Blinking LED stops blinking?

[rshayes]

Your second drawing is pretty close. The minimum input voltage to the 7805 should be about 8 volts. If the battery is at a minimum of 10 volts, and the series diode drops about .7 volts, the resistor should be no larger than 1.3 volts divided by .08 amp, or about 16.25 ohms. A 15 ohm, 10% resistor would be about right. The maximum power dissipation under normal conditions would be 15 ohms times .08 amp squared, or .096 watt.<p>The good news is that the power dissipated in the resistor is not dissipated in the regulator. With a 16 volt input, the regulator wil dissipate about
.8 watt.<p>However, there is a "gotcha" here. The peak power during a 200 volt transient will be about 2.6 kilowatts. This will probably only occur for a matter of a few microseconds, so the total energy is small. Film resistors (both metal and carbon) have very thin films as a resistive element and will probably heat heat enough locally to burn out. Carbon composition resistors have much more massive elements, and are better able to withstand short duration overloads. I would use a 1 or 2 watt carbon composition resistor, or possibly a wirewound resistor of similar or larger rating. Carbon comps are hard to get these days, but the wirewounds may be usable also.<p>Zener diodes are not held to very tight tolerances, and are not specified over a wide current range. The maximum input voltage for the 7805 is 35 volts. A 10% zener would have to be rated at less than 31 volts. On the other end of the scale, the battery voltage might go up to the 16 volt range under some conditions. This would imply a zener voltage greater than about 18 volts.<p>The rated voltage of a zener diode is measured at a specified test current. This current is usually different for each type of diode, and represents a power dissipation that is a fair fraction of the zener diode's maximum rating under steady state conditions. It actually starts conducting at a slightly lower voltage. To avoid useless dissipation in the zener diode, it should be chosen for a slightly higher voltage.<p>On the other hand, the test current is probably far lower than the current which will flow during a voltage transient. This will require reducing the maximum limit on the zener voltage.<p>In addition, zener diodes with breakdown voltages above about 6 volts have a positive temperature coefficient. On hot days, the voltage will be higher and on cold days lower. The data sheet voltages are measured with short pulses at 25 degrees centigrade.<p>I would choose the zener voltage at about the middle of that 18 to 31 volt range. A 22 or 27 volt zener would probably be satisfactory.<p>The peak power in a 22 volt zener with a 200 volt transient would be about 260 watts. Some data sheets give peak power ratings for zener diodes, and it appears that 1 watt units can withstand several hundred watts for very short time periods. I would probably choose a 5 watt zener, since they are not much larger or more expensive.<p>The zener diode give some reverse voltage protection even without a series diode. If the input voltage reverses, the zener diode acts like a forward biased rectifier diode, and limits the negative voltage to about .6 volts. The series resistor may take a beating, however.<p>With 1000 microfarad in there, you don't need the additional 10 microfarad capacitor. Some capacitance is needed near by to keep the regulator stable. This is mentioned on the data sheets in a fine-type footnote on one of the back pages. I would still use the .1 microfarad ceramic in parallel, since these are a much better capacitor above about 100 kilohertz.<p>Now that we've done it the hard way, you might consider a series resistor feeding 5 1N4007 type diodes in series. This will give about 3 volts, which is within the operating range of the LED. Another 1N4007 in series with the input can take care of the reverse polarity problem. The resistor would be chosen for .08 amp at the lowest input voltage voltage of 10 volts. A reasonable value would be 82 ohms. When the input voltage rises to 16 volts, this would dissipate 2 watts, so a 5 watt wirewound or two 2 watt carbon comps in series or parallel would be adequate. Advantages: cheap diodes, no regulator, no capacitors, and no heatsinks. Disadvantages: draws higher power at high battery voltage, resistors may get warm.

[Chris Smith]

Wasnt soo simple after all?

[Kookie]

Yeah, I know. Well, I learned something here and that's all good. Stephens great, I want to keep him, lol. It's still simple enough. I wanted to avoid putting together a board, but its better this way and I wont be replacing that blinky LED because it will be regulated and protected. GREAT, thanks soo much! I'm off to the store to get the zener, 7805, and a new blinky.