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Help with latching relay driver circuit

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rainman1

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The present design that i use in order to switch my single coil latching relay is:
The two 5Vs come from a single 5V-regulator, and Vcontrol-1,Vcontrol-2 come from two 3.3V GPIO legs of MCU.
Each time a different GPIO leg is turned on (while the other GPIO is off), the relay is switched.
**broken link removed**
The relay has 5V, 1W rated coil, and 25ohm (+-6%) coil resistance.
According to datasheet, it can be switched with only 75% of its 5V rating (and i use this fact in the design).

The problem with that design is that it consumes about 1A total the single 5V regulator i posses each time the relay is switched.

Can anyone help me to design a driver that wont consume that much current from the single 5V regulator?
Thanks.
 

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Got to go with the full H bridge. You're faking it now with a couple of resistors, that's the problem. You would replace R5 and R6 with drivers.

One sneaky way to do it might be to use P-channel MOSFETS. Put one in place of R5 with the gate hooked directly to Vcontrol1, the other on the other side the same way. Now when Vcontrol1 goes high, the 2N2222 is on, but the P channel mosfet is at -1.7V from the 5V rail, not enough to turn on. When Vcontrol1 goes low, the 2N2222 is off, but the mosfet gate is at -5V and turns on.

Ain't a real classy design, but it should cut that current down to .2A, and takes care of the "shoot through" logic you need for an H-bridge on a micro.
 
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I've used single coil latching relay as a auto power function for a Picaxe micro. By having a capacitor is series with the coil one can get the momentary switching action without the continuous current drain. Mine was a low current relay (13ma) so I didn't need any high current driver so my circuit is not a direct solution for yours, but the principle might be usable? The software makes the output pin high when first powered up manually, then when the program wants to power down it just makes the output pin low.


Lefty
 

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Hey! If that would work you could use just 2 transistors in a totem-pole arrangement, and cut the long-term power down to NOTHING! And free up a port pin! Will it work with just a single pulse like that, Rainman?
 
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"Will it work with just a single pulse like that, Rainman?"

It should work. It's mostly sizing of the cap so that it has a large enough charge to reliably energize the relay coil long enough for the latching magnet to hold the new state, some experimentation and testing will be needed.

The trade off is if too large a cap is used then it takes longer after pulse off for the resistor to bleed off the remaining charge on the cap so as to be ready in time for the next pulse on. So how short a time between setting and resetting and setting again is a factor to juggle the cap and resistor values. It's worth the effort, especially for battery powered applications where the continuous power drain directly effects battery run time, plus it's always nice to save a I/O pin when one can.

Anyway the concept worked great for me on my circuit.

Lefty
 
Thanks a lot guys.
Leftyretro
My Relay has 5V rated coil, therefore my GPIO pin in MCU can't drive the relay.
could you post here a design please of the totem-pole arrangement?
I could find a design thats related to that on web.

Thanks.
 
Using your existing circuit, since it is a latching relay, what if you pulse an output for just long enough (typically a few milliseconds) to trip the relay (either direction) and have a large enough bypass capacitor (say, 20 uF across the +5V power source) to supply the brief pulse current? Of course, you have to consider the power-up state of the MCU output pins.
 
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