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I know this is an old thread but searching "momentary switch with LED" brought me here. I have a question. At 3:30 in this videohe uses a 22ohm 150w resistor (I have a 15ohm 100W ) with an LED wired to a momentary switch to bleed the caps hooked to a 24v transformer. Makes sense. He used a switch because it keeps the caps from discharging during a weld to keep them charged up when you let go of the trigger. I can see the 2 wires on the resistor but can't see where the LED/ switch is in the system. Any idea on how that's wired?
My best guess would just be a LED with a series resistor across the caps. AS the caps discharge through the bleeder resistor the voltage across the caps will drop and eventually when the voltage is low enough the LED will extinguish.
The idea here is to bleed the voltage off the caps. This is normally done by placing what is commonly called a "bleeder resistor across a charged capacitor. The cap can be any value and charged to any potential. The LED and LED series resistor are across the cap (in parallel with the cap). The switch is in series with the resistor and when the switch is turned on it places the bleeder resistance across (parallel with) the cap. The cap discharges through the switched in resistor. The switch needs to be rated for the max voltage and current it will see. When the system is running the LED will be illuminated. When the system is shut down and the bleeder switch turned on the cap will discharge and when the voltage is low enough the LED will fade out. This is the simple and crude but effective way to discharge the cap(s). There can be more elaborate methods using more parts. That is why I mentioned "My best guess" above.
Reloadron Thanks again. I'm sorry for doing such a poor job of explaining my question. I understand the use of the bleeder resistor and the cap. The welder runs at 24V with a peak of 28-30. It is a 50V 100,000uf cap. The switch is rated for 120V or 220V.
When the system is running the LED will be illuminated. When the system is shut down and the bleeder switch turned on the cap will discharge and when the voltage is low enough the LED will fade out. Ron
Don't understand why the LED would be illuminated if the momentary switch isn't closed. If it's illuminated it is bleeding the caps which defeats the purpose of the momentary switch doesn't it? The goal is for the caps to stay charged until the switch is flipped illuminating the LED which bleeds the caps.
The LED Resistor is there to limit the LED current. When SW1 is pressed and held the capacitor will discharge through the Bleeder Resistor. Capacitor C1 can be a single cap or a bank of caps. I did not show the source. When in normal operation the LED will be lit, when source power is removed the LED will remain lit. When SW1 is pressed the cap(s) will slowly discharge through the Bleeder and when the voltage drops low enough the LED will extinguish.
Reloadron Thanks again. So, the difference in your diagram and what he describes in his video is that your circuit uses 2 resistors, one for a bleed and another for the LED? He doesn't mention a second resistor which is what confused me. Simple enough to do. What specs would the LED resistor be?
Is it not possible for the LED and bleeder resistor to be one in the same and just put the LED between the switch and the bleeder? I'm not doubting you. Inquiring minds just want to know. Your diagram seems to better serve the purpose as the light would remain lit whenever there is power in the cap. The way he described it the LED doesn't light until the switch is pushed. But if the LED is lit between pulls of the trigger isn't the cap being bled while it sits idle? The goal is to keep the caps as close to full charge as possible until they are bled down.
Here is what's going on. The video actually explains very little. A typical general purpose LED will have a Vf (forward working voltage) of about 2 or 3 volts and a If (forward current) of about 20 mA. LED circuits a resistor often refereed to as a current limiting resistor. Things for what I just mentioned work out like this. This is my typical 5mm LED data sheet.
Vsupply - Vfwd LED / LED Current so lets say Vsupply is 40 Volts across the cap(s). We get 40 - 2 / .020 = 1,900 Ohms or about a 2K Ohm resistor in series with my LED. The video mentions the bleeder resistor is about 22 Ohms 150 Watt which is a low resistance high wattage resistor. That would be typical but the video mentions nothing as to the LED configuration and the video is not very instructional at all.
The way I did the drawing is yes, when there is power on the cap(s) the LED will glow. A charged cap will, over time, discharge itself due to the internal resistance of the capacitor. Not a concern here. The idea is to rapidly discharge the cap, using a switch and dead short is not a great idea so we add a bleeder resistor and the cap can slowly bleed off the charge.
Reloadron OK. I'm getting there. Hopefully the last time I'll have to bug you.
Understood I need high resistance to drop the current down to 3V from the 24V feeding the caps. I used the Digikey Calculator using a 24V supply value and 3Vfwd that tells me I should use a 1K - 1.5k ohm resistor, rated at least .5W correct? Trying to use one resistor doesn't work because the bleeder needs to be low resistance to quickly discharge the caps and the LED resistor needs to be high to drop the voltage down.
Another option? Use a 24V muffin fan out of an old computer case across the caps as the bleeder with the switch and skip the LED, LED resistor and bleeder altogether. Hold the switch until I hear the fan stop spinning to discharge the caps?
Thanks again. It has been a very long time since I knew about this stuff. I was a Chemistry major.
Yes, that would be correct. 24V - 3V = 21V / .020 = 1050 Ohms
The LED series resistor is not a perfect exacting science as you may have guessed. You are not going to find an off the shelf 1,050 Ohm resistor so it is suggested you move up to the next available common size (off the shelf common stock). A 1.2K or 1.5K should do fine. Using your online calculator example the Power expressed in watts would be P = I^2 * R so we get (.020 * .020) = .0004 * 1050 = .42 Watts and we like some room so a 1.0 Watt resistor would do fine.
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What about using the fan instead? Is that an option?
Reloadron Can't tell you how much I appreciate the patient tutorial. I think I'll just use the fan since I have it already but the education was invaluable and know how to calculate and do it in the future. I've got a few things that could benefit from an LED. If you ever need the same kind of help regarding all things woodworking hop on over to our site lumberjocks.com.