Need help understanding how to use opto-isolator

Ok, I tried using the diode but I think my multimeter isn't fast enough to sample the data but I did see values ranging from 800mV to 1V this time (actually, I would see it ramp up from 80mV to generally topping out at 800mV, 900mV, and 1V).

What do you think? Does ~1V sound like a reasonable voltage across the buzzer? Are there any other tests I should try? I probably should invest in an analog multimeter. Is this where SWAG comes into play?

thanks again!

Reply to
mjohnson
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I was going to use an optical sensor to determine if the door was up or down so the device would be very close to the door. The garage door button is opposite from the door itself so it wouldn't be very close (13' away). Anyway, this project is just for fun. I just want to do it as an experiment to learn something from the whole process. Like I said I could go buy something but the goal isn't to solve a problem it's just for exploratory purposes.

Reply to
mjohnson

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What purpose will the output of the optical sensor serve?   That is,
how will it affect the operation of the system?
Reply to
John Fields

I think you clearly have AC, probably in the form of pulses, being applied to that buzzer transducer. Given that your meter loads that diode detector with a few hundred KOhms, ~1 V is believable.

A 0.1 uF cap across the meter would produce a value closer to what I mentioned, and provide a more accurate estimate of the peak voltage being applied across the buzzer.

At this point, if stranded on a desert island without more instruments, I would hook an optoisolator LED with a 1K series resistor, anode side, to the buzzer. Then measure the collector-to-emitter "resistance" with the Ohms function while the buzzer sounds. For one polarity of that measurement, you probably will see significant conduction when the buzzer runs. This is probably your best alarm pickoff circuit if you are still happy with the buzzer's audibility.

An oscilliscope is a good idea for the serious hobbiest.

That would be a last resort.

You're welcome.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.
Reply to
Larry Brasfield

I was thinking that I was just going to have an LED/Detector as the sensor which is coupled to the logic circuit with an AND gate. So when the detector is sensing the door in the up position and the buzzer is going off then pulse a signal to garage door remote.

Here's is an extremely high level diagram (i realize the AND function is more involved than what is represented).

+----------------+ [BUZZER]->|---\\ | | ON \\ |(activatation pulse) | AND---[555]---|-------------------->[REMOTE] | UP / oneshot | [SENSOR]->|---/ | +----------------+ LOGIC CIRCUIT
Reply to
mjohnson

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So if the door is up and the alarm clock goes off, the circuit will
pulse the remote, but if the door is down and the alarm clock goes off
the remote won't be pulsed?
Reply to
John Fields

Yup... that's the plan for phase 1. Plan for phase 2 (assuming phase 1 ever works) is to add another sensor to determine if there is actually a car (or movement?) in the garage. If there is no car (or if there is movement) then the remote won't activate.

Reply to
mjohnson

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For a solution for your phase 1 problem, click on this:

news:hc6871dqc4ir7ues829ntlac8h7p5cpepi@4ax.com
Reply to
John Fields

Wow! thanks for doing drawing up that schematic. That was very kind of you.

I do have a question, since the buzzer will be buzzing I need to "absorb" the transitions between the buzzer oscilations. Functionaly, I just need to know that the buzzer is going off. Is that the reason for the .1 uF cap between the alarm and the rest of the circuit, to absorb the oscilation and keep the voltage high for the duration of the alarm?

thanks!

ps: I have order my components and I'm waiting for them to arrive in the mail....

Reply to
mjohnson

--- No, that cap is to allow the pulses from the buzzer to get through to the + input of the first comparator while blocking any DC on the buzzer's output. With no input from the buzzer, The + input of the comparator will be pulled up to the +5V rail and will be at a higher voltage than the - input, so the output of the comparator will be quiescently high, and the 0.1µF cap on its output will get charged up to +5V through the 1 megohm resistor. Since the output of the first comparator is connected to the + input of the second comparator, the voltage on that input will be higher than the voltage on the - input, and the output of the second comparator will be quiescently high. The

100k ohm resistor on the output of the second comparator is used to pull the output up to +5V and to provide a DC return for the 0.1µF cap on the output of the comparator. the other 100k ohm resistor is used to pull the trigger input of the 7555 high and to provide a DC return for the cap. The 1 megohm resistor and 2µF cap are used to generate the ~ 2s wide output pulse from the 7555, and the transistor is used to turn on the relay when the output from the 7555 goes high.

How it works is that nothing happens until the alarm goes off, then when it does a low going edge at the output of the 0.1µF cap connected to the buzzer will pull the + input of the first comparator lower than the voltage on the - input, forcing the comparator output low. When the comparator's output goes low, it immediately discharges the 0.1µF cap on its output, pulling the output of the second comparator low. When that happens, the 7555's trigger input will be pulled low for about ten milliseconds, causing the 7555's output to go high and start the two second timeout period.

Now, as the pulse input from the buzzer continues to cause the + input of the first comparator to vary about the switching threshold, (the voltage on the - input) the output of the first comparator will alternately discharge the 0.1µF cap and then allow it to be recharged through the 1 megohm resistor. While it discharges the capacitor almost immediately, it will take much longer to charge back up to the switching threshold through the 1 megohm resistor, so as long as the buzzer is buzzing, the output of the second comparator will remain low and only one 2s pulse will be issued by the 7555.

When the buzzer stops buzzing, the + input of the first comparator will go high, the cap on the output of the first comparator will start charging, and when it gets to about 4.5V the output of the second comparator will go high, sending a positive pulse into the trigger input of the 7555. This positive pulse will be clamped to the 5V rail by the internal static protection diodes of the 7555 and, once that happens, the circuit will be armed and will trigger with the next output from the buzzer.

-- John Fields Professional Circuit Designer

Reply to
John Fields

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