Driving single coil latching relays with logic buffer IC's

You say you can run from 6.5 volts so you'll need a kind of H bridge with the top transistors replaced by resistors like this: -

If you read the data sheet you will see that you need to apply the following voltages to get it to operate: -

So, if your supply is 6.5 volts you can "drop" up to 2.75 volts across the collector resistors. Given that the coil resistance is 250 ohms, with (say) 4 volts across it the current will be 16mA hence, the collector resistors value can be (6 volts - 4 volts)/16mA = 125 ohms.

I'd probably go for 100 ohm resistors but don't go much lower or you might over-power the coil. Read the data sheet to get this detail.

A question has a risen over the inductive kick-back and my answer is this. If the current taken by the coil is 20 mA and the transistor switches off, 20 mA will want to flow through the collector resistors. Given that the collector resistors form a closed loop with the coil, the maximum fly-back voltage above the power rail (and at the collector) will be 20mA*100 ohms = 2 volts.

So there are no reasons to add flyback components - the resistors will protect.

If you like living dangerously, this circuit may work (it would use the 5V supply):

^{simulate this circuit – Schematic created using CircuitLab}

This is used to switch Fujitsu relays in a production design (yours draw about 1.5x the current so more edgy again).

Here is a link from the schematic. Credit to John Larkin of Highland Technology. The driver part:

The outputs are always connected to Vdd or GND so no flyback diodes should be required. Even if the transistors were off (for example due to very low supply voltage) there are still ESD diodes to the power supply pins (and if there is a cap as the 330nF shown - maybe use 1uF) that may not be an issue.