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-   -   Using a NAND gate; What's my mistake (http://forum.allaboutcircuits.com/showthread.php?t=87432)

wongojack 07-19-2013 07:04 PM

Using a NAND gate; What's my mistake
 
I purchased this item:

NTE7426 IC TTL Quad 2−Input High Voltage Positive NAND Gate

http://www.radioshack.com/product/in...uctId=20797106

datasheet: http://nte01.nteinc.com/nte%5CNTEMas...A?OpenDocument

Its also sold at Fry's

I'm giving the item because I'm thinking it may have something to do with my problem, and I want to find out if someone has tried to use it before.

I'm a database guy who likes to tinker around with new things and my latest discovery area is electronics. I'm mostly motivated by modifying and repairing old video game equipment, so I'm working through a prototype of a controller for the old Intellivision console. There's some good walk-thru's to follow and they require a NAND gate.

Anyway, I've connected the thing up in the simplest way that I can and I can't seem to get any kind of output voltage from the output pins. I must be doing something wrong. Here are the simplest steps that I've gone through to try and produce a result.

1) Insert 7426 into Breadboard across the 2 halves
2) Connect a battery power supply to + and negative rails and then connect + to Vcc (pin 14) and - to Gnd (pin 7)

From here I started complicated and got more basic. I'm pretty confident that I'm reading the pinout diagram correctly, so I don't think that's a problem. The NTE7426 doesn't mark pin 1 explicitly, but the "notch" does a good job of showing that pin 1 is just below it.

Ok, so my plan here was to simply connect wires to the + rail and create two high inputs. Then my experiment continues as I move the wires from + to - (gnd). If I'm thinking about this correctly then two wires connected between + and pins 1 and 2 should be the only scenario where pin 3 produces a low output.

BTW, I'm using a battery supply that has 4 AA batteries in it. I have batteries that produce different voltages available (different degrees of dead), but I was mostly using an output voltage of about 5.2V.

So I connect pins 1 and 2 to the + rail and pin 3 doesn't really do anything. I'm probing around measuring voltage supplied on pin 1 &2 (to ground), and I find the 5.2v on both. So the volts should be low on 3 - problem is that the volts on pin 3 when measured to ground are always low. No matter what combo of + or - connections that I make with my two wires on pins 1 and pins 2.

Its interesting that the millivolts detected on pin 3 do change, but it goes between something like 70mV down to 2mV

The other disconnected input pins show a voltage of 1.6v and the output pins have about 70mV

Just trying to think through what could be wrong, and I've come up with stuff like:
1) Not enough amps
2) Bad IC (although I've tried 2)
3) Some kind of noise

I've tried connecting with resistors instead of wire to pins 1 & 2 but that doesn't change anything. Next I was going to connect VCC with a resistor, but I don't understand why I'd need to. I've also contemplated wiring the unused inputs and outputs.

Anyone spot any common mistakes in my description here?

bertus 07-19-2013 07:09 PM

Hello,

The 7426 nand gate chip has open collector outputs.
You will need a resistor on each output to the + of your power supply.
A resistor between 1K and 4K7 would do for the experiment.

Bertus

Ron H 07-19-2013 07:31 PM

What does the output of your NAND gate connect to?

wongojack 07-19-2013 08:55 PM

Thanks for the replies guys. That was very fast.

I had read another post that talked about open controller outputs. I don't understand the core concept there, but I will try a resistor value in the range you are suggesting. To clarify, the resistor will connect the + to pin 14; correct?

Any links or material you can send me to that will assist my comprehension of the difference between open/closed?

As for what I want to connect the output to . . . I'm going to put output 1y as an input to another gate, same with output 2y they will serve as inputs to 4a and 4b. Any gotchas in doing that?

MrChips 07-19-2013 09:16 PM

There is no such thing as "closed collector".

Imagine you are holding the rope connected to a carillon or church bell.
The rope we call the "collector".
If the rope is not connected we have an "open-collector" with nothing to pull on.

You have to connect the collector to a load.

Ron H 07-19-2013 09:35 PM

Quote:

Originally Posted by wongojack (Post 629113)
Thanks for the replies guys. That was very fast.

I had read another post that talked about open controller outputs. I don't understand the core concept there, but I will try a resistor value in the range you are suggesting. To clarify, the resistor will connect the + to pin 14; correct?

Any links or material you can send me to that will assist my comprehension of the difference between open/closed?

As for what I want to connect the output to . . . I'm going to put output 1y as an input to another gate, same with output 2y they will serve as inputs to 4a and 4b. Any gotchas in doing that?

No gotchas, so long as the gates are in the same logic family.
You might want to look into 7400 NAND gates (7400, 74LS00, 74HC00, or 74HCT00). They don't need pullup resistors.

WBahn 07-19-2013 10:55 PM

Quote:

Originally Posted by wongojack (Post 629113)
Thanks for the replies guys. That was very fast.

I had read another post that talked about open controller outputs. I don't understand the core concept there, but I will try a resistor value in the range you are suggesting. To clarify, the resistor will connect the + to pin 14; correct?

What do you mean by "the + pin". The only pin that makes sense to call it that IS pin 14.

Open-collector means that the output is capable of actively driving the output to a LO level, but it can't pull it up to a HI level. The output looks like it is disconnected under these conditions. There are a number of reasons for using open-collector outputs. The easiest one to see is that you can directly tie the outputs of many chips together onto a shared signal line and they can never be in contention (two outputs fighting because one is trying to drive the line HI while the other is trying to drive the line LO, resulting in high current draw and an undefined logic level on the line). Any of the outputs can pull the line LO, regardless of the state of any of the other outputs. But the only way that the line can be pulled HI is for ALL of the outputs on the line to be HI. This allows one device to control the line as long as all of the other attached devices have their outputs HI.

But if none of the outputs connected to the line can drive the line HI, how does it become HI? That is the roll of the passive pullup device, typically a resistor. By connecting a resistor between the line that the output(s) are connected to and the logic power supply, the resistor pulls the line to a HI level when NONE of the connected outputs is driving LO. The smaller you make that resistor, the faster the output will change from LO to HI and the larger the load it can support. But it has to be large enough so that any one of the devices can sink enough current through it to pull the line LO.

You also don't want to leave logic inputs unconnected. This is particularly true for CMOS logic and less an issue for TTL logic, but it is still a good happen to tie unused inputs to well-defined logic levels (one of the logic power rails is generally used). Leave unused outputs disconnected.

Ron H 07-19-2013 11:10 PM

Quote:

Originally Posted by WBahn (Post 629156)
What do you mean by "the + pin". The only pin that makes sense to call it that IS pin 14.

Wongojack wrote:
Quote:

To clarify, the resistor will connect the + to pin 14; correct?
I think you misread it. It seems clear to me that he is saying that the resistor will connect between the +(vcc) and pin 14.

wongojack 07-19-2013 11:23 PM

Thanks for the clarification Ron. Rephrasing it would be something like "To clarify, the resistor will connect the +5v rail to pin 14; correct?"

From the context of the responses, I'm thinking the answer to that question is yes.

WBahn 07-19-2013 11:25 PM

Quote:

Originally Posted by Ron H (Post 629160)
Wongojack wrote:I think you misread it. It seems clear to me that he is saying that the resistor will connect between the +(vcc) and pin 14.

Pin 14 is ALREADY directly connected to + (Vcc).

Quote:

2) Connect a battery power supply to + and negative rails and then connect + to Vcc (pin 14) and - to Gnd (pin 7)
What is putting a resistor between + and pin 14 going to accomplish?

If he is asking if he should replace the direct connection between + and pin 14 with a resistor, then the answer is a resounding NO!

The recommendations that he received were to place a resistor between the OUTPUTS and Vcc. Pin 14 is NOT an output, it IS the Vcc pin.


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