You are charging a 6V lead-acid battery. It has 3 cells of about 2.1V in series (in rest, fully charged). For the Ic of the transistor to be 0, Ib needs to be 0. Let's calculate your minimum battery voltage. Your zener has a tolerance, but let's assume it's excactly 6.8V. Put the potentiometer at 0 (basically take it out of the schematic) to get your minimum battery voltage:
6.8V on the base of the transistor. Ube @ very small Ib = about 0.6V, so your minimum battery voltage is 6.2V. During float charge, the cell Voltage is about 2.25V, so 3*2.25V = 6.75V for your battery. That's higher than the minimum 6.2V of your schematic, so the current should indeed go near 0 if you give it enough time. (Never completely 0, as you keep float-charging your battery).
Maybe you wired something incorrectly? Did you check the B, C and E of the transistor?
Now, all that to the side:
- You are charging the battery incorrectly. Take a look at the battery university website how you should charge a lead-acid battery. If you use this method, the life span of your battery will be limited.
- If your battery is near empty, it's voltage will be very low, and you will burn out your transistor (even with infinite heatsink it can only take about 10W max.@Ta=50C and 1.5A), or your power supply.
- If you put your potentiometer all the way up (1k), your battery voltage will be 7.3V. That gives you a max. cell voltage of 2.43V, and you will over charge your battery until it boils, and destroy it in the process. Possibly there is also a danger of explosion, as the gasses from the battery are explosive.
So sorry, but I wouldn't reccommend building this schematic in a real-world scenario.