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Author Topic: Placement of resistors.  (Read 2177 times)
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No, but I have used a 12$ butane hot air pen successfully so I imagine that would work
tho for smd ics with more pins than a few I prefer to use the solder drag and wick method, hot air took a while
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I've tried the solder drag method before.  I ended up with all the pins connected to eachother.  I stopped there.  Was I supposed to wick away the excess after that?
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Yes, bridging is normal before that, the solder wick takes off the bridges leasving just enough under the pins for a good connection
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its funny, every time I try that theres never enough solder and I have to go back and reflow a pile of pins ... but if I want to remove an IC with wick, not one pin will pop loose.

not saying the method is invalid, just saying I must be the only one not doing it right
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Quote
but if I want to remove an IC with wick, not one pin will pop loose.
That is telling you that you did have enough solder in the first place.
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MMkay, well I think  I definatly have enough solder on there, haha.  I'll try to wick sometime.
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Btw, since we are on the topic of resistors, can anybody explain what pull-up and pull-down resistors are?  I knew at one point although since I've been broke, I haven't been able to really make any of my concepts and I've just forgotten.
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Pull up and pull down resistors set the pin at a logic HIGH and logic LOW respectively. They are used for inputs. You get it?
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No, Say I have a Vcc connected to a load, which also has a connection to ground at the same terminal.  I would put pull down resistors above the ground to make sure the Vcc doesn't go through to ground but rather my circuit?

Is that correct?  I think i'm getting confused with NMOS and PMOS transistors.
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Pull up/down resistors have nothing to do with that. I will try to find a link. I have to leave now. Try searching for more info.
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Yea I mean more so the logic i'm getting confused.  There aren't any resistors in a CMOS dealio but you get me.

still confused though, haha.
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When an input is not connected to anything it is known as "floating".  This is when there is no signal on it, so the input cannot know what logic level it is at.

When you connect the input (via a button, say) to either Vcc or GND, you are placing the input into either the HIGH or LOW logic state.

You want to avoid a floating input at all costs.  They can cause strange results in programs.  To avoid them you place the input into a "default" state - the opposite of the way you want your button to connect the input to Vcc/GND.  For instance, if your button connects the input to Vcc (HIGH), then you need to set a default state of LOW.  So you need to connect the input to GND for when the button isn't pressed.

However, you can't just connect it straight to ground, because then when you press the button there will be a direct connection between Vcc and GND, and that will basically be a short circuit.  The whole system will die, and in extreme cases batteries will blow up and wires / traces will melt.

So, you create the default state through a resistor.  The resistor can be quite high as you don't want much current to flow through it when you press the button, but at the same time it needs to be low enough that not too much voltage is dropped across it.  10KΩ is a typical value.

Now, when the button isn't pressed, the pin is connected to GND through the resistor and reads LOW.  You press the button, and the resistor is connected between Vcc and GND, and the input is connected to Vcc, so it reads HIGH.  No more floating.
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Hmmm I think I get it.  That would explain why there is a 10k ohm resistor on the IREF pin of a TLC5940, b/c you don't want your 5940 pulling unlimited current based off of the 0 resistence IREF. 
right?
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majenko made my tiny descriptions seem useless.... There is one more thing to add: if the resistor value is too high, the input may be suspectable to noise, and if it is too low, it will draw a lot of current, and waste power.
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So how would one figure out what is a good medium?  Just use 10K for now?
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