# ? Proper use of a zener diode?



## Steelwolf (Jun 1, 2001)

As I understand it, a zener diode is designed to be used as a voltage regulator. (or a current regulator depending on design). You place it in reverse polarity to the current flow, and it only allows a certain specific voltage drop across its terminals.

So if I had a memory cap I didn't want to fry, I would put the memory cap terminals in parallel with the terminals of the zener so that the voltage across the capacitor's terminals remain constant?

What would this do to the current? Can I model it as a proportional converter? Or more specifically:

I have a handcranked generator. The faster I crank, the higher the voltage output. I could control the voltage by maintaining a certain cranking speed, or I could put a zener diode. If I were to feel energetic, I could crank really fast. Would the extra voltage be converted to current to charge the capacitor or would the current be shunted away?

What about after I stop cranking. Will the zener then allow or prevent discharge across it? Or the zener doesn't matter after the charging is done?

Thanks in advance.


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## X-CalBR8 (Jun 1, 2001)

I have a question that is, in a way, related to the above question. If I were to use a super cap rated at 2.5-Volts, what would be the best, most efficient, way to crank that 1-2.5Volts up to the necessary voltage to run a LED at full brightness, such as about 4Volts 40ma. I know about Satcure, but it doesn't run the led at full brightness, from what I understand. Also, it would be very nice (almost essential) to have a circuit that could work from 2.5 volts to below 1 volt in order to suck every last bit of voltage out of the capacitor until it's completely drained. Is there a preconstructed voltage regulator that would fit the bill, or would I be forced to build my own? Does anyone know where I could get plans to build such a beast as this? Perhaps it would be possible to buy a voltage regulator that could be set to the exact requirements of the particular project. If anyone could help with ideas, info, or web links it would be very much appreciated. I've been racking my brain for several days now on this subject to no avail.


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## Steelwolf (Jun 1, 2001)

The simplest way I can think of is to use the builder's kit that ups the voltage of a 1.5V battery to 9V. Most of those kits floating around use the Texas Instruments TL496 IC chip. You can select to use a 3V input as well. The max output is 40mA. I would suggest such a circuit set up for the 3V input. You could probably even put 2 LEDs in series without any noticeable downsides. Of course everyone is saying that this is an obsolete IC, but until the kit comes out with the updated chip, or someone publishes a circuit to use the newer chips, this might be the simplest answer. It is brighter than the SatCure circuit anyway.

Now... what about that darn zener?


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## **DONOTDELETE** (Jun 2, 2001)

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.


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## **DONOTDELETE** (Jun 2, 2001)

Ok, now for the 2.5V cap:

If you could avoid using the 2.5V cap and use a 5.5V GoldCap instead, I would do this. It would be the easiest way and you would not have to build a step up converter.

If you WANT to use the 2.5V cap, you will have to build a step up converter with REGULATED output. You would have to use an IC, good ones would be the MAX1674, or the LT1308 fron Linear. But both are SMD devices.

You would have to order these ICs directly from the manufacturer.
I just found one IC not as SMD and available from www.conrad.com : LT1073-5, but it is and old type and not as efficient as the ICs above.
The next problem is the coil for the step up: I only know one ditributor selling good coils usable for these ICs: www.rs-electronics.com . But you could also look at the manufacturer: www.coilcraft.com 

So you see, you need some electronic knowledge to build an up-to-date step-up, and you have to hang a lot on the telephone before you get the parts, but it makes a nice project...

Another possibility would be to buy the 3to9volt-kit from radioshack, and then to redisign the circuit to an output of 4V. Have a look into the datasheet, this will be easy...


I hope, I could help you a little bit. If you have any further questions, please tell me.

Bye!


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## Steelwolf (Jun 2, 2001)

So... if I put a zener diode in reverse, parallel to the capacitor, I can protect the capacitor from over voltage, but I will end up wasting the extra amperage I might otherwise have been able to draw if I used a resistor of appropriate value in line.

Did I get that right?

So... is there anyway I can keep the voltage across the cap at the required 5.5V, but change the amount of amps going in according to how energetic I feel? Because, as I understand it, you can charge a capacitor faster by throwing in more amps (within reason, I guess), but have to keep the voltage within rated values to keep from frying it.


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## X-CalBR8 (Jun 2, 2001)

Phantomas2002: Thank you very much for all the helpful info. I learned a lot. I've got a lot to think about now. About the 2.5-Volt super capacitor, it's not really that I want to use a 2.5 volt capacitor, but all the 5.5 volt super caps that I've found for sale so far only go to 1 Farad and the 2.5 volt super caps rang from 50Farads to 2700Farads! I know that some company makes a 6.3-volt super cap, but so far, I haven't been able to find any for sale yet. That is why I was going to try to make a 2.5-Volt super cap work for me. It sounds like it would take a lot of electronics to make it work well though, so I may just wait until I can find one rated at a higher voltage for sale. Thanks again for all the helpful info.


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## X-CalBR8 (Jun 2, 2001)

Steelwolf: Yes. So long as you don't let the voltage go above what it's rated at, you can throw all the amps at it that you want (up to a very high point). It can handle many times more amps than any hand powered device could ever throw at it, but it is VERY important not to exceed the rated voltage, even for a short amount of time, because all it take is going too much over the rated voltage one time and poof you have a bad super capacitor on your hands. Going above the breakdown voltage (rated voltage) fries the dielectric in the capacitor by arcing through it, from what I've read about it. That's why I was trying so hard to get one of the super caps rated at 6.3 volts so that I would only charge it to 6 volts and would never get too close to the breakdown voltage. With a 2.5 volt capacitor, it would be hard to get enough useful voltage out of it without getting too close to it's rated voltage. With a 5.5 volt super cap, I wouldn't want to charge over 5 volts so that you would be sure to stay a nice safe distance from the breakdown voltage and should never have to worry about accidentally frying it.


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## X-CalBR8 (Jun 2, 2001)

Here is a page on building different types of voltage regulators. It is quite technical so it may take a while to fully understand it, but it will make for some very interesting reading. I just hope we can figure out how to apply this info in a useful way. Btw, It does contain info on Zener diodes which should be helpful to gaining a better understanding of them.


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## Steelwolf (Jun 2, 2001)

Thanks X-CalBR8. I'm checking through the webpage. Interesting. But I'm looking for a circuit that will not only limit the voltage, but will convert the extra to amps and allow it to charge the capacitor quickly and do the convertion automatically.

Slowly but surely, I will find the answer. Hopefully, one day, we will be able to post about the best way to go about building an emergency light that doesn't have rechargeable batteries that can go bad and can be charged quickly by hand or solar power.


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## **DONOTDELETE** (Jun 3, 2001)

Steelwolf:

I forgot to mention that, when a capacitor is not yet charged, it will short its two pins and then build up a voltage between them rising with the amount of charge flown through the cap.

What does this mean? 

When you have an empty cap in parallel with the 5.1V Zener diode and start cranking, the first component to limit the voltage is the CAP! not the Zener diode. The Zener will start to waste power when the cap would exceed the 5.1V which are reached at the end of the charging cycle and the cap is FULL. So the waste of power in the Zener diode would not have been able to be converted into useful charge as the cap is charged full already. It just keeps the cap in the safe voltage range.

Even with the most sophisticated electronics you would not be able to store more and more charge in a cap of a definded size. When you charged the cap fully and crank and crank and crank... the excessive power must be dissipated somewhere...

Got it?

XCalibr8:

If you want to use the 2.5V caps, just put two in serial. The capacity is only half if you use two equal caps, but the appliable voltage doubles. You will just have two make sure that the voltage is divided equally to each cap by adding two resistors in parallel with connectiing also between the two caps:

!
!
+-----+
! !
--- -
--- ! !
! ! !
! ! !
! -
! !
+-----+
! !
--- -
--- ! !
! ! !
! ! !
! -
! !
+-----+
!
!


The values should be in the some kOhms range so there is not much power wasted but charges can equal fast enough.

A better way would be to put a zener diode of about 2.4V parallel to each cap so protection is perfect...

Bye!


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## Steelwolf (Jun 3, 2001)

So, as I understand it now, I don't have to worry about excess voltage until the capacitor is charged. Only when that happens, then the voltage across the terminals will rise, then damage might occur. In that case, would there be any reason to keep the resistor that comes with the circuit? (They put a 10ohm resistor in-line in the original circuit. So the path runs from the generator to the resistor to the lamp back to the generator.)

Anyway, the idea would then be to just set up a zener diode across the capacitor to prevent overcharging. And probably a diode in line with the zener, but in opposite polarity to prevent discharge.

Hmmm.... the design is starting to get interesting. But to really get interesting, we'll need to go up to the 100Farad capacitors. Then it will be worthwhile to put in a solar charging circuit.





I wonder if I can put in a small, low power LED to indicate full charge.


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## X-CalBR8 (Jun 3, 2001)

Phantomas2002: I just wanted to thank you again for taking the time to give us advice on this project. Let me make sure that I understand fully. The 2.4 volt Zener diode would go between the + and - on each cap, right? So that when the capacitor tried to charge above 2.4 volts, the excess power would then be dissipated through the Zener diode, correct? So far I've found a 2.4-volt Zener diode rated at a half watt, would that be sufficient or would I really need to get the one rated at 1 watt. Also does it hurt a Zener diode to dissipate excess voltage for hours at a time such as a small trickle of voltage coming from a solar panel all day? I just thought of something else also, if either of the Zener diodes should ever fail on either capacitor, it would allow the cap to overcharge and be ruined wouldn't it? Is there a reasonable way to protect against this ever happening? One last question, would it still be necessary to put resistors between the series connection of 2 super caps if you already have the Zener diodes protecting against an overcharge? Thanks again for being so patient with helping me. I've still got many things to learn in order to have the knowledge to build all the neat things that I can dream up.


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## X-CalBR8 (Jun 3, 2001)

Steelwolf: Neato idea about the power charge indicator. I like that. I may have to give that a try also. Oh, I just thought I would mention, why not build a voltage limiter with Zener diodes if you still want to make your light powered with a hand generator? That way you could kill any very high voltage spikes before they ever get to the caps. Just seems to be a safer route to go when dealing with such expensive super caps. You could build the voltage limiter inside the base of a screw type bulb assembly (then run a wire out from it) so that you could screw it into and out of the hand-powered flashlight. That way you could still use the hand powered flashlight as a stand-alone or use it as a generator to charge your other flashlight. This project just has the potential to go so many different directions.


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## **DONOTDELETE** (Jun 4, 2001)

Hi guys!

The power dissipated by a zender diode is easily calculated:

Voltage across the zener x current = power

Example:

2.7V Zener diode rated 1/2 Watt:

2.7V x current = 0.5 Watt

current = 0.5Watt / 2.7V = 0.18 Amps

So the maximum continuus current is 180 mA.

By the way, I didnt find a 2.4 V zener diode, the nearest value is 2.7V, which could be a little much for the 2.5V cap. I would suggest to take a 1V and a 1.4V type in a row.

I dont know if 1/2 Watt is enough for the hand crank generator, because I dont know how much current it can deliver, but from estimating I would say it is NOT enough when Arnold Schwarzenegger is cranking



. Better take a 1 Watt type to be on the safe side.

For a solar panel it is easy to say: Just have a look at the max current of the panel...

And for the hand generator again: Of course you would have to put a diode in series with the positive lead of the generator to prevent discharging of the cap through the windings of the generator when you stop cranking.
And if you connect a LED to the output of the two caps (4.8V) you will have to use a current limiting resistor ( (4.8V-4.0V)/40mA = 20 Ohms ).

I thought about the resistors on the caps again and I think the best way is to put the zeners across each cap, so protection is perfect, no more resistors needed.
By the way, because a zener acts as a normal diode in the other direction, the maximum reverse voltage is limited to 0.7V on each cap which gives additional protection.

I think this will make an interesting project, please tell me if you have any results..

Bye!


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## Steelwolf (Jun 4, 2001)

OK. I'm starting to see now. Phatomas, thanks for all the explainations. Just a few more if you have the time.

I already have a bridge rectifier to straighten out the AC that the generator supplies. I've drawn up a circuit diagram and it seems to indicate that the diodes in the bridge would do the job of preventing discharge back through the generator coils. Did I get that right?

In addition, there were some mention in texts about putting a regular diode in line with the zener but in reverse polarity to prevent the capacitor from discharging. Using polarised storage like batteries and electrolytic capacitors, it appears that there would be no need for this diode either. Am I wrong?

Finally, from what I intepret your previous explainations about the behaviour of the cap, I should not require a resistor in the charging circuit as the capacitor will provide all the resistance needed to maintain the voltage drop at a suitable level and convert the rest of the energy to current to charge itself up. And this will happen until the capacitor is fully charged, whereupon, having no protection via a zener diode, the voltage drop across the capacitor will begin to rise and cause damage.

So the resistor will only be required in the lighting circuit to limit the current. I'm going to check if this resistor will detract from the brightness.

Again, Phantomas, thanks very much.


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## **DONOTDELETE** (Jun 4, 2001)

Hi Steelwolf!

If you have a bridge rectifier, you will of course not need an additional diode to prevent backflow.

And you will not need a diode additional to the zeners. This was a misunderstanding. I meant the zener itself will protect the cap from more than 0.7V voltage in the wrong direction (and of course 2.4V in the right one...).

The resistor is only needed in the output branch to protect the LED.

And, just if you are interested, the process of charging the cap:
When you start at 0V on the cap and start cranking, the cap will shorten the generator. So most of the voltage will drop in the generator itself. At this moment, the short-circuit-current of the generator will flow. This current is calculated by dividing the voltage generated divided through the internal resistance of the generator. With this current flowing through the cap, it builds up charge, which makes the voltage at the cap rising. The voltage built up at the cap subtracts from the voltage on the internal resistance of the generator, making the current going down.

There is no such thing as a conversion from voltage to current or vice versa.

The zeners are just to blow off any voltage over the safe voltage for the cap. Just like a overpressure valve at a steam bowl. 

To reain there: When you put steam (=current)into a bowl (=cap), starting at 10 bar (=volt), the pressure in the bowl starts at 0 bar and rises slowly. But the bowl is only safe up to 5 bar, so a safety valve (=zener) is attached, which opens at 5 bar. So you can add as much steam as you want, when the pressure reaches 5 bar, it is blown off completely by the valve to remain pressure constant.

Bye!


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## Steelwolf (Jun 4, 2001)

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>There is no such thing as a conversion from voltage to current or vice versa. <HR></BLOCKQUOTE>

Sorry Phantomas. That was just my way of thinking about it. If you have a 12V voltage source, then you will measure 12V 0A (or a small current because no ammeter is perfect). But if you put a 10ohm resistor across the terminals of your voltage source, then you will measure 12V dropped across the resistor and 1.2A line current.

If then you plug in your device and it acts like a 50ohm resistor, then you will measure 0.2A flowing in the circuit, 10V dropped across the device and 2V dropped across the 10ohm resistor.

I think the concept is right? But of course my way of thinking about it probably confuses everyone else.





Anyway, thanks for you help. I'm going off to build my emergency light now.



Complete with zener diode for protection and indicator LED for when it is fully charged.





X-CalBR8: Hope you can find all the parts you need. Remember to post when you start work on it.


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## X-CalBR8 (Jun 4, 2001)

Steelwolf: "Remember to post when you start work on it."

Sure thing. I've got a new picture host now so I hope to post pics as soon as I get to the point of doing anything interesting with it. I'm just not sure yet how long it will take to track down all the right parts and then figure out how to fit it all together in a way that I won't blow something expensive up. LOL. This is tougher than I first thought it would be. I could buy a premade step-up and voltage regulator (possibly the best way to go), but I want to be able to say that it was all hand made when I get done. Heck, I would even make the super cap myself if I only knew where to begin. Hehe. Looks like I'm gonna HAVE to build the stupid super cap myself if I ever get one like what I need.


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## X-CalBR8 (Jun 5, 2001)

Thanks again Phantomas2002. I now have a much better understanding of Zener Diodes and their uses. I did happen to find a place that sold 2 different Zener Diodes at 2.4 volts 1 watt, but they were out of stock of both. http://www.mouser.com That at least proves that Zener Diodes do exist at this size so it should just be a matter of time until we can find a place that has some in stock. I think we are getting closer and closer to getting this project to work. I can hardly wait.


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## zbermana (Jul 24, 2010)

i have same problem sir
I need 3,2 volt 5A and 10 A from 12 volt power source.

what kind of zener diode macth for that?

Pls someone help
Thanks a lot


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## Ray_of_Light (Jul 24, 2010)

You would like to use a zener to make a stabilised power supply from a 12 V source, while drawing 10 A?

You need a 0.82 Ohm 80 W series resistor, and a 3.3 V 35 W zener diode. Both the resistor and the diode need proper heatsinking.

There are less simpler, but much more effective ways to reduce the voltage from 12 to 3.2 V, by using a switching regulator.

Anthony


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## zbermana (Jul 24, 2010)

Ray_of_Light said:


> You would like to use a zener to make a stabilised power supply from a 12 V source, while drawing 10 A?
> 
> You need a 0.82 Ohm 80 W series resistor, and a 3.3 V 35 W zener diode. Both the resistor and the diode need proper heatsinking.
> 
> ...


==========

actualy i am very very novice at electronic, but i love to learn.
what you mean by swiching regulator. can i make by my self? or i have to buy that switching? do you have a simple schematic switching regulator?

thanks


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