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# Mystery voltage on input pins of 7408N (AND)

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In building another project I need to whip together some AND gate logic. Basically I want to produce high output if and only if my two inputs are high. Easy stuff, right? Everything seemed to be going well, but my output wasn't what I expected. In diagnosing the circuit (I stripped it down to the simplest circuit that reproduces the problem - see attached JPG), I noticed that there was something fundamentally weird going on. Here's the setup:

Circuit powered by 9V battery, running through a 7805CT +5VDC regulator (to power the 7408 gate). Nothing else interesting related to this problem. the IC is connected to two things: +5DVC and Ground. No other pins are connected to anything (take a look at the picture to see what I mean).

Here's the weird part: there is a roughly 1.65V potential (relative to ground) appearing at every A and B pin, and 3.6V at the Y pin (the 7408 is a quad gate IC, so there are 4 A's, 4Bs and 4Y's for the basic AND formula: AB=Y. I've included the spec sheet in case my explanation isn't clear.) So basically every input pin appears to be OUTPUTTING a voltage of 1.65V!!!

Of course, I would expect the input pins to be, well, input pins...not have a potential appearing at them. It took me quite a while to discover this mystery voltage, which appears at the input pins whether they are connected to anything or not (you can imagine what weird weird things it was doing to the rest of my circuit!).

Anybody ever seen this before? I'm baffled! I've also tried every other 7408 I have on hand (for a total of 10), same thing. I even tried a NAND to see if the mystery voltage appears at its input pins. And yes it does. So, is there some other circuitry I need? I figured I could just run two input voltages into the gate's inputs and get an answer. It's not that simple is it? Any help is appreciated! Thanks!!

7408-AND-49146.pdf

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The 7408 is an old fashioned (yours is more than 24 years old, I remember 1981) TTL logic IC. TTL means Transistor-Transistor-Logic. The input is the emitter of an NPN transistor that floats high if there isn't anything pulling it down to a low.

Each input needs 1.6mA to pull it down to a low logic level of 0.8V max. For noise immunity the inputs should be pulled down to 0.4V max.
What drives the inputs? You can't just leave them floating high. The input current is too high to be driven by an ordinary Cmos gate. A TTL gate's low output is at least 16mA to 0.4V max so can drive 10 TTL inputs.

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Thanks, guru. I'm recently getting back into electronics after a long break (since college - about 15 years). I discovered after I purchased 10 ea. of the 74xx series (AND, NAND, OR NOR, XOR and hex inverters) from Jameco that they're old, cantakerous TTL gates.

That being said, you're saying some stuff that I don't get.

The 7408 is an old fashioned (yours is more than 24 years old, I remember 1981) TTL logic IC. TTL means Transistor-Transistor-Logic. The input is the emitter of an NPN transistor that floats high if there isn't anything pulling it down to a low.

Each input needs 1.6mA to pull it down to a low logic level of 0.8V max. For noise immunity the inputs should be pulled down to 0.4V max.
What drives the inputs? You can't just leave them floating high. The input current is too high to be driven by an ordinary Cmos gate. A TTL gate's low output is at least 16mA to 0.4V max so can drive 10 TTL inputs.

I guess it's the "pull it down" comments that are throwing me. What do you mean? How do you "pull down" voltage? Throw it across a resistor? There's no CMOS gate involved in my circuit (that I know of).

Basically, each input to the AND is a 5V input. I figured (naively, I guess) that if both were 5V coming in, that the output would go high and 5V (give or take) would be output. That's what I want: two inputs AND'ed together to produce an "answer." That's  it.

I guess I need to research more about how these old 74xx ICs work so I can figure out how to get them to play nicely in my circuit. Maybe I need other circuitry to get them to cooperate.

Maybe I should order the basic logic series in CMOS form from Jameco and just cut my losses?

Thanks for your reply. I feel so lost!!
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That's correct. Your circuit needs something on its inputs to pull down the TTL gates' inputs to about 0.4V with 1.6mA. Ohm's Law calculates each resistor to ground to be 250 ohms, but 270 ohms will be OK.
Since your inputs are 5V, then each 270 ohm resistor will draw 18.5mA or a whopping 74mA for four of them.

If the gate doesn't drive much current into a load, a CD4081 Cmos quad AND gate package won't draw any supply current and works fine with 22M to ground on its inputs. Use 100k or 1M resistors to be reasonable. ;D

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Okay, I'm just going to have to play the "I'm an idiot" card here and ask exactly what the heck to do to wire up this 7408 AND gate. I've got +5 VDC going into Vdd, and Vss wired to ground. I want to use as inputs +5V or nothing. However, it doesn't seem to matter what I run in, since there is a mystery voltage on the input pins (even with NOTHING connected to them).

Audioguru suggests using a pull down resistor to pull the inputs to about .4V, and I'd be happy to do that, but don't know what to do. I've googled my fingers off trying to find a circuit diagram to show me exactly a properly wired (with all the "pull down" stuff in place) 7408 gate looks like.

So, my question is: what do I actually do on the breadboard to make this "pull down" resistor thing happen?

Thanks.

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what do I actually do on the breadboard to make this "pull down" resistor thing happen?

You decided to use old fashioned TTL with its high input current requirement. I told you what to do: Add a 270 ohm resistor from each input to ground.
Your +5V input signals must be able to supply a total of 74mA to the resistors.
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I got it now. Thanks for the help. The only reason I decided to use a 7408 is that's all I have on hand.

Thanks again to audioguru for the help. Somehow all of the great advice in guru's posts boiled down to "connect a 270 Ohm resistor between each input pin on the TTL Gate IC and ground."

I've posted a more detailed explanation with a picture on my blog. http://jstevenperry.com/blog/?p=37

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