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[ts-7000] Re: [electronic] 7-segment LED

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Subject: [ts-7000] Re: [electronic] 7-segment LED
From: Mike Dodd <>
Date: Mon, 31 Oct 2005 18:58:58 -0500
>Concerning resistors (outputs of 74LS47), I don't understand:
>
>The output pin (/a, ..., /g) are sinking current, so the current go from
>the LED to an output pin :
>
>5V -->---[LED|>]---->--- pin /a
>
>So this resistor (located after the LED) won't affect the lighting power
>of the LED, it's just for limiting the current entering inside 74LS47,
>am I right ?

No, you are wrong.

An LED is a light-emitting DIODE, which means it will pass as much current 
as you try to push through it (until it is destroyed). The purpose of the 
resistor is to limit the current to a safe level.

If you look at the spec sheet for the LED display, you will probably see a 
specification that describes "forward voltage drop." This is the voltage 
you can measure across the diode when current is flowing through it. I'm 
not sure what this is, but it's probably about 2V to 3V. This voltage is 
approximately constant, regardless of the current flowing through the LED. 
So to limit the current, you need a resistor in series. The circuit is:

5V --> RESISTOR --> LED --> pin /a

To calculate the value of the resistor do this:

1. Determine the supply voltage (let's say 5V).
2. Subtract the "forward voltage drop" from the spec sheet ((let's say it 
is 2V).
3. The result from #2 (3V) is the voltage that is dropped across the resistor.
4. Use Ohm's law R = E/I. E is the voltage from #2. Choose a current for I; 
let's pick 10mA (0.010A), which is a safe value. Therefore R = 3/0.010 = 
300 ohms.

A resistor in the range of 270 ohms to 330 ohms should be good, if our 
assumptions are correct. But the important thing is to READ THE SPEC SHEET 
TO DETERMINE THE ACTUAL FORWARD VOLTAGE DROP ACROSS THE LED. That will 
determine the voltage in step 2, and thus will determine the value of the 
current-limiting resistor.

>About voltage, how to know how many volts have we in pin /a; is 15V enough ?

The minimum required voltage is the forward voltage drop across the diode, 
plus a bit more. So if the diode drops 2V, you need at least 2.5V (through 
the appropriate current-limiting resistor). You could use 15V, but your 
resistor value would be much larger, and you'd be dissipating a lot of 
power in the resistor, so I suggest you go with 5V.

>V        Off-State Output Voltage /a to /g : 15V
>  O (off)
>I        On-State Output Current /a to /g : 24mA
>  O (on)

You are misinterpreting the specifications. These probably are MAXIMUM 
values, not what you want to use in normal operation.

If you have a display and a 5V power supply, get some resistors in the 200 
ohm to 470 ohm range, and connect one segment as I showed above, using 
different values for the resistors. Measure the voltage across the segment 
while it's lighted; this is the actual forward voltage drop. Calculate the 
resistor value, then try the display with the new resistor. You can reduce 
the value to make the display brighter, or increase it to make it dimmer. I 
suggest keeping the current below 15mA (0.015A).

Hope this helps.

Mike


------
Mike Dodd - Montpelier, VA
http://www.mdodd.com



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