Connect with us

any comments on this circuit?

Discussion in 'Electronic Basics' started by andy, Jul 22, 2004.

Scroll to continue with content
  1. andy

    andy Guest

    It's meant to send a 1 second pulse through an electromagnet at the end of
    every day, to start a plant watering system, as i said in an earlier post.
    I've tested it on breadboard, but with only one transistor and a single
    coil winding. I'm trying splitting the coil into 3 parts to get more
    current through it. It would be good to know what people think of this
    circuit before i build the final version - it's only the second circuit
    i've built.

    http://www.niftybits.ukfsn.org/electronics/daily-water.png
    http://www.niftybits.ukfsn.org/electronics/daily-water.gif

    andy.
     
  2. John Fields

    John Fields Guest

     
  3. andy

    andy Guest

    i've had it working with just one output transistor, so i'm not chucking
    it out just yet. it would help if you'd say what's wrong with it, or where
    i could look to do it right, rather than just trash it with one word.
     
  4. andy

    andy Guest

    I've been through it again, and worked out the component values for the
    second half of the circuit again. The new version is at:

    http://www.niftybits.ukfsn.org/electronics/daily-water2.png
    http://www.niftybits.ukfsn.org/electronics/daily-water2.gif

    If there's something wrong with the basic idea of how i've laid it out,
    i'd rather be told why than just have it written off.
     
  5.  
  6.  
  7. andy

    andy Guest

     
  8. Gordon Youd

    Gordon Youd Guest

    Would it not be better to use a 555 servo circuit, use the servo to open the
    valve, less parts, less current used.

    We are here to help not smack you down like some people.

    Regards, Gordon.

    ----------------------------------------------------------------------------
    ----------------------
     
  9. John Fields

    John Fields Guest

    ---
    OK, let's start with your output stage which, even after two
    iterations and comments from John Popelish, you don't have drawn
    correctly .

    Instead of three TIP3055's beasts being driven as emitter followers,
    why not use a single MOSFET which can handle a 13.3 amp load, (which
    is what your three windings will look like in parallel) and make it
    common source?




    +12>-------+
    |
    S
    VIN>-----G IRF4905
    D
    |
    +----+----+------+
    | | | |
    [L1] [L2] [L3] [1N4001]
    | | | |
    GND>-------+----+----+------+


    The only current you need to drive it with is what's needed to charge
    and discharge the gate capacitance, and with a channel resistance of
    20 milliohms and a 13.3 amp load, the IRF4905 will be dissipating
    about 3.5 watts for the 1 second it'll be on, which is nothing.

    Next, let's look at what you're using to drive the 2N3055's:



    +12>------------------------+
    |
    [50R]
    |
    +----+------+
    | | |
    C | +--[10R]--->TO 2N3055'S
    VIN>---[340K]--B Q1 | |
    E | |
    | C |
    +---B Q2 |
    | E |
    [2K2] | |
    | | C
    +----+-+---B Q3
    | E
    [1K0] |
    | |
    GND>-------------------+----+


    Notice that in order to keep the 2N3055's turned off, Q3 has to be
    turned on, which will be doing your 1.2AH battery no favors since
    it'll have to supply current through the 50 ohm resistor ***ALL THE
    TIME***. Assuming Q1, Q2, and Q3 are collector-to-emitter shorts, 12V
    through 50 ohms is 240mA, which means that the battery will be drained
    in about 5 hours after the circuit is connected. Period. End of
    story.

    But, if you change your driver and output stage to look like this:

    +12>-----------+----+
    | |
    [1K] |
    | S
    +---G IRF4905
    | D
    C |
    VIN>--[10K]---B Q1 +----+----+------+
    E | | | |K
    | [L1] [L2] [L3] [1N4001]
    | | | | |
    GND>-----------+----+----+----+------+

    You could use one of your 2N5551's for Q1, and then the only time
    current will be drawn from the battery is during the time when the one
    second pulse will drive the base of Q1 high.

    Assuming no self-discharge and a fully charged battery, with a
    capacity of 1.2AH and a load of 13.3A for one second every 24 hours,
    the battery would last for 0.09 hours, which is 324 seconds. That
    translates to 324 days... How big is your solar panel?

    which brings us to the next thing: Generating the 1 second pulse
    properly.

    But first, let's see your comments on the foregoing and whether you
    now understand _why_ your circuit is junk.
     
  10. John Fields

    John Fields Guest

     
  11. andy

    andy Guest

    i think you didn't see the version i posted after john popelish's comments:

    http://www.niftybits.ukfsn.org/electronics/daily-water3.png
    http://www.niftybits.ukfsn.org/electronics/daily-water3.gif
    like i said in my reply to john popelish, they aren't meant to be NPN
    emitter followers but PNP switches - I just bought the wrong part by
    mistake, and then copied the part number into the diagram without noticing
    it was the NPN version. my mistake, but that wasn't the mistake.
    paralleling the coils and using a bigger transistor is fair enough -
    that's just me not thinking. I used bipolar ones just to stick with what i
    know for this project - it's only the second i've built.
    That does sound like a better idea given the number of bipolars i've had
    to use to amplify the signal enough to drive the output stage.
    That's still the same mistake/misunderstanding over PNP/NPN transistors.
    yes i did draw it wrong, but the version i put up in reply to john
    popelish is right i think. in this version, that triple darlington is
    meant to be to switch /on/ a PNP switch, not switch /off/ an NPN follower.
    Like i say, i did think about this, and apart from the mistake with the
    part number (should have been tip2955 for the PNP design i drew), i think
    the original circuit was OK for this. What i was trying to do was keep the
    current through the two input stage transistors as low as possible (hence
    the high resistor values), but not worry so much about the output stage,
    because most of the current in this would be used by the coil anyhow.
    Thinking of using a 50mA panel, which should give a long term average
    current of about 1.8 mA minimum in uk winters, according to the table i
    have.
    that is what i wanted to know.
     
  12. John Fields

    John Fields Guest

    ---
    If you take 13.3 amps out of the battery for 1 second it should take
    7389 seconds to fill it back up at 1.8mA, but it'll take longer
    because there is no free lunch. Figure about 50% more, so that'll be
    about 11083 seconds, which is about 3.08 hours. Not bad, but that's
    not counting the quiescent current of the circuitry which, if it's
    1.8mA will keep the battery from charging in winter...
    ---

    ---
    You're welcome?

    Oh, well, it's a slow day, so here are some thoughts about the front
    end...

    You've got basically four approaches to choose from when deciding how
    to generate the 1s pulse:

    1. 7555/7556
    2. 4538/HC123
    3. LM393 or some other comparator array.
    4. Discretes/glue logic

    The 7555/7556 route looks attractive because of low quiescent current
    and a totem-pole output which would eliminate the wasted current drawn
    by a pullup resistor-open collector feeding the driver.
    Unfortunately, the slow-rising DC voltage from the LDR would have to
    be differentiated and AC coupled into the timer, which would require
    additional circuitry to do the job.

    The 4538 looks very attractive because of a totem pole output and DC
    coupled Schmitt trigger inputs, which could help to prevent
    retriggering because of noise on the output of the LDR circuitry.
    Unfortunately, the amount of hysteresis isn't specified. That is, I
    couldn't find it in the spec's. Still in all, there are two timers in
    each package, so the second one might be able to be used to hold off
    the first for long enough until it gets dark enough outside that there
    won't be a chance for a retrigger. That would also eliminate the need
    for hysteresis and you could use Fairchild's 4538BC part which has no
    hysteresis, but some other nice stuff.

    Using LM393's looks attractive because of the flexibility allowed, but
    because of no totem-pole output there'll always be a drain on the
    battery which can be avoided by using one-shots like 4536's. Also,
    LM393's have higher quiescent current requirements than CMOS does.

    Discretes... Unless you're doing this because you want to learn how to
    do it with discretes, forget it.

    So, my vote is for the 4538, and I'll post a schematic of how to do
    the whole thing to alt.binaries.schematics.electronic under "Water
    timer" in a few minutes.
     
  13. andy

    andy Guest

    The design i'm going for is like this:


    | :#: iron cored electro- |
    | :#: magnet. |
    | m |
    ========|X------O ballcock valve with |
    | magnet (m) on top. |
    | |
    | |
    | |
    | |
    | |
    | |
    | |
    | |
    | |
    | |
    | |
    | =============
    | | slow drain
    '--------------------------------' into drip hose.

    The tank is normally empty, with the magnet holding the ballcock shut
    against the electromagnet's core. Then a short pulse through the
    electromagnet should make the ball drop down. The tank fills quickly,
    the valve latches shut again, and then the tank drains slowly through the
    drip hose.

    the idea is:

    a) a short pulse to cancel the magnet's field should take less current
    than having to open a valve against friction. and much less than having to
    hold a solenoid valve open for the whole watering cycle.

    b) the system always comes back to a stable off (no watering) state by a
    purely mechanical process once the initial pulse has been sent. i.e. if
    the power goes, it can't get stuck open.

    c) it's easy to build out of common or garden parts - don't have to buy an
    expensive electrically operated valve.
     
  14. andy

    andy Guest

    why? where is the energy lost?
    that's why i was trying to keep that as low as possible. The final design
    i posted, this is about 0.3mA, I think.
    Doing it that way is probably better - i designed some hysteresis into my
    circuit, but a hold off period is more likely to work right i think.
    i was - is there any reason why the configuration i've drawn isn't a good
    way of doing it? i.e. the bit where the emitter of the transistor is
    connected to the wiper of the pot? I thought of doing it using a circuit
    from horowitz and hill for a differential amplifier, but the way i thought
    of seemed neater for something where the actual level it switches at isn't
    so important.
    the 4538 does look like it would be good for this - i was also thinking of
    adding a switch to select whether to water in morning, evening or both, so
    the rising and falling edge inputs could be useful. i'll have a go at
    working out a circuit with it, then compare to the one you've posted.
     
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.
Electronics Point Logo
Continue to site
Quote of the day

-