building usb-remote-cable

[PhyrePhoX]

okay, i played around a bit, managed to use a "remote switch" on both my a620 and s3is - it switches ON when it gets dark. i plan to use it in conjunction with a properly attached laser beam, so when the beam is interrupted, a shot will be taken. for highspeed photography, for example when you drop an egg, on its way down it interrups the laser beam.

i bought an extremely cheap solderless "experimenting board for kids".

however, i think i'm running 9V through my camera and i dont know why and how. attached are both the "circuit" plan from the book and the "real setup". i replaced the red diode (1.8 V) with two USB connectors, and then i found out that the whole 9V are on the contacts. so i replaced the 1K resistor with a 1M resistor: voltage drops down to around 4,5 - but that isnt enough for the cams to detect. so i replaced the 1M resistor with two 470k resistors (as in the picture), that triggers the cams but when i detect the voltage with voltmeter i still get 9V. either i'm dumb (long time ago i had physics in school about the exact same matter but i forgot...) or plain blind.
question: is it me using my voltmeter the wrong way or is there really 9V getting to my cams? what should i do to downsize that to 3V (dont wanna risk breaking something!).
the ldr is a light-dependent resistor (much light - 100 ohm, no light 1M ohm). the transistor is the "standard" transistor BC547.

maybe some of you more experienced users can shed light on this

edit: i know that the usb detection routine isnt that fast to detect a falling egg. but maybe i have to drop it from a much greater height

[jetzt]

I didn't check the schematics so far, but did you first connect the tansitor the wrong way?
Maybe it's gone than, but that's really only a "quick reply".

Edit1: How did you mesure the voltage, with camera connected?

Otherwise the full current has to go through the voltmeter and that's not how it should be?
I don't expect you mesured like that, but did you measure it like that?(pic1)
In that pic1 the value the voltmeter shows is only correct if the cam is connected.

(Remember your physic-edu-teacher: "You have to measure the voultage drop over an Resistor/consumer." )

However I would enjoy reading your results, and so I propose to put a potentiometer in to correct the daylight.
Like that you could adjust sensitivity.(pic2)

[GrAnd]

i replaced the 1K resistor with a 1M resistor: voltage drops down to around 4,5 - but that isnt enough for the cams to detect. so i replaced the 1M resistor with two 470k resistors (as in the picture), that triggers the cams but when i detect the voltage with voltmeter i still get 9V. either i'm dumb (long time ago i had physics in school about the exact same matter but i forgot...) or plain blind.

Hm... I'm not a hardware guy, but... If I'm not not mistaken, the using of a resistor leads to the limiting of the current. For voltage limiting the voltage divider is used. Or use a cascade of 6 diodes to reduce the voltage by 6x0.7=4.2V

[databoy]

Regarding cir.jpg.

The circuit you are using is just a simple light sensor circuit. Reduce the voltage to a six volt battery and reduce the 100k resistor to about 75k. The circuit is switching a led on or off. Do you really need the led in the circuit. If not, remove the led and put a 4.7 volt zener diode in its place. Increase the 1k resistor to a 5k resistor maybe 1 watt.

FOOTNOTE:   

If you want to reduce the voltage in USB cable circuits; it is best to use a series resistor in conjunction with a zener diode. You will need one below 5 volts. A 5 volt regulator is not the ideal solution. They need tantalum capacitors either side of the voltage legs as close as possible to the regulator. Otherwise the regulators are know to become unstable.

This page will show you the range of zener diodes available:

Jaycar Electronics

[PhyrePhoX]

I didn't check the schematics so far, but did you first connect the tansitor the wrong way?
Maybe it's gone than, but that's really only a "quick reply".

Edit1: How did you mesure the voltage, with camera connected?

Otherwise the full current has to go through the voltmeter and that's not how it should be?
I don't expect you mesured like that, but did you measure it like that?(pic1)
In that pic1 the value the voltmeter shows is only correct if the cam is connected.

(Remember your physic-edu-teacher: "You have to measure the voultage drop over an Resistor/consumer." )

However I would enjoy reading your results, and so I propose to put a potentiometer in to correct the daylight.
Like that you could adjust sensitivity.(pic2)

yeah i think i used the voltmeter the wrong way, i'm not used to this "electronics stuff". however i think that i had the camera connected. on the other hand, the camera itself isnt really a "resistor" or consumer, is it?
yeah, a potentiometer will definitly help, there just wasnt one included in the "starter kit".

i replaced the 1K resistor with a 1M resistor: voltage drops down to around 4,5 - but that isnt enough for the cams to detect. so i replaced the 1M resistor with two 470k resistors (as in the picture), that triggers the cams but when i detect the voltage with voltmeter i still get 9V. either i'm dumb (long time ago i had physics in school about the exact same matter but i forgot...) or plain blind.

Hm... I'm not a hardware guy, but... If I'm not not mistaken, the using of a resistor leads to the limiting of the current. For voltage limiting the voltage divider is used. Or use a cascade of 6 diodes to reduce the voltage by 6x0.7=4.2V

yes, you're probably right. will try to build something usable tonight

Regarding cir.jpg.

The circuit you are using is just a simple light sensor circuit. Reduce the voltage to a six volt battery and reduce the 100k resistor to about 75k. The circuit is switching a led on or off. Do you really need the led in the circuit. If not, remove the led and put a 4.7 volt zener diode in its place. Increase the 1k resistor to a 5k resistor maybe 1 watt.

FOOTNOTE:   

If you want to reduce the voltage in USB cable circuits; it is best to use a series resistor in conjunction with a zener diode. You will need one below 5 volts. A 5 volt regulator is not the ideal solution. They need tantalum capacitors either side of the voltage legs as close as possible to the regulator. Otherwise the regulators are know to become unstable.

This page will show you the range of zener diodes available:

Jaycar Electronics

i dont need the led, i replaced it with the usb connector. i do have a zener diode somewhere in the kit, will check it out, thank yoU!!!.
one question remains: i connected TWO usb connectors to it, but will not always actually USE two cameras. if i manage to get 3 volts into the "usb circuit" - it shouldnt matter how many usb connectors i "solder to it", right? i'm planning to actually connect as much as 4 or 5 cameras to it.

thanks for your replies guys, really helpful!

[databoy]

Technically the camera is detecting voltage not current. It should not matter how many cameras are connected. If you want to run multiple cameras; I would suggest that you use a larger current zener diode, 2 watt resistor and a higher power transistor.

Make sure the zener you put across the USB socket is over 3 volts and less than 5 volts.

Have a look at the Jaycar site; it will give you an idea of the voltage and current rating of the zeners. You are not looking at a big cost to upgrade to multiple sockets.

[a640]

is it possible to mod (usb) the phottix wireless remote since its small and cheap??

[databoy]

I do not know the brand name. Technically it is possible to mod any cheap remote doorbell. I found that the cheap doorbells contain a small board with an all in one receiver and chime chip. I could not find any technical details on the net for the chip. It is definitely an OEM manufactured for doorbell chime operation. Put a digital voltmeter across the speaker terminals and press the button. Record the voltage. Unsolder one wire from the speaker and measure the voltage. As long as the unit is under 5 volt operation you should have no problems. Be aware that most door bells/chimes have a chime which last from 5 to 8 seconds. For most purposes I found that it works ok. I used a scope to observe the waveform. The high end of the pulse waveform is enough to trigger the camera. I found that with the speaker in circuit the voltage is too low to trigger the camera.

The Kambrook unit is about $16 AUD. It is a 3 volt unit the size of a cigarette packet designed to clip on a belt. I found that the Australian 240 volt mains units were more trouble than they are worth. Too much work involved.

In Australia, Jaycar sell small remote novelty chimes. Very compact and cheap. Google Jaycar; search for novelty chime.   
   

[databoy]

For the hardware experimenters.

In Australia the motor vehicle alarms use all black wires with no number coding. The wire coding is at the end of the cables. It is chopped off by the installer as he connects the unit. This is for security reasons, so an intruder cannot immobilise the vehicle.

Within the alarm / immobiliser is contained a small daughter board with an RF receiver. The unit operates on a regulated 5 volt supply and generates a good square wave signal.

The wrecking industry cannot resell the units because without the connector coding they are useless. As one installer said it would cost more to trace the wiring than sell you a new unit.

If you try the immobiliser approach make certain you obtain the remote transmitter.

 

[photoscott]

9-Volts, not a good idea. Use some stacked button-cells or something else. Even 4 AAA NiMH batteries would be better, that will average at 4.8 to 5.2 volts total.

Wouldn't you only need 3 AAA batteries at 1.5v each?  I like Coutts idea. I just purchased the Canon RS60 remote trigger. I am going to try first plugging it in to the 3.5mm jack first to see if it works that way. I know the 3.5mm jack operates the remote capture  on the DiGiC II cameras from Canon. I saw a cell phone ear bud set's button to be the remote button. Remote shutter trigger for Digital Cameras - Instructables - DIY, How To, photography, tech
So, if the 3.5mm jack doesn't work then I will get the 3AAA battery holder from Radio Shack for $1.79 and wire and install it on the back of the housing of the RS60. Basically the same thing as Coutts design other than I would be using 3 AAA batteries and an actual Canon remote shutter release button.
I do know that when I plugged a 3.5mm tip into the 3.5mm jack on my A570IS that it does cause the screen to go blank. I am assuming because there is nothing to regulate the opening or closing of the circuit that is why the screen went blank(black).  So that tells me something is being affected when the 3.5mm jack is plugged into.  That is why I hope maybe the 3.5mm jack will work. I know that in the scripts menu for the CHDK that you can enable the remote capture via USB. I wonder if in DiGiC III's design that default remote input is the 3.5mm jack??? I will find out and post my results soon.
Thanks for this forum you guys.

[photoscott]

I am doing something similar to Coutts design. I bought the Canon RS60 remote shutter button for $20 online. It has the 3.5mm tip on it. I am going to try this in the 3.5mm jack first to see what it does. I plugged in the 3.5mm jack from my VoIP headphones into my A570 IS. No button to push, just a mute switch. I am assuming because the 3.5mm tip wasn't attached to a momentary switch it didn't do anything except black the screen out. I know that Canon's DiGiC II design let's you do a remote capture through it's 3.5mm jack. I saw a design that was made from a cell phone hands free ear bud piece.    Remote shutter trigger for Digital Cameras - Instructables - DIY, How To, photography, tech    The main component was the momentary switch in the mic area. So I will be trying the 3.5mm jack first to see what that does. If no success, I will buy a 3AAA battery holder from Radio Shack and wire to and on the RS60. From what I understand, a AAA battery is 1.5v.  If you do three(and the math) that is 4.5v. That seems to be enough according to all the people out there doing similar things.
I like the wireless garage remote one too! Though I already ordered the RS60 online. Same cost as the wireless design except the RS60 allows you to lock the button "on" for long exposure shots.
Thanks for this forum and to all who have contributed to it. 

[Barney Fife]

Deleted

[databoy]

Quite honestly people have purchased an expensive camera then try build a USB interface on the cheap.

I will add another word of caution. I would not trust the electronics guides on a large number of sites on the net. Most of the posters have very little electronics experience.

If you want a good starting point ePanorama.net
Be prepared to do a lot of reading and bread boarding.

A number of posts advocate using a 5 volt regulator. That approach will not work on the USB cable. You need a maximum voltage of a 4.7 zener diode across the + and - lines with a series resistor from the +ve feed. Your power supply will need to be regulated if it is above 5 volts. Voltage regulators need a minimum of 2 volts above the regulator output otherwise they will not regulate.

The simplest and cheapest approach is to buy a remote doorbell which operates on 3 volts.
   

[photoscott]

It's not too far-fetched that someone, someday, will find a way to tap into the microphone side of these cameras, and include things like a recording-volume meter, or ... more importantly, sound-triggered shutter capabilities.

Yeah like adding a condensed version of ProTools for portable, recordable media!

[photoscott]

It's not too far-fetched that someone, someday, will find a way to tap into the microphone side of these cameras, and include things like a recording-volume meter, or ... more importantly, sound-triggered shutter capabilities.

Yes, like how about adding a script that would enable recording with an external mic to get rid of all the noise generated from handling the camera or even to be able to record in stereo, not to mention mic placement which is key to a good audio recording.

Scott