Water top off

Discussion in 'The Projects Forum' started by erich_7719, Nov 13, 2009.

  1. erich_7719

    Thread Starter Active Member

    Oct 14, 2009
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    I have a water top off system that works, put it is using relays. I am wanting to replace the relays with a system that will still uses the float switches I have in place. Attached is a section of the relay system to show what I am wanting and also attached is a schematic that I assumed would have worked.
    My problem with the schematic is the nand input connected to the lower switch only stays high till the cap is discharged, but what I am is wanting is for the lower switch to stay high till the upper switch goes low.

    Any help would be greatly helpful.

    Thank you in advance.
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    No extensive analysis yet, but... in the left schematic:
    1) You're using 4011 NAND gates. You should use 4093 NAND gates instead, as they have Schmitt-trigger inputs.

    2) R7 is much too low in value. CMOS has a very limited current source/sink capability. R7 should be 10k Ohms.

    3) I'm not sure what you're trying to do with R3 and R4, but they would need to be swapped around to get IC1A's pin 1 up to a logic 1 level.

    4) Same thing with R5/R6 - for IC1A's pin 2.

    The right schematic is really hard to see any detail as it's so small.
    Could you re-post a larger version?
     
  3. erich_7719

    Thread Starter Active Member

    Oct 14, 2009
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    SgtWookie:
    I forget what sites I saw this at but R3/R4/C1 is to debounce the signal coming off of the float switch, same with IC1A Pin 2.

    I will most likely go to the 4093, I just had the 4011's on hand right now.
     
    Last edited: Nov 13, 2009
  4. erich_7719

    Thread Starter Active Member

    Oct 14, 2009
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    SgtWookie:

    I figured out what was wrong (Stupid mistake). Overlooked the fact of forgetting to put R7 on the bread board.

    I did now put in the 10K instead of 1K for R7. The circuit is working fine now.
    Thank you for your time.
     
  5. SgtWookie

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    Jul 17, 2007
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    OK, I see it now. R3/R4 and R5/R6 can stay in their original locations and work properly.

    I was quite tired when I replied yesterday evening, and I couldn't see the relay diagram.

    The 4093 Schmitt-trigger NAND gates really are a necessity for your circuit. 4011's may oscillate if their inputs are at an indeterminate level (not a full logic 0 or 1).

    Only one problem remaining; there is no definite discharge path for IC1A1 pin 1. Diodes D1 and D2 allow a logic 1 to be placed on pin 1, but the only actual discharge path is via reverse current through the diodes themselves.

    I suggest that you attach a 22k resistor or higher from IC1A1 to Vss. The value has to be fairly high, because R7 is already providing a load for the output of IC1B
     
  6. erich_7719

    Thread Starter Active Member

    Oct 14, 2009
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    Thank you for your reply.

    I never even though of that, it makes perfect sense though. The final implementation will utilize the resistor from IC1A1 to VSS and the 4093.

    Thank you again for your assistance.
     
  7. Wendy

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    Mar 24, 2008
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  8. erich_7719

    Thread Starter Active Member

    Oct 14, 2009
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    I've only made it through the 3rd page of the thread you suggested.

    It dose have some good ideas, only mine is not for a "Fish Tank", but for a Vivarium with 2 separate sumps, 3 pump, 1 Fogger, 1 Misting system, a Reverse Osmosis system, with a Water fall.

    I have attached Version 1 of my system that works great, and Version 2 Where I am trying to remove some of the reliance on relays, to lower Power consumption.

    If you would like to take a look you are more than welcome, Version 2 is still in the designing and testing stages.

    I will finish reading the other thread over the next couple of days though, while I have breaks in my day.
     
    Last edited: Nov 14, 2009
  9. SgtWookie

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    Jul 17, 2007
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    Erich,
    After looking at it again this morning, 33k might work OK, but it would actually be better if it were around 100k. The recommendation comes from the 10k resistor and diode in the current path from terminal B vs the 33k resistor, which will constitute a voltage divider network when terminal B is a logic 1, or 12v, and consideration for the Schmitt-trigger voltage levels of the 4093 quad NAND gate. The 4011 will be more sensitive to indeterminate levels than the 4093 is.

    The Schmitt-trigger inputs of the 4093 will have trigger levels of roughly 1/3 and 2/3 Vdd, or below 4v for a logic 0 in, and above 8v for a logic 1 in.

    10k + 33k = 43k across 12v; current from terminal B to Vss will be approximately 12v/43k=280uA, or 0.28mA.

    At a forward current of 0.2mA, a 1N400x series rectifier diode has a Vf of about 500mV, or 0.5v; so I'll subtract that from Vdd and get the resulting current:
    11.5v/43k = 267uA or 0.267mA.
    To get the voltage across the 33k resistor when B is high then, multiplying 0.267mA times 33k Ohms gives us 8.811 Volts. That's higher than the upper Schmitt trigger level, so it should work.

    However, increasing from 33k to 100k will allow the voltage to go even higher, and since capacitance is so low on the gate (that's why the resistor is needed in the first place) it will not appreciably increase the time it takes for the gate input to go low.

    11.5v/(10k+100k) = 0.105mA
    Vf of a 1N4002 diode at 0.1mA is about 0.46v; update & re-calc:
    11.54/110k = 104.91uA
    Now when point B is high (12v), the 100k resistor to ground will have:
    100k x 104.91uA = 10.49v across it. This is 87.5% of 12v instead of 73.4% like you'd get with the 33k resistor.
     
  10. erich_7719

    Thread Starter Active Member

    Oct 14, 2009
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    SgtWookie:

    Some details:
    The Diodes are 1N914 switching diodes.
    The LM7812 is regulating the power at 11.77V.
    The transistors are MPSA06.

    I know this will change the values some, what I have is:

    107uA before the Diode and 101.9uA after the diode (not sure how you calculated this "Vf of a 1N4002 diode at 0.1mA is about 0.46v"). Millimeter says it’s a .56V drop across the 1N914 so that would give:
    10.19V at B and 1.07V going to the transistor.

    The 4093 would be <3.92 for low and >7.84 for high; based on the 11.77V supply.
    Am I correct in this reasoning?

    I don't have any 1N002 Diodes on hand, but I do have some 1N001 diodes.

    I've updated the schematic in post #8 to reflect the changes.
     
    Last edited: Nov 14, 2009
  11. SgtWookie

    Expert

    Jul 17, 2007
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    That's all fine.

    Interesting; you're missing about 5uA in there. Must be leakage through the other diode.
    Actual measurements I took from a 1N4002 diode. There is some variance between even diodes of the same batch.
    Ahh, I thought you were switching IN your Vcc (11.77v) at point B from the relay?

    That sounds about right.

    Did you mean 1N4001? Those would work just fine. They would likely have a somewhat lower Vf than the 1N914's. They're slower to recover and have more capacitance, but in this application those don't really make any difference.

    That's good. Generally though, it's a good idea to leave schematic(s) as they were in the beginning of the thread, and post the updated version where it's relevant in the thread later; add a note by the earlier schematic where to find the next version to make it easier to follow the "trail of bread crumbs", so to speak. Besides, that gives you a good record of what was and what is, so if something suddenly stops working, you can refer back to the older schematic and see what changed.
     
  12. erich_7719

    Thread Starter Active Member

    Oct 14, 2009
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    Maltimeter says it’s a .56V drop across the 1N914 so that would give<- Actual reading.

    10.19V at B and 1.07V going to the transistor. <-- Trying to understand the math.

    I was now that relay is no more.

    Yes

    Corrected the schematic now.
     
  13. SgtWookie

    Expert

    Jul 17, 2007
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    OK.

    If you got rid of the relay at B, I don't know why you still have the other components. But actually, without the relay supplying voltage at B, you should never get an output at D.

    You don't have reference designators on many of the components in the schematic, so it makes it very tough to describe what part we're talking about. You say you get 1.07V going to the transistor - which one? If you're measuring Vbe, that would be about right for a transistor that's turned on. If Vbe were below 0.5v, it would be turned off.
     
  14. erich_7719

    Thread Starter Active Member

    Oct 14, 2009
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    Its a float switch at B. The relay I am in reference to was a "Hold" relay till the entire cycle has completed. IE...
    You have two float switches, one high in a sump the other low in the sump. As the water evaporates the top float switch makes a connection at B (this does nothing for now). The water evaporates more and make the lower float switch to make a connection at A, Causing the Relay/Circuit to trigger a NC solenoid (through D) to the open position. Once Solenoid is open the water level rises breaking the connection at A (the reason for the loop on the original relay and the loop from IC1B4 back to IC1A1) and the solenoid stays open until the water level fills high enough to break the connection on the higher float switch connected to B/IC1A2.

    Than the above cycle starts all over again.



    In post 1 the NAND.PNG is the portion of the total schematic I thought we were referencing. So the 1.07V is from BT1 to ET2. In that schematic I have a LED in place of the solenoid for test purposes only, that’s why I have the Darlington pair.


    The purpose is to lower the total power consumption of the entire system; this is just one step to accomplishing that.

    Hope that either I just misread one of your posts or I haven't confused you too badly.
     
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