All, I am wiring up my ‘removable’ lightning assembly - similar as discussed in previous thread. The wiring consisting follows this rudimentary, simple schema: 9V+ -> 41K resistor -> 1N4001 diode -> Pico 0402 Led -> 9V- When testing this schema the diode acts as intended but also seems to act like an additional resistor. The led light output is significantly lower with the diode in place compared to the resistor alone. Does someone have any idea what could be the cause of this and how to resolve this? Thanks in advance for any guidance! -Tiest Sent from my iPad using Tapatalk
Hi Tiest, I suspect your resistor is too large. I do have a few questions about the circuit to learn more. Is the voltage across the circuit 18v or 9V? Do you know the current and voltage ratings for the LED you are using? -Mark
Hi Tiest Why are you using the 1N4001 diode in series with the LED and resistor? You will have a voltage drop of about 0,6 Volt above the diode and with the very big 41K resistor, almost all of the remaining voltage will drop over the resistor. A "Normal" value of a dropdown resistor will be in the range of 200-400 ohm. The LED will use typically 2,5 Volt and typically 20mA. In your case you will have about 5,9 Volt drop over your resistor. You can use Ohms Law to calculate the resistor, R=U/I, like this: (9-0,6-2,5)/0,02 = 5,9/0,02=295 ohm. The standard E12 value to use will be 270 or 330 ohm.
Are you feeding it with 9 VAC? Seeing the diode as described above, I'm thinking you are rectifying your source to DC, something like as seen below, except with a 41,000 Ohm resistor?
Yes, a diode causes a voltage drop. No, there is no diode that doesn't, though some types cause more than others. Chopping half the cycles out without losing any power is too much to expect. Different diodes also have different resistance due to the materials they're made of. Rectifier diodes, for example, are silicon, which only moves electricity so well. But different diodes also differ in other properties. Rather than going with a diode with less resistance, you'd be wise to use the best diode for the job and adjust your resistor to suit.
Switch to a 2K resistor at 9V. The 1N4001 diode drops .7V so with a 41K resistor, you are driving the LED with 0.0002 amps. 0.2 milliamps is not enough, I like to use 2 or 3 milliamps for realistic brightness and long life, and manufacturers want you to use 20 milliamps so they burn bright and burn out. So if you were using 4.1K ohm resistor and the 1N4001 protection diode, you would be perfect for what I like to run LED's with my 12V accessory power. Alternately, if the 41K resistor was a typo, and you actually have 4.1K ohm resistors, just put one more in parallel with the first:
All, Thanks for the responses thus far. I just realized I did put in a typo, I am not using a 41K Ohm but a 4.7K Ohm resistor. Apologize for this mix-up. However my LED brightness is still impacted but per the responses; this is to be expected. Based on the feedback, I have two options: 1) Utilize a lower resistor ~2K Ohms for 9V accessory power 2) Increase the accessory power to 12V for usage with 4.7K ohm resistor. [mention]rray [/mention], what is the advantage of the two resistors in parallel? Thanks! -Tiest Sent from my iPad using Tapatalk
The effect on voltage is the same as using a single resistor, but both pass current, so amperage output increases.
Well, What actually happens when You have 2 equal resistors with the same value in paralell, the resulting value is half, like in your example 4,7 Kohm, the resulting value will be 2,35 Kohm. The brightness of the LED will increase.
Yeah, the reason I drew in adding a second resistor was because if you have a 4.1K resistor, you probably bought a pack of them so it's easier to half the resistance by adding another resistor in parallel to increase brightness instead of buying a 2.2K and waiting for it to be delivered. You can even parallel in 3 or 4 resistors, till you get the brightness you want, then calculate the new resistance, and order the desired resistors for your voltage source. Try to photograph your model, and see if the brightness is good, because camera's can see too much light easier than we can. I like to call it scale brightness, cause when you run LED's so bright that the walls and roof of a building glow, you went atomic and are doing the same thing as running your trains in full blast rocket mode, instead of using scale speed, scale brightness, and scale volume on a sound circuit. That is of course only valid you are Scale Modeling. Some people delight in Toy Trains, and that's where it's all about Lights, Camera's, Action! Bling, Bing, Bang! Haha! Somebody stop me! There's no right or wrong way, there's YOUR way!
To better understand what is going on, the 1N4001 diode is not acting like a resistor, it's just acting like a diode. When conducting, all diodes have a voltage drop. The current dissipated by the LED = [Vs (supply voltage) - Vd (voltage drop)] / R (resistance). Since the voltage drop is now the voltage drop of the LED plus the voltage drop of the 1N4001, you are increasing the voltage drop, which decreases Vs - Vd, and subsequently the current. I don't think anybody mentioned it, but if you are powering it with DC current and there is no chance of it getting a reverse voltage, then the 1N4001 is not needed.
[mention]rray [/mention] what type/brand of accessory power do you use? Thanks! -Tiest Sent from my iPhone using Tapatalk Pro
I use whatever 12V power supply I have on hand. When I started building my T Trak Z modules, I included a large bridge rectifier to prevent damage from anyone connecting the 12V DC accessory backwards: Or connecting a 16V AC accessory supply. So if I connect my modules to another club's TTZ modules, and their guy connects the wiring, my modules will not be harmed. The output polarity that my modules see is always the same, and always a DC voltage. Regardless of what the other club uses for accessory power, my modules will be safe with this device: So I recently made this 12V accessory power supply: It uses my 20V Black & Decker power tool batteries, and a 12V 5A buck boost converter to convert the 20V into 12V. In fact it will convert any battery from 9 to 36 volts DC, into 12V DC. I put the T Trak Z standard Anderson Powerpole connectors on it, so I can power my modules directly from power tool batteries. And as the batteries drain, and the voltage drops, the accessory voltage will stay at 12V until the batteries reach 8.75V, then the power shuts off. So, by using a 4K resistor in series with my LED's, I am driving a 20ma rated LED with 2ma. It's not super bright, but very realistic. This is assuming the accessory power is 12V. Suppose some club uses 20V AC as their accessory power, from an old buzzing Tyco Train Transformer, the LED's would still only see 3-4 ma and would still not burn out. And my module would still power up just fine!
[mention]rray [/mention] extremely helpful, will give it another go this weekend based on this information. Stay tuned. Tiest Sent from my iPhone using Tapatalk Pro
