Connect with us

Voltage sensing switch idea?

Discussion in 'Electronic Basics' started by Glenn Ashmore, Feb 4, 2004.

Scroll to continue with content
  1. I have two circuits A & B. Circuit "A" comes from a pressure sensor and
    varies from 0 to 5VDC. Using a couple of resistors as a voltage divider
    I can read the pressure on a cheap 200mV LCD meter. Circuit "B"
    operates a 12VDC relay that draws 42mA. I am looking for a way to
    switch off circuit "B" when the voltage on circuit "A" gets up to 4.5V.
    The solution must be cheap (like a dollar or three) and not effect
    the meter reading. Any suggestions?

    --
    Glenn Ashmore

    I'm building a 45' cutter in strip/composite. Watch my progress (or lack
    there of) at: http://www.rutuonline.com
    Shameless Commercial Division: http://www.spade-anchor-us.com
     
  2. CHRIS.WELSH

    CHRIS.WELSH Guest

    maybe a comparator circuit (lm311) driving a transistor switch for higher
    current.
     
  3. If you have two voltage sources, one at 5VDC, and one at 12VDC, you can do
    this with two bipolar transistors, a mosfet, and a couple of resistors.

    Use a PNP transistor, with the emitter connected to 5V. The collector is
    grounded through a 100k resistor, and the base is connected to your sensor
    through another 100k resistor.

    As the sensor gets to someplace near 4.4V, the transistor turns on, causing
    the collector voltage to rise (due to the 100k resistor.)

    The collector of the first transistor, Q1, is connected to the base of a NPN
    transistor through a 1MEG resistor. Call the second one Q2. Q2's emitter is
    grounded, and the collector connects to the 13V source through a 100k
    resistor. When the collector voltage of Q1 rises, it'll turn on Q2, which
    will then let current through. That will cause the 100k resistor to 12V to
    develop a voltage across it (ie, the collector voltage of Q2 will drop.)

    The final step is to use a power P channel mosfet (a 'high side' mosfet).
    Connect the gate up to the collector of Q2. That way, when it drops from 12V
    to near 0 (as Q2 turns on) this power mosfet will turn on. Connect the drain
    of the P-MOSFET to the relay coil, and the other side of the coil to gnd.
    The source is 12V. The relay coil is energized when the input goes to about
    4.3V.

    Here is the circuit (look at it using courier font:


    +------------+- 12V
    | |
    .-. |
    | |100k |
    | | |
    '-' |
    | |
    | |
    |---------||-+
    | ||-> P-MOFSET
    | ||-+
    | |
    5V | |
    - | .--------------.
    100k | | | |
    ___ |< NPN | | 12V |
    IN --|___|--| | | Relay Coil |
    |\ 1M | | |
    | ___ |/ | |
    -|___|-| '--------------'
    | |> |
    | | |
    .-. | |
    | |100k | |
    | | | |
    '-' | |
    | | |
    +--------+------------+
    ===
    GND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    Regards,
    Bob Monsen
     
  4. John Fields

    John Fields Guest

    ---
    It depends on how the relay's being switched, but basically:


    +12v>----------------+-------+
    | |
    | |
    0-5V>--+ +------+ |
    | | | |
    [R1] [R3] | [R4]
    | | | |
    | +-----|-\ | ___
    | | | >-----+ +---->OUT
    +---------+--|+/U1 | |
    | | | | | C
    | | +-----[R5]--+--[R6]---B 2N4401
    | | | E
    [R2] [CR1] | |
    | | | |
    GND>---+------+------+-------------------+


    The 0-5V is the output from your pressure transducer circuitry and R1R2
    is a voltage divider. CR1 is a shunt voltage reference like an
    LM385-2.5V and R3 is its current limiting resistor, U1 is 1/2 of an
    LM393 (the remaining 1/2 should have its inputs and output shorted to
    GND), R4 is the base current source limiter for the 2N4401 and R5 is to
    provide hysteresis around the comparator so its output won't chatter for
    slowly varying changes in the )-5V input.

    If you use an LM385-2.5V for CR1 it'll work fine with 1mA going through
    it, so you'll want to drop (12V-2.5V) across R3 with 1mA going through
    it, so that comes out to R3 = (12V-2.5V)/1mA = 9500 ohms, so just stick
    10k in there and it'll be fine too. For 10k it'll dissipate 9.5V*0.95mA
    ~ 9mW, so a 5% 1/4 watter will work just fine as well.

    Looking at the input divider, R1R2, what we want it to do is to generate
    a voltage at the junction of R1 and R2 which, when the input voltage
    gets to 4.5V is going to make the comparator switch. Since the - input
    of the comparator is sitting at the reference's 2.5V, that means the
    voltage at the R1R2 junction needs to be 2.5V when the output of the
    transducer circuitry is 4.5V. Not knowing how much current the
    circuitry can supply is a little bit of a problem, but if we assume
    100µA we can figure out R1 and R2. If we can have 100µA flowing through
    R1 and R2, then to make the hot end of R2 sit at 2.5V we can say R = E/I
    = 2.5V/100µA = 25k ohms. Assuming that you want to get pretty close to
    that 2.5V and that your reference is spot on, the closest standard 1%
    resistor is 24.9k, which is only 100 ohms (+/- the tolerance of the
    resistor, 250 ohms) off. Now, we can do the same for R1, and since we
    want to drop 2V across it at 100µA, that works out to 20k ohms, a
    standard 1% value!

    Next, I'm assuming that your relay switching circuitry looks something
    like this:

    +12V
    |
    +-------+
    |K |
    [DIODE] [COIL]
    | |
    +-------+
    |
    C
    ON>---[R]----B NPN
    E
    |
    GND



    And that by doing this:


    +12V
    |
    +-------+
    |K |
    [DIODE] [COIL]
    | |
    +-------+
    |
    C
    ON>---[R]--+---B NPN
    | E
    ___ | |
    OUT>-------+ GND

    We can pull down the base of the transistor turning the relay on,
    forcing the relay to open, no matter what.

    Combining the two circuits, we'll have:


    +12V
    |
    +-------+
    |K |
    [DIODE] [COIL]
    | |
    +-------+
    |
    C
    ON>--------------------------------[R]--+---B NPN
    | E
    | |
    | GND
    |
    |
    +12v>----------------+-------+ |
    | | |
    | | |
    0-5V>--+ +------+ | |
    | | | | |
    [20k] [10k] | [12k] |
    | | | | |
    | +-----|-\ | |
    | | | >-----+ |
    +---------+--|+/U1 | |
    | | | | | C
    | | +-----[R5]--+--[91k]--B 2N4401
    | |K | E
    [24.9k] [CR1] | |
    | | | |
    GND>---+------+------+------------------+

    What we'll want to do is divert the base current of your relay driver to
    GND, and to do that properly we'll need to saturate the CE juncton of
    the 2N4401, which will cause about 0.3V to appear across it, which
    would surely cut off the relay driver. Since the relay draws 42mA, if
    we assume a forced beta of 10 for its driver, that leaves us with 4.2mA
    to divert to GND. If we force the 2N4401's beta to 10 and it has to
    sink 4.2ma, then that means all we have to do to turn it on is force
    420µA into its base and we'll be done. But... there _is_ still R5 to
    figure out. It's there for hysteresis, and it serves to keep the
    comparator's output from chattering when its + input goes through the
    switching point slowly. The 2N4401 will only be required to sink the
    relay driver transistor's base current to turn it off, so if we set it
    up to sink 1mA and force its beta to 10, it'll only need 100µA of base
    current to turn on, so if we make the sum of R4 and R6 drop 1.3V with
    100µA flowing through them that'll do it. We need R6 because with the
    2N4401 fully turned on, its base to emitter voltage will only be about
    1.3V, which would never allow the comparator's + input to get to the
    2.5V switching point! So, R4+R6 needs to be (12V-1.3V)/100µA = 107k
    ohms. 100k is close enough, and R4 needs to be large enough to make
    sure the comparator's output transistor never comes out of saturation
    when it's on. It can handle 4mA, but if we keep it down around 1mA
    that'll make sure we have no problems, so R4 needs to drop 12V at 1mA,
    which makes it 12k. Then, since R4+R6 needs to be 100k, R6 needs to be
    88k. 91k is a standard 5% value, and it's close enough, so R6 will then
    be 91k, and 1/4 watt resistors will be plenty big for both R4 and R6.

    Now, the purpose of R5 is to source and sink a tiny bit of current into
    and out of the junction of R1 and R2 so that when the comparator turns
    off and on it forces the voltage on the + input to quickly rise and fall
    past the 2.5V switching threshold, helping to avoid any noise problems
    which would tend to make the comparator's output chatter. If we assume
    that 10mV of hysteresis would be enough, then we have to make the
    voltage across R2 rise by 10mV, and since R2 is 24.9k, that means we
    have to force an extra I = E/R = (2.51V-2.5V)/24.9K = 402nA through it.
    Since, when the comparator turns off, the base to emitter voltage of the
    2N4401 will rise to 1.3V, R4 and R6 will form a voltage divider with one
    end at 12V and the other at 1.3V, and the voltage at the junction of R4
    and R6 will go to:

    12V*R4
    E = -------- + 1.3V = 7.95V ~ 8V
    R4+R6

    Which is going to cause a problem in that it will cause the hysteresis
    to be asymmetrical about the switching point. Since the switching point
    is at 2.5V and the comparator's output goes to 0V when it turns on,
    pulling one end of R5 to 0V, we would like for that end of R5 to go to
    +5V when the comparator turns off, making the input to R5 symmetrical
    about 2.5V. That way, the hysteresis will also be symmetrical around
    the switching point.

    What we need to do to make that happen is to change the values of R4 and
    R6 so that the voltage at their junction will be 5V when the comparator
    is off. If, when the comparator is off, we allow 1mA of current to
    flow through R4 and R6 then, to make the voltage at their junction be
    5V, we need to drop 7V across R4. That means R4 needs to be 7k. The
    two 5% choices available are 6.8k and 7.5k, so let's choose 6.8k and see
    what happens. First off, if we want to drop 7V across 6.8k, the current
    through it has to be 1.03mA, which means that current is also going to
    be flowing through R6 and into the B-E junction of the 2N4401. The base
    is going to be sitting at about 1.3V, so that means R6 will have to drop
    5V-1.3V = 3.7V at 1.03mA, so it will need to be 3.7V/1.03mA = 2.98k. 3k
    is a standard 5% so, since that 1.3V is only approximate, 6.8K for R4
    and 3k for R6 should be OK.

    Now, looking at R5 again, we'll need to drop about 2.5V across it at
    400nA, so it'll need to be 2.5V/400nA = 6.25M. 6.2M is a standard 5%
    value, so that's pretty close, and here's your final(?) circuit:)

    +12V
    |
    +-------+
    |K |
    [DIODE] [COIL]
    | |
    +-------+
    |
    C
    ON>---------------------------------[R]--+---B NPN
    | E
    | |
    | GND
    |
    |
    +12v>----------------+--------+ |
    | | |
    | | |
    0-5V>--+ +------+ | |
    | | | | |
    [20k] [10k] | [6k8] |
    | | | | |
    | +-----|-\ | |
    | | | >------+ |
    +---------+--|+/U1 | |
    | | | | | C
    | | +-----[6M2]--+--[3k]---B 2N4401
    | |K | E
    [24.9k] [CR1] | |
    | | | |
    GND>---+------+------+-------------------+


    Interesting how something simple can turn out not to be huh?

    BTW, if your relay driver configuration is different from what I
    assumed, post back and we'll make everything match up.
     
  5. John Fields

    John Fields Guest

    On Thu, 05 Feb 2004 06:29:12 -0600, John Fields

    Errors, sorry...
     

  6. Now you got me thinking about the overall plan. This is one part of an
    interlock circuit consisting of 3 transistors and 2 switches in series
    on the low side of a relay. The relay is latched on by routing the
    ground of the coil through one pole of the relay and then through the
    interlock circuit. If the circuit is broken the relay unlatches.


    +-------------+
    |/ | +12V
    Pressure Switch -| | |
    |> | +-----+
    | o | |
    | / _|_ -
    | / |_/_|- ^
    | o/ o | |
    |/ | | +-----+
    Low RPM -| GND |
    |> |
    | |
    | |-------o |
    |/ | |=| ON Button
    High RPM -| | o |
    |> | |
    | | GND
    o |
    Oil Pressure '\ |
    (about 15') \ |
    o \ o |
    | | |=| Off Button (NC)
    +---------o | (about 15')

    Would it be better to have one "master" transistor at the relay which is
    held on and have the sensor output transistors drain its base like this
    when something gets out of spec?

    VDD
    | OIL STOP
    +-----+--------+
    | | | T T
    |/ | --- --- ---
    Pressure -| | GND -v \----o o----o o-+
    |> | |
    | | | T
    GND | _/ | ---
    | +-o/ o-------+--o o-GND
    +-----+ | _
    |/ | | | |
    Low RPM -| | +--|\|- +12V
    |> | |_|
    | | |
    GND |
    +-----+
    |
    |/
    High RPM -|
    |>
    |
    GND

    Also, what are the implications of having transistors and mechanical
    switches in series like this?

    --
    Glenn Ashmore

    I'm building a 45' cutter in strip/composite. Watch my progress (or lack
    there of) at: http://www.rutuonline.com
    Shameless Commercial Division: http://www.spade-anchor-us.com
     
  7. John Fields

    John Fields Guest

    ---
    Yes, but it looks like the transistors are wired backwards unless, for
    some reason, they're supposed to be PNP's.
    ---
    ---
    I don't know about 'better', but I like this way much better than the
    other because each transistor's emitter is being independently referred
    to ground and is being switched independently of the others without
    being daisy-chained emitter to collector to emitter to collector to...
    ---
    ---
    AFAIK, just force the transistor into saturation so it won't dissipate
    much power with the relay switched ON, and make sure the mechanical
    switch contacts have enough current flowing through them so that you
    won't wind up with high resistance contact crud problems. Also, if
    you're going to be doing this at sea, MAKE SURE any relays you use are
    sealed. Not just with a plasic cover over them, but _hermetically_
    sealed.
     
  8. This was my first attempt at this ASCII schematic program. I was lucky
    to get anything much less get it wired correctly. :)
    Thanks. I will adjust my schematic.

    The relays are sealed Omrons, the switches are waterproof membranes and
    I have a big can of conformal coating for the board.

    --
    Glenn Ashmore

    I'm building a 45' cutter in strip/composite. Watch my progress (or lack
    there of) at: http://www.rutuonline.com
    Shameless Commercial Division: http://www.spade-anchor-us.com
     
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

-