Newbie: Zerner diode voltage regulator ?

("Zener" diodes)

You've nearly got it right but coming in from the wrong angle. It is the Zener itself that controls the voltage to the load. A 13V Zener will supply 13V to the load. If you've a 10V load then you'll need a 10V Zener. Basic idea is that the Zener is set to -always- be conducting a small holding current (say 5ma) and it's kind of hovering on the cliff edge in a situation where it's running at it's Zener voltage but really wants to pass lots and lots more amps ("on the knee"). If allowed, yes it would quickly go short circuit but the dropper resistor will not allow this. The dropper is sized so that the load can take the maximum current it needs and also supply a small additional Zener current. Zener current is usually set at (a non critical value) 10% of the maximum load current.

Say you need a 5V 100ma load and only have a rough 12V power supply.

1] Get hold of a 5V zener. 2] Unpack the calculator. 3] Allow 10ma zener current. (10% of 100ma). That's a total of 110ma coming out of the 12V supply. 4] Dropper resistor value, is incoming supply voltage minus the zener volts, then divided by the total current. Which is (12V-5V) / 110ma. Which is 63.6ohms. Use nearest standard value of 68ohms. That's it. You've a working, Zener controlled, 5V supply!.

For good design, afterwards always check that the dropper and Zener are sized for the correct power rating. Zener power is at a minimum under normal running conditions and is 5V * 10ma =50mW (zilch). Zener power is maximum when the load becomes disconnected and all the design current runs through the Zener. It's then 5V * 110ma = 0.55W. A 1W rated Zener would be fine. Dropper resistor rating is (12V-5V) * 110ma = 0.77W. A 1W resistor would be OK. john

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Yes, that's about right. The zener just passes any extra current needed to keep the load voltage constant.

So, if the input is 12, and the output is 6, and the load varies from

100 to 200mA, the series resistor is 6/.2 =30 ohms. But you probably want to use something a little smaller so the zener is always conducting. Zeners need a minimum current to regulate well. You can look up the curve for the zener to determine the voltage at various currents. So, if you add a minimum zener current of say 10mA, the resistor is 6/.21 = 28.6 ohms. Round it off to 27 which is a standard value.

Then you need to make sure the zener doesn't overheat at minimum load. At that point, the zener will conduct 110mA at 6 volts which is 660mW so the zener should be rated above that, or around 1 watt. The series resistor power will be 36/27 = about 1 watt. A 2 watt size would give a little extra margin.

Now, if the input voltage also changes, the problem gets more complicated, since you need to consider the minimum and maximum input voltages, and also the min and max output currents.

-Bill

hdjim69 wrote: (snip)

Not quite. Think of the zener as a pressure relief valve that leaks current to keep the voltage from rising above its opening point. It can only act as a continuous regulator if its current is not allowed to go all the way to zero, because then it looses control of the voltage. Ordinarily, the upstream limiting resistor is chosen to make sure that at minimum input voltage and maximum load current, the zener current is just approaching zero. The zener wattage is chosen by evaluating how much power it has to dump with maximum input voltage and minimum load current, while the zener is passing all the excess current (at its breakdown voltage) that gets through the input limiting resistor and has no place to go except through the zener.

As mentioned by others, the zener diode is a shunt regulator -- that is, it shares the current coming from R with R(l). As load current decreases, the voltage at the node will rise. But as the voltage rises, the zener will instantly take more current, keeping the voltage fairly constant.

And if the load current increases, the voltage at that node will start to go down. But the zener will then instantly take les current, and the voltage again stays fairly constant. This effect is called shunt regulation -- current is shunted away from the load in such a way to keep voltage constant.

Without getting into details, zener regulators are holdbacks from the

1960s. A shunt regulator wastes the most power when there's minimum load current. If there's variation in load current or input voltage, you end up having to waste a lot of power to keep the shunt regulator within regulation. And their ability to regulate a changing load isn't anywhere near as good as a standard 3-terminal regulator like the LM7812, which uses an internal diode fed by a constant current source for its internal voltage reference. In short, I don't believe there are any advantages to using a 1N4742 with a power resistor over using an LM7812, except possibly saving a few cents for an application with a very steady load current and input voltage (and if that's the case, why not just use a resistive voltage divider?)

Learn it, learn how to design one, and then forget it -- it's doubtful you'll ever need it.

Good luck Chris

Hmm... this is quite different then what I had envisioned happening with the no current passing thru the zener until breakdown. I'm still trying to get a grasp on what's happening on all the interactions between components here. I wish there was an educational animation type program that showed you in SLOW motion exactly how the current flowed, voltage, reaction of components etc. Now that would be a great learning tool. Anyway, I'll go pound my head against the wall a few more times and I'm sure I will finally pound it in there. :)

Thanks everyone for their response.

J

Chris wrote:

Hi, J. A couple of things. First , take a look at the data sheet of a zener diode:

For this one, look at the representation/graph sketch "Zener Voltage Regulator" on the top right corner of p.2. That, along with the data sheet chart showing the parameters, is all you really need.

When the zener is forward biased, it acts pretty much like any diode. When it's reverse biased, it acts something like a regular diode until the reverse voltage approaches the Zener Knee Voltage (Vzk). As it gets closer to that zener knee voltage, the resistance of the zener diode starts to decrease in such a way that the current passing through the zener will imicrease dramatically as the voltage is slowly increased. This characteristic continues past the manufacturer-designated test current (at which the device is characterized), up to the point where the power being dissipated by the zener exceeds the package/die rating, when it is destroyed by overheating.

If you're designing with zener diodes, you want to set things so the zener current is between the test current (Izt) and the maximum current. If you're daring, you might go less than the test current, but definitely not low enough to get close to the zener knee -- you'll lose any voltage regulation there.

You don't need a .mov file, you just need a v-i curve.

Good luck Chris

(snip)

LTspice is a free circuit simulator that lets you see the current through and voltage across every component, as well as lots of other things.

It could easily simulate this regulator with a varying voltage input and/or a varying current load.

There is also a very active Yahoo discussion group dedicated to the use of this simulator where you can get some hand holding as you come up to speed on its use.

Forget the animations and sims. Way, way, way better to just connect a resistor and Zener across a power supply and look at the output voltage, while you vary the PSU. Feel the Zener body temperature, watch for the smoke.

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No, not at all. It's more like a calibrated spillway at the top of the dam that lets different amounts of water over to keep the level about constant.

In this case, the water level represents voltage, the water pressure at the bottom of the dam is the source for your load (say a turbine), and what falls over the spillway represents what gets dissipated in the series resistor. When your turbine needs more flow (called 'current'), less water falls over the spillway, and so on.

Hope This Helps! Rich