Need a latching circuit for DC motor

Discussion in 'The Projects Forum' started by Dimitris76, Sep 16, 2011.

  1. Dimitris76

    Thread Starter Member

    Jul 17, 2011
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    I have asked a similar question a couple of months ago but I need some more help now, as I need to eliminate all external switches and wiring.

    Here it goes:

    I have a 12v linear actuator (max 4 amp @ max. load) with built-in internal end switches that stop the motor at the end of it's travel. The extension/retraction is controlled simply by reversing the power supply's polarity.

    Now what I need is a circuit that with a push of momentary push-button will run the motor and extend the actuator's arm all the way out, then as soon as the end point is reached and the current flow through the motor is stopped, it would sense that and reverse the polarity in order to retract it all the in, until the other end point is reached and the motor stops again.

    Ideally holding the push button constantly in would repeat the whole action, but if it doesn't and I have to release and depress again - well... I can live with it... :)

    I though about using the 555 in bistable mode and sensing the voltage across an in-series to the motor load, small value resistor for the trigger, reset but I am sure there is a more elegant solution out there...

    Ideas, please?

    Dimi
     
  2. Bernard

    AAC Fanatic!

    Aug 7, 2008
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    My guess is that there are two diodes shorting end position NC switches to allow motor to start when polarity is reversed. Now for operation: Use an H bridge for logic control of reversing with a current sensing resistor in neg. leg of up side of bridge to detect stoppage of current, via comparator. diferentiate,& shape to use as reset for flip-flop. Set side is push button start., outputs go to oposing inputs of bridge. " just received some 4A, H bridges by Alegro UDN296 but not tried yet.
     
  3. Dimitris76

    Thread Starter Member

    Jul 17, 2011
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    I need something quick and dirty that I can understand though....

    I came up with this circuit myself:
    [​IMG]

    so I am using a DPDT relay to drive the actuator's motor and a 555 timer in flip-flop mode. When I press the push button the 555's output goes HIGH to the coil's one pole and the other pole is momentarily grounded. That changes the motor's polarity, which starts running and the current is causing a voltage gradient across the 0.3 Ohm, 5 Watt resistor that turns on the NPN transistor that takes over and keeps the coil grounded, until the actuator reaches it's mechanical end and the motor stops. That opens the coil's connection to ground the relay's contacts move (up in the drawing) again the motor polarity is reverse in order for the linear actuator to retract. During that phase the positive voltage to 555's threshold pin is resetting the output to LOW state.

    Do you think it could work?

    Dimitrios
     
  4. Dimitris76

    Thread Starter Member

    Jul 17, 2011
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    Yeah... I forgot the protection diode across the coil too...

    Dimitrios
     
  5. Bernard

    AAC Fanatic!

    Aug 7, 2008
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    After adding the relay coil diode, I see no reason that it should not work. I would add a 1/10μF cap in series with a 10Ω resistor across motor leads to protect relay contacts. The 2k base resistor seems a little high, maybe more like 100Ω depending on relay & transistor used. Just a quibble: also add 10k, pin 6 to ground so that pin 6 is never floating.
     
    Last edited: Sep 19, 2011
  6. Dimitris76

    Thread Starter Member

    Jul 17, 2011
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    Bernard,

    Thank you for your advice!

    The base resistor value is calculated according to the formulas listed here:
    http://www.kpsec.freeuk.com/trancirc.htm
    (scroll down to "Choosing a suitable NPN transistor")

    While I was looking at the 555 circuit though, I had an epiphany!
    What if I could get rid of it altogether?

    So I came up with this simpler circuit:
    [​IMG]

    Do you think it might work?

    Dimi
     
  7. Dimitris76

    Thread Starter Member

    Jul 17, 2011
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    1
    Bernard,

    about this capacitor and resistor - could you elaborate, please?

    The way I see it a 10 Ohm resistor across the motor's poles would create a 1.2Amp shunt @ 12 VDC!

    Also, what should the capacitor's other specs be (apart from the capacity you mentioned)?

    Dimi

    EDIT: I think I got it now! The capacitor in series to the resistor and both components across the motor poles. Thus no shunt with DC...
     
    Last edited: Sep 19, 2011
  8. Bernard

    AAC Fanatic!

    Aug 7, 2008
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    To figure resistor values would need to know relay resistance & motor operating current. As shown , when SW is closed, the thansistor is shorted, how about adding about 40 Ωs in one leg of SW?, will relay still pull in?
     
  9. Dimitris76

    Thread Starter Member

    Jul 17, 2011
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    1
    Bernard,

    do you mean tha pushing the momentary NO push-button will shortcircuit the transistor between it's collector and emiter?
    Why is this a problem?

    Dimi
     
  10. Bernard

    AAC Fanatic!

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    No problem- just a senior moment.
    PS: Be sure to look at the post by onlyvinod56 from yesterday on Projects concerning snubbers.
     
    Last edited: Sep 20, 2011
  11. SgtWookie

    Expert

    Jul 17, 2007
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    The circuit in post #6 is work-able, but I think you may run into problems with the current sense resistor being so low in value. The actuator might draw ~4A when starting up or under heavy load, but what is the current when running with a light load? What is the relay current requirement?

    Your circuit will need to keep the relay latched under both those conditions. The transistor won't start turning on until Vbe >= 0.63v, and with the 2k resistor in series with the base, you're probably going to need a lot of current through that 0.3 Ohm resistor to keep the relay engaged. 2A won't be enough.

    If you used a "window comparator" as a latch, you could also protect against the motor being stalled (too much current flow) in case something blocked the actuators' movement. Otherwise, the circuit would stay powered indefinitely (until something burned up).
     
    Last edited: Sep 20, 2011
  12. Dimitris76

    Thread Starter Member

    Jul 17, 2011
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    Thank you SgtWookie,

    the linear actuator I am using is this:
    http://www.ebay.com/itm/Linear-Actu...ltDomain_0&hash=item35aacdfa9f#ht_2670wt_1270

    and if you have a look at the specs in the tables it draws approx. 4 amp @ 300N push load. I have calculated the 0.3 Ohm resistor for this max load but if the load is lower I'll just raise the reistor value accordingly.

    Now the "window comparator" solution sounds very good but right now I have no idea what this is... I'll have to do some research. Do you have a schematic to share?

    Dimi
     
  13. Bernard

    AAC Fanatic!

    Aug 7, 2008
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    Window comparator is covered in Forrest M Mims's Engineer's Netebook II, also top of this page, with All About Circuits, upper right is a Find box; put in window somparator. ' think dual comparator ckt. would be best.
    Still plan on base current of 1/10 collector current. 4A thru .3Ω = 1.2V; 1.2V - .7V = .5V; R = .5V/ .005A = 100Ω, assuming 50mA coil current.
     
  14. SgtWookie

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    Jul 17, 2007
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    Sorry, Dimitris76 - I missed your reply. It's been busy on here and I've been pretty busy in my own life.

    I'll see about drawing up something for you.
     
  15. Dimitris76

    Thread Starter Member

    Jul 17, 2011
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    It's ok SgtWookie,

    in the meantime I took the liberty to draw something myself....

    [​IMG]

    What do you think about these resistor values?

    I want a relay that unlatches at current draw more than approx. 4 amps and as well as at current less than 1 amp. I am actually thinking of replacing R2 and R3 with a trim pot for "fine tuning"...

    the R5 value is allowed to let through approx. 1/10 of the current the coil needs, which is 33mA.

    Dimi
     
  16. SgtWookie

    Expert

    Jul 17, 2007
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    Well, I came up with fairly close to the same thing.
    Have a look at the attached. My values for R2,3, and 4 are different (I had them in a different order, too)

    It turned out to be a bit more complicated than I thought.

    R7, D2, D3 protect against exceeding the comparator input limits.
    ZD1, ZD2 suppress the spike from the motor when current is cut. Normally, that should be taken care of internally to the actuator, but if power fails while the actuator is moving, the spike needs to be snubbed.

    LTSpice doesn't have a pushbutton switch; that's what S1 and V2 are doing; acting as a pushbutton switch. It waits 50mS, then grounds the low side of the relay for 100mS. I used lower current limits than you did; trip current is ~3.63A (after all, if you set the thing to trip at 4A, it may just not trip!) and the minimum current is lower; about 12mA. We don't know what the no-load current will be. It might just be 400mA, but if it falls below 12mA, it'll drop out as well. If you want to increase R3, you'll need to decrease R2 or increase R4 to keep the high limit below 4A.
     
    Last edited: Sep 21, 2011
  17. Dimitris76

    Thread Starter Member

    Jul 17, 2011
    45
    1
    Thank you again SgtWookie!

    I think your circuit is up to the task so I'll go out to buy the extra stuff I need.

    One last question though:
    Does the current between the NPN transistor's base-emitter need to be 1/10 of the current I need at collector-emitter?
    And does this 1/10th rule apply for all the transistors as long as the current does not exceed the component's specs?

    I am trying to figure out what's the optimal resistor value for my 12V relay with a 360 Ohm coil....

    Dimi
     
  18. Bernard

    AAC Fanatic!

    Aug 7, 2008
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    My seat of the pants value of 2.7k vs The Sgt's calculated value of 3.3k for your 360Ω coil will both work- take your pick. The 1/10 base drive figure has been discussed on AAC before & I was finally won over with reason.
     
  19. Dimitris76

    Thread Starter Member

    Jul 17, 2011
    45
    1
    Bernard,

    Your 1/10th is straight forward and easy to remember, so I'll follow it.

    Thank you for your help!
    Dimi
     
  20. SgtWookie

    Expert

    Jul 17, 2007
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    Yes, either will work - but I chose the value for fairly specific reasons.

    Here's National Semiconductor's datasheet for the LM393/LM2903:
    http://www.national.com/ds/LM/LM393.pdf

    The LM393, LM339, and LM2903 all have a rather limited guaranteed sink current of ~4mA before the saturation voltage of the output can exceed 400mV. You really want to keep that saturation voltage below 500mV unless you want the transistor Q1 to start conducting.

    12v/4mA = 3k Ohms. 12v/3.3k Ohms = 3.636...mA, which is well under the worst-case limits stated in the datasheets.

    So, what's needed to turn Q1 on?
    12v/360 Ohms = 33.333...mA
    So, with Ib=Ic/10, 3.333...mA base current is needed.
    We subtract Vbe from the supply voltage to wind up with 12v-0.7v = 11.3v.
    11.3v/3.333...mA = 3,390.339.... Ohms.
    3.3k Ohms is the closest standard E24 value of resistance to ~3.39k Ohms.
    So, 11.3v/3.3k Ohms =3.424...mA, which places us neatly between the 3.636mA max target we were looking at, and just above the 3.333...mA we need.
     
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