# AND/Logic Questions

Discussion in 'Electronic Basics' started by [email protected], Oct 26, 2005.

1. ### Guest

Hi, I'm reletively familiar with electronics, and I'm trying to find
some information regarding implementing logic into a circuit I want to
build.

I am having trouble however actually learning how to actually USE an
AND gate. The IC I am using is: 74LS08, a data sheet can be found at:
http://pdf.alldatasheet.com/datasheet-pdf/view/12619/ONSEMI/74LS08.html

It is a simple QUAD 2 Input AND Gate. However, when I try to use it, I
am unable to make it work correctly. I am trying to use anywhere
between 6-9Vdc. The circuit I use basically has two switches, one LED,
and the IC. I have the IC hooked up to a positive and ground/negative
of my power supply, and each switch is hooked up to pins 2 and 3 on the
IC (the input gates of the logic switch), and they are hooked up on the
other side to a positive voltage. The LED is connected to the output of
the AND Gate. When both switches are on, the circuit is closed and the
light is on, however when one switch is off, the light still stays on.

is there anything i am doing wrong? or does anyone have some websites
that might help me? I have been googling this for a while but have had
no luck.

thank's

trevor

2. ### Anthony FremontGuest

1) Tie all unused gate inputs to ground or Vcc
2) Never allow an input pin to float (i.e. use a pull-down resistor {10K
should be fine} on the input pins that you are tinkering with)
3) Always use a current limiting resistor when driving LEDs

Your chip could be latching up or it could be that the input pins have
high enough impedance (like cmos devices) where the stored charge
continues to look like a logic 1 even though you turn the switch off. A
pull-down resistor will fix that.

3. ### John PopelishGuest

Take a look at the first page of the data sheet. The supply voltage
operating range is specified for 4.75 to 5.25 volts.

If you need a gate that operates up to 9 volts, you should look at the
4000 series CMOS family.

http://www.alldatasheet.com/datasheet-pdf/pdf/50864/FAIRCHILD/CD4081.html
Here is another problem. TTL gates are not high impedance voltage
inputs, like CMOS gates are. They are current operated inputs. They
contain an internal pull up current source (a resistor) and you have
to suck that current to very near ground before the input sees a logic
low input. Leaving the input open circuited allows that internal pull
up to bias the input positive. So you have a logic high input whether
your switches are open or closed.

Even if you use a CMOS gate, you will need to add a resistor to the
switched inputs to pull the voltage the opposite way that the switch
does, so there are no open circuit inputs at any time.

The LED is connected to the output of
You also need to make sure that the current to the LED is limited to a
safe value for both the gate and the LED.

4. ### John FieldsGuest

---
There are several things you're doing wrong.

1. The absolute maximum power supply and input voltage you're
allowed to use is 7V, so if you're running higher than that, all
bets are off. The recommended supply for LS is from 4.5 to 5.5V
in order for the device to stay within its spec's, so with your
supply cranked all the way down to 6V, if you put a diode in
series with the positive lead that would drop the voltage down
to about 5.3V which would be fine.

2. With LS, a floating input (which is what you have when you turn
either or both of the switches OFF) is automatically high, so
since the inputs never go low the LED always stays on.

3. The choice of an AND for this application is a bad one.
Not your fault, since you're just learning, so here's why:

LSTTL was designed with outputs that can sink 8mA but can only
source 400µA and still stay within their guaranteed switching
levels. Likewise, LSTTL inputs need to be able to source 400µA to
ground before they're considered to be at a logical zero.

What that means is that if you want to use the AND for your
application, both inputs have to be pulled down to ground with
somewhat less than 2000 ohm resistors in order for the inputs to be
less than Vih (0.8V) while sourcing 400µA each.

Unfortunately, there's nothing you can do with the output except
hook it up to the LED, since it can't source enough current to hurt
the LED (well, maybe that's fortunate

In any case, with an AND in there the circuit would look like this:

+6V>---[1N4001>]---+------+
| |
| O | O
S1| O S2| O
| |
| +--A
| | & Y--[LED>]-+
+---------B |
| | [R]
[1K8] [1K8] |
| | |
GND>---------------+------+---------------+

A good choice of a gate for this application, if you want to show
ANDedness by having the LED come on when S1 AND S2 were in the same
state would be to use the DeMorgan equivalent of the AND, the OR,
like this:

+6V>---[1N4001>]---+------+---------------+
| | |
[10k] [10k] [R]
| | |
| +--A |
| | & Y--[<LED]-+
+------|--B
| |
| O | O
S1| O S2| O
| |
GND>---------------+------+

That way, when S1 AND S2 are made the output of the OR will go low,
turning on the LED using the gate's sinking capabilities.

5. ### John FieldsGuest

On Wed, 26 Oct 2005 10:03:17 -0500, John Fields

DeMorgan equivalent of the AND, the OR, like this:

+6V>---[1N4001>]---+------+---------------+
| | |
[10k] [10k] [R]
| | |
| +--A |
| | & Y--[<LED]-+
+------|--B
| |
| O | O
S1| O S2| O
| |
GND>---------------+------+

Oops...

+6V>---[1N4001>]---+------+---------------+
| | |
[10k] [10k] [R]
| | _ |
| +--A _ |
| | _& Y--[<LED]-+
+------|--B
| |
| O | O
S1| O S2| O
| |
GND>---------------+------+