Continuing with MB2's thought, how about connecting the speaker leads to the input of a optoisolator, such as as a 4N25.
It has an open collector output, which should be able to directly trigger the camera.
Something like below:
(LED D1 gives a visual indication to help adjust the volume setting).
View attachment 161891

Thanks for providing just the circuit I was thinking about. My one concern is the camera trigger requirement versus the optoisolator's capabilities. And the resistor may not be needed, or possibly have a different value? some development work will be required.

 

My one concern is the camera trigger requirement versus the optoisolator's capabilities.

I think it's probably adequate but that would need to be determined.

And the resistor may not be needed

It's there mainly to avoid overdriving and blowing the LED input.
It's value may have to be adjusted to get the desired sensitivity.

 

Continuing with MB2's thought, how about connecting the speaker leads to the input of a optoisolator, such as as a 4N25.
It has an open collector output, which should be able to directly trigger the camera.
Something like below:
(LED D1 gives a visual indication to help adjust the volume setting).
View attachment 161892

Just a dumb question. Isn't the speaker output an AC signal?

 

I appreciate all this information and am even understanding some of it. I think I may have to go back to college to built this, though. I am NOT an EE.

In the circuit diagram in this image, https://www.digikey.com/product-detail/en/lite-on-inc/4N25/160-1300-5-ND/385762, excluding the LED addition, are you essentially saying that the speaker wires would connect at connectors 1 and 2 and the camera trigger at 6 and 5? Could it be that simple?

 

Isn't the speaker output an AC signal?

Of course.
Why do you ask?

excluding the LED addition, are you essentially saying that the speaker wires would connect at connectors 1 and 2 and the camera trigger at 6 and 5?

Close.
The camera trigger connection would be pins 4 and 5 (the polarity of the camera input would need to be checked so that its plus voltage goes to pin 5).
No connection to pin 6.
You also need the input resistor on pin 1 to connect to the speaker wires.

 

That seems simple enough. I'll do some checking on the camera to see if I can determine the polarity. If I can't figure it out, they may be some web posts I can research that has the needed information. I'll order some parts and see if I can make the thing work. Is there more I need to know about the resistor? I assume this needs to be 250 ohm. What wattage range should I be using?

Thank you for your help. I will get back once I've made some progress... or attemtped it.

 

Hello,

If you can not determine the polarity of the camera, you could use a H11F1 in stead of the 4N25.
That is also an optocoupler, but it has a FET as output.
The fet will work in the range upto ± 15 Volts.

Bertus

 

I'll do some checking on the camera to see if I can determine the polarity

Just measure it with a multimeter.
If you don't have one, just buy a cheap one.
I often use one I got free at Harbor Freight.

Is there more I need to know about the resistor? I assume this needs to be 250 ohm. What wattage range should I be using?

The value is somewhat arbitrary and may need to be changed to get the sensitivity you need, but around 200-300Ω should be a good value to start with.

The wattage can be any since it will be dissipating very little power, and only during the peak static noise.

 

Of course.
Why do you ask?

How does a 339 or such work when the input is AC? Thought putting an negative voltage on their input was a no-no?

 

If this is to be a "dedicated" thing, wouldn't it be better to take the triggering signal directly from the input of a AM receiver instead of the speaker output?

Did a quick search and seems like there are some 'lightning detectors' that aren't too expensive to buy, and might be a good place to start this project from. https://www.amazon.com/AcuRite-0202...ocphy=9015428&hvtargid=pla-313019908742&psc=1

And then saw this, "The lightning-radio frequencies are much lower than the frequencies used for the radio you listen to in your car; they range from 3 kHz to 30 kHz, when AM radio broadcasts at frequencies between 540 kHz to 1.6 MHz." from, https://www.popsci.com/article/diy/listen-lightning-strikes-live-citizen-scientist-map So AM doesn't seem to even be the correct frequency.

 

Yes, the really strong signals from lightning are at the lower frequencies, BUt there is plenty of energy at the higher frequencies and there is the double advantage of being able to select a specific, other wise quiet frequency, and having a receiver easl available and fairly cheap. AND, those receivers for the much lower frequencies also generally need a big antenna. Big antennas for low frequencies are seldom portable or cheap.

 

How does a 339 or such work when the input is AC? Thought putting an negative voltage on their input was a no-no?

Yes, it's a no-no.
So you just use a diode-resistor to clamp the negative voltages to near ground.

 

wouldn't it be better to take the triggering signal directly from the input of a AM receiver instead of the speaker output?

Why would it be better?
The speaker output has enough power to directly drive the LED of an opto-isolator.

 

Adding to the response in post#33, the input signal to the AM receiver is in the tens of MICROvolts, and from a very high effective impedance source, while the output that would be driving the LED device is a few VOLTS, from a very low impedance source, just a very few ohms.
About a million times more voltage from a much lower impedance source. Thus the output needs no additional circuitry, really, except to protect the LED from overvoltage.

 

Why would it be better?
The speaker output has enough power to directly drive the LED of an opto-isolator.

I guess I didn't describe my thoughts correct. I meant to take the signal from the AM detector circuit, before amplification to the speaker.

 

I guess I didn't describe my thoughts correct. I meant to take the signal from the AM detector circuit, before amplification to the speaker.

I understood what you said.
But you didn't answer my question as to why that would be better.

 

I guess I didn't describe my thoughts correct. I meant to take the signal from the AM detector circuit, before amplification to the speaker.

The power level at the detector in most rece3ivers is not great enough to drive a relay coil quickly, even a sensitive reed relay. Aside from that, it would require digging into the receiver package and making connections to a somewhat delicate section of the receiver. In addition, what benefit do you imagine would be gained by tapping in at the detector? That is what I want to know. I see no benefit at all.

 

In addition, what benefit do you imagine would be gained by tapping in at the detector?

I understood what you said.
But you didn't answer my question as to why that would be better.

My thinking was that it would both make the total item smaller and not be an annoyance to be around(not having to hear the sound). But guess I was wrong as usual.

 

not be an annoyance to be around(not having to hear the sound).

If that's a problem you can disconnect the speaker or replace it with a dummy resistive load.

 

If that's a problem you can disconnect the speaker or replace it with a dummy resistive load.

My thoughts were he wanted this in a small package that was connected with the camera. Like the ones you can buy.