14M2 time-lapse controller

parto

Member
I am making a long-term time-lapse camera controller, using as little power as possible--it needs to work on battery power for months.

I have the 14M2 sleeping until a real-time clock (DS3231) sends an alarm output to one of the 14M2's input pins. The 14M2 wakes up and using one of its pins switches on a mechanical latching relay. This provides power (11.1 V stepped down to 8.5 V) to the camera, which then turns on by itself. The camera is powered on for one minute to charge its internal supercapacitor (a coin "battery" which holds the date and settings and usually is charged with the camera's removable battery, which is not used in this setup). Then using a second pin and an opto-coupler (I read that this is better than a transistor since the two power systems are more isolated), the camera's remote shutter circuit is closed so an image is taken. The camera power is held on for another minute and then shut off. Then the 14M2 writes the new alarm time to the RTC. Then it goes back to sleep.

I see that the 14M2 project board kit I have (the AXE117) has an optional Darlington driver buffer chip that "enables you to connect higher power output devices (such as motors, solenoids, relays) directly to the board (current up to 500mA)". I am way out of my familiar territory here, but I am wondering whether I can use the Darlington driver buffer chip in place of the latching relay and if so, what would be the current draw through the AXE117 while the switch is held closed?
 
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Buzby

Senior Member
A non-latching relay needs power all the time it is 'on', and you need it 'on' for two minutes. It's obvious that a latching relay will not use anywhere near as much power, as it only needs a short pulse to turn it 'on' or 'off'.

But both of these power consumptions are probably a magnitude less than that used by the camera for two minutes.

Also, how is 11v reduced to 8.5v ?. If it's via a linear regulator then there is a huge power loss there.

The AXE117 power consumption I think would be insignificant compared to the relay, regulator, and camera.

Can you post a circuit diagram, then our resident 'flea power' experts will suggest solutions ?

Cheers,

Buzby
 

parto

Member
Hi Buzby, thanks for the reply. The converter is a buck converter--it's used only when the power is switched on for the two minutes. Yes, I was guessing that the power consumption for the AXE117 would not be much of a consideration--unless it draws power while the 14M2 sleeps. I don't know how to draw a circuit diagram but I can try. The main thing I am wondering is whether I can use the Darlington thing in place of the relay. And maybe in place of the opto-coupler too?
 

inglewoodpete

Senior Member
I don't know how to draw a circuit diagram but I can try.
Pen and paper is good enough if you do not have a software drawing package. Scan or photograph the drawing and post the picture.

When drawing the circuit, try to keep the 0v line at the bottom and the +ve supply at the top. Convention also has the input circuitry to the left and outputs to the right.
 

hippy

Technical Support
Staff member
I am wondering whether I can use the Darlington driver buffer chip in place of the latching relay and if so, what would be the current draw through the AXE117 while the switch is held closed?
I would guess it should work. One could possibly parallel Darlington outputs up to switch power to the camera if 500mA is not enough. It would be low-side switching, switching the 0V (-Ve) to the camera but shouldn't be a problem if the shutter connection is opto-isolated.

There is also some voltage drop across the Darlington transistors. That may be a problem or may prove advantageous.

The current drawn, and needed to be switched, would depend on what the camera uses when turned on, while charging its super-cap, and taking a shot. It might be best to measure that with a current meter in the relay operated circuit though it may be hard to get a reading of initial surge currents.

Each Darlington transistor can sink 500mA individually, but cannot support all transistors each sinking 500mA at the same time so you can't switch 8x500mA=4A. I cannot off-hand recall what the limit is.
 
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inglewoodpete

Senior Member
Each Darlington transistor can sink 500mA individually, but cannot support all transistors each sinking 500mA at the same time so you can't switch 8x500mA=4A. I cannot off-hand recall what the limit is.
Roughly (at temp rise of 70C!!!) you can get the following currents out of a ULN2803:
1 500mA (1 x 500mA)
2 800mA (2 x 400mA)
3 900mA (3 x 300mA)
4 1000mA (4 x 250mA)
5 1050mA (5 x 210mA)
6 1080mA (6 x 180mA)
7 1150mA (7 x 165mA)
8 1200mA (8 x 150mA)
Note that you'd have to switch all the appropriate pins on or off with the same command. Eg Let pinsC=#11100000

I'd be using discreet Darlington transistors or a half decent MOSFET rather than many Darlingtons from the same chip.
 

parto

Member
Thanks for the replies! I am still learning more about the suggestions for the Darlington chip; I am not finding much for practical-use examples. My initial question is, I think it can control voltages other than the 4.5 volts that the microcontroller uses? So, the 11.1 volts for the camera is routed through the Darlington chip?

As for the MOSFET, I would really like to know more about an option like that. I need to keep the camera power on for two minutes at a time (16 times a day). I think the camera uses about 200 ma--that's the highest current of the system. Then I need to snap the shutter with a second switch. A third switch applies power to the clock during the time the 14M2 communicates with it. If the latching relay uses less power overall than the MOSFET, I would use the relay, even though it takes up more space and requires more messy wires.

One problem I had with the opto-isolator as the camera switch is that the camera requires specific resistances for idle and for shutter snap. With the opto-isolator in the circuit the resistance seemed to vary even when I changed the range of the multimeter. I understand the latching relay--the resistance doesn't change. So to get the project done for now, I am using the latching relays for each switch. In using a transistor as a switch, I am concerned that I am getting myself into unknown territory. I like to save the space, but I need to cut down on the learning time and get the project done. If there's something straightforward I'll take a look though!
 

westaust55

Moderator
The ULN2803 Darlington transistor array is a set of 8 low side switches.
Each channel is in effect open circuit when off and connects the output pin to ground when active.

So for your 11.1 V supply then sequence would need to be.

11.1V battery - voltage regulator — camera.
Then the ground of the voltage regulator and the ground of the camera to a drarlington output pins.

You will lose around 1.2 volts across the Darlington transistors.

If that volt drop is too much then consider a DMOS alternative chip. See details in my 2017 post here:
https://picaxeforum.co.uk/threads/uln2803a-udn2981a-drop-in-dmos-alternatives-to-consider.29775/
Those devices will halve the volt drop between output pin and common ground.
 

parto

Member
Thank you westaust55. The voltage regulator is adjustable and I need to end up with 8.5 V for the camera--so it seems that could make up for the voltage drop. The grounds for both the voltage regulator and camera go to a single Darlington output pin? Also, I wonder what would be the power loss considerations: Power through the Picaxe/Darlington to keep the pin connected to ground, and power leakage when the circuit is open (when the Picaxe chip is asleep and waiting for the next alarm interrupt). Between shots and for a month before the shots begin the system needs to retain as much power as possible in the both the Picaxe and camera batteries. Then the camera shots go on for four months.
 
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