windozeuser

No it's not Bill Gate's ghost traveling around to assume the posistion of the anti-Christ. :lol:

I'm talking about the high impedence inputs of logic gates. What is meant by "they are floating", and can anyone please explain a pull up and pull down resistor? Also, if someone could help to explain impedence better, I feel a little shaky on it's definition. I know it's the total opposition in an AC circuit taking in account Inductance, and capacitance. However, why is it still measured in ohms?

Thanks

JimB

An input is said to be "floating" if it has no obvious connection to another external point. If it is a high impedance input, it can "float" to any voltage it likes due to stray current through high resistance leakage paths.

A "pull-up" resistor is used to pull an input (or sometimes an output) up to the supply voltage. It is just a resistor connected between the supply rail and the input. It will hold the input at logic 1.

A "pull down" resistor connects the input to the 0v line, so holding it at logic 0.

Impedance is the vector sum of resistance and reactance.

Z = sqrt(R^2 + X^2)

Resistance is measured in Ohms, so is reactance, so impedance cant help but be in Ohms!

JimB

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Papabravo

I'd like to expand on Jim B's post just a bit. Impeadnace is a complex quantity, not in the sense of difficult to understand, but in the sense that it is a complex number. Complex numbers have a real part and an imaginary part. The imaginary part is a real number multiplied by the square root of -1. It is usually written as the letter i by mathemeticians or j by electrical engineers.

As Jim B mentioned the units of impeadnace are ohms and when you take the square root of the sum of the squares of the real part (R) and the imaginary part(X) you are calculating the MAGNITUDE of the impeadance. But wait there's more. Impeadance also has an angle associated with it. The angle is the inverse tangent or arctangent of the imaginary part divided by the real part.

The following are all 50 ohms impeadance, can you figure out why?

upand_at_them

Impedance is a complex number? I didn't know that. Interesting.

Mike

JimB

PapaBravo

Are they Elbonians in your avatar?

Only an Elbonian could have a negative resistance (-35.36 + j35.36), I assume you mean (35.36 - j35.36).

JimB

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Papabravo

Jim B

Yes they are Elbonians. I thought they were kinda cute when I saw them in the strip for the first time, and I still think they're an amusing concept.

It often happens that the mathematics can be correct even when the physics is counter intuitive. There are devices which exhibit a negative resistance characteristics.

So

35.36 - j35.36 = 50 angle -45

Right?

audioguru

A floating input on an IC can cause it to oscillate. A floating input on a 74HCxx IC can cause the output transistors to draw a high current and over-heat because they might get biased halfway and both be turned on.
74HCxx ICs have a max switching time rating of 1us unless they are Schmitt-Triggers.

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Uncle $crooge

JimB

OK, I was thinking of "simple" components rather than some semiconductor devices, also not following the maths closely. I think you are correct on all counts.


Audio, I once left a couple of inputs to a PIC floating, it took me ages to suss out what was causing the lock-ups/resets/errors/ridiculous behaviour every few minutes!
A couple of pull down resistors fixed it instantly.

JimB

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RadioRon

Here's a floating input story. I built my first computer starting in 1977. It was a couple of S100 board kits on a home made backplane board. The system expanded slowly over about two years to the point where I had about 48K of memory and an 8 inch floppy disk drive. I used to use it for word processing a lot, like working on school reports for example. Sometimes the computer would frustrate me by crashing for no apparent reason, but being impatient I just learned to "save" my files very often so not much work would be lost. But the inevitable happened when I was about 7 hours into a report one day without having done a backup and the system crashed. I was livid and vowed there and then to finally debug this occasional crashing. Well, the next day I started poking around with an old oscilloscope but didn't find anything. I continued to poke at it one way or another for about a month. I finally found the problem in that my I/O board (not designed by me) had a bus driver gate connected to the backplane (on a control line of some sort) and the input to the driver gate was floating. The designer used LSTTL and assumed the input would always appear high because there was no input current. This was true most of the time, but when the gate heated up after a few hours, it would swing the other way and crash the system. The moral of the story is always terminate your inputs, CMOS or otherwise.

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RadioRon

RadioRon

Here's a floating input story. I built my first computer starting in 1977. It was a couple of S100 board kits on a home made backplane board. The system expanded slowly over about two years to the point where I had about 48K of memory and an 8 inch floppy disk drive. I used to use it for word processing a lot, like working on school reports for example. Sometimes the computer would frustrate me by crashing for no apparent reason, but being impatient I just learned to "save" my files very often so not much work would be lost. But the inevitable happened when I was about 7 hours into a report one day without having done a backup and the system crashed. I was livid and vowed there and then to finally debug this occasional crashing. Well, the next day I started poking around with an old oscilloscope but didn't find anything. I continued to poke at it one way or another for about a month. I finally found the problem in that my I/O board (not designed by me) had a bus driver gate connected to the backplane (on a control line of some sort) and the input to the driver gate was floating. The designer used LSTTL and assumed the input would always appear high because there was no input current. This was true most of the time, but when the gate heated up after a few hours, it would swing the other way and crash the system. The moral of the story is always terminate your inputs, CMOS or otherwise.

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RadioRon

mstechca

The posters above have your answer, but to help you understand further, consider these points:

1. a "floating" input is an input connected nowhere. Don't get it confused with a high impedance input. In simplest terms, impedance is resistance.
You can tell that an output is in an impedant state when an LED lights dimly when it is connected the right way between output and ground.

2. In digital terms, a pull-up and a pull-down define the default logic values for a given section. For example, if you connect a resistor between +ve and an input of an inverter (this is a pull-up resistor), that inverter has a default value of 1. If you instead connect a resistor between -ve and an input of the inverter (pull-down resistor), the default value is 0.

"Default" is the value that is assigned to the logic gate, should no input be defined. To override a default value, connect an input to +ve, or -ve.

3. When dealing with pull-up or pull-down resistors, you need to choose a wise value. Anything between 1K and 5K is ok. If you go too low, then current consumption of the circuit will be high when the logic input is the opposite of the default value, since the resistor is then connected between +ve and -ve. If on the other hand, the value is too high, the logic value might not be defined correctly.

In some cases, a logic "1" is defined between certain voltage levels, and a logic "0" is defined between another set of voltage levels.

I suggest a 1K pull-up or pull-down resistor because It works well with alot of logic gates.

Before you start taking everyones suggestion, you need to think about what you want to do with logic gates and why. In some cases, pull-up and pull-down resistors may not apply.

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audioguru

A 74xx TTL logic low input is a max of 0.8V at 1.6mA which needs 500 ohms. A low input level of 0.4V is more desirable which needs 250 ohms.
Why just guess? Look it up.

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Uncle $crooge

windozeuser

Thanks for all help I think I understand it now(atleast better).

ljcox

By -ve, he means ground or 0 Volt. -ve implies a negative voltage, ie. a voltage that is more negative than the ground potential.

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Len

mstechca

Because I don't know which logic family he intends to use (74xx or cmos).

Yes. The reason why I use -ve more is because I am referring to the -ve terminal of the battery, which can be considered as ground in logic circuits.

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