - do you know any relays that will operate from that low of a voltage? I don't.

    Why don't we go back to the beginning, and have our TS/OP describe what it is exactly they are trying to do; what is the power source, and what is it that they are trying to control?
     
    Last edited: Mar 13, 2015
  15. ebeowulf17

    Thread Starter Active Member

    Aug 12, 2014
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    Thanks to everyone for the great input so far.

    This was intended to be a timer that would repeatedly close a contact briefly at regular intervals, emulating a never ending string of button pushes on another device (the relay contacts would be wired in parallel across the other device's switch pins.) The problem is that there are a variety of possible "other devices" that it might need to work with, with varying operating voltages.

    I originally planned to just make a 5V (or maybe higher) circuit which would be completely independent of the other device, with the two connecting only at the relay outputs. However, the person I'm designing this for requested variable input voltages so that the other device and this timer could share a single power supply instead of needing two separate power supplies, one voltage for the timer/relay circuit, and one for the other device.

    I don't think a buck converter is worth the effort for this particular project, and it doesn't look like there's going to be any other good way to handle the entire 3-12V range effectively. I think now that I might compromise by including an LDO regulator and designing around a different minimum voltage (say 4.5 - 4.75V circuit off of LDO from 5V minimum supply) so that the circuit could handle most of the original intended range on a shared power supply of anything from 5-12V, but then if the other device needs less than 5, two separate power supplies would be needed. That feels like a pretty fair compromise.

    That said, I still need to re-think my Darlington choice, as has been pointed out by several people above, and get my head on straight about how much voltage I'll lose through whatever transistor I choose, and how that effects relay performance.

    Unless that sounds like a bad idea, I'll get started on reworking the plan and post a new schematic sometime soon.

    P.S. Special thanks to MCU88 for going back and answering the original question. I really did want to understand those calculations better, even if it turns out not to be useful for this particular circuit!
     
  16. MCU88

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    Mar 12, 2015
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    Huh? I guess you say what you say because (I) is proportional to (V) and inversely proportional to (R) -- the zener and (R) has to dissipate / waste this energy as (V) rises. This is true that such an setup is very inefficient. But could be insignificant if the current drain is low. There always has to be at least 5mA to keep the zener in regulation.

    I personally would only ever use an zener diode for crowbar protection circuits.
     
  17. MCU88

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    Mar 12, 2015
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    No problem. Ohm's law is not simple when you have half a dozen or so variables to consider.

    Members try to transpose this ...

    RZ = (VDC - ZV) / (IZ + I)

    (properly decompose it and write it out in algebraic)

    Q1: Transposition for IZ?

    Q2: Transposition for I?

    Q3: Transposition for VDC?
     
  18. MrChips

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    Oct 2, 2009
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    Bob is correct. I will demonstrate with actual numbers why the zener and the calculation is (almost) totally useless.

    The zener regulation circuit provides constant voltage and constant current. The zener and its current limiting resistor is designed for a given load current. What current the load does not consume has to be taken up by the zener.

    Let us assume that the load current in this situation is 100mA, that is, 20mA for each LED and 60mA for the relay.

    Assume the supply voltage Vs is 12V and the zener voltage Vz is 3V.
    The voltage drop across the resistor is VR = Vs - Vz = 12 - 3 = 9V
    Hence the required resistor R = (Vs-Vz)/I = 9V/0.1A = 90Ω
    The power dissipated by the resistor = V x I = 9V x 0.1A = 0.9W
    The power dissipated by the zener when it has to conduct all of the current = 3V x 0.1A = 0.3W

    Now, if the supply voltage Vs increases slightly above 12V, the zener will still regulate the output voltage to 3V but has to dissipate more power.
    If the supply voltage Vs drops below 12V, the zener no longer conducts (for an ideal zener) and the output voltage falls below 3V.

    So let us design the circuit for a minimum supply voltage of Vs = 4V
    Required resistor R = (4V - 3V )/0.1A = 10Ω
    Power dissipated by the resistor = 1V x 0.1A = 0.1W
    Power dissipated by the zener is the same as before = 3V x 0.1A = 0.3W

    If the supply voltage drops below 4V, the zener no longer regulates the voltage.
    If the supply voltage rises to 12V, the current through the resistor is (12 V - 3V)/10Ω = 0.9A
    The power dissipated by the resistor = (V x I) = 9V x .9A = 8W
    The zener would have to dissipate 3V x 0.9A = 4W

    To design the circuit for a supply voltage of 3V, the current limiting resistor would have to be zero ohms, i.e. no current limit. Your zener either has to dissipate infinite power or it will blow as soon as the supply voltage rises above 3V (for an ideal zener).

    In summary, zener regulation circuit is handy when the load current can change from 0 to Imax.
    It is not useful when the supply voltage swings over a wide range.
     
    BobTPH likes this.
  19. MCU88

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    Mar 12, 2015
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    Yes Mr.Chips makes points with current drain of 100mA. (IT) Current total. But if the current required for the circuit is much less than this then it would be a different story. Perhaps the circuit only needs say 6.66mA?
     
  20. MrChips

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    Oct 2, 2009
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    100mA is just an example.

    You can do the same calculations using 10mA as another example.
     
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