I am not sure I can properly "teach" a 70 year old codger about zener diodes. Not even with videos showing happy electrons marching around the circuit. You need to get some basic concepts nailed down, such as Ohm's Law and Kirchoff's current and voltage laws and how to use them to analyze circuits . And those concepts have NOTHING to do with quantum physics. Also, understanding a V-I graph is probably the most essential "visual" concept. The graph I posted in post #17 has just two active quadrants: I and III. Nothing happens in Quadrants II and IV. The lines on the graph represent the current through a diode (any diode, including zener diodes) on the vertical axis at the voltages applied on the horizontal axis.
In a thousand words or more...
You have positive currents and negative currents, positive voltages and negative voltages. These are plotted in pairs: for every voltage on the plot there is one and only one current plotted. Positive voltages represent forward bias of the diode. Positive on the anode and negative on the cathode causes positive current. Negative voltages represent reverse bias of the diode. Negative on the anode and positive on the cathode causes negative current. You start out at zero voltage and zero current and either apply a positive forward bias or a negative reverse bias and measure the resulting current to create the graph. Make sure you have a fresh battery in your digital multimeter when taking measurements.
If you substitute a resistor for the diode, and plot current at various positive and negative voltages applied across the resistor, the plot will yield a straight line that passes through the origin, rises in a positive direction for positive voltages (Quadrant I), and falls in a negative direction for negative voltages (Quadrant III). The slope of this line is the ratio of the current at any given point to the voltage at that point, which by definition is the conductance (I)/(V) =1/(R). Thus, lower value resistances (higher conductance) will produce steeper slopes approaching a vertical line, whilst higher value resistances (lower conductance) will produce shallow slopes approaching a horizontal line. Easy peasy. Now on to actual diodes. See graph in post #17 for visual aid.
You will note that the line for forward-biased diodes is a curve, going sharply upward to the right of the "knee" in the curve in Quadrant I. For reverse-biased diodes, the "curve" is almost horizontal at a very small current (leakage current), until either the zener voltage is reached (zener diodes) or the break-down voltage is reached (regular diodes). In both cases (zener diode or regular diode), the negative current increases sharply at a certain reverse voltage. Don't worry about how or why this happens. It just does. This reverse current is then limited only by the external resistance in series with the diode.
The same is true for forward-biased diodes: the current is limited only by the external resistance in series with the diode. However, for forward-biased diodes the transition from low current to high current is spread over a much wider range of forward bias voltage.
Some people will talk about "ideal" diodes where the curve in Quadrant I is just a vertical line, parallel to the vertical (current) axis, and intersecting the horizontal (voltage) axis at 0.7 V. From the origin to +0.7 V the "ideal" diode conducts zero current, behaving like an open switch. Then suddenly, at +0.7 V, the diode behaves like a closed switch, conducting with a voltage drop of 0.7 V but with current limited only by whatever external resistance is in series with the diode.
When a reverse voltage is applied to this "ideal" diode, zero current flows and the "curve" is a straight line parallel and coincident with the horizontal axis, extending from +0.7 V back through the origin in the negative direction... until the zener voltage or the breakdown voltage is reached. At that point, the "curve" dives straight down, parallel to the veritical axis and, again, the current is limited only by the external series resistance.
So, is this "ideal" diode model of any practical use, or is it like the "ideal" projectile fired from a rifled gun barrel that takes zero time to reach a target at any distance and without any bullet drop caused by gravity? Sure would have liked to have one of those "magic" bullets, whether bustin' varmints or bigger game! Never gonna happen with real bullets, but the answer is YES for real diodes. The "ideal" diode model
is useful. It allows you to make circuit calculations that are "gud enuf" for circuit design, You can calculate the power that will be dissipated in the zener diode when it is reverse-biased and conducting its rated current. You can calculate the ohmic value of the necessary current-limiting resistor and you can calculate the power that will be dissipated in that resistor and in the zener diode. And this is
all you need to calculate in order to use a zener diode.
In most circuits, a zener diode is
always reverse biased with sufficient voltage to cause conduction at the zener voltage, using a power supply that provides
more than the zener voltage. With any less power supply voltage, the zener does not conduct any current at all. The difference between the power supply voltage and the zener voltage is the voltage drop across a series current-limiting resistor.
So, using the "ideal" model, you first determine how much current you want in the zener diode when it is doing its thing (conducting at the zener voltage, which we assume is a constant. It is a good idea (for reasons explained later) to make this current the same as the maximum current that will be drawn by the load that you attach in parallel with the zener diode. If this load current is insignificant, as it is in the circuit using the TIP122, then choose any small current that is less than the maximum current rating of the zener diode. For the TIP122, the load current is the base current and it will be only about 1/1000th of the emitter current. This is important because the current drawn at the TIP122 emitter is NOT the load current of the zener. So, if the emitter is charging a battery at, say, 2 A then its base current will be something on the order of 2 mA. Pick a zener current of, say, 30 mA and you should be okay.
Next, pick a zener voltage appropriate for your circuit. As
@Audioguru has mentioned, this is a horrible circuit for charging a battery. Using a 12 V zener, it will not charge a 12 V lead-acid battery which requires at least 13 V (or more) for even a "trickle" charge. Worse than that, the two-diode voltage drop from base-to-emitter of the Darlington-connected TIP122 will result in more than 1.4 V
less voltage on the emitter output than is present on the base input, i.e., the zener voltage. That means the output will be on the order of (12 - 1.4) = 10.6 V. This may be sufficient to charge a 9 V NiCd battery, but there is no way it will charge a lead-acid 12 V battery.
There are other reasons why this is a horrible battery charger, but let's not dwell on that while I am trying to explain how zener diodes work. You need more than 13 V to charge a 12 V lead-acid battery, so add 1.4 V to that to allow for the base-emitter voltage drop, arriving at a minimum zener voltage of 14.4 V. So pick a
15 V zener diode to stand any chance of getting the circuit to charge a 12 V lead-acid battery.
So now you have a 15 V zener drawing 30 mA and providing a voltage to the base of the TIP122. That means the zener must dissipate (15)(0.03) = 0.45 W = 450 mW of power. That's a bit on the high side for your typical 400 mW glass-encapsulated diode, so up-size to 1 W to be "safe". Can you identify a 1 W zener diode by looking at it? No. You must test it to see how hot it gets while dissipating one watt. This may result in destruction of the test sample, so have a few spares on hand. You can "guess" how much power a salvaged diode will safely dissipate by comparing sizes, but it won't be an educated guess. It will be a pure rectal extraction.
One simple way to measure the power dissipation capability of a diode is to
forward bias it and apply a variable DC voltage source to it through a current-limiting resistor. Adjust the power supply voltage to vary the current while measuring both the current and the voltage drop across the diode. Why forward bias? Because you want a fairly constant and low voltage across the diode whilst adjusting and monitoring the current. Be quick on the 4-banger calculator to multiple voltage measured
across the diode and current measured
through the diode to arrive at the power dissipated in the diode. Two meters, one for voltage and one for current are highly recommended.
Monitor the diode case temperature. If it gets too hot too touch, that's too much power. Assuming you can find a zener diode that operates continuously at one watt dissipation without excessive heating, don't assume that every diode with the same size and shape (unless it is identical) will also operate at one watt. There is a lot of variation in manufacturing, so even threaded stud-mounted diodes you would think could handle a boatload of power may turn out to be wimps. It all depends on the internal construction. If you can get a part number and the corresponding datasheet, you can use that information instead of performing your own power dissipation tests. But where's the fun it that? And someone might accuse you of practicing engineering if you go that route.
Now that you have found a 15 V zener, and it is capable of dissipating one watt of electrical power. What next? Calculate the value for the current-limiting resistor you connect is series between the zener diode and your DC power supply. That value will be the found by subtracting the zener voltage from the power supply voltage and dividing the difference by the zener current. Your posted circuit claims a DC power supply voltage of 19.4 V. This will vary with the mains supply voltage, the amount of current drawn, and perhaps the phase of the moon. Treat this number as a starting figure to get you into the ballpark. Subtracting 15 V gets you 4.4 V across the current-limiting resistor. Dividing that by 0.03 A gets you a value of about 147 ohms, which will dissipate (4.4)(0.03) = 0.132 W. A 1/4 watt resistor would be "safe" to use there.
Life gets more interesting when you add a load connected in parallel with the zener diode. Assuming the power supply voltage doesn't change with this load, and the zener voltage is also constant, then the current through the current-limiting resistor remains constant no matter what the load is! What does change is the current through the zener diode. It decreases as the load current increases, because the sum of the zener current plus the load current is equal to the current through the current-limiting resistor... and that current is constant if the power supply voltage doesn't change.
Now why would you get a much larger voltage out of your "battery charger" when you used the big, black, diode instead of zener diode? The simple explanation is you have inserted an ordinary rectifier in a reverse-biased direction and it is NOT conducting. The entire power supply voltage, nominally 19.4 V, is applied to the base of the TIP122 and approximately 1.4 V less than that appears on the emitter of the TIP122. The unloaded output soars to about 18 V. To test this theory, just remove the big black diode but leave the resistor connected between the base and the power supply.
Finally, why is this circuit so horrible? Even with a 15 V zener it is still horrible. Lead-acid cells should be charged with a variable constant-current power supply while monitoring the battery terminal voltage and reducing the charging current to a trickle current as the open-circuit battery terminal voltage approaches the full-charge voltage. A constant-voltage power supply, which is what this circuit attempts to provide, does not do this. There is NO current-limiting resistance between the battery and the power supply. The battery charging current will be determined by whatever the effective series resistance happens to be, a complicated function of transformer selection, rectifier bridge selection, "smoothing" capacitor selection, wire gauge and connecting length... but not the phase of the moon. You are lucky you are not using a larger transformer in your bread board setup. BTW, I haven't seen an actual bread board setup using real wooden boards since the 1950s. Congratulations on keeping this tradition going, even if you did cheat by screwing a proto-board to the bread board.
