Connect with us

Rain Sensor Status

Discussion in 'Sensors and Actuators' started by chopnhack, Sep 9, 2014.

Scroll to continue with content
  1. Harald Kapp

    Harald Kapp Moderator Moderator

    9,908
    2,100
    Nov 17, 2011
    chopnhack likes this.
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,363
    2,758
    Jan 21, 2010
    They're nice and easy to use. And they're rugged.
     
    chopnhack likes this.
  3. chopnhack

    chopnhack

    1,573
    352
    Apr 28, 2014
    LOL

    I get into too much trouble with just two transistors! Thank you Harald, that is an excellent post, but above my comprehension at the moment. I ordered some LM2576, about 45 cents US each.
     
  4. chopnhack

    chopnhack

    1,573
    352
    Apr 28, 2014
    If I am reading this chart correctly, the inductor that is needed for the basic regulation circuit is dictated by the current usage of the circuit and Vinput.
    upload_2014-9-10_22-17-53.png
    If I have about 28v coming into the circuit and the circuit consumes about, maybe 100mA(unsure?):
    Would I need a 1000μH inductor?
    What power rating should I select? 1/4 W or higher?
    How does one know when they have inductors or resistors if they use the same color code?!

    Reading through the datasheet, I noticed that the regulator will operate in discontinuous mode if under 300mA - do I need to design the circuit differently?

    upload_2014-9-10_22-42-34.png
     
    Last edited: Sep 11, 2014
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,363
    2,758
    Jan 21, 2010
    I believe that circuit gives you the maximum current vs inductance. Sure, you could use 1000uH (1mH), but you could also get away with 100uH, or anything in between.

    No problems with discontinuous mode. It will give a slightly lower frequency ripple, but that is unlikely to be an issue.

    edit: the important thing about the inductor is the saturation current. You won't want to use one that looks like a resistor as they're chokes and don't have the low loss characteristics you're after. Ensure that the inductor you choose (toroids are very commonly used) is rated for a current in excess than the current you'll ever need. For 100mA, try to get one rated for 250mA or so -- more is OK, it will just be larger and ore expensive.
     
    chopnhack likes this.
  6. davenn

    davenn Moderator

    13,561
    1,856
    Sep 5, 2009
    also, dropping from 28V to 5V is going to cause a LOT of heat generated by the poor 7805
    eg 23V drop x 1A current = 23W of heat !!
    in you last example .... 23V drop x 300mA = 7W .... its still gonna get hot

    and you are pushing it past its limit ... most datasheets state 25V max input to a 7805
    so in reality you don't want to be anywhere near even 25V to give the 7805 a long life expectancy ;)

    Dave
     
    chopnhack likes this.
  7. chopnhack

    chopnhack

    1,573
    352
    Apr 28, 2014
    eep... LOL I will update the schematic, its supposed to read LM2576 - switch mode supply.
     

    Attached Files:

  8. davenn

    davenn Moderator

    13,561
    1,856
    Sep 5, 2009
    LOL well there's always a good reason for these things ;)

    interestingly ... all I saw was the 7805 label .... didn't even notice the extra 2 legs on the device :rolleyes:

    D
     
    chopnhack likes this.
  9. chopnhack

    chopnhack

    1,573
    352
    Apr 28, 2014
    I thought chokes and inductors were synonymous? Shame too, I don't seem to find an assortment of toroidal style or coil style inductors on e-bay. I did find the resistor type. Will this type not work at all?

    Thank you


    No worries, I do worse... have you seen some of my schematics? :eek::p
     
  10. davenn

    davenn Moderator

    13,561
    1,856
    Sep 5, 2009
    OK ... next point

    your diag. says 29.2V on secondary, so 29.2VAC rms x 1.414 = 41.3 Vpk (peak) - say ~2V drop through the diodes
    you are going to have 39-40V DC going into the regulator. Not the 28V you have stated ;)

    Ohhh and I would up C1 to at least 1000uF. 100uF just aint enough :)

    Dave
     
    chopnhack likes this.
  11. chopnhack

    chopnhack

    1,573
    352
    Apr 28, 2014
    I need to clarify this, I just couldn't find an appropriate component in eagle. The input is really 29.2VAC as measured on my DMM. It's coming out of a 24V wall adapter.

    upload_2014-9-11_0-26-35.png
     
  12. davenn

    davenn Moderator

    13,561
    1,856
    Sep 5, 2009
    ok so it really is 29VAC and you need to take that into account regardless of what the label on the adaptor says ;)
     
  13. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,363
    2,758
    Jan 21, 2010
    Looking at the entire circuit, if you're just using the power supply to power a couple of LEDs, a switchmode power supply is going way over the top.

    You could probably run the whole lot from the rectified and filtered AC, possibly using some transistors capable of 60V.

    Unless you actually need a source of 5V for a particular reason...?
     
    chopnhack likes this.
  14. Harald Kapp

    Harald Kapp Moderator Moderator

    9,908
    2,100
    Nov 17, 2011
    Those transformer based wall adaptors typically drop in voltage as they are loaded by a current. Measure the output voltage of the adaptor at no load and at max. load (max. current drawn). The voltages you measure give you the input range for which the regulator needs to be designed.
    As you go from >20V to 5V this should be no issue hre - just wanted to notice.

    Another issue: If you measured 29.2V AC you probably used the AC range of your meter. This means 29.2V are RMS. The peak value of a 29.2V RMA voltage is 29.2V*sqrt(2)= 41V! This is the peak voltage you will see across C1!
    This means you would need the HV version of the LM2576 as the standard version is rated only up to 37V input voltage. Build a mockup consisting only of transformer, rectifier and Capacitor and measure the real voltage across C1.
     
    chopnhack likes this.
  15. chopnhack

    chopnhack

    1,573
    352
    Apr 28, 2014
    I had 7805's on hand so that seemed the easiest way to produce the 5v. Why 5v? - I figured with two LED's I might need the Vdrop - but now realize I would only need about 3v since they would never be in series.

    I had given thought to a capacitor fed p.s. - but remembered that the wall adapter might give varying voltage (under load I have seen several volts dropped with two different adapters). Unless the voltage is consistent cap-fed supplies are not suitable since they have no v-reg (LED's would burn out). I could be wrong, but that is what I thought from my limited experience.

    Can you elaborate on what you are thinking of? Rectified and filtered I get (bridge rectifier and caps) - that would still have us at a peak of ~39VDC. What would the transistors do?
     
  16. Harald Kapp

    Harald Kapp Moderator Moderator

    9,908
    2,100
    Nov 17, 2011
    You may be best off if you use another wall wart that has a 5V output.
     
  17. Harald Kapp

    Harald Kapp Moderator Moderator

    9,908
    2,100
    Nov 17, 2011
    Alternatively you can reduce the voltage after the rectifier by putting an impedance in series with the rectifier. The impedance will create a voltage drop such that the output voltage of the rectifier is lower. If you put the impedance before the rectifier, it will see an AC current. You can then use a capacitor for impedance. This has the advantage that it dissipates much less energy as heat than a resistor. Here's a simplified example:
    [​IMG]

    Green is the input voltage (+-30V from the transformer), blue is the output voltage across C2. By varying the value of C1 you can adjust the output voltage. Note that R2 (crossed red) is for simulation purposes only and should not be part of a real circuit.
     

    Attached Files:

    chopnhack likes this.
  18. chopnhack

    chopnhack

    1,573
    352
    Apr 28, 2014
    The VDC looks pretty stable there! I didn't think it would be because of the source...
    Am I correct in assuming then that with a X2 or Y2 rated cap for c1, I can build a suitable supply for the two transistors and leds without a regulator?
     
  19. chopnhack

    chopnhack

    1,573
    352
    Apr 28, 2014
    It got cut off on the bottom, but 3.3v and 33mAis the output across and through R4 if I use 4.7μF for C1, C2 1000μF and 100Ω for R4. The schematic shows 2200 - I was playing with values to see what they do and forgot to change it back. What are your thoughts on this?

    upload_2014-9-11_22-14-36.png
     
  20. Harald Kapp

    Harald Kapp Moderator Moderator

    9,908
    2,100
    Nov 17, 2011
    1) you don't need an X and/or Y rated capacitor for C1. C1 is on the low voltage side of the transformer so (almost) any capacitor is suitable.
    2) you sjould use a regulator anyway to acommodate for variations in input voltage, load, temperature etc. By lowering the input voltage to the regulator via C1, you reduce the power loss in the regulator.

    Looks pretty much like my schematic with diffferent values. Of course you have to change the values to adpat the circuit to your needs (in term of voltage and current).
    Note that C1 sees AC. You cannot use a polarized capacitor (elko) for this one.
     
    chopnhack likes this.
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Similar Threads
Loading...
Electronics Point Logo
Continue to site
Quote of the day

-