Pull Down Resistors

What value is the pot?

 

Oh, pity. It would have been easier to calculate the exact values for you knowing that. Do you have a multimeter to measure it's resistance? (Measure across the outer terminals, with the pot out-of-circuit.)

 

It means it's 10 kilo ohms, 10,000 ohms, (in any context). That's a fairly standard value. If you measured correctly, it will be 5K from one end pin to the centre pin, 5K from the other end to the centre pin, and 10K from end to end.

It's actually handy to number the components, to make them easier to refer to.
I could explain the theory now, but I'll just do some quick calculations and then I can do it better, with voltages at different points.
Edit: I'll be a little while - have to eat lunch first.

 

Interesting. Maybe I'm missing something, but the way I see it LED1 will always be on, if your circuit is identical to the drawing.
Even with the pot wiper all the way to the left, there will still be a little under 1V available at the base of T2, (which will clamp at 0.6V), turning it on, and 2.5V available at the base of T1, clamped at a little over 0.6V, also turning it on. It might be dimmer than when the wiper is in the centre, but still on.
There will be roughly 1.9V at the right-hand side of the pot in this position, so LED 3 will be very faintly lit too. LED 2 will be fully on.
If you say that LED1 only lights when the pot is in the centre, then obviously I've made a mistake somewhere, but that's how I see it.

Similar situation when the pot wiper is to the right.

Someone else want to check this out and see what I overlooked?

 

If it could turn 360 degrees, yes. (At least I got that right.)

 

It's OK, I made a silly mistake. In reality, (with the pot wiper to the left), the base of T2 will be at about 0.35V and it will be off, not on, so the LED could not be lit. And the right-hand side of the pot will be at about 0.7V, so LED3 will be fully off. R3 / R5 act as a voltage divider, dividing the 0.7V at the right of the pot in half, supplying the aforementioned 0.35V to the base of T2. R5 is not a 'pulldown' in this situation. Similarly, with the pot in this position, R2/R4 are a voltage divider, supplying 2.5V to the base of T1, which cannot conduct because T2 is off.

When the pot is to the right, the situation is reversed.

I miscalculated the current through the pot and the right-hand leg.

Edit: So you're not going blind, LED1 is definitely off when the wiper is at the ends.

 

My problem now, though, is when the pot is in the centre, I get 0.4V at the base of both transistors, so they should be off and LED1 still not lit! ???

 

A pulldown resistor holds an input low, until a high is applied to the input. And see my last post.

 

I just did a quick LTSpice simulation of the wiper in the centre of the pot. (Attached, and screenshots below.)

The circuit:-


The trace, green is Vb1, (T1), and blue is Vb2, (T2):-

 

Hmmm. At 410mV, both transistors should be off, and LED1 unlit.
You've got me beat.

It's hard to see in your pics, but carefully double-check that it's wired the same as the schematic.
If so.....you win. I give up.

 

Before I disappear, if someone who has LTSpice wants to load the schematic above and double-check what I've done, please do so. This is a mystery. I'd love to know what's going on here.
Edit: They're just generic NPN transistors, but that should be all that's needed here.