ChevyBoy95 Posted October 7, 2016 Report Posted October 7, 2016 linear regulator? same sort of deal as a pot as far as wiring in, ground leg, input leg, and output leg. https://www.arrow.com/en/products/mcp1702-1502eto/microchip-technology?gclid=CIvnz7Hgx88CFQ-raQod3aQH7A Issues people talk about are with using resistors are voltage regulators are with the loads that vary along with the voltage sources not being regulated well, so your voltage that the LED (or output) might see will vary and could cause the LED to fail. Best Score to Date : 160.5 dB Outlaw (47Hz)[4 XM 15's & 2 Taramps Bass 12k's] BL : http://www.stevemeadedesigns.com/board/topic/147800-chevyboy95s-4-15s-7krms-wall-1533-db-on-half-power/YouTube: http://www.youtube.com/hitemwiththeflex/
DLHgn Posted October 7, 2016 Author Report Posted October 7, 2016 linear regulator? same sort of deal as a pot as far as wiring in, ground leg, input leg, and output leg. https://www.arrow.com/en/products/mcp1702-1502eto/microchip-technology?gclid=CIvnz7Hgx88CFQ-raQod3aQH7A Issues people talk about are with using resistors are voltage regulators are with the loads that vary along with the voltage sources not being regulated well, so your voltage that the LED (or output) might see will vary and could cause the LED to fail. That looks awesome. So, with that putting out 1.5v and my LED being wired to it as well, wouldn't the LED be seeing less than 1.5v?
Krakin Posted October 7, 2016 Report Posted October 7, 2016 I still have absolutely no idea why you wouldn't use a resistor. If you are wanting 20mA from a 5V lets say it has around 1.5V dropped across the IR LED so you have 3.5V across the resistor which means you should have roughly a 180Ω resistor. I don't see why a 5V line couldn't be regulated more than needed by the PSU. Krakin's Home Dipole Project http://www.stevemeadedesigns.com/board/topic/186153-krakins-dipole-project-new-reciever-in-rockford-science/#entry2772370 Krakin, are you some sort of mad scientist? I would have replied earlier, but I was measuring the output of my amp with a yardstick . . . What you hear is not the air pressure variation in itself but what has drawn your attention in the two streams of superimposed air pressure variations at your eardrums An acoustic event has dimensions of Time, Tone, Loudness and Space Everyone learns to render the 3-dimensional localization of sound based on the individual shape of their ears, thus no formula can achieve a definite effect for every listener.
DLHgn Posted October 12, 2016 Author Report Posted October 12, 2016 UPDATE: So, after talking to my bestfriend's dad (has been the top electrician at his company for many years now), I have decided to just go with the resistor. It's super simple and should get done what I need done. How I figured out the resistance needed: I know my source is 5v. The specs for the LED are 1.45-1.65 Vf (forward voltage) and 20 mA. For my purposes I am just rounding to 1.5Vf. Ohm's Law: V=I*R rewritten as R=V/I to find resistance The voltage going across the resistor (as Kraken mentioned) is 5v - 1.5v = 3.5v. This is the voltage that will be used in the above equation. Convert 20 mA to A by dividing by 1000. So 20/1000 = .02 Now, R = 3.5v / .02A = 175 ohms This means that I need a 175 ohm resistor Unfortunately, I couldn't find a 175 ohm resistor that I could use for my purposes on amazon so I had to settle with a value that I could add up to ~175. I ended up going with a 51 ohm resistor. To achieve my 175 value I need to put 3 resistors in series which are in series with 2 resistors in parallel. For series resistors, Rt = Sum(Rn) from 1 to n For parallel resistors, Rt = [sum(1/Rn) from 1 to n] ^ (-1) Let's calculate the value of my two resistors in parallel first (1/51 + 1/51) ^ (-1) = 25.5 ohms We can treat that 25.5 ohm value as a standalone resistor that is in series with the other 3 resistors. To calculate the final resistance 25.5 + 51 + 51 + 51 = 178.5 ohm This would get me 1.43v seen by the LED. It's a bit lower than the spec sheet but all that means is that my LED will be slightly dimmer which, for my situation, is okay. But keep in mind, this assuming each resistor is exactly 51 ohms. The resistors I got have an error rating of +- 5% which amounts to as low as 48.45 ohms to as high as 53.55 ohms. With these new values I can calculate my minimum and maximum voltage seen by the LED. My calculations (using same equations as above) say that I should be somewhere between 1.25 and 1.61v. Again, it being less than rated just means that it will be dimmer. My main concern was the upper limit as too much voltage can blow the LED. Thankfully the upper range is still below my 1.65v spec sheet limit. Anybody see any mistakes I made or wrongful assumptions?
Krakin Posted October 13, 2016 Report Posted October 13, 2016 Nope, just the same as I said before. I just said 180Ω because they don't make a 175Ω resistor. Can you post what LED you have just so it can be easier to find the datasheet. Krakin's Home Dipole Project http://www.stevemeadedesigns.com/board/topic/186153-krakins-dipole-project-new-reciever-in-rockford-science/#entry2772370 Krakin, are you some sort of mad scientist? I would have replied earlier, but I was measuring the output of my amp with a yardstick . . . What you hear is not the air pressure variation in itself but what has drawn your attention in the two streams of superimposed air pressure variations at your eardrums An acoustic event has dimensions of Time, Tone, Loudness and Space Everyone learns to render the 3-dimensional localization of sound based on the individual shape of their ears, thus no formula can achieve a definite effect for every listener.
DLHgn Posted October 13, 2016 Author Report Posted October 13, 2016 Nope, just the same as I said before. I just said 180Ω because they don't make a 175Ω resistor. Can you post what LED you have just so it can be easier to find the datasheet. It's the Chanzon 5mm LED IR Emitter 940nm. I bought them from amazon: https://www.amazon.com/Chanzon-100pcs-Infrared-Emitter-Emitting/dp/B01BVGIZGC/ref=sr_1_1?ie=UTF8&qid=1476374926&sr=8-1&keywords=IR+LED&refinements=p_89%3ACHANZON
Krakin Posted October 13, 2016 Report Posted October 13, 2016 You should be fine. Krakin's Home Dipole Project http://www.stevemeadedesigns.com/board/topic/186153-krakins-dipole-project-new-reciever-in-rockford-science/#entry2772370 Krakin, are you some sort of mad scientist? I would have replied earlier, but I was measuring the output of my amp with a yardstick . . . What you hear is not the air pressure variation in itself but what has drawn your attention in the two streams of superimposed air pressure variations at your eardrums An acoustic event has dimensions of Time, Tone, Loudness and Space Everyone learns to render the 3-dimensional localization of sound based on the individual shape of their ears, thus no formula can achieve a definite effect for every listener.
DLHgn Posted October 13, 2016 Author Report Posted October 13, 2016 I just got all of the components in today. I'm running into the problem of needing to connect the LED system to the power/ground run in the USB. I was thinking about cutting the USB and running all data/power wires to a distro of some sort then going from there. That would allow me to power the camera and the LED at once. Thoughts?
ShadeTreeMechanic Posted October 13, 2016 Report Posted October 13, 2016 You could get one of those cheap USB splitters and crack the plastic case off. Then just solider your power wires to it. 91 C350 Centurion conversion ( Four Door One Ton Bronco) 250A Alternator (Second Alternator Coming Soon) G65 AGM Up Front / Two G31 AGM in Back Pioneer 80PRS CT Sounds AT125.2 / CT Sounds 6.5 Strato Pro component Front Stage CT Sounds AT125.2 / Lanzar Pro 8" coax w/compression horn tweeter Rear Fill FSD 5000D 1/2 ohm (SoundQubed 7k Coming Soon) Two HDS315 Four Qubes Each 34hz (Two HDC3.118 and New Box Coming Soon)
DLHgn Posted October 13, 2016 Author Report Posted October 13, 2016 You could get one of those cheap USB splitters and crack the plastic case off. Then just solider your power wires to it. That would be a good idea but I don't have a soldering iron. I'm thinking about straight ghetto riggin' it with a screw as a distro block
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