> I'm probably stupid, let's summerize:
> pin CA : is common anode from my 7-segment LED
>
> Keith said :
> /-->---[LED|>]-- RESISTOR --->--- pin /a
> +5V -- pin CA --+-->---[LED|>]-- RESISTIR --->--- pin /b
> \-->---[LED|>]-- RESISTIR --->--- pin /c
>
> Mike said :
> /-->---[LED|>]--->--- pin /a
> +5V -- RESISTOR -- pin CA --+-->---[LED|>]--->--- pin /b
> \-->---[LED|>]--->--- pin /c
No, I didn't mean one resistor for the entire display. You need one
resistor per SEGMENT, as Keith said. My example was a general
discussion of how a resistor is needed to limit the current through
one LED.
> That's true, I usually see limiting current resistor placed before a LED
> and not after. Two much current entering the LED with burn it.
Well, not exactly.... :) In a series circuit, it doesn't matter
whether you put the resistor before or after another component; the
current through all components will be the same. These two generalized
circuits are equivalent:
+5V --> RESISTOR --> LED -->
+5V --> LED --> RESISTOR -->
The difference here is that the display contains _eight_ separate LEDS
(7 segments and one more for the decimal point). You must treat each
segment as a separate LED, with its own resistor. Since the display
has a common anode, the only choice is to add a resistor to each
segment's cathode pin.
> For LTS-546AP:
> - Foward Voltage Per Segment : typ=2.1V max=2.6V (test cond Ir=20mA)
> - Reverse Voltage Per Segment : 5V <=== what's that ?
>
> * Who is right ?
They are both right - they mean different things.
The forward voltage per segment (typical 2.1V) is how much voltage is
"dropped" (appears) across the LED when the stated current is flowing
through it (20mA in this case). Thus, if you apply a voltage through a
series resistor, such that a 20mA current is flowing through the
circuit, then connect a volt meter across the LED, the meter will read
about 2.1V.
The reverse voltage per segment (5V) is the MAXIMUM voltage you can
apply if you connect the LED backwards (i.e., with the positive
voltage to the cathode). Anything higher will destroy the LED.
> * Can I connect the common anode to 3.3V or it must be fixed 5V ?
Since the LED's forward voltage is specified at 2.1V, you may use 3.3V
to power the display.
To calculate the series resistors (one per segment), subtract 2.1V
from the available 3.3V, leaving 1.2V to drop across the resistor.
Ohm's Law then gives a value for the resistors (assuming a 10mA
current, which should make the display bright enough):
R = E / I
R = 1.2V / 0.01A
R = 120 Ohms
So you will need eight 120 Ohm resistors. If you want a brighter
display, choose a higher current (but less than 20mA) and recalculate.
If I may suggest.... As you are discovering, there is a lot to learn
about how electronic components work, and how to understand data
sheets. It would be worthwhile to read an introductory book on basic
electronics and logic-level circuits. This stuff isn't hard, but it
can be very confusing if you jump right in the middle of it.
Regards, Mike
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