Circuit for 2 coil latching relay

Thread Starter

carto

Joined Feb 3, 2014
22
Hi, I am using a 5-V 2-coil bistable latching relay (the RT424F05) with a ATmega328P (at 2.5-5.5 volts) chip to switch 12-V power on and off once an hour. I think that between the chip and the relay I need to have a transistor and a diode, but I am having trouble figuring out the specifications required for those, and where to put them in the circuit. Can someone help with this? If more information is needed please let me know. Thank you!
 

AlbertHall

Joined Jun 4, 2014
9,873
You will need two transistors (suggest 2N2222A with 470 Ω base resistor) and two diodes plus 2 output pins from the ATmega to drive that relay, one set for each coil.
 

Thread Starter

carto

Joined Feb 3, 2014
22
You will need two transistors (suggest 2N2222A with 470 Ω base resistor) and two diodes plus 2 output pins from the ATmega to drive that relay, one set for each coil.
OK, Thanks very much! I am a beginner at this; what about where to put what?
 

Thread Starter

carto

Joined Feb 3, 2014
22
Have you looked at logic level Mosfets?
2n7000 etc.
Max.
MHR, thank you for the suggestion. I will look into those. My application requires extremely low power consumption of the system. The chip will sleep for 1 month, then wake up only every 90 minutes to turn on the 12 V power for 2 minutes, and then back to sleep--and continue this for 4 months--all on three AAA batteries. The latching relay uses just a bit of power each time it's switched--I'll compare that with what the logic-level Mosfets can do. If they would work, they would save some space!
 

Thread Starter

carto

Joined Feb 3, 2014
22
MHR, maybe I am confused--were the logic-level MOSFETs for switching the 12 V, or for controlling the relay? I really don't know much about this.
 

MaxHeadRoom

Joined Jul 18, 2013
20,700
MHR, maybe I am confused--were the logic-level MOSFETs for switching the 12 V, or for controlling the relay? I really don't know much about this.
Just a suggestion to use them in place of a Bi-polar version especially where you use a high impedance signal to switch a relay etc.
Max.
 

rlp123

Joined Apr 7, 2009
10
"wake up only every 90 minutes to turn on the 12 V power for 2 minutes, "

Let me get this straight. You say

Sleep 1 month (not a relay issue...)
Wakeup
activate relay coil 1 (set)
apply 12v to load for 2 min
activate relay coil 2 (reset) remove 12V
wait 88 minutes
activate relay coil 1 (set) again
apply 12v to load for 2 min
activate relay coil 2 (reset) remove 12V
wait 88 minutes
turn on again and continue this cycle 24hrs a day for the next three months

Is this your intention? Also, what is the 12V load being driven for 2 minute increments?
Robert
 

Thread Starter

carto

Joined Feb 3, 2014
22
Hi Robert, Thank you for the reply. Yes, you have it right. And the 12V load is a camera. The specs say max 2.8 W.
 

rlp123

Joined Apr 7, 2009
10
It seems you wouldn't have a real startup surge current for the camera (& drive platform?) You don't really need that 8A relay. You could probably get by just fine with TXS2-L2-4.5V Panasonic relay. It's a couple bucks cheaper than the RT424F05 and takes a whole lot less current (15.6mA). In fact, if you were courageous, you could possibly get by without a driver transistor since the atmega can sink about 20mA per I/O pin with a drop (VOL) of about 0.5V with Vcc 4.5V. That leaves 4V for the coil operation and the coil has a required must operate voltage of 3.6V. Current would have increased to operate the coil but probably not exceed 20mA. In this case a LOW out turns the relay coil on and a HI out removes drive from the coil. You must of course still have the 2 back diodes across the 2 relay coils but you could save two transistors and two or 4 resistors and the board space. Only by testing the circuit would you verify Panasonic's quality control testing of it's relays.... At any rate, if you want to be safe and use the driver transistors or skip them and direct drive the relay, you are still only consuming about 1/7th of the current necessary to acuate the RT424F05 relay if you switch to the Panasonic relay.

Robert
 

rlp123

Joined Apr 7, 2009
10
One more note: I don't know what your continuous current draw might be for your circuit but to maximize battery life and retain operating voltage for the relay, you would definitely be better off with the driver transistors for the relay coils as long as you pick a device with a low saturation voltage. A good one for this would be the ON Semiconductor KSC945YTA . It's only 20 cents for one at Mouser. The saturation voltage would be around 0.1V with Ic=20mA. That lets your battery pack drop to 3.7V (1.23V per cell) and still operate the relay.
Robert
 

AlbertHall

Joined Jun 4, 2014
9,873
For that kind of useage you ditch the relay completely and use the ATMega output to drive a MOSFET directly. For the load current you have you could use a 2N7000.
 

rlp123

Joined Apr 7, 2009
10
Al

That would certainly be the physically smallest and least cost method if you change carto's approach entirely. Not knowing what else is in his circuit and whether or not he has done any battery life analysis, leaving the atmega alive for a couple minutes every 90 minutes versus a 20mS pulse twice every 90 minutes , may or may not adversely affect his desired battery life running for at least 90 days. That's an unknown when no other information is available to me. However, I would strongly opt for your method if there is insignificant change in battery life. Nice call.
Robert
 

Thread Starter

carto

Joined Feb 3, 2014
22
I was thinking that powering the MOSFET all the time would take more power overall than would activating the the latching relay. I guess that's where the capacitor idea comes in--there would be no current flow through the MOSFET between wake-ups? I need the system to be as reliable as the relay approach, but am all for saving space and simplifying. I'll look into these suggestions; thanks!
 
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