Author Topic: Function of a zener diode (Read 7957 times)

ZooT · « **on:** August 09, 2006, 08:50:15 AM »

I've got a special project going on that requires 4.8vdc from PV's.(to charge 4-AA Nicads)

I've got a small stack of 6-10 vdc@1 watt x 100ma coming in for this project.

But I don't see any info about 6vdc charge controllers/shunts floating around anywhere.

Looking at the redrock site I see a simple circuit called "shunt1" which should serve my purpose but is rated for 12vdc.

If I were to exchange the 12vdc zener diode for a 5.1vdc zener diode, would this act as a usuable shunt regulator.

I've uploaded the picture of the schematic for "shunt1" to my folder here, but am not posting it here, because I don't know if that's against board rules or not and am not going to risk getting anyone, particularly me in trouble for doing something naughty....

commanda · « **Reply #1 on:** August 09, 2006, 04:05:16 AM »

You need to also lose the other 3 diodes in series with the zener. Each adds 0.6 volt (12 + 3 * 0.6 = 13.8 volts as specified on the drawing). This, unfortunately ignores the 0.6 volt Vbe of the transistor. Oh well.

Zener diodes have a characteristic called dynamic resistance. The effect of this is that they are not as on/off as the simple theory books would lead us to believe. The voltage across the zener does vary somewhat as a function of the current going through it. If you're going out to buy the 5.1 volt zener, also get a 4.7 volt and maybe some 3.3 volts and a few regular diodes so you can SOT (select on test).

Personally, I'd use a TL431 shunt regulator and a pot to make an adjustable zener. But that's just me.

Amanda

Opera House · « **Reply #2 on:** August 09, 2006, 04:20:12 AM »

Those circuit changes would be ok. I'd reduce the number of diodes to two, remember the transistor is another .65 volts. The load resistor value should also be increased to reflect the output of the panel. If you don't, all the power will be dissipated in the transistor. I prefer to use a LM341 adjustable zener to obtain the exact voltage desired. These can often be found in an old PC power supply.

dinges · « **Reply #3 on:** August 09, 2006, 04:23:22 AM »

You could also look on the web for a very simple adjustable power supply, based on the LM317 voltage regulator IC. This thing is adjustable, so no need to bother with zeners.

It's a voltage source though, which in theory is not ok for your application (you need a current limit too), but, since you have only a few & small PV panels, the output current is limited by your panels. Here are some example schematics:

http://www.tkk.fi/Misc/Electronics/circuits/lm317.html

http://www.national.com/pf/LM/LM317.html

http://www.uoguelph.ca/~antoon/circ/vps.htm

Now, to be really honest; I think you can take a shortcut here, and skip the regulator all together, since you've got such a limited PV supply: max 100mA. When you charge 4 nicads at 100mA, if they're modern nicads of, say, 1000mAh, you need to charge for 15 hours. But, at such a low charge current (1/10 of the capacity), you can almost permanently let them be charged; It's called trickle charge.

Many of the simple & cheap battery chargers do just that: supply a current of 1/10C, and expect the user to stop the charge after the required time. And even if you forget to stop it, no great harm is done to the batteries.

In that case, all that would be needed would be a diode, to prevent the batteries unloading themselves via the PV panel, during the night...

That's the shortcut, and the one I'd take. But, if you want to do it the proper way, build a small charge controller or voltage regulator. If only because it's fun and you'll learn a lot about electronics...

I know curiosity killed the cat, but what's the special project?

Norm · « **Reply #4 on:** August 09, 2006, 06:39:35 AM »

I took 6 solar units from those solar lights,

each one used to charge 2 AA rechargable batteries

...put them in series and charge my 14.4 cordless

drill packs...open voltage is 25-30 volts

and they work just fine...voltage under load

starts out at12-13volts gets up to 15.75 and

after that 8 hrs or so may get up to 15.83

never any higher....like trickle charge I guess

it would stay right about there no matter what

never gave a thought to regulating the voltage

wouldn't know how if I wanted to....

or at any rate it would be a real challenge.

( :>) Norm

ghurd · « **Reply #5 on:** August 09, 2006, 09:24:55 AM »

I'd just use a 50 cent LM7805.

Solar + to Vin, Battey + to Vout, all 3 grounds together.

Meaning, personally, I don't think I would even bother with the 2 capacitors that are usually shown in 78XX circuits. Unles the first one failed.

G-

ZooT · « **Reply #6 on:** August 09, 2006, 12:10:15 PM »

It's just a general purpose PV powered 4.8vdc battery charger/power supply.

The reason why I say general purpose is because I've got multiples of at least four different 4-AA devices I'm planning on using this thing with.

But....it has to be cheap because it's going to be at a location where I have experienced vandalism and theft in the past so that counts out a "real" PV array or anything that can't be replaced for under 10 dollars or so per unit.

I can encase these little $6 panels in glass and mount them on the roof or in the skylights(prefered) or simply duct tape them into the corner of a window and not draw much attention to them, but still have "some" light and other rustic niceties.

ZooT · « **Reply #7 on:** August 09, 2006, 01:00:30 PM »

Thanks Ghurd

The LM7805 costs $1.59 at my local radio shack, nevertheless I think that's what I'm going to use.

Found a page from Iguana labs that tells about using the LM7805 along with a 220uf capacitor(also at my local radio shack)to build a 5vdc power supply and THAT looks like something I can handle.(under $3.00 for parts per unit from radio shack and I need a dozen units)

Now would I still need to use a blocking diode with this?

As you can tell, I'm not an electronics guy 8^P

willib · « **Reply #8 on:** August 09, 2006, 03:00:09 PM »

Peter

I couldnt agree more

just hook the panel up to the batteries , at 100mA they are not going to over charge !

the panels do have a reverse diode installed dont they?

altosack · « **Reply #9 on:** August 09, 2006, 05:21:31 PM »

Buying locally is great (if you can call a RadioShack "local"), but since you're getting some quantity, I thought I should tell you that I just bought (

7805s for $0.25 each at futurlec.com; they don't have a minimum order as far as I know.

Since you're paying shipping, you might as well throw in the capacitors and whatever else you think you might need. All in all, I think you could save some bucks.

Dave

ZooT · « **Reply #10 on:** August 10, 2006, 01:18:57 AM »

Yeah......I think I'll do that(order them)

But I'm still going to buy enough to build a couple at radio shack tomorrow so as to have something to play with this weekend laughing

The reason why I was interested in a charge control system is that these panels produce 6-10 voc(stickers on the back of each say the lowest produces 11.2voc and the highest 12.0voc)and upon testing one (they just came today and I work till after dark) it produced 5.6voc under a 40 watt flourescent light, and I was/am a bit worried about overcharging.....

ZooT · « **Reply #11 on:** August 10, 2006, 01:31:30 AM »

No, they don't have "anything".....just a couple clips to solder leads to.

ghurd · « **Reply #12 on:** August 10, 2006, 08:21:50 AM »

The 220uF is way overkill!

Maybe 0.1uF on the output. The batteries act like a cap anyway.

The input is DC (instead of AC from a transformer), so the input cap is probably not at all needed. But 0.1uF should be plenty just the same.

Standard data sheets only call for 0.1uF on the output, 0.22 or 0.33uF on the input.

Avoid the blue tin can electrolitic type. The little tiny flat orange ones would be good.

G-

commanda · « **Reply #13 on:** August 10, 2006, 08:58:09 AM »

The 0.1uF caps on the input & output are for stability. They should be placed close to the device. It has nothing to do with how DC the incoming dc is, nor the fact that the battery appears to be a really big capacitor. You normally only need large output capacitors like 220 uF when driving a load which draws a lot of heavy current spikes, and even then you still need the 0.1uF in parallel.

Amanda

ghurd · « **Reply #14 on:** August 10, 2006, 09:43:20 AM »

I guess I was trying to say I don't care for the circuit he found.

(no input cap or even a mention of one, 220uF elco for the output)

ZooT · « **Reply #15 on:** August 11, 2006, 11:14:37 AM »

Ok, I've built a LM7805 based circuit, as shown in the picture below(?)

Right now I have it hooked up to an 12v/4watt ICP amorphous VW panel.

On the input side I'm getting 13.x volts mainly because it's a bit cloudy today, and that's not an issue.

On the output side though, right now when connected to 4-AA NI-MH batteries I'm getting readings of 5.83vdc....

Is this going to be a problem?

ghurd · « **Reply #16 on:** August 11, 2006, 01:42:52 PM »

I have seen that with 78XX's and no load or no output cap.

It usually does fine with a load.

Some brands or types are different than others. One batch I had needed a load in addition to the battery it was charging. A high value resistor, seems like 10 or 20K, from output to ground fixed it.

Maybe someone else will chime in.

I would sure like some info on that new style VW PV.

Did the grey box come connected to the PV?

Is the highest voltage you have seen 13.8V?

Does it say "ICP" on the panel?

Does it say "4W", and thats all? Typical for ICP.

Are there any other specs on the PV?

G-

ghurd · « **Reply #17 on:** August 11, 2006, 02:01:00 PM »

Afterthought...

The controller in the grey box, if there is one, may be screwing around with the 7805. I think I would use a input cap on the 7805. Maybe 10uF?

And I would put a 24K resistor between the grey box output (or 7805 input, depending on how you look at it) and the ground. A half a mA should keep the PV controller running. Just guessing out loud.

ZooT · « **Reply #18 on:** August 12, 2006, 01:09:10 AM »

Yeah there is a controller circuit in the grey box of the ICP panel....chip on it says H431CL

Tomorrow I don't work, so I can put a "production unit"(the actual components that I want to use) out on the porch with some brand new nicads for testing.

Think I'm going to connect four panels in parallel and see if it'll charge 4-AA nicads in 4 hours

ZooT · « **Reply #19 on:** August 12, 2006, 01:36:02 AM »

Yes the grey box came connected to the ICP panel, and no there is nothing on the ICP panel except a sticker that says "powered by ICP solar" on the back it has the model number and a VW logo molded into it along with the usual CYA warnings.

In fact the only reason why I have it is because it's handier to play with then my 15 watt panels(12"x36") and is advertised to make more power than the panels I want to use in this project......

Another person here mentioned 15 hours to charge 4-AA nicads using a 100ma panel.....so I guess I'm going to have to connect 4 in parallel to bring the charge time into an acceptable range(4 hours?) for this project.....

On another thread here someone mentioned "lightning bug" 4-AA flourescent flashlights.

Well I didn't find "lightning bugs" but I did find 4-AA flourescent flashlights with another socket that holds a nice bright conventional bulb, which I can also use for $1 each

I've got a bagful(two dozen) of these, a box of 100 brand new AA nicads, as well as two dozen of those little 6-10vdc 100ma 1 watt panels, and I can get more of all of them so I kinda' want to do this right 8^P

ghurd · « **Reply #20 on:** August 12, 2006, 08:59:34 AM »

You might want to look at some tech data. Eveready has some nice info. 4 hours is kind of fast with a simple circuit.

I have had very good luck with C/10 max. C/6 in some stuff, where its not charged often. C/4 I feel could be hard on the batteries.

There is a point where additional current(panels) won't help at all. The batteries will reach 5V, then the current is reduced to 'not much'.

Like it is charging at 20ma, with 1 PV or 50 PVs.

I made a C/12 (C/15?) charger (voltage and current limited, 12V powered) for a friend, maybe 5 years ago. He has 2 sets of "D" Ni-Cds batteries, charges a set about twice a week, meaning each set gets charged once a week. Last I knew, it is still using the original batteries! That's about 250 cycles.

G-

SamoaPower · « **Reply #21 on:** August 12, 2006, 10:31:57 AM »

I don't believe you've mentioned what the capacity, in mAh, of your nicads are. They're available over quite a range and that capacity determines charge time and impacts charger design.

Peter (dinges), in an earlier comment, had the right idea. Nicads really want a current source for charging, not a voltage source. The impedance of the batteries is very low compared to that of the panel so the voltage will be dragged down to the battery voltage in any case and that voltage will be dependant on state-of-charge. Using a voltage source like you're currently doing, will extend the charge time as Ghurd pointed out.

If you really want to charge in a significantly shorter time than a C/10 rate will allow, either close monitoring or a charger with a lot more smarts will be necessary if you don't want to either cook or undercharge your batteries.

Charging batteries with solar is a pretty iffy proposition at best, because of the time-variable nature of insolation. If you want to get reasonable lifetime from your batteries, your're going to have to do a lot more than what you're doing now.

ZooT · « **Reply #22 on:** August 16, 2006, 12:25:56 AM »

In the first circuit shown at at

http://www.tkk.fi/Misc/Electronics/circuits/lm317.html

I notice that R1=240 and the value of R2= (192 x Vout) - 240

Is the value etched in stone or is it a relative value and used only in the context of explaining the circuit where is the plugged in value of R1 = 250 then

R2= (192 x Vout) - 250

At the national semiconductor site I found a .pdf named LB-35 that discusses the LM317's possible application and the circuit shown and described is Fig #1 of that document also lists a resistor, this time R2 =240

And the only reason why I'm asking this is because I can't seem to find a 240 ohm resistor.

Another question that comes to mind is what wattage rating should these resistors have?

And I'm sorry if these are "dumb" questions.....but I'm a machinist not an electronics engineer.......maybe someday....but not yet 8^P

commanda · « **Reply #23 on:** August 16, 2006, 12:51:44 AM »

The LM317 maintains 1.2 volts between output and adjust. The resulting current in R1 also flows through R2. The output voltage is the voltage across R1 (1.2 volts) plus the voltage across R2 (which is R2 * 1.2/R1).

The value of 240 for R1 is not etched in stone. I often use 120 ohms myself.

Amanda

dinges · « **Reply #24 on:** August 16, 2006, 01:48:14 AM »

The power (wattage) rating is not important either; anything should do, a 1/10 W or 1/8W should be enough.

A 240 ohm resistor has colourbands red-yellow-brown.

120 ohm should work to, as Commanda explained.

But, if you can get 2 120 ohm resistors and put them in series (one end connected to the other) like this:

--=--*--=--

you 'magically' get a 240 ohm resistor.

120+120 = 240, at least it was when I checked it last time.

dinges · « **Reply #25 on:** August 16, 2006, 02:20:35 AM »

That IC of yours (H431CL) is very likely (in fact, I'm certain) a TL431; see the datasheet:

http://www.fairchildsemi.com/ds/TL%2FTL431A.pdf

It would probably be easiest to simply modify that for a 5V output, instead of the current output of 12V or whatever it is.

Take a look at fig. nr. 10 in that datasheet, and check whether it's the same as in your PV panel/charger. If so, you could adjust resistor R1 to get 5V output...

Maybe best thing to do would be to post a picture of the innards of that regulator of yours (the H431CL), along with the values of the resistors in there.

I think modifying it would be the simplest (unless it's SMD...), for only the cost of a resistor. You wouldn't need extra boxes, wires running between them, and extra parts (7805s). Unless you want your panel/regulator to remain original.

But, to do that, we need more information.

ZooT · « **Reply #26 on:** August 17, 2006, 12:19:37 AM »

Ok, as per our discussion I'm planning on doing an SOT for R1 while replacing R2 with a 5K pot so that I can adjust the thing via multimeter without using a soldering iron for each change.

Several LM317 circuits show the resistor that's labeled as R1 from the adjust pin and Vout and use a 5k pot in place of the resistor that's labeled as R2 between the adjust pin and -Vout.

Even the national semiconductor data file labeled PB-35 shows a pot to allow adjustments rather than a fixed value resistor between the LM317 and the - Vout

On another site....I think it's by a "tony" something or other that has gobs of circuits listed mentions that AA Nicads like to be charged to 1.42 vdc.

Using the 7805 circuit even with a resistor to hold the voltage to 5 vdc leves me lacking but without the resistor I really have no idea of what the top voltage that these batteries could be charged to is....after all under bright sun, the single panel that I've been using out of the stack produces 11-12 vdc.

Not enough to make it a 12 volt panel but too much to be the 6 volt panel that they were sold as

Now ghurd mentioned using a high value resistor(10-20k) between the ground and the output of the 7805 to limit the output to 5vdc....wonder if replacing the resistor with a pot would allow some sort of adjustment in that circuit?

commanda · « **Reply #27 on:** August 17, 2006, 01:27:49 AM »

Now ghurd mentioned using a high value resistor(10-20k) between the ground and the output of the 7805 to limit the output to 5vdc....wonder if replacing the resistor with a pot would allow some sort of adjustment in that circuit?

NO.

What ghurd is referring to is that the 7805 tends to be unstable with no load. The resistor is just to give it a load and draw a fraction of a milliamp.

The pot with the LM317 is standard practice. There is no need to do an SOT (select on test for those who haven't seen this particular TLA (three letter acronym) before), for R1. Just pick a value between 120 ohms and 270 ohms, and select a pot to give the maximum voltage required.

If R2 was 120 ohms, the current in the resistors will be 1.2 volts / 120 ohms = 10mA. (Now you can see why I use 120 ohms, it makes the math easy.)

So, to get a maximum of 12 volts out, the voltage across R1 will be 12-1.2 = 10.8 volts. So R1 needs to be 10.8 volts / 10 mA = 1.08K.

A 1K pot will give an output from 1.2 volts to just short of 12 volts.

HTH (theres another TLA :-)

Amanda

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Author Topic: Function of a zener diode (Read 7957 times)