OK, got it. --->Infrared LEDs. Vf=1.5 x 3 series = 4.5V
Fed by 3 series lithium for 11 V at full initial 700mA; 11V - 4.5V divided by 0.7 A = use 10Ω in series. Current will diminish while cells discharge to their rated minimum recommended.
For 12V lead-acid; 12V - 4.5V divided by 0.7A = use 11Ω to 12Ω
Or, your 'puck driver' set for 4.5Vf / 700mA instead of resistors.
Check their time on until Li cells reach their minimum voltage (2.5V ¿?) with or without your 'puck driver'