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Basic setup question, trying to use Or Gate

W

WW

Jan 1, 1970
0
Hello,

I have a bread board and a 4xAA battery pack, I'm just getting started,
learning about electronics and wanted to try playing around with some
simple gates and switches.

I have connected the red lead to the top line of the breadboard, and
the black lead to the bottom, testing with a multimeter i get about
4.99v.

I then put the DM74LS08N chip on the board, it's a national
semiconductor chip, so i went to their website and found the chip
datasheet, it says pin 7 is ground and pin 14 is VCC so i assumed that
pin 1 was the pin under the little circular marker and set power and
ground accordingly.

For fun, i thought i'd test setting the output from pin 11 to an Led
but the Led lit up. I didn't expect this, i checked with the
multimeter, at pin 14 the voltage is 5.03, at pins 13 and 12 (which
aren't connected to anything), the voltage is 1.70v at pin 11 the
voltage is 4.30 (this is one of the gate output pins), which is turning
the Led on (I figure)

If I connect pins 13 and 12 directly to ground the gate behaves as
expected, but why is there voltage on these pins when the breadboard
holes aren't connected to anything.

Should the pins have voltage when not directly connected to anything?
Am i missing something in my circuit setup?

I tried hooking up a switch, it has 8 switches on it , i set the first
two input to Vcc and their outputs to the input pins of the DM74LS08N
but when the switch is outputing 0v the input pin to the chip still has
1.7v (the output pin from the switch has 1.7v too, but if i disconnect
it from the chip it has 0v.

I hope my explaination makes sense,
if there is a better forum/group to ask these questions in please let
me know,

Thanks,
Wilma
 
L

Lord Garth

Jan 1, 1970
0
WW said:
Hello,

I have a bread board and a 4xAA battery pack, I'm just getting started,
learning about electronics and wanted to try playing around with some
simple gates and switches.

I have connected the red lead to the top line of the breadboard, and
the black lead to the bottom, testing with a multimeter i get about
4.99v.

I then put the DM74LS08N chip on the board, it's a national
semiconductor chip, so i went to their website and found the chip
datasheet, it says pin 7 is ground and pin 14 is VCC so i assumed that
pin 1 was the pin under the little circular marker and set power and
ground accordingly.

For fun, i thought i'd test setting the output from pin 11 to an Led
but the Led lit up. I didn't expect this, i checked with the
multimeter, at pin 14 the voltage is 5.03, at pins 13 and 12 (which
aren't connected to anything), the voltage is 1.70v at pin 11 the
voltage is 4.30 (this is one of the gate output pins), which is turning
the Led on (I figure)

If I connect pins 13 and 12 directly to ground the gate behaves as
expected, but why is there voltage on these pins when the breadboard
holes aren't connected to anything.

Should the pins have voltage when not directly connected to anything?
Am i missing something in my circuit setup?

I tried hooking up a switch, it has 8 switches on it , i set the first
two input to Vcc and their outputs to the input pins of the DM74LS08N
but when the switch is outputing 0v the input pin to the chip still has
1.7v (the output pin from the switch has 1.7v too, but if i disconnect
it from the chip it has 0v.

I hope my explaination makes sense,
if there is a better forum/group to ask these questions in please let
me know,

Thanks,
Wilma


The notch in the IC package denotes the pin 1 end. The pins are numbered
sequentially counter clockwise from the notch.

There is no case in this logic family where an open circuit is defined as a
logic zero. An open input is undefined and not allowed even on unconnected
gates.

The allowable input voltage is 5 volts plus or minus 5% so 4.75V to 5.25
volts
is the correct range. Four 1.5V batteries is 6 volts.

Finally, the necessary current to drive your LED should be compared to the
current that is available from the gate. Typically, the transistor that
sinks
current to ground, in the totem pole structure, is larger than the
transistor
that pulls the output high. This means that a logic zero is often used to
light
the LED. A driver is required when the LED requires more current than can
be supplied directly.
 
J

John Popelish

Jan 1, 1970
0
WW said:
Hello,

I have a bread board and a 4xAA battery pack, I'm just getting started,
learning about electronics and wanted to try playing around with some
simple gates and switches.

I have connected the red lead to the top line of the breadboard, and
the black lead to the bottom, testing with a multimeter i get about
4.99v.

I then put the DM74LS08N chip on the board, it's a national
semiconductor chip, so i went to their website and found the chip
datasheet, it says pin 7 is ground and pin 14 is VCC so i assumed that
pin 1 was the pin under the little circular marker and set power and
ground accordingly.

For fun, i thought i'd test setting the output from pin 11 to an Led
but the Led lit up. I didn't expect this, i checked with the
multimeter, at pin 14 the voltage is 5.03, at pins 13 and 12 (which
aren't connected to anything), the voltage is 1.70v at pin 11 the
voltage is 4.30 (this is one of the gate output pins), which is turning
the Led on (I figure)

If I connect pins 13 and 12 directly to ground the gate behaves as
expected, but why is there voltage on these pins when the breadboard
holes aren't connected to anything.

The inputs to the LS series of gates (or any other variant of TTL) is
not an open circuit, but includes an internal pull up current that
makes open circuit inputs act as if they were pulled up to logic
one's. So you gate is acting normally. If you decide to experiment
with 4000 or HC or HCT any of the CMOS kind of gates, their inputs do
behave as floating lines when left unconnected, and tiny leakage
currents or stored charge, so those inputs are indeterminate when left
hanging. They also draw lots more supply current if the input
voltages are not clean one's or zero's, so operating them this way is
usually a bad idea. But your gate is perfectly happy with its inputs
unconnected, as long as they are not subjected to strong noise that
might bump them down to somewhere between a one and a zero.
Should the pins have voltage when not directly connected to anything?
Am i missing something in my circuit setup?

You would have to look at the internal circuit to see where that
positive voltage is coming from. Take a look at page 2 of:
http://www.datasheet4u.com/html/S/N/7/SN74LS08_TexasInstruments.pdf.html

The LS08 type has a 20k pull up resistor feeding a transistor base and
there is a diode in series with the emitter, raising the base voltage
above the normal 1 diode drop. This voltage gets out through the
input pin through another diode.
I tried hooking up a switch, it has 8 switches on it , i set the first
two input to Vcc and their outputs to the input pins of the DM74LS08N
but when the switch is outputing 0v the input pin to the chip still has
1.7v (the output pin from the switch has 1.7v too, but if i disconnect
it from the chip it has 0v.

You activate the inputs of TTL chips by grounding the input, detouring
the internal pull up current out through the input pin. A logic high
in acts just like a disconnected pin, since both do nothing to the
internal pull up current.
I hope my explaination makes sense,

Me too. :)
if there is a better forum/group to ask these questions in please let
me know,

This is the right one.
 
J

Jasen Betts

Jan 1, 1970
0
I have a bread board and a 4xAA battery pack, I'm just getting started,
learning about electronics and wanted to try playing around with some
simple gates and switches.

I have connected the red lead to the top line of the breadboard, and
the black lead to the bottom, testing with a multimeter i get about
4.99v.
I then put the DM74LS08N chip on the board, it's a national
semiconductor chip, so i went to their website and found the chip
datasheet, it says pin 7 is ground and pin 14 is VCC so i assumed that
pin 1 was the pin under the little circular marker and set power and
ground accordingly.

For fun, i thought i'd test setting the output from pin 11 to an Led
but the Led lit up. I didn't expect this, i checked with the
multimeter, at pin 14 the voltage is 5.03, at pins 13 and 12 (which
aren't connected to anything), the voltage is 1.70v at pin 11 the
voltage is 4.30 (this is one of the gate output pins), which is turning
the Led on (I figure)

If I connect pins 13 and 12 directly to ground the gate behaves as
expected, but why is there voltage on these pins when the breadboard
holes aren't connected to anything.

even though the pins are designated as input there are indirect connections
to the VCC and ground pins inside the chip.
Should the pins have voltage when not directly connected to anything?
Am i missing something in my circuit setup?

the pins should always be connected to something, input pins left unconnected
(or "floating") tend to pick up signals from other parts of a device and
cause wierd effects.
I tried hooking up a switch, it has 8 switches on it , i set the first
two input to Vcc and their outputs to the input pins of the DM74LS08N
but when the switch is outputing 0v the input pin to the chip still has
1.7v (the output pin from the switch has 1.7v too, but if i disconnect
it from the chip it has 0v.

when it's turned off it's not connected to anything, if you want the chip to
see 0V put a 10K (or less) resistor between the chip input pin and ground
and then connect the switch..

the resistor is called a pull-down resistor because when the switch is off
it pulls the voltage on the pin down (close) to 0v,

a different configuration would be to put the resistor to +5V
and the switch between ground and the input pin. this time it would be a
pull up resistor.
I hope my explaination makes sense,
if there is a better forum/group to ask these questions in please let
me know,

I can't think of one.


4xAA cells probably isn't the best way to power a 74LS series chip
they're pretty fussy about getting the right voltage,

generally this means using a higher input voltage (say 8-12V) and a
LM7805 regulator and apropriate filter capacitors.

The 74HCxx and 4xxx series chips are more forgiving about voltage,
but even fussier about having all their inputs connected.

Bye.
Jasen
 
J

Jasen Betts

Jan 1, 1970
0
The allowable input voltage is 5 volts plus or minus 5% so 4.75V to 5.25
volts
is the correct range. Four 1.5V batteries is 6 volts.

and four 1.25V rechargable cells is pretty close to 5V when they're
reasonably fresh, but still probably not a good idea.

Bye.
Jasen
 
E

ehsjr

Jan 1, 1970
0
WW said:
Hello,

I have a bread board and a 4xAA battery pack, I'm just getting started,
learning about electronics and wanted to try playing around with some
simple gates and switches.

I have connected the red lead to the top line of the breadboard, and
the black lead to the bottom, testing with a multimeter i get about
4.99v.

I then put the DM74LS08N chip on the board, it's a national
semiconductor chip, so i went to their website and found the chip
datasheet, it says pin 7 is ground and pin 14 is VCC so i assumed that
pin 1 was the pin under the little circular marker and set power and
ground accordingly.

For fun, i thought i'd test setting the output from pin 11 to an Led
but the Led lit up. I didn't expect this, i checked with the
multimeter, at pin 14 the voltage is 5.03, at pins 13 and 12 (which
aren't connected to anything), the voltage is 1.70v at pin 11 the
voltage is 4.30 (this is one of the gate output pins), which is turning
the Led on (I figure)

If I connect pins 13 and 12 directly to ground the gate behaves as
expected, but why is there voltage on these pins when the breadboard
holes aren't connected to anything.

Should the pins have voltage when not directly connected to anything?
Am i missing something in my circuit setup?

I tried hooking up a switch, it has 8 switches on it , i set the first
two input to Vcc and their outputs to the input pins of the DM74LS08N
but when the switch is outputing 0v the input pin to the chip still has
1.7v (the output pin from the switch has 1.7v too, but if i disconnect
it from the chip it has 0v.

I hope my explaination makes sense,
if there is a better forum/group to ask these questions in please let
me know,

Thanks,
Wilma


I'll mention two points. First, IC pin orientation. You correctly
identified pin 1. It's always the same - heres a diagram: (Imagine
the pins pointing away from you and into the monitor screen)

-------------
|
) Any DIP IC
|
-------------
| | | |
Pin1



Next, your breadboard supply: If you want to power the breadboard
from the AC supply to save batteries, get a 12 volt DC wall wart,
and an LM7805 voltage regulator in a TO 220 package. Wire it up like
this:

------
+12 ---+---Vin| 7805 |Vout---+--- +5 volts
| ------ |
[C1] | [C2] C1 = .33 uF
| | | C2 = .1 uF
Gnd ---+---------+-----------+


_______
| o |
|_______|
/_______/|
| ||
| 7805 ||
|_______|/
| | |
| | |
Vin Gnd Vout

Have fun!
Ed
 
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