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Related

Wireless switch for power

Former Member
Former Member over 11 years ago

  Hello......

      New to this board and also new to this subject. I am trying to power a Ledex B12 solenoid and be able to switch the power on and off wirelessly. I have to try to fit the components into a box which is 2.5"x2.5" and about 1" deep. I am using the solenoid to control a latch/lock.......if a solenoid is not the best or smallest way to accomplish this, please make any suggestion you may have. Your help is greatly appreciated. Keep in mind I have no experience in this field......only an idea on how to use it.

 

        Thanks, Keith

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  • D_Hersey
    0 D_Hersey over 11 years ago

    Meant to type: DC solid-state relay, getting sleepy.

     

    So we only want to do 1 of 2 things at a time with this particular opto. It isn't designed to operate in a linear (continuous) mode.

     

    -not energize the input led

    -energize the led with 8.75 ma of current, in the proper direction.

     

    If one were to give a kid a bargain-bag of LEDs and a drawer-full of miscellaneous batteries, and he caught on to the two-terminal nature of the sources and loads, different things might happen:  He could destroy a given LED in a flash of light.  He could destroy a given LED without a flash of light.  He could hook up a battery without result.  He could steadily illuminate a LED.  If this last happened, the battery was nearly exhausted.  Its Thevenin equivalent resistance grew large, as its voltage fell.  If the voltage falls below the threshold voltage of the LED, still no soap.  The first-order model of a battery is an ideal voltage source with a series resistance.  They have to put this resistance in at the factory, else a clumsy mechanic could destroy the universe.

     

    So the chip implies that we must insert a resistor between the front-end of our opto and a driving potential (voltage) source.  We gotta make our voltage supply look like a dying battery to the LED or it will fry.  LED not linear in effort/flow.  Double the woltage, bring up the current by a factor of ten.  All diodes are LEDs.  Some are cruelly encapsulated in dark epoxy.  Some are so low in frequency (and now high!) that they are hard to see.  LEDs tend to be monochromatic, and wavelength is inversely proportionally to the drop.  Red LEDs have about 1.2 V drops (xout voltages), Violet, about twice that. We are figuring upon about a volt and a quarter of drop, cause its late, and when its late, I like to keep the math easy.  If our logic supply was 3.3V, and our outputs were CMOS, we could probably take our choice of turning on the LED when we go to the positive logic true state, or vice versa.  Pulling down to activate might be slightly better electrically, but not worth fretting over.  If our driver is TTL or nMOS, we wanna pull down.

    Anyway, our LED has a potential of one-and-a-quarter, say.  This means (forward mode) in order to operate but not too much we expect such a drop.  We need to subtract this from the supply voltage, then whip out hoary old Ohm's law to wind up with our R's value.  R=V/I (think about it, more effort, more flow)

     

    --Is that right, darn it's late!  3.3-1.25 = 2.05! ***! R=2.05/8.75*1000  = 234and a quarter, about!--

     

    We're not done, bucko!  We still need to reckon the power we intend to dissipate in the ballast R in the active phase:  Power =IIR =(7/8/100)^2*234 (circa)

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  • D_Hersey
    0 D_Hersey over 11 years ago

    Meant to type: DC solid-state relay, getting sleepy.

     

    So we only want to do 1 of 2 things at a time with this particular opto. It isn't designed to operate in a linear (continuous) mode.

     

    -not energize the input led

    -energize the led with 8.75 ma of current, in the proper direction.

     

    If one were to give a kid a bargain-bag of LEDs and a drawer-full of miscellaneous batteries, and he caught on to the two-terminal nature of the sources and loads, different things might happen:  He could destroy a given LED in a flash of light.  He could destroy a given LED without a flash of light.  He could hook up a battery without result.  He could steadily illuminate a LED.  If this last happened, the battery was nearly exhausted.  Its Thevenin equivalent resistance grew large, as its voltage fell.  If the voltage falls below the threshold voltage of the LED, still no soap.  The first-order model of a battery is an ideal voltage source with a series resistance.  They have to put this resistance in at the factory, else a clumsy mechanic could destroy the universe.

     

    So the chip implies that we must insert a resistor between the front-end of our opto and a driving potential (voltage) source.  We gotta make our voltage supply look like a dying battery to the LED or it will fry.  LED not linear in effort/flow.  Double the woltage, bring up the current by a factor of ten.  All diodes are LEDs.  Some are cruelly encapsulated in dark epoxy.  Some are so low in frequency (and now high!) that they are hard to see.  LEDs tend to be monochromatic, and wavelength is inversely proportionally to the drop.  Red LEDs have about 1.2 V drops (xout voltages), Violet, about twice that. We are figuring upon about a volt and a quarter of drop, cause its late, and when its late, I like to keep the math easy.  If our logic supply was 3.3V, and our outputs were CMOS, we could probably take our choice of turning on the LED when we go to the positive logic true state, or vice versa.  Pulling down to activate might be slightly better electrically, but not worth fretting over.  If our driver is TTL or nMOS, we wanna pull down.

    Anyway, our LED has a potential of one-and-a-quarter, say.  This means (forward mode) in order to operate but not too much we expect such a drop.  We need to subtract this from the supply voltage, then whip out hoary old Ohm's law to wind up with our R's value.  R=V/I (think about it, more effort, more flow)

     

    --Is that right, darn it's late!  3.3-1.25 = 2.05! ***! R=2.05/8.75*1000  = 234and a quarter, about!--

     

    We're not done, bucko!  We still need to reckon the power we intend to dissipate in the ballast R in the active phase:  Power =IIR =(7/8/100)^2*234 (circa)

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