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Pull Down Resistors

nintendoeats

Oct 18, 2013
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I have only recently started to fiddle with electronics seriously and set myself a simple task. I had a potentiometer (taken from a 360 controller) connected to two LEDs, using the pot to switch between them. I decided to try making a third light come on when it was centered. After much fiddling I created the circuit below, which works exactly as I wanted. However, I don't fully understand what I did and was wondering if somebody could please clarify. Specifically, the way the transistors are connected. I have a vague notion that the resistors between the transistor bases and ground are pulldown resistors, but I don't fully understand what that means and I don't understand exactly what's going on in that whole area to my satisfaction. Sorry for the messy diagram, my handwriting is quite poor.

Thanks!
Pot_circuit.jpg
 

nintendoeats

Oct 18, 2013
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I really don't know. As I say, I took it out of an Xbox 360 controller and I haven't been able to find any info on it.
 

Old Steve

Jul 23, 2015
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I really don't know. As I say, I took it out of an Xbox 360 controller and I haven't been able to find any info on it.
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.)
 

nintendoeats

Oct 18, 2013
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If I've done this correctly then it is 10 kΩ. What exactly does that mean in this context?
 

Old Steve

Jul 23, 2015
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If I've done this correctly then it is 10 kΩ. What exactly does that mean in this context?
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.
 

nintendoeats

Oct 18, 2013
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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.
Sorry, I wasn't really clear, I known what 10k is :). Say I turn the pot 90 degrees to the left (assuming it rotates 360), Would that mean that the left now had 2.5k of resistance and the right had 7.5k?

I'll just put on some numbers and reupload the image to the original post.
 

Old Steve

Jul 23, 2015
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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?
 

Old Steve

Jul 23, 2015
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Sorry, I wasn't really clear, I known what 10k is :). Say I turn the pot 90 degrees to the left (assuming it rotates 360), Would that mean that the left now had 2.5k of resistance and the right had 7.5k?
If it could turn 360 degrees, yes. (At least I got that right.)
 

nintendoeats

Oct 18, 2013
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Here are some pictures in case I drew it incorrectly. As you can see, it does work as described. The transistors are BC547B Which do appear to turn on at about 0.6V. Well, I'm glad to at least have produced an interesting mystery.



I have to head to bed soon, so if I disappear that's what happened. Thanks!
 

nintendoeats

Oct 18, 2013
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I just double checked those resistors. The 1.5Ks actually come out at more like 1.25kΩ. I'm colourblind and can't really read the bars so I sorted them into compartments a while ago, and I guess I miscategorized those ones as 1.5KΩs. Since this is the bit of the circuit I really don't understand I don't know if that explains the discrepancy.
 

Old Steve

Jul 23, 2015
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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. :oops:

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

nintendoeats

Oct 18, 2013
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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.

Aaaaah, ok. That makes perfect sense now. If that's not a pull down resistor, then what is?
 

Old Steve

Jul 23, 2015
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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! ???
 

nintendoeats

Oct 18, 2013
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A pulldown resistor holds an input low, until a high is applied to the input. And see my last post.
Ah thanks, ok that makes sense. I can see where the confusion was.

Hmm, I double checked all the resistors, and aside from the ones I already mentioned they are correct. Is there anything in the circuit I could measure that might shed light?
 

Old Steve

Jul 23, 2015
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I just did a quick LTSpice simulation of the wiper in the centre of the pot. (Attached, and screenshots below.)

The circuit:-
Centred CCT Screenshot.JPG

The trace, green is Vb1, (T1), and blue is Vb2, (T2):-
Centred Trace Screenshot.JPG
 

Attachments

  • Pot circuit centred.asc
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Old Steve

Jul 23, 2015
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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.
 
Last edited:

Old Steve

Jul 23, 2015
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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.
 
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