i'm using an arduino to interface with a canon a650 point and shoot digital camera. i currently have the arduino turning on and off the camera but it cannot determine if the camera is currently on or off.
i could solder in a couple of leads to the power light on the camera but i'd prefer not to modify the camera.
i have the camera on an ac power supply. is there a way i could examine the dc power being supplied to the camera to determine if it is on or off?
Could you glue a phototransistor over the camera's "power on" light? Or some other LED that lights up when it's on?
i'm exploring that idea as well but i don't like the idea of affixing something to the camera like that - could be a point of failure.
My approach would be to cut one of the power supply wires and use a current sensor in series with this to measure how much current was being drawn. There should be a measurable difference between when the camera is on and when it's not.
i've determined -
the camera draws 0 amps when off
the camera draws .07 amps, minimum, when on
if i put an LED in line on the power supply, pressing the camera power button turns on the LED but not the camera. (if i can do an inline led then i can definitely hook that up to my arduino)
does anyone know what it would take to have an inline LED in the power cable?
What kind of voltages are we talking about at the power supply? DC? AC?
4.3v dc
This could be a good starting point...might require some tweaking:
very interesting. i haven't used a comparator before so i'm trying to figure out how this works. the comparator is comparing two currents which will determine if the LED turns on. somehow the different sets of resistors establish a current level for pin 2 which will be compared to pin 3 determining if current will flow through pin 6 allowing the led to light.
hopefully someone else can add a better explanation.
You may be able to use a much simpler solution. If you connect a one ohm resistor in series with the camera power supply then the voltage across the resistor will be 0 when the camera is off and 70 millivolts when on. 70 milllivolts will produce an analog reading of around 14 using analogRead.
So, if you connect one end of the resistor to ground and the other end to both the negative supply to the camera and to an analog input pin, you can tell if current is flowing by checking the value of analogRead. If the value is 10 or more then the camera is drawing at least 50 ma is therefore on; less than this value and the camera is off.
Yes there is a simpler solution, but in the sprit of understanding that circuit:-
the comparator is comparing two currents
No the comparator compares two voltages not currents. It is really a differential amplifier which has a very high gain. A differential amplifier amplifies the difference between the two inputs. If one input has a higher voltage than another the output is so big it hits the rail (supply).
somehow the different sets of resistors establish a current level for pin 2
Remembering it is a voltage not a current the some how is a simple potential divider.
R6 develops a voltage across it, the more current that flows the bigger is this voltage. It is the increase in this voltage that feeds the potential divider R2 & R3 less voltage than it had before so making the voltage at pin 2 smaller.
This is the negative input so when this negative input is more negative than pin 3 the output (pin 6) goes negative to the ground. This allows current to flow from the LED thus lighting it up.
it works! have to use a 1 Ohm resistor, even 2.2 will prevent the camera from booting.
my symbol for the power supply is incorrect, as thats the plug end.