Slotted optical switch

[matty matt tmatty]

I've bought some optical switches for use in a project:

https://www.amazon.co.uk/gp/product/B01G6PE5NU/ref=oh_aui_detailpage_o02_s00?ie=UTF8&psc=1

foolishly thinking it would be as simple to connect as a reed switch!

I've tried following various examples online but I am getting nowhere. All I am aiming to do is something very simple - trigger an output (lcd glass) to come on/off when the gate is open/closed.

Can anyone help?

 

[Minder]

They are slot opto's and very easy to use, how do you have them connected now and what supply voltage and what input is the output connected to?
M.

 

[73's de Edd]

Sir matty matt tmatty . . . . ...( hmmmmmmmmm . . . . .now ? did I fully get all of the alliterations of Matt in ? )

Try this known circuitry to initially get your unit working. Use “ Diode test mode” of a DVM to locate the infrared diode half / section.

It will be showing the junction voltage of the diode, while the series C-E junction of the detector transistor half, might not register at all, if not being of a much higher reading.

Then depend upon the 220 series resistors current limiting, in not “blowing” the IR diode section, by holding its current feed at approx 20 ma. .

Use a video camera or your phone in monitoring the slot of the diode section in being able to then “see” if otherwise invisible IR light is coming forth from the slot, when you put power to the IR emitter diode.

Then take DC metering in > 5VDC range to monitor across from ground to the bottom of the 10K load resistor to confirm if there is a LOW-HIGH logic shift as you move the shutter in and out of place. (Or turn the LED on and off if NO shutter has been set up yet.)

I saw NO polarity / nor lead markings / nor case markings on your info supplied by the supply link, unless it was on a separate sheet of paper, enclosed within the shipment.

TEST HOOK-UP . . . . .



73's de Edd

 

[matty matt tmatty]

Thanks for the responses!

The switch does have markings on the top for the diode on one side and a H and Y on the other.

My supply voltage is ideally 9V or 6V batteries but I can get this down to 5V by using resistors?

My output is a piece of LCD shutter glass that turns black with a voltage of 3V:

https://www.aliexpress.com/item/6-o...lgo_pvid=ba5b2245-55e7-406c-bd5d-3a1debc16040

I've had it working with a 3V supply, a reed switch and a 100Ω resistor across the lcd to discharge it immediately when the switch is open (otherwise the LCD slowly fades away).

Given Edds suggestion this is my proposed circuit. R1 and R2 will divide the voltage down to 5V. Do you think the LCD will be happy with 5V?

Thanks

 

[Bluejets]

Forget the voltage divider idea......

Your circuit is wrong anyhow as you have a short across the 100 and LCD ,it's in the wrong place as well and I have no idea what the 100 is supposed to do either.

Your LCD runs on 3v so you will need a supply to match that plus the approx. 0.7v drop you get across the open collector transistor in the slot opto.
Also, get into the habit of drawing circuits in a conventional manner, much easier to read.

I'll draw you out a diagram similar to Edd's.
One of a few different ways to approach this.

 

[matty matt tmatty]

Thanks Bluejets. I put the 100Ω resistor on there to ensure the lcd turns off instantly when the switch is closed, otherwise it takes a while to turn transparent. I got the idea from the following video,
The relevant bits are at 4:11 and 7:21

Although he uses 1K resistor, I tried a few different ones and they all worked. Do you think there is a better way to do this?

 

[AnalogKid]

No matter what the values of R1 and R2 are, a 10K / 100 ohm divider means 1% of the system voltage will be available to the LCD (after the other errors are corrected). I can't draw schematics this week, so here is a text suggestion:

Diode side:
9 V
470 ohms
anode
cathode
GND

Transistor side:
9V
1K
collector
emitter
10K
GND

3.3V zener diode and LCD in parallel with the 10K

ak

 

[matty matt tmatty]

Thanks ak, updated diagram below. Does this look ok?

 

[73's de Edd]

Sir matty matt tmatty . . . . .

Back at my post one . . . . . at that time, I had ZERO idea of WHAT was wanting to be accomplished.

I just knew that the circuit I posted would positively get you introduced to the basic circuit familiarity to let you expand upon its basics.

After seeing the video, and what he had accomplished I now need to know what exactly you are wanting to accomplish, as you show us as having a large LCD plate instead of the two individual small eye size units that the other person was using..

Therefore, basically, I see zero need for the optical shutter, as replicating his effect of turning on and off of the LCD panel eyepieces only requires an application of voltage to darken the display and elimination of that applied voltage to make the display transparent again.

WITH one little caveat, that he discovered, which is the need of a bleeder resistor shunted across the display element to SPEED up the deactivation of the blackened display, in going back to its transparent state.

When he utilized that backwards installed electrolytic, he created a bleeder resistor by virtue of the excess DC leakage thru the capacitor by that reverse polarity action.

Sooooo . . . . .we need a full fill in on your full application, in order to see if that left optical interrupter/ shutter IS EVEN NEEDED.

Another question is if your display is being CLEAR on its no power state or does there seem to be some degree of tinting.

ON HIS UNIT . . . . I know that his units incorporate 90 degree offset polarization filters that are alternatively switched on and off at a speed greater than the persistence of vision of the human eye . . . . .specifically, MUCH FASTER . . . . due to the frame refresh rate of a LCD TV display.

ALERT !
(Crystal Ball prompt #1 is now glowing and coming in to me) . . . . . If this projects end result is being related to making a fast reaction shutter for welding “goggles “ the degree of opacity and spectral wave,length suppression, in no way approaches what is being needed for that capacity.

Here is the correction on the last schematic of yours, with its actual final state being depending upon what action you are trying to complete.

SCHEMATIC REPLOT :

The 470 gets the right current to the IR emitter with a 9V supply.
The 1k value should adequate to get adequate power needs thru the bottom loop
The Zener sets / limits the end supply voltage level going to the LCD panel
The 10K speeds up the drain of quiescent retentive voltage from the LCD panel . . . after its power off.
( Experimant and lower its resistance value incrementally, if decay time is being too slow.)



73’s de Edd
.

 

[AnalogKid]

Thanks ak, updated diagram below. Does this look ok?

Nope. 73's schematic is what I was suggesting. With the optocoupler datasheet you can calculate the max current available to the LCD before the zener is starved and the output voltage sags below 3.3 V.

ak

 

[matty matt tmatty]

Thanks to you both. As you might have guessed I'm new to all this. I posted the video as it was the only reference I could find to illustrate the need to bleed the LCD glass.The application requires the LCD glass to turn on and off rapidly (like the opposite of a strobe), so there is a need to bleed! I plan to control/sync the turning of the glass on and off using a slotted disc between the optic switch.


This way, for every 30 degrees of rotation of the disc the LCD is on most of the time and off for a short flash at the end.
I hope this all makes sense! The disc will make a full rotation about once a second. I will pick up some components at the weekend and give it a try!

 

[matty matt tmatty]

Just wanted to say the circuit worked perfectly. Thanks to everyone who helped out!

 

[matty matt tmatty]

Hi!
The DC motor that runs the disc is currently operated using a model train controller box, so I can speed up and slow down the motor as required.
Would it be possible to run the motor off the same 9V battery that powers the LCD circuit? I've tried connecting the motor directly to the 9V battery and it works just fine, the motor can take up to 24V but 9V appears to be enough for my needs.
I was hoping I could just start a new circuit off the battery (parallel to the LCD circuit) that has a mosfet and 100k pot to control the speed of the motor? Like this:

Here is a link to the motor:
http://www.maplin.co.uk/p/mfa-12-24v-small-301-single-ratio-motor-gearbox-n96bn
The mosfet:
http://www.maplin.co.uk/p/2n7000-mos-fet-transistor-to92-case-uf89w
and the pot:
http://www.maplin.co.uk/p/omeg-100k-andohm-linear-single-gang-potentiometer-fw05f

 

[duke37]

The 2N7000 is a diddy transistor in a to92 case and may not be strong enough for the job.
Most fets need a considerable voltage source to gate to turn on. With a 9V supply you may not get more than 5V out to the motor. A Darlington may be better.

 

[Minder]

The 2n7000 is a logic level fet equivalent to a darlington, Siliconix originally called it a FETlington.
Here is the original spec sheet.
M.