I presently run an 8 camera system as follows:
Camera power supply
Power is provided, for all camera, by an ebay purchased 20A/5V psu. The psu voltage can be varied by +/- 10% (4.5V - 5.5V), I have it turned down to 4.5V.
DC V+ of this psu is connected to a blade fuse distribution board consisting 8 number 2A fuses. A power diode is then connected to each terminal of the blades before connection to the positive supply terminal at the camera (V_Batt+).
Voltage across each camera (V_BATT+/V_BATT-) is therefore approximately 3.8V (they would normally each operate with 2 AA batteries but I have had no issue at this higher voltage).
Camera Focus/Shutter release mechanism
Using a spare 20A/5V psu (which is turned up to 5.5V), camera autofocus is achieved on all cameras by applying 5.5V to pin 1 terminal of their mini female usb port (USB_P1).
I presently do this by way of a mechanical switch (SPST) connected to DC V+ of this spare psu. The output of the switch then fans out and is individually connected to each camera USB_P1.
When the switch is closed a small current flows into USB_P1 and exits at V_BATT-.
DC V- of both psu's (camera power and switch) are shorted with thick wire.
With the switch closed and 5.5V now applied between USB pin 1 and V BATT-, current flowing into USB pin 1 is measured (via DMM) as 0.42mA for each camera (3.36 mA total)
I have therefore modelled this high resistance switch path for each cameras as a
R = 5.5 / 0.42^10-3 = 13.1 KΩ resistor
With 8 cameras the equivalent parallel resistance is modelled as a 1.6375 KΩ resistor
Subsequently, opening the SPST switch causes the shutter mechanism to operate and exposure occurs.
Requirement (if possible)
I'd like to automate this switching action.
I believe the situation described requires High side switching. I imagine a logic level P channel MOSFET, operating in linear "triode" region, connected to Arduino digital i/o (with suitable current limiting resistor) to replace the mechanical switch.
has an RDS(ON) stated as typical 1.3Ω and maximum 1.5Ω (VGS = -5 V, ID = -300 mA).
Assuming, for now only, that I can operate the Arduino from this same 5.5 V supply already described, there is no IDS/VDS characteristic shown for VGS = -5.5V.
Is it safe to assume that this MOSFET will operate with VGS = -5.5V and that the RDS(ON) at that level will be better (i.e. lower) than 1.5 Ω ?
Now, if I do understand that correctly, assuming no worse than 1.5Ω operation and 8 camera switching
total series resistance is (1637.5 + 1.5) = 1639 Ω
I(tot) = 5.5/1639 = 3.356 mA
VGS = 3.356 mA x 1.5 = 5.034 mV
V USB_P1/V_BATT- is 5.5 V - 5.034 mV = 5.4949 V
Are my calculations a correct interpretation of the intersection of the load line with the IDS/VDS characteristic?
Now, with only a single camera turned on the same calculations are:
total series resistance is (13100 + 1.5) = 13101.5 Ω
I(tot) = 5.5/13101.5 = 0.42 mA
VGS = 0.42 mA x 1.5 = 0.63 mV
V USB_P1/V_BATT- is 5.5 V - 0.63 mV = 5.499 V
Is that also a correct interpretation of the intersection of the load line with the IDS/VDS characteristic?
I think anything higher than 5.4 V across V USB_P1/V_BATT will operate the camera focus and shutter action correctly.
However, the following is stated in the Characteristics for
VGS Gate threshold voltage
Min = -0.5 V, Typ = -0.7 V, Max = -1.0 V Conditions VGS = VDS, ID = -1.0mA
Zero gate voltage drain current
Max = -1 mA Conditions VDS = 0.8 MAX RATING, VGS = 0V, TA = 125C
Do these two parameters mean the switch won't work as I expect for the case when only 1 camera is powered up and requires switching?
If so, is there another MOSFET better suited to my application?
Any comment appreciated.