 |
|
|
|
|
|
| |
| |
|
|
|
|
| |
| |
|
|
>> I always thought of it as logic high meaning "connected to the +V
>> rail" and logic low meaning "not connected to anything".
>
> Logic "high" and "low" are just names given to a certain agreed range of
> voltages.
Indeed. You can use any two states you want, so long as it's easy to
distinguish them, and all the components are using the same rules.
> The difference between "not connected to anything" (floating)
> and "logic low" is very important when working with data busses.
There is that...
>> From the looks of things, it seems that actually logic low means
>> "connected to the -V rail". So it's like a bipolar setup...
>
> But don't assume a "logic low" output will be exactly -V, it is only
> guaranteed to be within the range of voltages given for "logic low" in
> the data sheet.
Heh. It gets weirder and weirder...
>> Yes, that's the other fun thing. Depending on which type of gates
>> you're using, they supply about 20 mA. But some of the LEDs I'm
>> looking it draw 40 mA...
>
> What you mean is, they draw *up to* 40 mA?
No idea. The spec sheet just says "forward current".
> How much they actually draw
> depends on what value series resistor you put in there. Using Ohm's law
> and the V-I curve of the LED from the datasheet you can work out exactly
> what resistor size you need for a given current.
Heh. And here I was thinking it's just a case of "if the voltage is less
than X, you don't need a resistor"...
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
Hey, check this out:
http://www.maplin.co.uk/Media/PDFs/N24AZ.pdf
content of this datasheet means! :-D
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
On 5/19/2010 6:58 AM, Invisible wrote:
> scott wrote:
>>> "74LS series: Inputs 'float' high to logic 1 if unconnected."
>>>
>>> Oh crap. Well that would explain a thing or two! >_<
>>
>> You should never be leaving any inputs as floating anyway, that's just
>> asking for trouble (for example if you use a different IC series
>> later). Best to always tie your inputs to whatever you want them to be.
>
> Sure. But I'm sitting with the IC between my fingers, poking the pins
> with the end of a wire. Not so easy to tie it all down just for a quick
> test.
So, you put it on a breadboard with each pin tied to a pull-up or
pull-down resistor, and probe with the opposite sense. If your resistor
is pull down (i.e. connected to GND) then use VCC to probe (or better
yet, just get a dip switch ...) If the resistor is connected to VCC,
then use ground to set the signal to 0
> So it seems that every input pin is supposed to be connected to one rail
> or the other. (Or an output, which I guess is going to be the most
> common case...)
Yes. 5v is high, ground is low, not connected is neither of these,
though TTL does float high, a non-connected input on a CMOS chip can
lead to erratic behavior, or even destructive oscillation. It also
leaves the chip wide open for ESD damage. FETs are interesting in that
they don't sink or source any current on their inputs. without anywhere
for the voltage to go, a charge can stay on that input. That charge can
go dangerously (for the chip) high can cause a dielectric breakdown of
the gate's insulating layer. Some CMOS gates will have a high-value
resistor tied to ground on the input to bleed away some of the stray
voltage as a safety measure for ESD. Don't rely on that.
> The page also claims that you can configure an output pin to drive an
> LED so that it lights when the pin goes *low*... Man, electronics is weird!
LED anode connected to VCC, cathode connected to out. ;) Not too strange
when you think about it. A logic output is either connected to VCC or GND.
One other thing to be cautious of: If a TTL chip has an open collector
output, make sure to use a pull-up resistor before connecting it to a
CMOS chip. High is floating on an open collector output.
Oh! and by the way: with TTL chips VCC is *critical* due to the nature
of the bipolar transistors used in the gates, they must have a regulated
voltage. Look at Falstad's circuit simulator to figure out why. They
have a resistor set to a particular value to bias the transistor. If the
supply voltage goes out of range, the transistor is no longer properly
biased, which will cause misinterpretation of signals.
CMOS doesn't require a bias, so can run on any voltage (within limits)
--
~Mike
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
clipka wrote:
> Texas Instruments has a nice collection of 74xx data sheets in a single
> proper PDF file at <http://focus.ti.com/lit/ug/scyd013b/scyd013b.pdf>
> ("Digital Logic Pocket Data Book"), apparently covering all their
> current products (as of 2007) in the 74xx line, from single-gate 5-pin
> thingies like the 74x1G00 (single NAND gate) to 100-pin monsters like
> the 74x32543 (36-bit registered bus transceivers with 3-state outputs).
Ah, neat!
Most of the books I've seen only cover the 7400, 7401, 7402, etc., and
don't mention things like the 741G00. Could be useful to know. (Except
that Maplin don't seem to sell this stuff... *sigh*)
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
While we're here, does anybody comprehend what VCC, VSS, VDD and related
symbols mean?
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
On Wed, 19 May 2010 16:16:28 +0200, Invisible <voi### [at] dev null> wrote:
> While we're here, does anybody comprehend what VCC, VSS, VDD and related
> symbols mean?
http://en.wikipedia.org/wiki/IC_power_supply_pin
--
FE
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
>> What you mean is, they draw *up to* 40 mA?
>
> No idea. The spec sheet just says "forward current".
It's probably under a section titled "typical operating conditions" or
"absolute maximum ratings". There should also be a graph showing a curve of
current against voltage, from zero up to the absolute maximum current
rating. Usually with LEDs the light output is fairly linear with respect to
current, but certainly not with voltage.
>> How much they actually draw depends on what value series resistor you put
>> in there. Using Ohm's law and the V-I curve of the LED from the
>> datasheet you can work out exactly what resistor size you need for a
>> given current.
>
> Heh. And here I was thinking it's just a case of "if the voltage is less
> than X, you don't need a resistor"...
If you want to run the LED at 10 mA you look up on the V-I curve in the LED
datasheet what forward voltage corresponds to 10 mA. Say for example 2.8 V.
If your logic output is 5 V (and the other side of the LED is connected to
0V), then you want 2.8 V across the LED and hence 2.2 V across the series
resistor. So your resistor needs to drop 2.2 V with a current of 10 mA,
that makes it a 220 Ohm resistor.
Connecting the LED directly with no series resistor to a voltage source is
usually a bad idea (even if you think the voltage is correct), because even
slight changes in the voltage cause large differences to the current the LED
draws which might damage your circuit.
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
> content of this datasheet means! :-D
I can send you a longer one if you want :-)
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
scott wrote:
>> the content of this datasheet means! :-D
>
> I can send you a longer one if you want :-)
Why thank you, you're too kind. :-}
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
| |
|
|
>> Heh. And here I was thinking it's just a case of "if the voltage is
>> less than X, you don't need a resistor"...
>
> If you want to run the LED at 10 mA you look up on the V-I curve in the
> LED datasheet what forward voltage corresponds to 10 mA. Say for
> example 2.8 V. If your logic output is 5 V (and the other side of the
> LED is connected to 0V), then you want 2.8 V across the LED and hence
> 2.2 V across the series resistor. So your resistor needs to drop 2.2 V
> with a current of 10 mA, that makes it a 220 Ohm resistor.
>
> Connecting the LED directly with no series resistor to a voltage source
> is usually a bad idea (even if you think the voltage is correct),
> because even slight changes in the voltage cause large differences to
> the current the LED draws which might damage your circuit.
With my old electronics kit, if I'm remembering this rightly, you
connect through a resistor if you're using the 3V power source, and skip
the resistor if you're using the 1.5V source.
Oh, and *don't* connect the LED to the 9V source at all. I did that once
by mistake... that particular LED doesn't do anything now.
Ah, life seemed so much simpler back then. :-(
Post a reply to this message
|
|
| |
| |
|
|
|
|
| |
