# Pull up/Pull down

#### Surendar

Joined Sep 12, 2005
18
What is the purpose of Pull up and Pull down resistors and how the absence of

these will affect the circuits!!

Thanks

Surendar

#### Papabravo

Joined Feb 24, 2006
21,225
Originally posted by Surendar@Apr 25 2006, 02:45 AM
What is the purpose of Pull up and Pull down resistors and how the absence of

these will affect the circuits!!
Thanks

Surendar
[post=16502]Quoted post[/post]​
The purpose of these resistors is to force an output or input to go to a defined state. This may happen for lots of different reasons. The alternative is to let that point in the circuit float. If it is an input then sometimes it will be a zero and somtimes it will be a one. Clearly an undesireable situation.

Some outputs are called "open collector" or "open drain". In these cases the pullup resistor is essential for establishing the high state. Without them the output will be stuck at zero(low).

#### SparkMan

Joined Apr 27, 2006
8
Pull Up/Down resistors are very important for optimized logic operation.

For example, TTL Logic:

Logic low is defined as 0.8 volts or less, but not negative
Logic High is 2.0 volts to 5 volts

Voltages higher than 0.8 and lower than 2.0 are "Undefined" or Illegal.

TTL Logic Integrated Circuits are pretty much transistors. Well, the definition is Transistor-Transistor Logic.

The inputs, when left untied to any connection, are not always at the same voltage from one to another. Meaning, however the transistor logic is designed inside the chip, determines what the voltage on the unconnected input will "float" at.

Some definitions:

Active low - For an input, a Low is required to cause a change. For an output, When the output is activated, it will be a low

Active High - For an input, a High is required to cause a change. For an output, when the output is activated, it will be a high

For example, a "AND" gate has 2 inputs and one output. To activate the output, both inputs need to be high (Active High). When the output is activated, it will be high.

So the output of the AND gate is only high when both inputs are high.

So where do the resistors come in?

Lets say you want to use 2 Normally open push button switches to trigger the AND gate inputs.

You would connect one end of the switches to the +5V Power supply, and the other ends to the AND inputs.

So theoretically when you push both buttons, the AND output will be high.

But that might not happen. Because the switches are normally open, you have left both inputs in an open circuit, thus they are free to go to the voltages they float at.

So if one input happens to float at 4.2 volts, then it is high, even without pussing the button, thus the logic you want, will not be what you want.

So you use "Pull-Down" resistors, and you connect them between the inputs and ground. They close the circuit and drop the voltage on the inputs to less than .8 volts, or a low.

When you hit a button, the voltage on the pin rises to above 2 volts, enough to be a high.

You also dont use 100 ohm resistors, because the current draw would be too much. You use 1k or higher, typically 5-10k or more.

Pull up resistors do the opposite, they force the inputs to be a high, or greator than 2 volts.

The same applies to outputs as well.

I know that was a lot of talking, so I'll try to simplify things:

When you are not activating any inputs, or doing anything to the circuit, basically when it's in "Standby" mode, you have designed the logic to work a certain way.

In Logic, we have whats called the "Normal" or "Standby" state of the inputs or outputs. The Pull resistors are there to help keep the inputs/outputs at the state they are supposed to be during standby.

What happens when you dont use pull resistors is you get erratic or incorrect logic operation.

#### Mazaag

Joined Oct 23, 2004
255
Originally posted by SparkMan@Apr 27 2006, 12:41 AM

But that might not happen. Because the switches are normally open, you have left both inputs in an open circuit, thus they are free to go to the voltages they float at.

So if one input happens to float at 4.2 volts, then it is high, even without pussing the button, thus the logic you want, will not be what you want.

So you use "Pull-Down" resistors, and you connect them between the inputs and ground. They close the circuit and drop the voltage on the inputs to less than .8 volts, or a low.

[post=16596]Quoted post[/post]​
Sounds great... but I have two questions if you don't mind SparkMan

First, you say that when there are no inputs, the voltages they float at. I don't quite understand what you mean by that.. ? could you give me a schematic as an example.

Secondly, (lets just condiser one input) lets say that the "floating" voltage of one of the inputs is around 3 volts. How does connecting this node to a resistor and to ground cause the voltage at tha tnode to drop down to below 0.8 ? wouldn't it just cause whatever floating voltage we had to drop across the resistor and thats it?

Thanks

#### Papabravo

Joined Feb 24, 2006
21,225
Originally posted by Mazaag@Apr 27 2006, 12:51 PM
Sounds great... but I have two questions if you don't mind SparkMan

First, you say that when there are no inputs, the voltages they float at. I don't quite understand what you mean by that.. ? could you give me a schematic as an example.

Secondly, (lets just condiser one input) lets say that the "floating" voltage of one of the inputs is around 3 volts. How does connecting this node to a resistor and to ground cause the voltage at tha tnode to drop down to below 0.8 ? wouldn't it just cause whatever floating voltage we had to drop across the resistor and thats it?

Thanks
[post=16622]Quoted post[/post]​
No. A TTL input is actually the emitter of a multiple emitter transistor. It wants to source current into some kind of load. The amount of current is a "standard TTL load" of 1.6 mA. The biggest pulldown resistor you could use would be 0.8V/1.6 mA or 500 ohms. A smaller resistor would pull the voltage to a lower value. A larger resistor would allow the input to exist in the no-mans land between 0.8V and 2.0V Another kind of current sinking load would be an NPN transistor in an open collector gate or the NPN transistor in the bottom of a totem pole output. One problem with TTL totem pole outputs is that they have a hard time sourcing current and the have a hard time getting all the way to the 5V rail.

The drawback of a pulldown in TTL is that it wastes power. It is much more common to pull TTL inputs up. Then the emitters on the inputs are not able to source current.

It's all different with CMOS. Those inputs are high impeadance; they neither sink nor source any appreciable current.

#### Gorgon

Joined Aug 14, 2005
113
Originally posted by Papabravo@Apr 27 2006, 08:55 PM
It's all different with CMOS. Those inputs are high impeadance; they neither sink nor source any appreciable current.
[post=16626]Quoted post[/post]​
The pullup/down resistor is just as important here (or more in fact) to define the logic level. If you let a cmos input float it may do that to the midpoint between + and -. In this state it 'opens' both the positive and negative drive path inside the circuit, resulting in a much higher than normal current flowing through it.

It is always important to read the datasheet for the different parts, to see what to do with unused inputs. Normally you need to terminate all inputs one way or the other for the normal logic parts/ families.

TOK

#### SparkMan

Joined Apr 27, 2006
8
Originally posted by Mazaag+--><div class='quotetop'>QUOTE(Mazaag)</div><div class='quotemain'>Sounds great... but I have two questions if you don't mind SparkMan

First, you say that when there are no inputs, the voltages they float at. I don't quite understand what you mean by that.. ? could you give me a schematic as an example.

Secondly, (lets just condiser one input) lets say that the "floating" voltage of one of the inputs is around 3 volts. How does connecting this node to a resistor and to ground cause the voltage at tha tnode to drop down to below 0.8 ? wouldn't it just cause whatever floating voltage we had to drop across the resistor and thats it?

Thanks
[post=16622]Quoted post[/post]​
[/b]

<!--QuoteBegin-Papabravo

No. A TTL input is actually the emitter of a multiple emitter transistor. It wants to source current into some kind of load. The amount of current is a "standard TTL load" of 1.6 mA. The biggest pulldown resistor you could use would be 0.8V/1.6 mA or 500 ohms. A smaller resistor would pull the voltage to a lower value. A larger resistor would allow the input to exist in the no-mans land between 0.8V and 2.0V Another kind of current sinking load would be an NPN transistor in an open collector gate or the NPN transistor in the bottom of a totem pole output. One problem with TTL totem pole outputs is that they have a hard time sourcing current and the have a hard time getting all the way to the 5V rail.

The drawback of a pulldown in TTL is that it wastes power. It is much more common to pull TTL inputs up. Then the emitters on the inputs are not able to source current.

It's all different with CMOS. Those inputs are high impeadance; they neither sink nor source any appreciable current.
[post=16626]Quoted post[/post]​
[/quote]

Papabravo has added something I forgot to mention: the maximum currents, and he also mentioned "sinking" and "sourcing".

~Tries to remove dust off of the TTL theroy I learned~

If I remember correctly, sinking means current flows into an input/output, and sourcing means current flows out of the input/output.

Pull resistors are properly calculated using these values.

For inputs, 1.6mA is the sinking, and 40uA is sourcing

For outputs, 800uA is sinking, 16mA is sourcing.

I am still dusty on TTL, so there is a lot more to the calculations, but I think this will give you somewhat of an Idea.

When I said "when there are no inputs, the voltages they float at" I mean that the inputs are an "open circuit", meaning that electrically, there is nothing connected to them. When this occurs, the voltages on the inputs are floating (and nobody knows execatly where).

This is a schematic representation of an OR gate:

Why and where the voltages float, I don't fully remember, but as you can see, the inputs go directly to emiters of transistors, just like Papabravo said.

I son't know if this would be a correct analogy, but if you know anything about transistor amplifiers or transistor circuits, the pull resistors act kinda like biasing.

Hope that clears some things up better for you

#### skypher

Joined Apr 16, 2006
16
Originally posted by SparkMan@Apr 28 2006, 03:30 AM
If I remember correctly, sinking means current flows into an input/output, and sourcing means current flows out of the input/output.

Pull resistors are properly calculated using these values.

For inputs, 1.6mA is the sinking, and 40uA is sourcing

For outputs, 800uA is sinking, 16mA is sourcing.
How can you have two different values for each current direction? How do you set the direction? Resistors won't do that.

Isn't this about whether the resistor is placed in series with ground ("sinking") or Vcc ("sourcing")?

#### hema.alla@incat.com

Joined Nov 6, 2006
1
How do we caliculate value of pulldown resitor?

#### antseezee

Joined Sep 16, 2006
45
Pull Up/Pull Down resistors are placed into circuits for the purpose of setting a high or low, rather than letting the pin float. Sometimes, a chip will not operate if a pin is floating because there is not a definitive voltage value. No defined voltage value means the chip receives no logic. Some pins can be left floating; depends on the data sheets.

Usually you place a VCC above the resistor, with the other end of the resistor connecting to the pin. Sometimes DIP switches or a pushbutton are placed nearby to ground. A good setup is to link ground to a DIP switch, then place a pull-up resistor right of the switch with VCC above it. This way, you can alter between a HI voltage or a LO quite easily without requiring a STDP switch.