Transistor help request

Hi, I'm trying to run a 700mA LED load through a one-shot 555 timer circuit, using a hall effect sensor as the trigger and with a 12V power source (schematic attached). I can't make my TIP122 transistor switch properly - it opens the current even when the hall sensor is not triggered.

How do I determine the necessary base current? The TIP122 datasheet (https://www.electro-tech-online.com/custompdfs/2013/04/TIP122.pdf) suggests that the DC current gain is 1000...so I think I need a 7mA base current? Assuming 11.3V gets through the timer: 11.3V/0.007A = 1.6K base resistor?

Any help GREATLY appreciated I'm going crazy. I can make it work with a 2N2222, but I believe that transistor is not rated for more than about .8A and I need 1.4A here.

Thank you!

Do you want to light the leds while the 555 is triggered? If so, the emitter of the Darlington is only going to pull to [(Voh of the 555) - (Vbe of the Darlington)] - the drop across the base resistor. The base current into the Darlington decreases as the emitter voltage increases, so you cant get it to pull high enough.

I would suggest flipping the Leds and their resistors to the collector side of the Darlington, emitter grounded. This puts 12V-Vcesat = 10V across the Leds and resistors. The resistor that supplies the Darlington base current has 8.8V ={11.3 - Vbe(sat)} across it. 1K would drive the darlington with 8mA, which should do it.

Amazing! Thank you.

You should be able to use the circuit in your fist post without any base series resistance at all. That puts the transistor in an emitter follower configuration. The emitter pin will be a voltage source about 1.2 volts less than the output voltage of the 555. The load current will be controlled by the resistors in your LED strings.

This configuration will also be a bit more energy efficient than a grounded emitter circuit, since the base current flows through the load, instead just going to ground.