A Few Small Questions

1. Jan 2, 2010

RestlessMind

I am constructing a solar powered nightlight with http://i538.photobucket.com/albums/ff341/the_cake_is_a_lie/ElectronicsSchematicCOMPLETECUSTOM.gif" [Broken] circuit, and I have a few questions.

1. I do not know what amount of ohms the resistor designated "R4" should be. I think I can calculate it with: (ohms needed) = (battery voltage) - (LED voltage) / (LED amperage). The package of my battery says that it is 3.6V, and the LED package states "FW current: 25mA, FW supply: 3.3V (typical), 3.6V (max.)". Therefore, is it correct to calculate the needed ohms of R4 like this?

3.6 - 3.3 / 0.25 = 1.2 ohms

Also, do 1.2 ohm resistors exist? If not, what should be used?

2. A solar nightlight tutorial recommends use of http://www.anybodyburns.com/pathlight/images/batteries.jpg" battery that says "Cordless Telephone" on the package, but is also a NiCad rechargeable battery (at 3.6 volts; there are actually three batteries in series in the case). Is this okay, or is there a special property of the tutorial's batteries that were meant for solar applications that mine don't have?

3. Does the circuit, given the above confirmations, seem that it will be functional to you (of course I will test as well)?

Last edited by a moderator: May 4, 2017
2. Jan 3, 2010

vk6kro

25 mA is 0.025 amps so that resistor would be 12 ohms. That is a common value.

3. Jan 3, 2010

RestlessMind

Ah! A decimal point off. Thanks very much for correcting me!

Does anybody know anything about the battery issue?

4. Jan 3, 2010

RestlessMind

Another little wondering; does this configuration...

...work the same as this configuration?

5. Jan 3, 2010

vk6kro

No, the extra battery holder won't make any difference.

6. Jan 3, 2010

RestlessMind

Good to know.

Do you know anything about question 2., perhaps?

7. Jan 3, 2010

8. Jan 3, 2010

RestlessMind

Oh no, the voltage is not an issue.

And they are simply three AA NiCad rechargeable 1.2 volt 600mAh batteries wired in series, in a plastic package. My question is; they are not specified for solar applications, but seem identical to the ones in solar applications anyways. Are they?

9. Jan 3, 2010

vk6kro

They are very low capacity NiCds.

Current NiMH like Eneloop ones are 2000 to 2500 mA-H.

Yours are good enough to try though.

10. Jan 3, 2010

RestlessMind

Yes, I got them at 600mAh because the schematic I am working off of says they need to be ~550mAh and I don't want a very long charging time. Since I am using an 80mA solar panel it will already take 550/80=6.875 hours to charge.

11. Jan 3, 2010

vk6kro

You don't need anything else to test them.

Just connect the cells across the solar panel and put it in sunlight. Check the charging rate with an ammeter.

This will tell you if you have enough voltage to charge those batteries.

12. Jan 3, 2010

RestlessMind

The solar array gives 6 volts, the battery array is 3.6 volts, and the diode is relatively nothing. Should this not charge the battery fine? There will be ~2 volts "excess."

13. Jan 3, 2010

robert2734

The circuit has a belt and suspenders quality to it in that both Q2 and R4 protect the LED from overvoltage/overcurrent. Anyways Vce of Q2 should be at least 0.2V and will actually be whatever voltage the LED doesn't "want". So R4 should be less than 12 ohms to allow some voltage to be dropped across Q2.

I would expect Q2 to be used as a current source (You know Vbe is 0.7V so a resister in parallel with Vbe will draw 0.7V/R amps so voila current source) but they aren't doing that.

You realize a bigger battery never hurts you. Say you have a full 2400mAh battery at dusk and the night light draws 600mAh during the night. Then it will only take 600mAh to fill back up. Or say the battery is empty at dawn and it is charged 600mAh during the day. The battery is one quarter full but you still have 600mAh of juice in the tank.

The reason I reccommend the bigger battery is that it will be harder to destroy a bigger NiCad battery by overcharging. NiCads don't like being trickle charged. I would think a lead acid chemistry battery would work better and lead acid cordless batteries exist tho they might be harder to find. But then it would be at 4V (lead acid is 2V a cell) and all your numbers would need to be recalculated.

14. Jan 3, 2010

robert2734

You do not want 2V of excess voltage across a 3.6V battery array. Not safe for the battery. If you have built this circuit can you tell us what the voltage across the batteries actually is during charging?

15. Jan 3, 2010

RestlessMind

No, I did not design this circuit. I did modify it (with the assistance of vk6kro). Here is the original: http://www.anybodyburns.com/pathlight/images/schematic1led.gif"

Yes, I know. I added in R4. What is Vce? The package of Q2 (and Q1) says "Vcbo: 75V, Vceo: 40V, Vebo: 6V".

Less than 12 ohms? What's the closest ohm rating that is less than 12 ohms?

What do you mean? Will this cause problems?

Ah. I'm not sure if I've seen any NiCad batteries that big, though. I did see a NiMH that was 2400mAh, though, but I don't know their weaknesses and strengths when it comes to charging and draining.

I don't want to use a lead acid battery. I'm short on time and available resources and my batteries need to fit in a rather small space.

Last edited by a moderator: Apr 24, 2017
16. Jan 3, 2010

RestlessMind

What voltage do I want across a 3.6V battery array? No, I have not yet build the circuit.

17. Jan 3, 2010

vk6kro

Q2 is not there to protect the LED.
The circuit charges the battery during daylight hours and then turns on the LED after dark.
Q2 does this switching

Solar panels become current generators if short circuited. This one can only give 80 mA which is a reasonable charging current for a set of NiCds.
This circuit does exactly that with only the series diode D1 between the solar panel and the battery.
It does not have any provision to stop overcharging the battery and, as you suggest, this is more likely to happen with low capacity cells.

There is a possible problem with the battery having enough voltage to actually turn on the LED. White LEDs can need 3.5 volts to turn them on. Below that, you get no light output at all.
After charging the batteries fully, you can try putting the batteries across a LED/ resistor series combination.

18. Jan 3, 2010

robert2734

The original circuit http://www.anybodyburns.com/pathligh...ematic1led.gif uses a 4.5V solar array. I would suggest not substituting a 6V solar array for the 4.5V solar array. You can hurt your batteries. No, I do not know the exact voltage you can put across the battery without causing trouble.

If you already have a 6V solar array I would suggest bleeding off the extra 1.5V by putting two diodes each a with forward drop of 0.7V in series with the battery.

The 12 ohm resister you added will do nothing except make the LED slightly dimmer. There's no particular reason to starve Vce of Q2 of voltage so I would suggest zero ohms. If you must have a resister there, there is no magic value, but the bigger the resister the dimmer your LED will be.

19. Jan 3, 2010

robert2734

20. Jan 3, 2010

RestlessMind

Okay, this is confusing.

First of all, http://www.anybodyburns.com/pathlight/flow.htm" is what the site I got the schematics from says about how the circuit works in the light and dark. From what I understand the LED does not turn on in the light.

But the other circuit only uses two batteries, with a total voltage of 2.4 volts. My modified circuit: http://i538.photobucket.com/albums/ff341/the_cake_is_a_lie/ElectronicsSchematicCOMPLETECUSTOM.gif" [Broken] Uses 3 batteries with a total voltage of 3.6 volts.

Earlier, in another thread, vk6kro seemed to say that ~2 volts "excess" voltage wouldn't hurt anything. Why do you say that it will?

R4 is meant to protect the LED.

Last edited by a moderator: May 4, 2017
21. Jan 3, 2010

vk6kro

Sorry no. Q2 and the LED are on during the day. They are powerred by the solar array during the day. Maybe you can't see the LED glowing in daylight. That doesn't stop the solar array from also charging the batteries.

During the day, Q1 is conducting which removes the forward biasing of Q2. This is the function of the circuit.

This is why the original circuit doesn't have R4 which only dims the LED.

Putting a resistor in series with a LED is not to dim the LED. It is there to protect the LED from being destroyed by excess current. A NiCd battery can deliver adequate current to destroy a LED, even if it only has a voltage slightly greater than the turn-on voltage of the LED.
A fully charged NiCd cell will have a voltage of about 1.35 volts, so this set of 3 cells could have a voltage of about 4.05 volts.

22. Jan 3, 2010

RestlessMind

In that case, do you think that a 12 ohm resistor is appropriate for R4?

Also, I think I am going to stick with my 600mAh batteries. However, what is this that robert is saying about the voltage from the solar panels being too high for the batteries? The solar panel outputs ~6 volts and the batteries are ~3.6 volts (again, the diode is relatively inconsequential due to its very low voltage drop).

23. Jan 3, 2010

vk6kro

In that case, do you think that a 12 ohm resistor is appropriate for R4?

I think I would put 47 ohms in there for a start. However you need to do some experiments to see what charging currents you can get and what voltage turns on your LEDs.

Also, I think I am going to stick with my 600mAh batteries. However, what is this that robert is saying about the voltage from the solar panels being too high for the batteries? The solar panel outputs ~6 volts and the batteries are ~3.6 volts (again, the diode is relatively inconsequential due to its very low voltage drop).

We did discuss this before. The voltage of the Solar panel doesn't matter as long as the panel can only deliver a limited charging current to the batteries.

I built up a version of that oscillator circuit we were discussing. I had no trouble getting 20 volts DC from a single NiMH cell. Might do some work optimizing it for best efficiency, though.
I used transistors and an inductor I had on hand, so it is probably not as efficient as the original.

24. Jan 3, 2010

RestlessMind

Why 47? The equation you showed me says 12 is needed.

The package of the LEDs I bought say "FW supply: 3.3V (typical), 3.6V (max)". Doesn't that already tell me how much voltage they need?

Well, subtracting the 3.6V and ~0.4V of the batteries and diode from the solar array, there will be 2.6V extra going to the battery in optimal sunlight conditions, 2V in typical conditions. I suppose you are saying that this is too much? Robert proposed using diodes to bring the voltage down. How much would it need to go down? Does the voltage need to be something like only ~0.5V higher than the battery + diode?

25. Jan 3, 2010

vk6kro

Why 47? The equation you showed me says 12 is needed.

The package of the LEDs I bought say "FW supply: 3.3V (typical), 3.6V (max)". Doesn't that already tell me how much voltage they need?

A fully charged NiCd cell will have a voltage of about 1.35 volts, so this set of 3 cells could have a voltage of about 4.05 volts.

I said I would put 47 ohms in there. Until you do some measurements, you don't know the voltage of the LEDs or the fully charged voltage of the NiCds. This 1.35 volts is only there for a short time after you remove the cell from charge, but you need to allow for it.

Well, subtracting the 3.6V and ~0.4V of the batteries and diode from the solar array, there will be 2.6V extra going to the battery in optimal sunlight conditions, 2V in typical conditions. I suppose you are saying that this is too much? Robert proposed using diodes to bring the voltage down. How much would it need to go down? Does the voltage need to be something like only ~0.5V higher than the battery + diode?

No, I would say there is no problem. You don't need more diodes.