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Mid Position Valve

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  • 26-02-2021 1:00pm
    #1
    John.G
    Registered Users Posts: 3,468 ✭✭✭


    The above is used in Y system heating systems, it can supply water for DHW or CH or both. It achieves its mid position by applying a reduced voltage of ~ 69V to the actuator motor to keep the valve in its mid position against spring tension.
    It sends 230V through a diode to convert to DC and then through a 13Kohm resistor, the motor then stalls against the spring tension in the mid position.
    A de magnetising AC current is then passed through the motor at other times but this has caused problems on some gas boilers as they won't shut down if the valve has been in CH mode only and the roomstat is satisfied as 230v is then supplied through the de magnetising resistor to the switched live on the boiler.

    My query is why can't the required reduced mid position motor voltage be supplied through the resistor as 69V AC.


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    #2
    aido79
    Moderators, Regional Abroad Moderators Posts: 5,374 Mod ✭✭✭✭aido79


    John.G wrote: »
    The above is used in Y system heating systems, it can supply water for DHW or CH or both. It achieves its mid position by applying a reduced voltage of ~ 69V to the actuator motor to keep the valve in its mid position against spring tension.
    It sends 230V through a diode to convert to DC and then through a 13Kohm resistor, the motor then stalls against the spring tension in the mid position.
    A de magnetising AC current is then passed through the motor at other times but this has caused problems on some gas boilers as they won't shut down if the valve has been in CH mode only and the roomstat is satisfied as 230v is then supplied through the de magnetising resistor to the switched live on the boiler.

    My query is why can't the required reduced mid position motor voltage be supplied through the resistor as 69V AC.

    A resistor doesn't reduce voltage. It limits the current but doesn't change voltage. A voltage divider circuit would be needed to reduce the voltage.


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    #3
    John.G
    Registered Users Posts: 3,468 ✭✭✭John.G


    The motor is ~ 8W so its resistance is 6613 ohms, the resistance in series with it is 13000 ohms so total resistance = 19613 ohms.
    voltage drop across resistor = 230*13000/19613 or 152V so volt drop across motor, 230-152 or 78V
    or volt drop across motor = 230*6613/19613 or 78V


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    #4
    aido79
    Moderators, Regional Abroad Moderators Posts: 5,374 Mod ✭✭✭✭aido79


    John.G wrote: »
    The motor is ~ 8W so its resistance is 6613 ohms, the resistance in series with it is 13000 ohms so total resistance = 19613 ohms.
    voltage drop across resistor = 230*13000/19613 or 152V so volt drop across motor, 230-152 or 78V
    or volt drop across motor = 230*6613/19613 or 78V

    There is no way the motor alone has a resistance of 6613. How are measuring or calculating that?
    It sounds like there is a resistor in series with the motor to make a voltage divider circuit with the other resistor to drop the voltage to the required voltage for the motor.


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    #5
    John.G
    Registered Users Posts: 3,468 ✭✭✭John.G


    I just calculated the running resistance from R = V^2/watts, 230*230/8, 6613. I do know that the volt drop across the motor is 68/80V.
    Attached is the diverter valve.


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    #6
    Hoagy
    Registered Users Posts: 707 ✭✭✭Hoagy


    Here is a fairly clear explanation of how it should work for anyone (like me) who isn't familiar with the system.

    http://www.diyfaq.org.uk/plumbing/controls/midpositionvalve.htm


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    #7
    John.G
    Registered Users Posts: 3,468 ✭✭✭John.G


    Its design is very ingenious, its just my electrical question of the need for a DC supply.


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    #8
    Cerco
    Registered Users Posts: 2,034 ✭✭✭Cerco


    If you run AC down the SW2 leg without the diode the motor would just continue moving the valve towards A. Would be the same as keeping SW1 connected to white.
    The diode gives a pulsating dc output through the series 13k resistor.
    From the description I expect the pulsating dc moves the motor in short steps towards A till it reaches midpoint when SW2 is switched. The series 13k reduces the load voltage to fine tune the steps That’s my understanding of the operation.
    This is a clever way of using a unidirectional motor where a bi-directional would be intuitive.

    It would be interesting to see it operating


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    #9
    Bruthal.
    Registered Users Posts: 209 ✭✭Bruthal.


    DC to these synchronous type motors cause them to hold the current position. That is likely the setup here.


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    #10
    Wearb
    Moderators, Home & Garden Moderators Posts: 6,228 Mod ✭✭✭✭Wearb


    https://www.youtube.com/watch?v=F5m8f6HN2Us&t=5s

    John Ward has a good explanation above.

    Please follow site and charter rules. "Resistance is futile"



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    #11
    John.G
    Registered Users Posts: 3,468 ✭✭✭John.G


    There is 240V only on the white wire when mid position is called for (HW&CH), The motor starts moving due to 240V applied vie SW1, a reduced DC voltage is also fed to the motor via SW2 and the diode/resistor. As the valve moves further across SW1 will switch to the grey wire but this has no effect as 240V not present if the HW cylinder stat is not satisfied, the only power to the motor is then ~ 69V from the white (roomstat not satisfied) 240v via the diode/13K resistor and the motor stalls in the mid position.
    All this is fine but my query still stands, if there was no diode and with a resistor of sufficient value in the supply to the motor would it still stall with a reduced AC only voltage??. All motorized valves, including 2 port ,stall at "full" AC power when they move to their end positions and draw ~ 8W. In other words can a AC motor be made to have reduced stalled torque with a reduced AC voltage only?, it does stall at full torque with 240VAC applied.


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    #12
    Bruthal.
    Registered Users Posts: 209 ✭✭Bruthal.


    John.G wrote: »
    All this is fine but my query still stands, if there was no diode and with a resistor of sufficient value in the supply to the motor would it still stall with a reduced AC only voltage??. All motorized valves, including 2 port ,stall at "full" AC power when they move to their end positions and draw ~ 8W. In other words can a AC motor be made to have reduced stalled torque with a reduced AC voltage only?, it does stall at full torque with 240VAC applied.

    It would be impractical to try balance a reduced supply to the motor against the return spring to stop in a desired position, when simply supplying a hold current is far simpler. The diode applied to the circuit will hold the motor from moving either way via being DC, like an electric brake, which is operated in an easily selected position by virtue of the switch design setup.

    When moving to the full position, they encounter a hard stop which they cant move past.


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    #13
    Cerco
    Registered Users Posts: 2,034 ✭✭✭Cerco


    John.G wrote: »
    All this is fine but my query still stands, if there was no diode and with a resistor of sufficient value in the supply to the motor would it still stall with a reduced AC only voltage..

    If you take this to a logical conclusion and reduced the AC voltage to zero then do you think the motor would stall? It would continue under it s own momentum. The way I see it, the PULSATING DC voltage has a juddering effect on the motor causing it to stall.


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    #14
    Bruthal.
    Registered Users Posts: 209 ✭✭Bruthal.


    Cerco wrote: »
    If you take this to a logical conclusion and reduced the AC voltage to zero then do you think the motor would stall? It would continue under it s own momentum. The way I see it, the PULSATING DC voltage has a juddering effect on the motor causing it to stall.

    If you reduce the AC voltage to zero, the spring returns the valve to the rest/de-energised position.

    If you apply DC, whether it is half wave or completely smooth, the motor holds its position, not just stops, it electrically brakes.


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    #15
    Cerco
    Registered Users Posts: 2,034 ✭✭✭Cerco


    Bruthal. wrote: »
    If you reduce the AC voltage to zero, the spring returns the valve to the rest/de-energised position.

    If you apply DC, whether it is half wave or completely smooth, the motor holds its position, not just stops, it electrically brakes.

    Removing the AC will allow the spring to eventually return valve to rest position.

    This article seems to support Bruthal’s analysis . It answers the Op’s question.
    :
    https://electrical-engineering-portal.com/methods-braking-motor


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    #16
    Bruthal.
    Registered Users Posts: 209 ✭✭Bruthal.


    Cerco wrote: »
    Removing the AC will allow the spring to eventually return valve to rest position.
    However imho it is the combination of pulsating dc voltage pushing motor against the spring gives the motor locking effect, not the dc voltage alone.
    DC will hold them in position without any spring. The position of rest is set according to when the DC is applied.

    If you visualise a permanent magnet inside a winding, no comutator, you have a synchronous motor.

    The permanent magnet will follow the AC magnetic field as it reverses, in sync with it, with the permanent magnet rotor N pole following the winding S pole as it swaps from one side to the other.
    (and vice versa)

    Apply a DC current, you now have a winding with N/S poles fixed, and a permanent magnet with fixed N/S poles. The rotor will stay fixed with its N pole at the winding S pole, going nowhere.


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    #17
    John.G
    Registered Users Posts: 3,468 ✭✭✭John.G


    I get the gist of that I think but if the spring was removed while valve is in mid position I dare say the valve would continue to move over against its stop?.
    The other resistance I think provides a de magnetizing voltage with HW only supplied but at 270K & 0.25W implies a volt drop of 260V ?.


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    #18
    Bruthal.
    Registered Users Posts: 209 ✭✭Bruthal.


    John.G wrote: »
    I get the gist of that I think but if the spring was removed while valve is in mid position I dare say the valve would continue to move over against its stop?.
    The other resistance I think provides a de magnetizing voltage with HW only supplied but at 270K & 0.25W implies a volt drop of 260V ?.

    The point about the spring was that the DC alone will hold these motors in position.

    If the inertia overshoots the DC hold position, the SW2 changes over, disconnecting any power, so the spring will pull it back, SW2 changes back again, so the DC reconnects, so the motor will settle to the same position anyway. The spring is needed if that happens. But the DC alone holds the motor position once its in its mid position stopped.

    If the spring is removed while the valve is stopped mid position by the DC, it will not move. The gearing in these units is high, so it is easy for the very small synchronous motor to move the mechanism, and also so it can easily start.

    So any stopping power the motor applies while DC powered, will be hard to move against.


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    #19
    John.G
    Registered Users Posts: 3,468 ✭✭✭John.G


    Fantastic! Thanks.


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    #20
    Bruthal.
    Registered Users Posts: 209 ✭✭Bruthal.


    A common place where we see a slightly different electric braking in action, is battery drills. When you release the trigger fully, it bridges the windings, so the inertia spinning motor is acting like a generator, so when it is bridged, it is like shorting this generator, thus magnetically stopping it almost instantly.

    Usually a position can be found with the trigger where the drill is not powered, or braked, so inertia keeps it going for a few seconds. Release the trigger the rest of the way during this, and it can be seen how it stops it instantly.


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