I really appreciate all the replies. I guess what I'm trying to figure out here is why a battery would cause a different response than the power supply...

Post #5.

Add a 0.1uF capacitor across the new 4.7K resistor. That will keep the base of Q1 low long enough for the supply to settle down.

Post #5.

ak

 

It is a bistable circuit. There is nothing in the circuit to determin its state when the power is applied. Add a 0.1uF capacitor across the new 4.7K resistor. That will keep the base of Q1 low long enough for the supply to settle down.

So I initially misread your post and put the 0.1uF across R2, as I had removed the 4.7K from base Q1 to GND, and it's now working as expected. Do you think it's possible that there was some noise coming off of emitter Q2 on initial powerup, and the capacitor now in place is using that noise to charge instead of making it all the way to the base of Q1?

 

So I initially misread your post and put the 0.1uF across R2, as I had removed the 4.7K from base Q1 to GND, and it's now working as expected. Do you think it's possible that there was some noise coming off of emitter Q2 on initial powerup, and the capacitor now in place is using that noise to charge instead of making it all the way to the base of Q1?

In the original circuit, there was nothing in the circuit to determine which state it started out in. When Q2 was off, the base of Q2 was in a high impedence state, with the only reference to circuit ground being through the leakage current of the turned off LED. In that state, any noise appearing on the base would turn on Q1 and flip the state of the bistable to on.

By adding the 4.7K resistor from the base of Q1 to ground, it gave the base a solid reference that is more immune to noise. In that condition, which state the circuit would power up in was still anyone's guess. By adding the capacitor, on power up, the base of Q1 is held low until the capacitor charges up by Q2 collector leakage current through R2. By then, the bistable has settled in the off condition. When the button is pushed, the capacitor will quickly charge up and the base of Q1 will become positive enough to flip the state of the bistable to on.

I hope you can follow this explanation. Post if you have more questions.

 

In the original circuit, there was nothing in the circuit to determine which state it started out in. When Q2 was off, the base of Q2 was in a high impedence state, with the only reference to circuit ground being through the leakage current of the turned off LED. In that state, any noise appearing on the base would turn on Q1 and flip the state of the bistable to on.

By adding the 4.7K resistor from the base of Q1 to ground, it gave the base a solid reference that is more immune to noise. In that condition, which state the circuit would power up in was still anyone's guess. By adding the capacitor, on power up, the base of Q1 is held low until the capacitor charges up by Q2 collector leakage current through R2. By then, the bistable has settled in the off condition. When the button is pushed, the capacitor will quickly charge up and the base of Q1 will become positive enough to flip the state of the bistable to on.

I hope you can follow this explanation. Post if you have more questions.

Thanks for the explanation. So in the now working circuit I'm using, there is no 4.7K from Q1-B to GND, just the 0.1uF in parallel to R2. Still the same theory though? Capacitor absorbs noise preventing unintended biasing of Q1?

Also, what would have made it work using the 5V breadbord PSU instead of the battery?

Thanks again, you guys are super helpful.