# Pull Up & Pull Down Resistors Explanation

Discussion in 'General Electronics Chat' started by BobaMosfet, Jul 1, 2009.

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1. ### BobaMosfet Thread Starter AAC Fanatic!

Jul 1, 2009
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I'm trying to understand exactly how a resistor can "pull up" on voltage? For that matter, a "pull down" resistor only underscores the fact that neither pull-up or pull-down resistors do anything. They are just their to prevent a short.

Why are they called "Pull-up" and "Pull-down" resistors, and how can Ohm's law prove they are useful for anything other than preventing a short to Ground?

I've looked high and low for a real answer on this, bought books, and more. Will someone *please* explain this so my new-to-electronics mind can understand this.

Jul 17, 2007
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3. ### Wendy Moderator

Mar 24, 2008
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Basically it has to do with digital electronics. With a pull up (or a pull down) you can have a 1 and a zero with a simple switch, be it mechanical or one transistor. In many cases it is the voltage that is important, not current capacity.

There is another benifit to pull up or pull downs. If you have several switches going to ground (which is similar to what open collector does), you can form a logic function with that one resistor. In this case a AND function.

4. ### Papabravo Expert

Feb 24, 2006
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Inputs which are pulled up or pulled down can be driven to the opposite state by an external device as might be required on a bed-of-nails tester. There is no problem with Ohms Law, since with zero current through the pull up or pull down, the voltage must be the same on both sides of the resistor. Any deviation from equality REQUIRES current to flow and the presence of a voltage drop across the resistor.

5. ### BobaMosfet Thread Starter AAC Fanatic!

Jul 1, 2009
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I appreciate all the instantaneous responses. I looked at wiki, and the problem is, you are all way over my head on this. You all have told me what it does, but not really how it does it at the electron and voltage level.

A resistor in series drops voltage, not current. A resistor in parallel drops current, not voltage.

I want to be able to really understand this, not just spout what I'm told.

I need a real example with real values for everything and a "walk through" from Vcc to Ground on each leg to understand when the resistor is doing something and conversely when it is not.

You can apply all the voltage you like across a resistor that connects to an open (floating) pin. There will be no current flow and hence no voltage across the resistor, no matter what it's value is. This effectively eliminates the resistor. The open end of the floating pin will go high (to full voltage) simply because there is maximum attraction from the power-source, trying to make current move. So again, the pull-up resistor is only there to prevent a short-circuit.

The fact that voltage goes high on an open pin (as described above) is in direct contrast to Ohm's Law, unless this is a result of the resistances within the power-source interfering:

E = I * R
E = 0 * (Near Infinity + Pull-Up Resistor)
E = 0

Help me get my mind in line with yours, please, because I keep coming back to one perspective, which means I'm not being given enough discriminatory information (new information).

Thank you very much for your patience and help. If I'm wrong, don't be afraid to say so, and explain WHY. My Qwest for understanding far outweighs my pride.

Apr 5, 2008
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7. ### SgtWookie Expert

Jul 17, 2007
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Hang in there - some concepts can be tough to wrap your head around.

The resistor creates a voltage drop due to it's internal resistance to the electron flow. It's like putting a kink in a garden hose; the pressure at the far end of the hose goes down, and the pressure near the head end stays high.
Let's get this straight.
If you place a 1 Volt potential across a 1 Ohm resistor, you will get a current flow of 1 Ampere.
If you place a 1 Volt potential across two 1 Ohm resistors that are wired in parallel, each resistor will have a 1 Ampere current flow for a total circuit current of 2 Amperes.

Well, any time the voltage on both sides of a resistor is the same, it isn't "doing anything".

One thing you need to realize is that the input pins to MOS/CMOS/MOSFET devices actually appear (electrically speaking) more or less like a small capacitor. If you connect a resistor between a voltage source and a capacitor, current will flow through the resistor from the more negative potential to the more positive potential until the voltage at the capacitor matches the voltage at the voltage source; then the current flow stops.

Not precisely true; the open (floating) pin actually has some value of capacitance.

You're actually interpreting Ohm's Law incorrectly.
E = IR should be read as:
"The voltage measured across a resistance is equal to (the current in Amperes) times (the resistance in Ohms)"
So, if E=0 (the voltage across the resistor is zero) thus the current and/or the resistance is equal to zero.

8. ### beenthere Retired Moderator

Apr 20, 2004
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In the case of either an input or output pin on a logic IC, you are incorrect in your statement -
That might be the case for perfect models, but real-life IC's have extremely imperfect inputs and outputs. Current will always flow into or out of the pin. The pullup/ -down resistor will always work as advertised.

The effect is pronounced enough in old TTL that a 4.7K resistor is the best value for an input pullup. For CMOS, something more like 20K works well. A 4.7K resistor will hold the input high and cause the logic to malfunction.

You might be interested in this TI paper - www.focus.ti.com/lit/an/scba004c/scba004c.pdf

9. ### BobaMosfet Thread Starter AAC Fanatic!

Jul 1, 2009
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All

Thank you so very much. Both for correcting my thinking and explaning the capacitance issue with the pin, as well as my misuse of Ohm's law. I took time to digest what you said and work it out.

Now I get it. Atlast. And that ROCKS!!!!

Thank you!

10. ### Tree1 Member

Nov 30, 2010
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I am in the same situation! Have red many books and explanations online and still can't understand the reason of pull-up resistor in a transistor other than preventing a short in it's on state. Before I comment any further, let me read this thread. Let's see if I can finally understand this.

11. ### Tree1 Member

Nov 30, 2010
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Does this "capacitance issue" also apply to an open collector transistor? When a transistor is in cut off, the collector to emitter junction is like an open switch ( open circuit). The collector side( without a pull up resistor ) is still tied to VCC( Supply Voltage). So why would you not measure VCC at the collector ? Why is it still floating ?

Nov 30, 2010
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13. ### sreenivasulu New Member

May 19, 2016
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pull up resistors are just resistors connected in between the supply and required pin/circuit.
these are mainly used for making the circuit to make high level (enable) and is always connects the pin/circuit to the supply voltage.

14. ### BobaMosfet Thread Starter AAC Fanatic!

Jul 1, 2009
748
182
All--

I'm very much farther along in this now than when I originally wrote this. Pull Up and Pull Down Resistors now make perfect sense. My problem at the time, was an as-of-yet imperfect understanding of how current and voltage behave. No further answers are necessary. Thank you very much for your responses.

15. ### hp1729 Well-Known Member

Nov 23, 2015
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Application may be clearer than theory. Take any logic gate. Apply power with no connection on the input. What voltage do you find? The specific voltage may vary with the family (CMOS, TL, LS TTL). It will not be a valid high or low, but a "float".
Specifications for the logic state a given voltage and current for a valid high or low. TTL, <0.6 V, 1.6 mA for a low. >3.5 V, 400 uA for a high. (Or what ever fits the family in question.) If the driving input only pulls low you need a resistor to provide a high of that 400 uA, or what ever.
They could be calculated, and I'm certain engineering students will say they should, but the answers fit a large range using 1 unit load to 10 unit loads and the answers are always the same.
It doesn't mean that the resistors MUST stay within the designed boundaries. For TTL 1K to 10K. for CMOS 10K to 100K. Unused inputs can be tied directly to VCC or ground, so the calculations are just for the classroom.

16. ### HenryTj New Member

May 18, 2016
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Not likely technically precise, but might provide a mechanical means, or actually hydraulic, way to conceptualize this. At least it's the one I use. Imagine you have a plumbing situation. A valve with a pressure sensor to turn it on or off. But the seals aren't perfect. Somehow a tiny bit of water leaks to the other side.of the valve. The water will slowly build up. So you put a drain that lets a tiny amount drain out, so only the proper surge of valve trigger flow will trip the sensor and turn the valve on. You don't want the water leaking in and building up to trip the sensor. That would be one way of thinking about a high ohm "pull down resistor." It puts a small drain on that part of the circuit so that voltage/current doesn't leak in and build up.

A "pull up" resister is sort of the opposite where a small line lets a tiny trickle of water in to keep the maintain the pressure on the sensor, otherwise a leaking fitting will slowly let the pressurized water leak out and the sensor will trip when it's not supposed to.

At least that's my way of conceptualizing it. There are interesting similarities between electronics and plumbing.

17. ### WBahn Moderator

Mar 31, 2012
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Glad you got your answers. It sounds like you would prefer this thread be closed to further responses.