Zener Diodes / Voltage Regulation

I'm having some confusion regarding zener diodes which I hope somebody much brighter on the subject can clarify somewhat, or at least point me in the right direction!

The basics of diodes has always been "current can only flow one way", and beginners are left to assume this means they'll always do just that. But obviously that's not the case, since normal diodes will stop doing such a thing if they're pushed too far, I understand. And as a result, people take advantage of this effect more reliably with the zener diode apparently, by making them "fail" at much lower rates, it would seem.

So while this is possibly out of the scope of asking on a newsgroup, what sorts of applications could they be used for reliably? I hear they can be used as regulators, but only if the load is fairly constant. So, for something like an electronic circuit, with chips going on and off and constantly changing the load, would this be of any use there? I've seen them used in situations such as when one is drawing power from the PC parallel port. I'm assuming this isn't related to actually getting that power, but keeping it from going too high.

I'm curious because, despite my lack of knowledge regarding the details of analog circuits, I have a pretty decent understanding of digital ones, and find myself making and planning new things in that realm all the time. I'd like to power some of these sorts of things off of batteries sometime instead of a 7805 on an ac adapter, and am interested in the best way to go about it. One project in particular involves an 8052 microcontroller with an lcd which I'd like to make portable, for example.

On a side note, how possible would it be to boost the voltage from a couple of AA's to run 5v logic reliably? I can't really think of how such a circuit would be made, though I know they surely exist. I assume it uses capacitors and an oscillation of some kind, but I dunno. When I think of changing voltages, I mostly just think of transformers.

Anyhoo, any help or pointers would be greatly appreciated!

Reply to
FyberOptic
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FyberOptic wrote: (snip)

That rule applies below reverse break down voltage, and zeners are operated above reverse breakdown.

I'm not sure what you are trying to get across with that, but the difference between normal, rectifier diodes and zeners are that normal, rectifier diodes are designed to have a minimum guaranteed reverse breakdown voltage and be operated below that guaranteed minimum, while zeners are designed to have specific reverse breakdown voltages and to be operated with in a reverse breakdown situation.

Rectifier diodes have forward current ratings based on the heat they can get rid of while dropping a volt or so in the forward direction. Zener diodes are rated for the power they can get rid of while leaking reverse current (and dropping their rated zener breakdown voltage.

Any application that keeps their die temperature below the rated temperature. These include signal clamping, and reference voltage generation and supply regulation.

They act as pretty fair regulators as long as the current through them does not go too low (or the breakdown process gets sloppy) or too high (and the zener overheats). The zener shunt regulator must carry a minimum current when the load current is at maximum (to avoid the first case) and must carry that and the difference between the maximum and minimum load current without overheating. In addition, variations in the unregulated voltage that is current limited by a series resistance into the parallel combination of the zener and the load, must not push the zener operation into either of those extremes through its expected variation.

It depends on whether power is cheap or expensive and how much the raw voltage is expected to vary, and how much current variation the load will produce. there are some pretty large zener diodes produced. But the most common types are axial leaded devices with power ratings of 5 watts or less. If the load power is much less than that, a zener might be a practical regulator.

The power available through a parallel port is quite limited, so a small zener is quite practical as a way to limit the voltage from that source. Of course the limited power also implies that the load is draws only a small current.

Since zener regulators must consume some raw supply current, just to keep them regulating, they are poor choices for most battery applications. 3 terminal linear regulators, like the 7805 have pretty low minimum current requirements, and very much better (in that department) regulator types are available. But best efficiency for higher current loads involves switching the battery voltage off and on, as completely as possible, and then filtering those pulses to extract their average voltage. There are no intentional losses in that process, so the power out equals the power in, except for unavoidable losses. in many cases, the output current exceeds the input current, because the voltage has been converted down. The key words for this sort of thing are [buck converter].

It is often done the same way that spark voltages are generated by contact points and a coil. inductors have the property that they generate voltage to fight sudden variations in their current. You connect the battery across an inductor and its current ramps smoothly up, as it produces a voltage that almost holds back the battery voltage. Then you switch that current off, and the inductor produces a voltage that tries to keep its current going. If the battery had been applying positive voltage to the inductor, when the current is interrupted, the inductor will produce a negative voltage to try to keep sucking that current through the switch that is blocking the current. If you provide an alternate path for that current driven by the inductive so called flyback voltage reversal, you can connect that negative voltage to a storage capacitor that charges to an arbitrarily high negative voltage. If, instead, you add a second winding to the inductor, making it into a transformer, that flyback voltage can be reversed (by swapping the winding ends) and you can use the diode to connect the output to a capacitor that is charged to an arbitrarily high positive voltage. Or add several windings and produce several proportional voltages, either positive or negative, all at the same time.

You also need a control circuit that varies the current ramp up time , each cycle to produce the desired output voltage.

The key words for this sort of thing are [boost converter].

There are ways to charge a pair of capacitors, in parallel, with the battery voltage, and then reconnect them, so that they are in series, doubling the battery voltage, and connect that voltage to an output storage capacitor to act as a power source while you go back and repeat the cycle. The key words for this sort of thing are [charge pump].

Here is a basic tutorial for the buck and boost converters:

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Reply to
John Popelish

A zener usually isn't used directly as a power supply regulator, per se. It may be a good voltage reference, though, as input to a comparator (which doesn't require much current.) If a zener is used for supplying some current, a BJT voltage follower is often added (the base doesn't require much current, once again.)

Take a basic circuit:

R1 / Z1 +Vsupply ----/\\/\\--+--|---+-----| NPN | Vref---|+/ |>e | |/ | | | | Vout | gnd '-------+----+----- LOAD ---gnd | | | | | | | \\ --- C1 | R1 / --- | \\ | | / | | | | | | gnd '-------------------------------+ | | \\ / R2 \\ / | | gnd

Here, you can see that an opamp is used to compare the Vref (output of the constant current version of the zener regulator I just mentioned above) with a portion of the output voltage (Vout.) The opamp then drives Q1 as needed so that the difference is nil.

[That opamp might only be a differential pair of BJTs with some additional BJTs for things like a constant current at the pair's emitter or in the collector of one side as another constant current to 'stiffen' (make it much higher gain) it so that only one diff pair is needed instead of, say, two pairs.]

Of course, you can just buy an IC that does all this and more for you.

Jon

Reply to
Jonathan Kirwan

Here's a little challenge for the OP. Given that

Q=CV C(series) = 1/(1/C1 + 1/C2) (alternatively C1C2/(C1+C2)) C(parallel) = C1 + C2 Qt(series) = Q(C1) = Q(C2) Qt(Parallel) = Q(C1) + Q(C2)

Then take two capacitors with a ratio of 10:1 (say 1uF and 0.1uF). Hook them up in parallel and charge them to some voltage (say 5V)

Disconnect them and reconnect in series (a thought experiment, really) and calculate the total voltage across the pair. You might be surprised at the result.

Cheers

PeteS

Reply to
PeteS

Could you perhaps try a post with a couple of straight questions ?

Graham

Reply to
Eeyore

Dammit, you KNOW that this is news:sci.electronics.basics yet you just have to slam the newbies.

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Michael A. Terrell
Central Florida
Reply to
Michael A. Terrell

In the old days, voltage regulation wasn't common. There were a few things that could make use of it, and you'd see voltage regulation applied to that stage rather than to the whole equipment. Only exotic lab equipment would have fully regulated power supplies.

So in the tube days, that voltage regulation would be done with VR tubes, ie Voltage Regulation tubes. There'd be a current limiting resistor from the power supply, and then the two terminal tubes would kick in when the voltage went too high. Change the load, and you'd need to change the current limiting resistor.

Transistors came along, and so did zener diodes. They were like those VR tubes, except you could now get zeners that would regulate at low voltage.

IN the early days of transistors, you would see those zener diodes used basically like those VR tubes, regulating specific stages rather than whole power supplies. They were lousy for general supply use, since they tended to be low current devices and you had to adjust the current limiting resistor as the load changed.

But they disappeared mostly after a decade. People made the transisition to solid state devices, and realized things weren't quite like the days of tubes. Tubes used such low current at high voltage that the power supplies were pretty high impedance output.

But solid state needed high current at low voltages, and the old high impedance output power supplies didn't really work so well. You'd see people adding and adding large value electrolytics to the output of the rectifier(s) in the solid state supplies, trying to get that low impedance output. And it was never a complete success.

You'd start to see zener diodes used to supply a constant voltage, rather than real power, so they'd be used with transistors that would pass the actual current. And then you'd see supplies where there was feedback, so no matter what the load on the supply, the voltage was constant. Gradually these came into force, so fewer and fewer solid state supplies had anything but a regulated supply. But they weren't regulated because the devices had an absolute need for an exact and constant voltage, but because the regulating state allowed for that low impedance output that the devices wanted.

It was also a lot easier to make up a solid state regulator than a tube type regulator, since large numbers of transistors would fit in the space of a single vacuum tube.

Then a few years later, the discrete regulators basically disappeared. IC regulators became common, especially 3 terminal regulators. It became so easy to use them, it was hardly worth not using them. No need to figure out the current limiting resistor, and no need to adjust it every time the load changed. It was great.

You now rarely see zeners. They are relegated to where a low current voltage source is needed. And where the drain of a 3 terminal regulator is prohibitive.

Michael

Reply to
Michael Black

Is the challenge to find the error in these equations?

Reply to
John Popelish

Needs some interpretation. Q is the charge transfered by a change in voltage V.

True but not terribly applicable to placing two charged capacitors in series and measuring their open circuit voltage.

True, and useful in determining how much total charge it will take to produce a given change in voltage across the parallel pair.

What does that have to do with the open circuit voltage of an arbitrary string of precharged capacitors?

The question is applicability of those somewhat simplified equations to this problem.

I assert that if you charge a 1 uF cap and and a a 1 nF cap, each to 12 volts and place them in series, the total voltage across the pair can be either zero volts or 24 volts, depending on the relative orientations. How do your generalities show this to be true or false?

Reply to
John Popelish

On Sat, 21 Jul 2007 22:01:28 -0700, in message , FyberOptic scribed:

That's a bad assumption, and not the "basics" of diodes at all. Maybe a Cliff Notes version.

Start understanding better by using the proper term, which is not "fail," it is "breakdown."

Not at all, your question is exactly what this newsgroup is for.

Actually, as long as the load and the supply are reliable within a fairly large range, the zener can be used pretty safely, as long as the regulation does not have to be tight. Zeners have quite a large +/- tolerance range, on the order of 15%.

The best way to go about it is to choose a battery value that most closely matches the need of the load. Any voltage regulation using a zener will waste power from the battery, and there are many, many batteries available on the market to pretty much match any requirement.

It's best done, as above, by choosing the battery voltage you need. Any voltage conversion, up or down, again, will require some possibly significant amount of power just for the conversion. Since getting 5V with a battery of AA cells is not practical without conversion, due to the voltage variation of differing types, you want to find a marketed 5V rated battery.

To choose the right one, figure out the amperage requirement of your project, then decide how long you want to run the project reliably. You mentioned wanting to use AA batteries, which are in the range of 2000-2500 mAh for NiMH rechargeables. Say you are going to use 1A for 4 hours. That's 4 amp-hours. Multiply by 125% and you arrive at the requirement for a 5 amp-hour (5000mAh) battery, which is twice the typical capacity of the NiMH AA. Now find a 5000mAh 5V (typically 4.8V, 1.2V per cell) battery pack, which can be had for around $30 US.

Which, of course, don't work well with DC. Hence the prevalence of AC in power transmission.

Reply to
Charlie Siegrist

Snort

Do tell. Let's see

Q = CV. I think we can agree on that. It was figured out a long time before we were born. Series capacitance; reciprocal addition is the same (for two devices) as product over sum. Parallel capacitance is simply the addition of the capacitances. The charge on any number of caps in series of any capacitance is identical The total charge on any caps in parallel is the sum of charges.

I don't think I said anything different above.

Cheers

PeteS

Reply to
PeteS

On Sun, 22 Jul 2007 15:48:36 -0400, in message , John Popelish scribed:

That's the question of the day. I'm thinking the answer depends highly upon unstated conditions, and then after the stating of said conditions, lots of sammy-the-snake squiggly lines and cuss words such as "dv/dt" and "logarithmic graph paper." Either that, or magic.

Reply to
Charlie Siegrist

This is not a binaries newsgroup. You can post it to news:alt.binaries.schematics.electronic by attaching it as a GIF, JPEG, or other common image file. Then, post a message here giving the name of the post. If you post a binary file here, most people will never see it.

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Service to my country? Been there, Done that, and I\'ve got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida
Reply to
Michael A. Terrell

The four resistors would only "waste" 15 watts if they were pulling current all the time. The fuel injectors only open very briefly to allow a spray of fuel in.

The resistors may also serve another function - like modify the inrush current or help absorb flyback current.

15 watts is a drop in the bucket when it comes to auto efficiency - even if the injectors stayed open 100% of the time. Synthetic motor oil, intelligent driving, tire pressure, etc., will all make a greater impact.
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Reply to
default

Just thought I'd mention that Jaycar (and others I presume) sell a device that does this - ie a portable USB supply

David

Fyber>

Reply to
quietguy

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