Why worry about 1 milliamp being used by the resistor? That's easy: because it's wasted power. The power being used to drive the LED is not wasted because it's doing useful work -- or at least, the power that is wasted by the LED producing heat is beyond the control of the designer, but the power being wasted by the resistor is within the control of the designer, so it's a valid target for improvement.
Don't ask "why try to make it work better?", ask "why not try to make it work better?" "Why not?" is the question that drives progress.
Wow, I have been looking for a solution to my problem for three days. While this is an older thread and you brought it back from the dead I might not have seen it otherwise. I just wanted to say thanks to everyone that posted all the helpful information here.
I am working on a project using a fade-in/fade-out circuit for LEDs. I am building some repulsor props for a friend of mine who has an Iron Man costume. He wanted to use some reed switches to activate the LEDs but when I tried one it got stuck in the closed position. I am not sure what caused it. The rating on the switch is 500mA. My circuit draws about 50mA so I don't think that it was excessive current. Anyways, I thought that isolating the switch from the circuit through a relay was the answer. I was looking at some relay circuits when I came across this forum/thread. I am just a beginner working with circuits and components. I would not have thought about using a Mosfet instead of a relay. So thanks again for all the help given in this thread. I am going to start some experiments now.
Well, I did some more experiments and unfortunately using a mosfet as a switch won't work with my circuit. My circuit needs to maintain a connection to the ground in order for the capacitor to drain through the LEDs and produce the fade-out effect. I am not sure if it is possible to connect the mosfet in this manner.
This is my circuit in case anyone wants to take a look over it and tell me if they see a way around this issue.
My power source for this circuit is 4 AAA batteries at 6v (6.5v w/fresh batteries). I am using a BC548 at Q1 and a BC327 at Q2. I am also using a different power transistor, 2N5296, at T1 but I don't even know if I need that. The max current for the LEDs will be 240mA - 720mA max depending on the LEDs I use.
It was mentioned earlier in this thread that a DPST reed switch would be a good idea but I don't quite understand why or what the purpose of that would be. I would love for someone to go over that with me.
Then, I am just wondering if there is a way to use a mosfet in this circuit or if there is something else I need to do to protect the reed switch from being damaged. It is possible that it was a defective switch but I am afraid of messing up another one.
My friend already has SPST reed switches and I have to activate this LED circuit and a sound module at the same time. I was thinking of looking for a DPST or DPDT relay to activate both at the same time. Otherwise I would have to use two mosfets, right? The sound module will be running off of the same power source; but I am not connecting power to it, I am closing the switch to activate the sounds. I know a relay would do that but I don't know about the mosfet. It seems that the power across the switch has to be at a certain voltage for the IC to play the sound. I can't just send power from the battery to the other side of the switch. I have to send power from one side of the switch to the other. Would the mosfet 'corrupt' that power/voltage signal to play the sound?
One thing I did find while looking at how to protect the reed switch was that most companies selling them recommend using a relay and on DC circuits to wire a diode (1N4004) parallel with the relay. I don't understand the mechanics of what is happening there or why the resistor/mosfet setup in this thread works without it.
Any help would very much be appreciated. Thanks, Mike