I just purchased this thiel equalizer, but the green power indicator led was broken off. I powered the unit up and measured the voltage at the leads and get 20vdc. I go to Ratshack and thumb thru the Digikey catalog, almost all LEDs are 2 or 3v. Maybe 1 out of 100 is 12v. No 20v. I end up getting this part
http://www.radioshack.com/product/index.jsp?productId=2062551&cp
2.1v/25mA/6.3mcd. I'm not interested in a direct replacement, any brightness will do. This equalizer is about 15 years old, are 20v leds not made anymore? Can I use the RS part? (I'm suspecting not).
http://www.radioshack.com/product/index.jsp?productId=2062551&cp
2.1v/25mA/6.3mcd. I'm not interested in a direct replacement, any brightness will do. This equalizer is about 15 years old, are 20v leds not made anymore? Can I use the RS part? (I'm suspecting not).
Attachments
Found this looking for info on LEDs, might be of interest to some.
http://www.smithappens.com/picture/542/Leds-and-resistors-circuit-sex-positions.html
http://www.smithappens.com/picture/542/Leds-and-resistors-circuit-sex-positions.html
Hi 
The 2-3Volt you see is the voltage drop caused by connecting the led in a circuit.
You can put almost any voltage through a led... As long as the current doesn't exceed around 20-25mA.
The led you identified would do the job ! Just make sure the LED's longuest lead is connected to + and shortest one to -
Good luck !
The 2-3Volt you see is the voltage drop caused by connecting the led in a circuit.
You can put almost any voltage through a led... As long as the current doesn't exceed around 20-25mA.
The led you identified would do the job ! Just make sure the LED's longuest lead is connected to + and shortest one to -
Good luck !
Is there a resistor in series with the LED? That LED looks quite wide. What is its diameter?
You could substitute any normal LED that will fit as long as you use an appropriate resistor in series. You could also customise it's brightness by changing the resistor.
Start with about 5KOhms and reduce or increase until you get the desired brightness but don't run it too hot or you'll be replacing it soon.
You may find that 20volts will burn your chosen LED if used without a resistor. The average LED cannot handle even a 9volt battery for very long unless you add a resistor in series to limit the current!!
If all els fails then you find these types useful (obviously not in this quantity though!): http://cgi.ebay.co.uk/24V-LED-CHROM...yZ108854QQssPageNameZWDVWQQrdZ1QQcmdZViewItem
You could substitute any normal LED that will fit as long as you use an appropriate resistor in series. You could also customise it's brightness by changing the resistor.
Start with about 5KOhms and reduce or increase until you get the desired brightness but don't run it too hot or you'll be replacing it soon.
You may find that 20volts will burn your chosen LED if used without a resistor. The average LED cannot handle even a 9volt battery for very long unless you add a resistor in series to limit the current!!
If all els fails then you find these types useful (obviously not in this quantity though!): http://cgi.ebay.co.uk/24V-LED-CHROM...yZ108854QQssPageNameZWDVWQQrdZ1QQcmdZViewItem
Did not know that bit about the longer/shorter leads, thanks. If I place my meter on the broken led so that I get the +20v indication on the meter, then the leg with the red probe is + and the black probe is -, right? Not sure if the board is marked.
I also found this led at RS that says is frequently used for power indication. I see the mcd is 620 and the mA is 30. Seems like a significant difference in mcd, which would do a better job, you think?
I don't know if its common knowledge around these parts, but if you walk into a RS and they don't have the part you want you can pay for it right then and there and they will mail it to your home at no extra cost, no need to go back to the store.
I also found this led at RS that says is frequently used for power indication. I see the mcd is 620 and the mA is 30. Seems like a significant difference in mcd, which would do a better job, you think?
I don't know if its common knowledge around these parts, but if you walk into a RS and they don't have the part you want you can pay for it right then and there and they will mail it to your home at no extra cost, no need to go back to the store.
The LED's that are rated for 12/24V usually have a resistor built in already.
A quick Goodle search and 10 minutes if your time will teach you everything that you need to know about LED's and how to use them. Here's a link to get you started: http://www.theledlight.com/LED101.html
And here's a handy calculator that tells you what series resistor to use for your application: http://led.linear1.org/1led.wiz
There are quite a few of these calculators available on the web.
Google is your friend!
A quick Goodle search and 10 minutes if your time will teach you everything that you need to know about LED's and how to use them. Here's a link to get you started: http://www.theledlight.com/LED101.html
And here's a handy calculator that tells you what series resistor to use for your application: http://led.linear1.org/1led.wiz
There are quite a few of these calculators available on the web.
Google is your friend!
If the LED is broke, then most probably it is open circuit. Even if there is a resistor in series with this LED, u will read the higher voltage that is supplied. In this case it is 20Volts. Check, there may alredy be a resistor that is installed on the pcb. So all u have to do is connect any color LED of your choice/liking.
Gajanan Phadte
Gajanan Phadte
OK, I think I got this figured out, thanks for the tips and links. What I need to do is remove the existing LED and check its resistance. The new one is only 1 ohm. If the old is the same as the new, then I can just insert it, there is resistance somewhere else to keep the current to acceptable levels.
If the resistance is in the old LED (the lamp part is sheared off, the base is intact), then I will determine what resistance I need based on the following:
20v supply voltage - 2.1 voltage drop of new LED=17.9v
17.9v/.025A = 716 ohms-- select 1,000 ohm resistor
Calculate current w/ this resistor: 17.9v/1,000 ohm = .0179A
Calculate wattage of resistor: 2.1v/.0179A = .0376, therefore 1/4watt resistor is fine.
Thanks, looks like I'll be opening my own repair shop soon. I'll use some of the figurines from the LED figure link above as my van-side logo...
If the resistance is in the old LED (the lamp part is sheared off, the base is intact), then I will determine what resistance I need based on the following:
20v supply voltage - 2.1 voltage drop of new LED=17.9v
17.9v/.025A = 716 ohms-- select 1,000 ohm resistor
Calculate current w/ this resistor: 17.9v/1,000 ohm = .0179A
Calculate wattage of resistor: 2.1v/.0179A = .0376, therefore 1/4watt resistor is fine.
Thanks, looks like I'll be opening my own repair shop soon. I'll use some of the figurines from the LED figure link above as my van-side logo...
DreadPirate said:
An LED is a diode, diodes don't obey Ohms Law. The forward voltage (nominally 2.1V) changes by only a fraction of a volt for currents from under 1mA to over 30mA.
Your maths looks OK for finding the resistor, but another thing about LEDs is they'll glow quite adequate for an indicator on 2mA or less. But there's bound to be a suitable resistor already in the circuit. Easy way to check is to fix a 1k across where the LED was and measure the voltage drop across the 1k. Knowing there's 20V altogether you can get the existing resistor value.
With round LEDs there's usually a flat on the rim indicating the cathode lead (which connects towards ground); handy to know if the leads have been cropped.
"If there is no current flowing, there will be no voltage drop".
Most voltmeters have a high 'impedance' (formerly called "Ohms per Volt") so they don't load the circuit under test. So--if you measure a 20v source in series with an LED current limiting resistor, but without the LED in the circuit, you will still read 20v. Without the LED, it's an open circuit. The resistor will not have any significant current flowing, therefore (by Ohms Law) it will drop no voltage.
Likely there is already a limiting resistor in the circuit, so adding another resistor would lower the current to the new LED.
Most voltmeters have a high 'impedance' (formerly called "Ohms per Volt") so they don't load the circuit under test. So--if you measure a 20v source in series with an LED current limiting resistor, but without the LED in the circuit, you will still read 20v. Without the LED, it's an open circuit. The resistor will not have any significant current flowing, therefore (by Ohms Law) it will drop no voltage.
Likely there is already a limiting resistor in the circuit, so adding another resistor would lower the current to the new LED.
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