single coil latching relay

Presumably this is the type of relay that needs a reversal of coil current to change over. You need some kind of bridge drive, the simplest I can think of is to put a resistor from each end of the coil to the power rail and an open collector driver to each end of the coil.

Bo schrieb:

The simplest way is a push-pull driver of whatever kind and a capacitor in series to the coil.

--
Dipl.-Ing. Tilmann Reh
http://www.autometer.de - Elektronik nach Maß.

Perhaps I misunderstood:

I don't see how you can do what you describe-- take for example, a 5V relay

40mA coil.

If I put even a small resistor, 50 ohm, on the coil ends as you describe-- once the driver is turned on, the 40mA across the R is going to drop the voltage too much and the relay will not switch at 3V. Am I to assume that I would need for example, 7V on the relay coil to account for the V drop across the opposite coil end resistor? Or perhaps a very low resistance of

5-10ohms?

I like the idea a lot, but am having trouble visualizing how is would function reliably. If feasible, I could use 2 outputs from the A6832 on either coil end, right?

Thanks,

Paul

>

I have done this for 9 (actually 8, but the matrix was 3 x 3 anyway)

5V relays. I used 3 rows and 3 columns, each driven by a 1/4 74HCT125; the tri-state controls of all the 6 bufers are separately driven, while there is just a single bit for the state of the rows and one for the columns - that is, you set the direction using the logic inputs (2 bits - or use one with an inverter) and then you select for some time the row and column you want using the tri-state controls.

Dimiter

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Lets suppose your relays will switch at 4 v and needs 40ma, your resistor needs to drop 1 v at 40ma so it needs to be 25ohm. This means your driver has to sink 240ma when operated. A typical operate time for a latching relay is 1-2 ms so your driver only needs to be on for that time and so power consumption is still very low unless you are switching at high repition rates. If you can run from the higher pre regulator voltage that will ease your power problems.

You could use an active pullup (assuming your Allegro chip has open drain outputs). Here's a Spice-like netlist (easier than complex ASCII schematics):

# Node list: # 0 GND # 1 +5V # 2, 3 Relay coil # 4, 5 Open-drain outputs from driver IC # D1 A=2, K=4, D1N4148 # Diode of your choice D2 A=3, K=5, D1N4148 Q1 E=2, B=4, C=1, QBC547 # Small NPN transistor Q2 E=3, B=5, C=1, QBC547 R1 1, 4, 10k # Non-critical resistor R2 1, 5, 10k

The problem with that circuit is if both drivers are on together, ie software fault or power up condition, you short out the power supply possibly causing damage.

I don't see this problem. Perhaps I'm misunderstanding the netlist.

If both outputs are on, both transistors are on and applying 5v to both sides of the coil? This solution seems OK to me, though perhaps more parts than your suggestion, but does not have the voltage drop issue.

The A6832 relay driver I would like to use in conjunction with either of these circuits, is open-collector--so I think it would work with either.

Thanks for the suggestions!

Paul

With both drivers on there is 5 volts to each side of the coil. Isn't there a diode from the coil to the driver?

It would not be a problem - you do need the diodes to 5V and to GND at both sides of the coil, but they will only conduct when "discharging" the relay coil. However, using open drain outputs and resistors is an archaic approach. You do get HCT chips capable of driving both sides with enough current for small relays for decades now, and you do get sot-23 individual buffers which are small enough to hide beneath any relay I have seen (Teledyne included) for about a decade.

Dimiter

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Yes I had it drawn wrong, and yes it does need some extra diodes. A driver chip is the best solution but the op wanted something cheap and easy.

I'm confused.

Where are extra diodes needed? and why?

Paul

If your relay coil is between two buffers, you need two diodes at each side of the coil. One of the diodes has its cathode connected to the coil and its anode to GND, the other has its anode connected to the coil and its cathode to +5V (this circuit at both ends of the coil). A pair of these diodes will conduct only while discharging the coil - which pair depends on the direction you have been driving the coil before switching off. If you do not understand why the diodes are necessary I strongly suggest you just put them there, this will save you burned buffers etc. Perhaps looking up something like "flywheel diode" could help (I have not tried Google on it, though).

Dimiter

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When you turn off the driver the coil current has to find another path, it can only go 1 of 2 ways, either it breaks down the emmiter base diode or it breaks down the driver. If you put a diode from each end of the coil to the power rail it can flow that way instead.

Didi,

Thanks for the extra details. I understand the need for the diodes that have their anodes tied to the coil. I do not understand the diodes between coil and ground (with anode=gnd, and cathode tied to coil). In what circumstances are these diodes forward biased? The word 'flyback' catches my attention. Is this the 'std' diode that goes across a coil to snub inductance spikes from the coil?

Paul

Paul, when the diode with its anode to the coil is forward biased (the one you understand), the diode at the other side of the coil with its anode to GND (cathode to coil) is also forward biased, thus closing the current path.

Dimiter

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That depends on the drive signals. If you always pulse one side low, while holding the other high, then the lower diodes are not needed.

If you pulse high, while holding low, then the upper diodes are not needed, and the lower ones may be inherent in the driver.

Again, depending on the driver, but if you drive actively in both directions, you can omit the diodes, as the ON driver resistance (now shorting the coil) provides the current path.

If you are unsure, then put the diodes in, it is cheap insurance.. :)

-jg

I beg to differ. With both IC outputs on (ie low), both transistor bases are pulled low, so both transistors turn off. No current drain, except a small amount through the base pullups. Likewise, with both outputs off (high), everything pulls high, both relay terminals are high: again, no current drain.

I do agree with later posters about flywheel diodes. Because of my isolation diodes (between emitter & base) I wouldn't count on those on the driver IC. Provide your own: upper & lower. To minimise voltage drop, you might choose Schottky diodes for D1, D2.

Relay Drivers from ON Semi NUD3112DM for example

gm