It cant be this hard...

[AxleD]

Hi All,

I was trying to create a simple latching switch but despite my best efforts i cant seem to get it to work. In theory it makes complete sense why this circuit should work, but when building the circuit the LED lights up and stays on as soon as power is applied. Not sure what what I'm doing wrong here as it's a pretty simple circuit (see attached). Q1 is a 3906 and Q2 is a 2219 and built on a breadboard.

Any suggestions on what the issue could be would be appreciated.

Thanks

[Whales]

Dislcaimer: guessing out of my arse, little analog experience.

Surely the impedance to ground for the PNP's base (with the button pressed) is lower than the impedance to V+ for the NPN's base.  That would ensure that the PNP is always in a better situation than the NPN when you let go of the button and they race to latch. 

[Zero999]

The transistors don't have enough base current to turn on properly.

Reduce R1 and R2 to something more sensible and it should work.

[Seekonk]

I think it is a rittle reaky.  Put a 10K base resistor on Q1 E-B.

[dmills]

The problem is you are using an LED as a load...

Even at very low currents the LED Vf is going to be very much greater then the Vbe of Q2, so Q2 will be biased slightly on, and from there it is pretty much game over.

The emitter/base resostor on Q1 will help as will shunting the LED with a few K so there is a load that exists right the way down to 0V.

The 100K resistors are probably excessive if you actually want to handle reasonable current by the way, a few k would make for more reasonable base currents.

Regards, Dan.

[Mechatrommer]

when you apply power, pnp base is floating not so high up to V1 level, so she conducts in linear mode pulling top of R2 higher, then the npn base goes high turning him on, maybe slowly in linear mode too, then he will pull down pnp base even deeper down, then the she will pull up the he base harder even further up to the climax or V1. congratulation now you have fully conducting pnp & npn regardless you push the switch or not. maybe you may add pull up resistor to pnp base to satisfy her up during the start, i'm not sure esp if shes in moddy condition.

[Zero999]

Of course I didn't read the text properly, just looked at the schematic with a 10R load on it.

I think it is a rittle reaky.  Put a 10K base resistor on Q1 E-B.

Yes and if that doesn't work, put a B-E resistor on Q2 as well.

[LA7SJA]

I think someone can benefit from this video.
EEVblog #262 – World’s Simplest Soft Latching Power Switch Circuit

Johan-Fredrik

[AxleD]

Thanks I see what the issue could be.

Given the explanations as to the possible reason it looks like the resistor across Q1's B and E could be the solution.  I will try and report back.

LA7SJA, this circuit is from that video 

[Seekonk]

Noise pickup can also be an issue in these circuits.  Q2 amplifies any spike it might see on the base should you have some long leads on a proto board.  A little momentary spike could get the whole thing running.  If you continue to have problems place a capacitor from B-E on Q2 to form a time delay.

[bugs]

Try  shielding the LED from the light and see if it still switches on.
Possibly the LED is working as a photo-diode and it's current flows trough the base of Q2, switching everything on.

[mrkev]

It's funny how many different ideas are perfectly plausible in this scenario.
Basically, you've got very strong positive feedback and since the base current of Q1 is determined by the col. current of Q2, overall amplification is multiplication of both Betas of each transistor (could be well over 9000  ).

Keep in mind that even 1nA at the Q2 base will make about 0,2uA at the Q1 base and that's gonna switch on whole circuit.
It doesn't really matter if it's load, the C-E capacity of Q1 (or Q2),  Q2 base picking up some rf (or even mains) signal, etc.

The solution is the same,  to break the high gain loop. Probably best way with resistor between B and E of the Q1, something like 10k - 47k should do it. It's gonna limit the minimum current going through Q2 that is able to switch on whole circuit (about 0,5V/47k - B-E voltage drop over the BE resistor).