Sharpie is quick and easy. I bought and use a laser printer for the latter method, matter of fact from the same website you show. I still need to get some decent software and learn how to use it. I've some inqueries on usenet for just that process.
Then take a look at this:One problem with something like 7805, if you draw .5A then you are disipating well over 3 Watts. This will likely fit in a cigarette lighter adaptor, with the fuse being the center pin, [...]
As to getting the job done or getting experience, I'd have to answer yes to both.
Well, it should be a short matter to build one and give it a test drive to get some hard facts on it rather than just assuming. It's so easy to critisice something you haven't experienced, but not much harder to get knowledgeI don't like that 'black regulator' link... I don't buy the 93.1% efficiency..
If you completely re-draw the schematic, I don't see how?I may redraw it as I interprete it and post it here. Does this run into copyright issues?
Shooting off an eMail to ask for pe... PS: The Smiles button isn't working either.
I did a quick simulation with your circuit, using available models in my LTSPice library. The CR4 1N4001 might not work correctly. I often seen Schottky or fast recovery diode used in this location. Also beware that a load current of 2A would means diode current will also be 2A or more so 1N4001 is out of the question. Thus I replaced it with a Schottky diode of 3A current rating.Well, the son of a gun didn't work on several levels. CR4 was wrong, I had to turn it around.
The 150Ω base resistor of MJE2955 might be too high and thus the transistor will operate in linear mode instead of going into Off/full saturation alternatively. Obviously you don't want this. I have reduced its value to 22Ω instead. I also reduced the LED series resistor to 2K2 to try to saturate 2N2222 more.The reason I say this is I'm getting the same current in as the current out, so no conversion is taking place, and Q1 is getting pretty hot. This is with 14.8VDC in at 2.1A.
I tested the above simulated circuit with and without this base-emitter resistor(R20) for Q1 and the instantaneous power dissipation in Q1 appears to be the same in both cases. The red trace is with the 900Ω resistor in circuit.Hi Bill, Q1 is probably getting hot because its base is floating when Q2 is turned off. A tiedown resistor across Q1 b-e should cure this.
by Jake Hertz
by Duane Benson
by Duane Benson
by Aaron Carman