How to I replace the 12 volt incandescent bulb in my Yaesu G-800DXA rotor
control unit with an LED bulb? The incandescent bulb is a fuse looking
bulb with wires out each end. I tried putting an LED in, it worked for a
while then went out. Do I need a resistor in the circuit?
Joe,
Yes, the value of series limiting resistor is actually highly depended on
the color of the LED. A typical red LED requires between 15-20 mA to
achieve full brightness and a corresponding forward voltage drop of about
1.7 VDC. On the other hand, some super white and blue LEDs have forward
voltage (Vf) drops near 4.0 VDC. I would first measure your 12VDC supply
and ensure it is really 12 VDC and not some other value. Once this value is
known, you are ready to proceed with your computations:
To compute the value needed, subtract the Vf (forward voltage) from the
measured supply voltage. For a red LED, this value would be close to:
12.0V-1.7V = 10.3 VDC (This represents the voltage drop across the resistor
you are solving for)
Next, using Ohm's law, divide the resulting resistive voltage drop by the
current required to achieve full red illumination:
10.3V/20 mA = 10.3V/0.02 A = 515 ohms. Since 515 ohms is not a standard
value for 5% tolerance resistors, pick one close -- like 560 ohms. You now
have a properly operating LED that will last nearly forever, provided the
supply voltage remains reasonably constant.
But what if you want a super bright white or blue LED? Typically, the Vf
value will be close to 4.0 VDC. So, let's solve for the correct resistor
value needed for use with a super-bright LED:
12.0V - 4.0V = 8VDC.
Next, using Ohm's law, divide the resistive voltage drop by the current
required to achieve full white illumination:
8.0V/20mA = 8.0/0.02 A = 400 Ohms. Again, find a 5% resistor close to 400
Ohms -- like 390 Ohms.
But you may ask, how do I know the real Vf value for the LED? Well, you can
experiment and find it or more easily...when you purchase an LED from Radio
Shack, Mouser, Allied, Newark, Digi-Key, etc., the Vf value will always be
stated on the blister pack or the manufacturer's data sheet. Use that Vf
value.
Now, you may also be asking about the resistor wattage size. Let's use
Watt's law to calculate the power dissipation of the red LED's
current-limiting resistor...
If we choose a 560 Ohm resistor, we can closely compute the dissipation by
squaring the current (20 mA), and multiplying it by the resistor value
(recall the target value was 515 Ohms, but we substituted 560 and that's
close enough). Here goes:
20 mA = 0.02A. Then, 0.02^2 x 560 = 0.22 watts. At a minimum, a 1/4 watt
resistor should be used, or even a 1/2 watt resistor since were getting to a
marginal break point for the 1/4 watt resistor.
If your plan is to use more than one LED to illuminate the meter face, then
you can use these examples above for each LED. On the other hand, it is
possible to stack LEDs in series and use one current limiting resistor but
you will need to re-work all the calculations. It should actually be an
interesting exercise to go through the computations and compare and contrast
both ways.
That's it. To summarize:
1) Measure the true supply voltage;
2) Acquire the proper Vf voltage from the LED's data sheet;
3) Always observe correct LED polarity;
4) Choose a resistor close in value;
5) Choose a wattage of resistor that is at least equal to the amount being
dissipated by the resistor. Going up in wattage value in marginal cases is
good engineering practice.
Paul, W9AC
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