Pull Down Resistors

[nintendoeats]

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

I'm inferring that reality is disinterested in our petty mathematics. Would it help if I gave you EXACT measurements for each resistor? Since the difference is so small, is it not possible that the minor variances in resistor values from the "standard numbers" might be enough to account for it?

 

[Old Steve]

I'm inferring that reality is disinterested in our petty mathematics. Would it help if I gave you EXACT measurements for each resistor? Since the difference is so small, is it not possible that the minor variances in resistor values from the "standard numbers" might be enough to account for it?

No, absolutely precise values aren't necessary.

 

[nintendoeats]

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.

And what's the UNO doing and where's the pot exactly???

I've double and triple checked and as far as I can tell that wiring is correct. The Uno is just a power source in this project. The pot is that little box on the left. An analog stick is actually two spring-loaded potentiometers, one for each axis. The reaosn you can see my hand in the second two pictures is that I have to hold the stick left or right to make LED2 and LED3 come on.

 

[Old Steve]

I've double and triple checked and as far as I can tell that wiring is correct. The Uno is just a power source in this project. The pot is that little box on the left. An analog stick is actually two spring-loaded potentiometers, one for each axis. The reaosn you can see my hand in the second two pictures is that I have to hold the stick left or right to make LED2 and LED3 come on.

Let's wait until someone else runs the LTSpice simulation. If they get the same results as me, and agree the circuit is accurate, then you have it built differently to the schematic. I've spent enough time on this for now. I want to get back to my project.

 

[nintendoeats]

Let's wait until someone else runs the LTSpice simulation. If they get the same results as me, and agree the circuit is accurate, then you have it built differently to the schematic. I've spent enough time on this for now. I want to get back to my project.

I've just run through it twice more and your diagram really does look correct. I'll post a good clear pic of the breadboard in the morning when I have proper light. Thanks for all your help!

 

[nintendoeats]

Quick update before I knock off, I just plugged a more traditional (still 10k) pot into the circuit and it works perfectly. So we can eliminate that as a concern.

 

[nintendoeats]

EDIT: Sorry, I put a wrong thing here. I thought I had found a discrepancy, but I was mistaken. The mystery stands.

 

[nintendoeats]

Ok ok ok, I should go to bed. Last thing. At center the voltage from the pot to ground on each side is 1.5V. Since we have matching resistors in the voltage divider ( I did some reading ) we would expect that to be split in half at the base of the transistor, so about 750mv. I measured base to ground that was about 700 mv, so that checks out. The issue must therefore lie with calculation of the potentiometer itself. I got the same results with two different 10KΩ pots, so we can eliminate malfunctioning hardware. Since a pot is just a voltage divider I would expect 2.5V on each side, but I'm guessing the large resistance affects that. I'm not totally clear on how, but as far as I can make out that HAS to be the problem area.

BED.

 

[duke37]

If the transistors take no current, then the base voltage of Q1 will be 5*1.5/8 = 0.94V
The voltage shown of 0.41V indicates that the transistor is taking base current. The normal rule of 0.6V or 0.7V is only a very approximate value based on considerable current draw. The relation between voltage and current is logarithmic.

 

[nintendoeats]

If the transistors take no current, then the base voltage of Q1 will be 5*1.5/8 = 0.94V
The voltage shown of 0.41V indicates that the transistor is taking base current. The normal rule of 0.6V or 0.7V is only a very approximate value based on considerable current draw. The relation between voltage and current is logarithmic.

Ah, so you're saying that at very low current draw the base voltage drops considerably, enough to account for the confusion?

Anyway, here are pictures as promised. This is a link to a video of the circuit in action.