Questions about pull-up/pull-down resistors

Greetings group,
First-time poster and EE by education, but often has a lot of questions, like this one concerning pull-up and pull-down resistors:
I will use the following link as a reference circuit: http://en.wikipedia.org/wiki/Pull-down_resistor
I wondered what they were for, in reading a datasheet about a MIL- STD-1553 chip, looked it up, and gathered that the purpose of each resistor is to pull any "line floating voltage" to the specified level (down to GND or up to X VDC), the mechanism of course being the voltage differential on either side of the resistor causing a current to flow.
What I don't really grasp is, at what "threshold" or level does the pull occur? In the circuit diagram, on the input to the 7407, if, as the article states, unconnected IC inputs tend to float "high" (some voltage), how high does it need to be before the pull-down action starts? I assert the answer is: almost immediately, because any differential above 0V GND (assuming a solid ground) will cause current to flow.
So, how can the input in this case ever "float high"? After all, it is still unconnected according to the diagram.
In other words, maybe my question is, "Does pull-down mean 'hold-down' or 'force-down', or is it really a voltage-activated pull?"
Also, if a line were to "float high" (possess a positive voltage), and I admit it's been many years since I've been doing EE, how would the presence of resistor cause the voltage on the input line to change? I know there's a "voltage drop across the resistor" from the float voltage down to ground, but how would the act of dropping a resistor in there actually drop the line voltage down to zero? All I know is, a little current would flow and the resistor would dissipate power in the form of heat, and that's that. Can someone straighten me out? Thanks.
Mike
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The reason you pull-up or pull-down is to prevent RF noise from triggering the input. The floating leads act as antennas. Pull-up or pull-down is used depending on what the rest of your circuit is doing. If you have, for example, a four input nand gate and you only use two of the inputs, you would pull the floating, or not used ones, up. Those two pull-ups would give you logic highs. So your other two inputs would use the gate as a two input nand.
Al
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Zerex71 wrote:

What you need to look at is the equivalent circuit of the IC, eg:
http://focus.ti.com/lit/ds/symlink/sn7400.pdf
Generally, an unconnected ic input pin acts like a little aerial connected to a high input impedance amplifier. Any signals, such as stray radio waves, data on adjacent lines coupled to the pin by stray capacitance, or whatever, can be picked up by the pin and be taken as data input.
Studying the equivalent circuit diagram shows that, typically, the normal voltage on an unconnected input pin is neither at supply or ground but somewhere in between - the undefined "floating voltage". This voltage will be nearer the threshold voltage (half supply rail voltage, typically, for CMOS) than would be the case if the pin was pulled to either rail by a resistor.
Using a pull up/down resistor does two things. Firstly, it moves the voltage on the pin as far away from the threshold voltage as is possible. Next, it lowers the input impedance of the pin, so that more signal power is needed to move the pin voltage across the threshold.
The combination is otherwise known as increasing the noise immunity of the input. A resistor ensures that relatively small amounts of noise won't cause erroneous transitions. The received noise has to be around half supply voltage or greater whilst driving the much lower input impedance produced by the resistor and IC input, in parallel. Without it, a noise signal of a few tens of mV can cause false triggering, in some cases.
--
SUe







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Hi and thanks to both of you for responding so quickly. That is an interesting bit of information that I never had described to me. Also, thanks for not being nasty in your response - I have had a run of that lately just for asking sincere questions.
Okay, so these answers seem to answer the "why do we use the resistors" question, but I didn't seem to pick up on the mechanism of how the pull-up/down works, as I was describing. In other words, if I hooked up a scope to the terminal, minus R, I should see the floats you descibe. Then let me drop the resistor in there and the voltage differential on either side causes a current draw from the unconnected input terminal down through R to ground. Power is dissipated, but how in the world does that force the terminal to drop from float voltage to zero? [Which raises another question: How can you suck current out of a 'current-less' IC down through the resistor? or maybe I'm just a dummy...]
Here's another question: If we want to force a voltage instead of letting it float...why not just tie it directly to the voltage we want? (I realize as soon as I write this the problem becomes when you actually do connect the input, but you may see why I'm asking it.) What good does dropping the resistor do for us? I am probably missing something in the way you are explaining this to me.
Mike
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Zerex71 wrote:

It's all down to impedances - the IC pin has a floating voltage on it fed from a very high impedance. It also has a resistor connected between it and a power/ground rail that presents a relatively low impedance. So the voltage on the pin is pulled to the voltage on the other side of the resistor - if the resistor is relatively low in value.
For example, let's say that the "floating voltage" is floating at 5v and has a source impedance of 10 megohm - that has an equivalent circuit of an ideal 5v supply, in series with a 10 megohm resistor - the other side of which is the pin.
Connect a 1000 megohm resistor from the pin to ground. This could be the typical surface resistance of the circuit board. The voltage on the pin will drop - but not by much. It's new voltage will be 5 x 1000/1010 volts. Very nearly 5v.
Replace the 1000 megohm resistor with a 10 megohm one. The voltage will now be 5 x 10/20. Down to 2.5v
Replace the 10 megohm resistor by a 1000 ohm one. The voltage will now be 5 x 1000/10001000 - very, very nearly 0v.
A "current less" IC still has currents flowing - it is just that they are very very tiny.

It's back to the equivalent circuit of the IC. In some case you *can* tie the pin directly. But often that would result in too much current, or an unnecessarily high current flowing, either normally or in a fault condition. In some cases it is important that there remains some small voltage difference between pin and rail, or the operation of the IC will be compromised. eg, if there are intrinsic diodes built in to the design that have to be kept reverse biased.
Ultimately, what you have to do is to get the equivalent circuit of the IC and work out how it works and work out what the effect of an external pull up/down resistor will have. You can see what the effect of using a very high pull up resistor or a very low one would be. But, ultimately, the chip designer has already done this for you and will recommend a value to use.
--
Sue


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wrote:

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Don Lancaster's TTL Handbook says two reasons to use a resistor instead of directly tying TTL inputs to +5 volts is 1. Some variants of TTL are sensitive to transients on the +5 bus and tying through a resistor helps prevent these transients from affecting the inputs and 2. If the input is tied to +5 through a resistor, it can be easily and safely pulled down to logic 0 for troubleshooting.
I recall doing labs using an LS TTL flip-flop and left inputs floating, thinking they'd float up to logic 1 and be disabled - only to find that the resulting breadboard was very sensitive to stray noises (the circuit worked differently depending on where I held my hand) - pulling up the inputs through 1 k resistors made the circuit predictable and work like it was intended to.
Bill
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says...

Yes, the original 74xx series was sensitive to Vin>Vcc. Any reverse current was a reliability problem waiting to happen. Since the inputs were emitters, one could get reverse transistor between the emitters. This also brings up the point that one should never bias TTL inputs with resistors (other than pull-ups). SOme really funky things can happen between those emitters. Of course the Schottky variants of TTL didn't have this problem because they're really SDTL so don't have the input emitters.

This is a good idea that's not limited to TTL designs. It makes board test easier too.

Never leave *any* input float. I didn't think anyone beyond EE 101 made that mistake. It only takes once.
--
Keith

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Without the resistor, when the input you would like to be connected is connected, it will be shorted to ground. With the resistor in place, the resistor will "resist" current flow more than the wire would and there will then be a voltage across it. As long as the resistor is sized properly, there will be voltage observed at the input pin at the logic "high" level. The wire by itself will brown out the power supply because it has a limited power dissipation capacity. Hopefully this is helpful.
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http://www.interfacebus.com/IC_Output_Input_Pullup_Resistor_Values.html

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