Ok, Steelwolf, first for the Zener diode:A Zener diode is in a usual diode when used in flow direction: it drops about 0.7 V of voltage. Where it is diffenrent from a regular Silicon diode is when you put a voltage on it in reverse (stopping) direction. Then it has a specified breakthrough voltage that remains exactly constant (the Zener voltage, available types from 1V to about 56V). If you apply a voltage above the zener voltage, a current flows through the diode. If you would virtually have a ideal power supply with no internal resistance, you could read the current flowing from the U-I-diagram of the diode. This U-I-diagram makes a sharp turn at the breakthough voltage of the diode. What means: There is almost no current flowing when below the zener voltage, then the current steeply increases.Ok, so what does it mean for a generator of variing voltage (the crank-powered one) and a zener parallel to a cap?If the generator delivers less than the zener voltage, all of the current is delivered into the cap and charging it. When you come over the zener voltage, the diode begines to bypass as much current as necessary to keep the voltage down to the zener voltage. This means: The diode pulls current that causes a voltage breakdown at the internal resistance of the generator (dc-resistance of the copper wire). So this current is shunted and generates heat in the zener diode. This is why you should use a type of appropriate wattage (I would use not less than 1 Watt, otherwise you could easily blow it).From the cap it looks this way: He gets a stable voltage source in height of the zener voltage which doesnt vary with the drawn current (if the cap draws more current, the zener doesnt need to draw as much to keep the voltage down (both the zener current and the cap current add together flowning through the internal resistance of the generator)).So if you use a 5.5V cap to power a white LED it doesnt make sense to use a 3.8V zener, because then the cap could charge only to this voltage and store less energy. If you use a 5.1V zener, the cap can take much more charge. You then have to use a resistor to keep the current of the LED down.Ok, if you have any more questions, or if I wrote some nonsens (happens sometimes), please tell me.
Luiz Vieira wrote:
> I would like to carry the final workbench tests that you suggested, but I have no precision "meter", just one multimeter that I repaired once ...
Hello Luiz, Well, to get a clear picture you want to use a Tektronix oscilloscope, a mainstay back at the secret underground laboratory of Captain Zener. It helps to see exactly what's cooking with the waveforms. The charger has got a bit of 3rd harmonic distortion for example, and also, to exactly peg the peak current in my 'torture test' that made sure the zener wouldn't melt
:o(
:o)
Luiz Vieira wrote:
> About the "final" Captain Zener solution: if not available, what risks one will face if using a 6.2V, 1W zener diode that is not the 1NA?
What matters here is the 1 Watt most especially. After that, something around 6.2V is good. If you drop down in voltage (like a 5.6V or a 5.1V) at some point your going to get some leakage even at the idle battery voltage (2.8V). I'm not sure which other zener you have in mind, because there's not that many zener families in circulation, one of the most common alternative families is a 1/2 watt series... dont go there.
Again, again, again, let me dwell on something about the family of zener diodes that comes from a semiconductor factory: They have a very sharp transition at 6.2V, and don't leak any current if they are at or below that optimal 6.2V threshold. But if the diode is adjusted by dopants so that it clamps lower, for its manufacture, specification, and sale, like the ones specified at 5.1V or heaven forbid 3.3V (those are ghastly) then the thing is so "moooshy" that it will leak current almost no matter what voltage you put onto it. That's why we are, in the end, reaching for the 6.2V units, to exploit the sharpness of the threshold. I would probably breathe easy with a 5.6V but expect trouble even with a 5.1V (it will tend to draw down the battery gradually due to leakage in such a zener).
Luiz Vieira wrote:
> if batteries are O.K. and in place, the zener diode is transparent to the circuit because battery voltage will not be over (nominal voltage plus 15 to 20%), right? If voltage across the zener diode does not reach the test-voltage value, then it will not conduct. I never considered the fact that the zener diode would leak battery current. Is it a wrong analysis?
Hello Luiz,
Yeah, you are drifting over to the 'dark side' here (sorry, old Star Wars laserdiscs tonight). Your comment is true for the 6.2V unit because it transitions so sharp and below 6.2V does not leak hardly even a few microamps. At 3.3V no problemo.
But if you go and grab for example, a 3.3V zener and figure it wont conduct at 3.2V, you are in for heap-plenty disappointment. A 3.3V zener will still be conducting crazily at 2 1/2 V. And that unacceptable leakage is present even on a 3.9V or a 5.1V zener, because of the qualities of the semiconductor doping.
Basically a zener diode "likes to be around 6V" they can be built to run very well there with a sharp threshold. When they fill the purchase-orders for diodes with lower voltages, the excessive dopants create a mooshy and gradual transition, which isn't really what people want. They want abrupt conductivity at the specified zener voltage, and no conductivity below it.
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BTW, for purposes of navigation, zeners specified HIGHER THAN 6V (8.2V, 10V etc) are also quite reasonably sharp. But the 6V zeners are sharpest of all. The data sheet chooses values spaced logarithmically so 6.2V is the typical number called out (like for 1NA).
Luiz Vieira wrote
> The other thing: isn't the charger based on a half-wave rectifier?
Yes it is. For the torture test, I happened to choose 2 zeners in "back-to-back series" so that the transformer has a more symmetrical loading. Asymmetric loading (one diode) is a bit distasteful. For purposes of charging the batteries, the single diode, obviously, was enough. For purposes of observing heat at the zener, the torture test chosen was fair enough, and worse than the actual situation for which we are applying it .
Luiz Vieira wrote:
> Should the peak voltage be considered 14V or 7V?
Not 7V. It is 14V. You must always have one lead of the transformer as a ground reference, and there is a peak at +14V and another at -14V, and nothing else. It is true that due to the single diode, HP only uses the positive half-peak at +14V.
Luiz Vieira wrote:
> I once thought about driving a transistor to form a "higher-current" zener. As it will only act when batteries are not O.K. or absent
It is common to utilize transistors as pass elements when a zener is not available at a high-enough wattage. Such techniques can also 'synthesize' a multi-component zener that behaves better with low thresholds (that 3.3V specified zener problem). But now you are into all sorts of tedium to actually specify and design the transistor and its circuit. You would only do that if necessary; here its not so use a 1NA.
Luiz Vieira wrote:
> I had a few doubts (not about your posts, about my own concerns) because I did not have the necessary equipment
"Have no fear, Captain Zener is here."
:o)
- Norm Seattle area
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