Trying to build a supercapacitor powered solar light - Page 1
Forum section: electronics / beginners

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 09, 2017, 09:28:56 pm
content: I would like to build a small solar light for indoor usage powered by a single supercapacitor.I stumbled across this:https://github.com/njsharkracer/infinitysolarjar, which seems to be exactly what I am looking for.Parts list:https://github.com/njsharkracer/infinitysolarjar/blob/master/parts%20list.txtSchematic:https://raw.githubusercontent.com/njsharkracer/infinitysolarjar/master/schematic.jpgMain components that I plan on using:Polycrystalline solar panel 5 V 81 mAGREEN-CAP 400F 2,7V (Samwha DB5U407M35060HA)SparkFun 3.3V Step-Up Breakout - NCP1402LED 55° 1.800 MCD 5MM Neutral-White (Nichia NSPL515DS)My unfinished test circuit looks like this (falstad.com):http://tinyurl.com/j8frugkHow can I automatically turn the LED off, when the solar panel output voltage reaches a certain value?Edit: Fixed the GitHub link

author: Audioguru
author_rank: Super Contributor
timestamp: January 10, 2017, 02:36:35 am
content: Your first link does not work because it has a comma at its ending.Instead of using the NCP1402 you should use the IC from a solar garden light that charges the battery (your capacitor) when there is light and turns off the LED. When lighting is low the output from the solar panel is also low then the IC lights the LED from the battery (your capacitor). My solar garden lights use a cheap little AAA Ni-MH battery cell, not a super-capacitor.It is wasteful to adjust the amount of light by connecting a resistor parallel to the LED. Instead control the amount of light and current from the battery (your capacitor) by changing the value of the inductor as shown in the datasheet of the IC.

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 10, 2017, 08:41:22 am
content: First of all, thanks for your reply.When you say "IC from a solar garden light" you mean something like thishttp://www.flytron.com/electronic-parts/240-ana608-solar-charge-and-led-controller.html?Q1:Why is it a bad idea to use a NCP1402 chip for this setup?QuoteI know that my solar light is not going to be very economical. I'm just building it for the fun of it.Edit 1:I've attached a schematic that shows my current setup. Am I using the zener diode correctly?Q3:Can I use something like this as a switch to turn the LED on/off based on the solar panel voltage output?MOSFET Fairchild Semiconductor J175_D26Z 1 P-Kanal 350 mW TO-92-3Edit 2:Works great.

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 12, 2017, 09:30:31 pm
content: Can someone please take a look at the circuit I posted above?Am I using the Zener diode correctly?

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 13, 2017, 05:46:48 am
content: I would have two concerns about your use of the Zener.First, the zener breakdown can be up to 3.78 volts according to the data sheet. After the drop from your Schottky diode you will be well above the rated voltage of your supercap. While many will survive this abuse, it is a poor design practice.Second, you have no current limit in the zener. I didn't look up the solar cell you are using but it seems quite possible that you will exceed the power dissipation of your zener diode in bright sunlight. The most common way to prevent this is to add resistance in series with the zener, if you take the output to your Schottky diode after the resistance you will still have good voltage regulation. Of course this has a negative effect on the efficiency of your circuit. Trades like these (and looking for other implementations that allow regulation with efficiency) are the essence of engineering.

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 13, 2017, 10:17:00 am
content: The datasheets of the components I am using right now are attached to this post.The Zener diode is rated to break between 2.2 and 2.6 Volts, which seems to be about perfect for my use.I currently have a 47 ohm resistor (R1) in series with the Zener diode, and a 150 ohm resistor (R2) in series with the LED.Edit 1:I will try to make a picture of my breadboard circuit later today.Edit 2:I think I calculated the value for the R2 resistor incorrectly. It needs to be around 5 ohm.

author: tszaboo
author_rank: Super Contributor
timestamp: January 13, 2017, 04:06:51 pm
content: Did you calculate the thing? I've held a supercapacitor in my hand, it was half a liter volume, and contained the same amount of energy as a AA battery. Sure, it can be charged and discharged fast, but the energy capacity...

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 13, 2017, 08:27:46 pm
content: @NANDBlog, The super capacitor should be able to power my LED for at least 10 hours when charged to 2.5 V.@CatalinaWOW I've attached a picture of my circuit as it is right now.I measured ~ 20 mAh going through the Zener diode with a 4 V power source and the super capacitor disconnected.And the super capacitor stops charging at ~ 2.4V.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 14, 2017, 12:08:02 am
content: My first circuits professor always said that a "Working circuit is a thing of beauty". Congratulations. A fun project and a bit of learning.One of the projects on my back shelf, waiting for time, motivation and all to come together is a "Perpetual light". Not practical, just fun. The idea would be to combine solar, wind and perhaps solar thermal to keep a small light going "forever" in something the size of a yard ornament. Since that project first got put on the shelf there have been many technologies which have come along to make it easier. Super caps are one. Low density energy storage, but also pretty low loss and few restrictions on charge and discharge. Integrated switching converters are another.Glad to see you implementing part of the vision.

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 14, 2017, 03:00:24 pm
content: One last thing I don't understand yet.The LED forward voltage is 3.2 V, the booster circuit outputs 3.3 V and I am using a 10 ohm resistor in series with the LED.I expected to see ~ 10 mA flowing through the LED, but I can only measure ~ 6 mA.Edit 1:I just realized that there is a +-3% measurement tolerance for the forward voltage.That means the actual forward voltage is 3.24 V in my case?Edit 2:Ambient temperature also influences the forward voltage. I should have read the datasheet more carefully before asking this question.

author: Audioguru
author_rank: Super Contributor
timestamp: January 14, 2017, 03:49:16 pm
content: The datasheet for the Flytron IC is written in Chinese so I do not know anything about it.1) The LED forward voltage can be anywhere from 2.65V to 3.5V so it IS NOT 3.2V. Some of the LEDs might be 3.2V.2) How can the LED light with only 2.5V from the fully charged capacitor and much less voltage as its voltage runs down?3) The P-channel Jfet conducts all the time. Its gate must be a few volts higher than its source for it to turn off which never happens in your circuit. Your schematic shows a Mosfet, not a Jfet.4) The zener diode wastes away much of the capacitor charge.

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 14, 2017, 04:51:05 pm
content: QuoteThe boost circuit turns on at 0.9 V and runs down to 0.3 V. It steps the voltage from the super capacitor up to 3.3 V.QuoteThe LED (in my breadboard circuit) goes completely off once the input voltage rises high enough. 4 volts are definitely enough, but I think even less will work.QuoteI guess there is no simple solution for this problem? What would be the alternative?I'm open for any suggestion on how to improve my breadboard circuit.

author: Audioguru
author_rank: Super Contributor
timestamp: January 14, 2017, 05:00:58 pm
content: Quote from: 0-8-15 User on January 14, 2017, 04:51:05 pm

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 14, 2017, 06:37:23 pm
content: I tried my best to draw a schematic that shows what I currently have.Edit 1:The Zener diode orientation is wrong in the schematic.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 15, 2017, 03:00:45 am
content: One change to your circuit you might consider is moving the Zener diode. In its current location the capacitor voltage is limited to approximately 2.1 volts (the Zener voltage less the Schottky diode drop). This costs about 10% of the available voltage with your selected Zener. If you parallel the Zener with the super cap it will charge to the full 2.4 volts. The current limiting resistor will still protect the Zener from overcurrent while it is shunting excess solar current.

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 15, 2017, 10:09:22 am
content: In order to measure the voltage limit I can just replace the capacitor with a multimeter that measures the voltage right? The reading I get will be the "target" voltage of the capacitor?Doing this in my current circuit gives me a reading of 2.55 V when using a 4 V battery pack as power source. I think I'm doing something wrong there.I will parallel the Zener as you said and measure again.

author: BrianHG
author_rank: Super Contributor
timestamp: January 15, 2017, 12:13:51 pm
content: What part number of zener are you using? Don't most zeners waste/conduct/consume a little current even before they reach their nominal voltage?Try using an N-Channel Jfet, as a 0v drop voltage regulator, with a 1-2 diode adjustment jumper on the gate & GND, between the solar cell and super-cap instead of a zener diode shorting out your solar cell. This will also allow support for higher input voltages, though, your max charging current will be limited by the Jfet. You can also use a P-Mosfet with NPN transistor & diodes at it's gate to create a high current 0v regurator, and you can get rid of the series protection diode. This will however not begin to conduct charge from the solar cell to the super cap until the solar-cell is something like 1.2v higher than the charge in the super-cap.

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 15, 2017, 01:31:38 pm
content: Quote from: BrianHG on January 15, 2017, 12:13:51 pmThis one:http://www.conrad.com/ce/us/product/1263089Type: BZX79C2V4

author: BrianHG
author_rank: Super Contributor
timestamp: January 15, 2017, 03:48:29 pm
content: Quote from: 0-8-15 User on January 15, 2017, 01:31:38 pmHave you measured the reverse bias current below the rated 2.4v? Say, through a 2.0v through 2.2v source with an amp meter in in series? I know according to NXP's data sheet that the current begins to sky-rocket beginning at 0.8x it's rated value, but, it you zoom into the .pdf data sheet graph, you see it always draws about 1ma though this is not well shown since the scale is 20ma per division and the line sort of curves as you get further away from the 0.7x. I know solar cells don't always give too much current and every drop counts.

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 15, 2017, 05:59:40 pm
content: Quote from: BrianHG on January 15, 2017, 03:48:29 pmI measured 0.2 microAmp flowing through the Zener diode in reverse direction, when charging the capacitor with a source voltage of 2.8 V.I don't have a bench power supply for in-depth testing. But from what I can tell I lose hardly any energy from the Zener diode in my current setup. I might be wrong of course.I also measured the amount of current that flows into the capacitor when it is being charged by the solar panel. It only changed by a few percent when I removed both the Zener diode and the R1 resistor.Quote from: 0-8-15 User on January 15, 2017, 10:09:22 amPlacing the Zener diode in parallel with the capacitor increased the voltage by 50 mV.Edit:I think the biggest problem is the P-Channel JFET (Type: J175_D26Z), which I am using right now. It limits the current through the LED too much. I can't seem to get above 6 mA. When I removing the JFET I get 22 mA through the LED.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 15, 2017, 11:52:46 pm
content: Quote from: 0-8-15 User on January 15, 2017, 05:59:40 pmI would have expected a bigger difference, but measurement trumps theory always. I suspect that the reason is that the Schottky drop is falling as current goes down and eventually lets the voltage climb to the levels you are seeing. If you carefully watched the voltage over time it would quickly rise to to the 2.1 volt I was expecting, and then climb ever more slowly to the voltage you observed.In your application this gradual rise doesn't hurt a thing. Assuming my new theory is right. It is always good to have a model for what is happening. It helps to diagnose faults and guides future designs.

author: BrianHG
author_rank: Super Contributor
timestamp: January 16, 2017, 02:00:15 am
content: Can you try a P-Channel Mosfet instead of the J-fet, something like this: SSM3J328R. You will need to move the mosfet to the other side of the LED in your circuit. Or, even better, place the mosfet inbetween the super-cap and the boost converter. You will waste 0 current charging when the sun is up. This will give you a ridiculous low Rds on of 0.09 Ohm when the solar cell is 1.25v less than the charge in your super-cap. You can shrink, or enlarge this with a series diode or 2 on the gate with a 20Meg resistor. Adding a cap on the gate and source can filter out quick changes in light, like when a person moves past the solar cell casting a shadow, delaying the turn on and turn off...

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 16, 2017, 09:11:12 am
content: Thanks for all those great suggestions. I will try all of them later today.QuoteI only have access to N-Channel MOSFETs and P-Channel JFETs right now.Yesterday, I was able to increase the current flow through the LED by placing the JFET between ground and the voltage booster.I can't seem to make the P-Channel JFET conduct (it seems to always add a big resistance to the circuit). If I read the datasheet correctly I would need a higher drain-source voltage differential to reduce the resistance?

author: BrianHG
author_rank: Super Contributor
timestamp: January 16, 2017, 02:09:49 pm
content: Ok, which N-channel Mosfets?Any with a Vgs of 1.2v?

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 16, 2017, 02:15:55 pm
content: Type: 2 N 7000 ONS

author: BrianHG
author_rank: Super Contributor
timestamp: January 16, 2017, 03:33:35 pm
content: Quote from: 0-8-15 User on January 16, 2017, 02:15:55 pmI'm sorry, but that device requires a good 4 volts to turn on completely. That's too high for your circuit. The mosfet I mentioned earlier turns on to the same degree with only 0.7v on the gate, compared to 4v on the gate of the 2N7000.Though this might be done with 2N3904/2N3906 NPN & PNP transistors, they will eat current when operating, drop 0.5v on your circuit and you would need 2.Does your Sparkfun 3.3v step-up converter have an enable/inhibit input?What's the minim operating voltage?

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 16, 2017, 07:23:22 pm
content: Quote from: BrianHG on January 16, 2017, 03:33:35 pmNo, only IN, OUT and GROUND.Quote from: BrianHG on January 16, 2017, 03:33:35 pm0.3 VQuote from: BrianHG on January 16, 2017, 03:33:35 pmI haven't been able to find a similar MOSFET in a German shop yet.I solved it using two N-Channel MOSFETs (Type: 2 N 7000 ONS). The new schematic is attached to this post.The LED is now fully on (~20 mA current), when no input source is connected. And completely off when I connect an input source with ~ 2.5 V or higher.The new circuit works extremely well. Replacing the 2.2 kiloohm resistor (R3) with a 1 megaohm one dropped the current through the boost circuit down to 70 micro-ohm when the super capacitor is being charged.QuoteWorks great! I'm using a 6.3 V 470 microfarad capacitor (Type: EEU-FC0J471), but I think I will go for an even bigger one.QuoteI think I don't understand that.

author: BrianHG
author_rank: Super Contributor
timestamp: January 16, 2017, 07:56:23 pm
content: Quote from: 0-8-15 User on January 16, 2017, 07:23:22 pmSorry, my schematic concept is different than the one you just posted. If the current circuit works to your liking, then, don't change it.The super low on resistance mosfets, many exist, means when you turn them on, they act like a shorted wire instead of a 5 ohm resistor. The very low gate turn on voltage means that when placing 1 volt at the gate, is will already be a short. For the 2N7000, placing 1v at the gate, it will act like an open circuit. Placing 3 volts at the gate, the 2N7000 will act like a 10 ohm resistor.My design place the LED and 10 ohm resistor on the other side of the mosfet, using it as a switch, not a source follower as well as some other alterations. Wrapping my head on how your current schematic actually charges the supercap has baffled me.

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 16, 2017, 08:06:57 pm
content: Quote from: BrianHG on January 16, 2017, 07:56:23 pmI just spotted a mistake in my schematic. The two wires next to the R2 resistor should haven been connected.

author: BrianHG
author_rank: Super Contributor
timestamp: January 16, 2017, 09:57:38 pm
content: Ok, now that makes sense...Next, why is the LED and R2 connected to the Source of the mosfet instead of being on the Drain side, with the Source connected to direct to GND?

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 17, 2017, 11:56:57 am
content: I have updated the schematic once again to reflect the circuit I currently use (this time hopefully without any major mistakes).The LED on/off circuit works flawlessly. I'm very happy with it.But I still have some questions (and things I don't understand):* Why does the C1 capacitor voltage increase when I lower the R1 resistance?* Do I need the R1 resistor at all? Are there other ways to reduce the stress on the Zener diode?* What options do I have to improve the charging efficiency? MPPT?

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 17, 2017, 07:21:58 pm
content: C1 voltage increase as R1 decreases because you are increasing current through the zener diode. zener diodes do not have a perfectly flat voltage vs current response beyond the breakdown. Some are closer than others.You may or may not need R1. Find the current in your zener when the solar cell is in full sun. Measure the voltage there too. The product of voltage and current is power, and if that power is below the rated dissipation of your diode you don't need any protection. Since it is a good chance that this protection is needed, a better approach is to measure current and voltage as R1 is decreased. This way you can know when you are approaching the limit, instead of finding out by blowing your zener diode up.I have taken the liberty of redrawing your circuit in a more conventional form (signal or power flow from left to right/top to bottom with negative at the bottom and highest supply along the top.) It makes it easier for me to see what is going on, it may also help you. Also added another protection method for the zener which will buy you a little voltage on C1.You are on track for another method to protect the zener with your turn on/turn off function for the LED. Think about what you have done and see if it doesn't give you some ideas.

author: BrianHG
author_rank: Super Contributor
timestamp: January 17, 2017, 07:40:25 pm
content: CatalinaWOW, there is a mistake in your schematic. Q2's output should be tied to the gate of Q1. The way you have it, it is shorting out the 3.3v source.+ Your N-Mosfets are drawn as P-Channel and Q2 is upside-down. The Source should be connected to the GND, the Drain should be connected to the Gate of Q1, switching Q1 off when the solar cell pulls up the Gate of Q2.0-8-15 User, though you write 'N-Channel mosfet' text in your schematic, you have also drawn it as a P-Channel, and your Q1 is also wired backwards.If you are using a 2N7000, it is an N-Channel Mosfet.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 18, 2017, 01:21:43 am
content: Quote from: BrianHG on January 17, 2017, 07:40:25 pmProof that haste makes error.Here are corrections.

author: BrianHG
author_rank: Super Contributor
timestamp: January 18, 2017, 02:17:28 am
content: Error again,The output of Q2 goes straight to the Gate of Q1. R3 stays tied to the +3.3v pullup.The second solar schematic is still wrong.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 18, 2017, 03:30:10 am
content: You found only one error. Here are both corrected. Maybe all, but I am still not putting much brainpower into this.

author: BrianHG
author_rank: Super Contributor
timestamp: January 18, 2017, 05:21:52 pm
content: Quote from: CatalinaWOW on January 18, 2017, 03:30:10 amTake a look at your schematic I modified below...I'm still not sure why 0-8-15 User has configured Q1 as a 'Source Follower' instead as a switch. When producing these, each 2N7000 will generate slightly different brightness on the LED. Placing the mosfet beneath the LED and resistor will generate a brighter LED light which will be more consistent over many different Mosfets, relying on R4 to control the LED's brightness.The second schematic is how I would wire it...

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 18, 2017, 05:39:48 pm
content: Thanks a lot to both of you for being so patient with me.Quote@CatalinaWOW, that helped me a lot. Your schematic is much easier to read than mine.Today I tested a few circuit variations in full sunlight (behind a window).R1 (47 ohm) - Shottky behind Zener -> Vout: 2.80 VR1 (47 ohm) - Zener behind Shottky -> Vout: 3.05 V~ 70 mA charging speedR1 (100 ohm) - Shottky behind Zener -> Vout: 2.55 VR1 (100 ohm) - Zener behind Shottky -> Vout: 2.65 V~ 50 mA charging speedIt looks like I can't take full advantage of the solar panel with my current setup.What do you think about using something like a LM317 voltage regulator instead of the Zener diode?QuoteThere is no particular reason for this. It was simply the first solution to the problem, which came to my mind. Remember, this is my first electronics project.QuoteI will try to understand that.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 18, 2017, 05:45:54 pm
content: Quote from: BrianHG on January 18, 2017, 05:21:52 pmI agree with both your changes. The first actually explains why I redrew the circuit. I wasn't thinking about how the circuit worked, but was really have trouble seeing what the OP was doing laid out as it was. I botched the transposition because I was not mentally latching onto what was there, and didn't think about what it should be.

author: BrianHG
author_rank: Super Contributor
timestamp: January 18, 2017, 05:55:17 pm
content: Quote from: 0-8-15 User on January 18, 2017, 05:39:48 pmI noticed you had a N-Channel Jfet earlier, have you tried using it as a regulator. It may already be 2.5-3v out.It wont consume any power as a regulator and it has a 0v drop.Combined with a 2N3904 in an emitter-follower configuration in place of your current diode, you would get an up to 350ma charge current right up until the regulation voltage.To test your J-Fet's voltage, just connect the Gate to the GND, The drain to your power source & the Source should have a fixed voltage output. When testing, give it a little load to make sure.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 18, 2017, 10:53:03 pm
content: Quote from: 0-8-15 User on January 18, 2017, 05:39:48 pmI am a little unclear what you are measuring. If the measurement is the voltage across C1, you have exceeded its maximum voltage rating with several of these values. By more than 10% in one configuration. I don't know about this particular product, but early versions of supercaps were notorious for having little margin in the specs. You really should not be trying for more voltage here, instead you need to scale back a bit. As I read the data sheet 2.7V is OK if you keep the temperature below 60C. That shouldn't be a problem if you don't place your capacitor in direct sun or in a box which sits in the direct sun.If the measurement is at the output of the Sparkfun device it implies a defective part, since the data sheets say that regulation should be better than that.

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 19, 2017, 09:47:11 am
content: Quote from: BrianHG on January 18, 2017, 05:55:17 pmI only have P-Channel JFETs lying around.I like your suggested voltage regulator concept, but does that also mean that a LM317 would be inappropriate for my usage?Quote from: CatalinaWOW on January 18, 2017, 10:53:03 pmI replaced the super capacitor with a multimeter that reads the voltage.Quote from: CatalinaWOW on January 18, 2017, 10:53:03 pmI don't intend to achieve higher voltages (2.5 V would be just fine), I only want to increase the charging speed.It's probably a bad idea, but I placed two 2.4 V Zener diodes in parallel to increase the charging current. It works great so far.Vout is below 2.7 V in full sunlight.And the charging current is around 90 mA.I measured a current difference of 0.3 mA in full sun light with no load connected.Current through ZD1 ~ 27.7 mACurrent through ZD2 ~ 27.4 mA

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 19, 2017, 04:17:43 pm
content: Adding another zener is a creative solution. It helps two ways. It reduces the voltage rise for a given total current, allowing greater charging current without exceeding the capacitors voltage rating. And it splits the power distribution between the two diodes. Based on your current measurements they are dissipating 0.027 A times 2.7 V, or a little less than 90 mW. This has a very good chance of being within the dissipation limits of your zener diodes.There are often many ways of skinning a cat, and old hands often get trapped in their tried and true solutions. Don't be ashamed of your explorations.

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 20, 2017, 09:34:50 am
content: The charging speed is much better now, but still too slow for my taste. It took 3 hours of direct sun light (through a window) to charge the capacitor from 0.42 V to 1.45 V.I would like to make it three times as fast. That means I need to replace the solar panels with a bigger one and rethink the voltage regulation part.A 5 V 3.0 W solar panel (16 cm x 14 cm) should do the trick. It would replace the two 5 V 0.4 W panels (6 cm x 6 cm) I currently use in parallel.I also thought about buying a small 12 V 5.0 W solar panel (same price as the 5 V one), but that would probably be an overkill and much harder to work with.What do you think about using a step-down voltage regulator like this:https://www.pololu.com/product/2841

author: BrianHG
author_rank: Super Contributor
timestamp: January 20, 2017, 01:30:01 pm
content: try replacing the diodes with a good npn transistor, the emitter should go to the cap, the collector to the solar cell +, place a 2.7v zenner on the base with a 100 ohm pullup and see what happens.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 20, 2017, 04:29:14 pm
content: You are attacking the problem the right way, but it is worth stopping to do some math to understand some details. It can save component purchases and test time.Your current .4W cells should provide 80 milliamps in full sun, or 160 milliamps for the two in parallel.Voltage is Q/C and Q is current times time. Your 3 hour charge should then provide a charge of 3 x 60 x 60 x 0.16 or 1728 Coulombs. Dividing that by 400 says that starting from zero volts you get 4.3 volts.You are not getting nearly this performance. One reason is that the cells are rated for "full sun". That means straight overhead with clear air with the cell oriented directly toward the sun. You are probably not steering your cell, so over three hours will likely see a few percent loss in output. If you are not pointing the cell directly at the sun you can easily lose another 20-30 percent from the ideal figure. Finally, dirt, absorbtion and reflections in and on your window can cause loses ranging from 5 to 50 percent.It would take losses on the high end of all of these factors to get the performance you are reporting. It is worth checking to see if you have other loss factors, and to see if you can reduce the ones which are actually there.

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 20, 2017, 06:31:59 pm
content: Quote from: CatalinaWOW on January 20, 2017, 04:29:14 pmYes, but they are not supposed to meet those ratings under load, as far as I understand. I measured their open circuit voltage at around 6 V and their short circuit current was around 82 mA.The highest charging speed I measured so far was 80 mA, but that was with both cells connected and no window in between the cells and the sun.The charging speed in the location I want to use the solar light in is much slower than that (~40 mA peak charging speed).Quote from: CatalinaWOW on January 20, 2017, 04:29:14 pmCan I really calculate it like that even though I have a series resistance of ~ 50 Ohm in between the solar cell and the capacitor?Quote from: CatalinaWOW on January 20, 2017, 04:29:14 pmI will try to optimize the position of the solar cells some more, but I think there is no way around a bigger cell. I think I will try some amorphous thin film cells instead of the polycrystalline ones I am using right now.The questions I ask myself right now are:* Should I go for low voltage solar cells to keep the voltage difference between the solar cell output and the target voltage as small as possible?* Or should I better shop for solar cells with the lowest price per watt and use a buck converter (like the Pololu one) to make use of the extra voltage I get from the solar cells?I will have to shop around for some more parts to play with anyways. Both you and BrianHG have suggested interesting things I would love to try out.Edit:I would have calculated the charging speed like that (assuming that we are not current limited by the solar cell).RC (time constant) = 50 ohm * 400 farad = 20000Vs (supply voltage) = 2.7 voltI takes t = 3382 s (~1 hour) to charge the capacitor from 0 V to 0.42 V:Vs * (1 - e^(-t / RC)) = 0.42 VIt takes t = 15402 s (~4.3 hours) to charge the capacitor from 0 V to 1.45 V:Vs * (1 - e^(-t / RC)) = 1.45 VThat means it takes ~3.3 hours to charge the capacitor from 0.42 V to 1.45 V.Quote from: BrianHG on January 20, 2017, 01:30:01 pmI don't have any of those. I will try to get my hands on some more parts next week.

author: BrianHG
author_rank: Super Contributor
timestamp: January 20, 2017, 06:51:42 pm
content: My suggestion is an emitter-follower regulator. Even a 2N3904 will deliver more than 300ma charge current if your solarcell is generating that current if it were shorted to the cap, minus the 0.7v drop between the base and emitter. If you can find a germanium transistor, that would be around 0.3v. Another suggestion would be a very low RDS on mosfet switch who's gate would be disabled once the charge reaches your Zener Voltage.Note that I once found a super-low dropout voltage linear voltage regulator KF25B from ST. 0.4v dropout with 500ua quiescent current, 500ma drive, and maybe you wont need the reverse diode protection with it simplifying your design and only getting a 0.4v drop from solar cell to s-cap. It only consumes 50ua when disabled.http://www.digikey.com/product-detail/en/stmicroelectronics/KF25BDT-TR/497-4202-1-ND/725503

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 20, 2017, 08:10:34 pm
content: QuoteI will definitely try it out.QuoteThat sounds like it could be a very efficient solution.QuoteI could get my hands on a LF25CDT-TR, even though I would prefer stuff I can through-hole mount.But won't all linear voltage regulators burn at least half of the energy in my case?I thought something like the Pololu 2.5V, 500mA Step-Down Voltage Regulator D24V5F2 would increase the efficiency in my current setup a lot.Example when stepping down from 5 V to 2.5V with an efficiency of 80%:(5 V / 2.5 V) * 0.8 = 1.6That would mean 60% more current and I could remove the Zener diodes and the resistors in between the solar cell and the capacitor.Or am I making a mistake here?

author: BrianHG
author_rank: Super Contributor
timestamp: January 20, 2017, 08:37:31 pm
content: Yes, a switcher is more efficient, but, remember, you are sending 100% of the solar cell's current into the cap until it reaches the desired 2.5v, then, you are finished charging. The real question is whether having the same 5v solar cell at full sunlight switched down to 2.5v, will the solar cell drop in voltage you get anyways when you want to consume as much current as possible VS a similar drop connecting the solar cell directly to the cap over optimal lighting conditions. I have no doubt that if you has a 10-12v solar cell at full light, the switcher would be the best choice, at 5v, we are sitting at the edge of which solution is best with the current you are getting.In the case of my 2N3904 regulator, it would make your current 100 ohm series resistor, if you use the same 100 ohm resistor pulling up the base, with a gain of over 100, the s-cap will now see a 1 ohm connection to the solar cell, with a silicon diode drip in voltage, while the new 2.7, or 3v zener (3v approx -0.6v drop in transistor between base & emitter) still only sees a 100 ohm pullup so it wont blow.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 20, 2017, 08:47:47 pm
content: The 80 mA number is based on the assumption that you are operating the cell somewhat near the operating point that the vendor used in rating the cell for power. Obviously there is no power at either the short circuit condition, or at the open circuit condition. Since your cells are rated 5 V the assumption is that power is rated at that voltage. The resistor affects whether you are operating the cell at that point. The fact that you are getting only 80 mA short circuit current it implies one of two things. Either your location provides less than one standard sun of illumination (which is common this time of year) or the vendor was very optimistic in his ratings (another fairly common event) or some combination of the two. It isn't a bad first order approximation, but needs further work if you want answers with more than one significant digit.I believe it is a coincidence that your time constant calculation gives an answer so close to your observation. Until your capacitor charges to 2.7 volts the supply voltage is effectively your solar cell output, not the 2.7 volt zener voltage. (Assumes that the reverse current through the zener is negligible). There are a number of analysis scenarios which depend on how loaded your cell is operated as you are using it. Perhaps it is near 2.7 volts and that explains why your calculation is working so well.

author: edavid
author_rank: Super Contributor
timestamp: January 20, 2017, 09:19:19 pm
content: Low voltage zeners are a poor choice for protecting a supercap. They have wide tolerance and poor regulation, which is a recipe for cooking your supercapSeries regulators such as the LM317 or an LDO are also a poor choice, since they waste too much energy.You should use a better shunt regulator. If your solar cell max current is under 100mA, you can use a TL431. Otherwise you need a TL431 + transistor.Also, with a 5V solar cell, you would be better off with 2 supercaps in series.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 21, 2017, 03:45:22 am
content: Quote from: edavid on January 20, 2017, 09:19:19 pmSupercaps (or any capacitor) in series can be a problem also. If they are not well matched they won't share the voltage equally, leading to failure of one, followed by failure of the other. Also, as you have measured, the open circuit voltage of your 5 V nominal cell is 6 V which could fry your supercaps even if they are well matched. There are a number of ways to address these problems, all with their own drawbacks.The best thing about this is exposing the OP to a variety of ideas, along with the trades in selecting between them.

author: edavid
author_rank: Super Contributor
timestamp: January 21, 2017, 05:07:56 am
content: Quote from: CatalinaWOW on January 21, 2017, 03:45:22 amWell, if he insists on using that panel, he has to do something, right?3 options I can think of:1. panel -> buck converter -> supercap2. panel -> 2 supercaps in series, with individual TL431 clamps3. panel -> 3 reasonably matched supercaps in series

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 21, 2017, 04:35:36 pm
content: Quote from: edavid on January 21, 2017, 05:07:56 amThere are no perfect solutions. I personally like your solution 2, but there are other options with other trades. He could use a resistive divider to balance the voltage, at a cost of continual power lost.The trades have different outcomes in different situations. As you have seen in this thread, the OP rightly rejects some technically sound solutions because he doesn't have the parts available. I have seen some very strange high production designs which apparently occurred because someone discovered a nearly free warehouse full of some component. Others have resulted from the in-laws ownership of a company. As an engineer your job is to find a creative solution that meets all of the constraints. Some of which may be very strange.By the way, another trade which may be appealing to the OP is to add a concentrator to increase the output of his cell. This can be something as simple as a few sheets of white paper, though a concentrator that simple is unlikely to provide the performance boost he wants.

author: BrianHG
author_rank: Super Contributor
timestamp: January 21, 2017, 04:42:11 pm
content: Quote from: CatalinaWOW on January 21, 2017, 03:45:22 amMAGIC-TRICK, place 2 normal green LEDs, 1 on each super cap. At the charge currents and voltages, they will prevent one cap from overcharging, basically balancing. LEDs draw in the low pico-amp range until they get close to their operating voltage. I use them as zener diodes for my 4v to 250v down to 3v regulator in my telephone line powered apps where when you are on the hook, you cannot draw more the 5 ua since a zener diode draws over 10x that just at a volt.

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 21, 2017, 04:58:26 pm
content: Quote from: edavid on January 20, 2017, 09:19:19 pmI like that idea a lot, but then I need a more complex circuit to drive the LED, unless I miss something obvious here.Quote from: edavid on January 20, 2017, 09:19:19 pmLike those?TL431ACZT (http://www.conrad.com/ce/en/product/155636)LM431AIZ (http://www.conrad.com/ce/de/product/1014147)Quote from: CatalinaWOW on January 21, 2017, 04:35:36 pmI will stock up next week.I'm currently reading through this:http://www.linear.com/solutions/4545Just found this:https://www.physicsforums.com/threads/charging-a-capacitor-with-solar-panel.713123/#post-4519778Quote from: CatalinaWOW on January 20, 2017, 04:29:14 pmI think one of the main reasons why I get such a slow charging speed is that I operate the solar cell way below the optimal voltage.

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 27, 2017, 10:45:47 am
content: New parts have just arrived.The 2.5 V buck converter outright doubled the charging speed.Quote from: edavid on January 20, 2017, 09:19:19 pmThe TL431 works great! The charging speed is almost constant until the super capacitor gets close to 2.5 V.Maybe I can combine both charging methods?1. Initial phase: Buck converter2. Final phase: TL431

author: BrianHG
author_rank: Super Contributor
timestamp: January 27, 2017, 07:38:35 pm
content: Can you provide us with the 2 charging setup schematics. I'm trying to come up with a method of switching from switcher to linear regulator automatically.Do you have a preferred voltage of when you should be doing the switch?

author: edavid
author_rank: Super Contributor
timestamp: January 27, 2017, 10:52:11 pm
content: Quote from: 0-8-15 User on January 27, 2017, 10:45:47 amDoes your buck converter have an output voltage control? If so, that should be effective at limiting the output voltage, and you don't need the TL431, except for extra protection.

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 28, 2017, 09:10:14 am
content: Quote from: edavid on January 27, 2017, 10:52:11 pmThe buck converter (D24V5F2) has a fixed output voltage of 2.5 V. Input voltages have to be between 3 V and 36 V.When using the buck converter the charging speed is very fast at the beginning but then declines almost linearly towards the end.When using the TL431 the charging speed is noticeably lower at the beginning (compared to the buck converter), but it stays almost constant until the end of the charge.My idea was to charge the capacitor up to ~ 2 V through the buck converter and then switch to the TL431.But I do not really understand what is causing the different charging curves.Quote from: BrianHG on January 27, 2017, 07:38:35 pmI have attached schematics for both charging setups. And a picture of the buck converter breadboard circuit.I have measured the time it takes to charge a 3 F super capacitor from 0 V to 2 V on all three circuits:1. Zener diodes:160seconds2. Pololu 2.5 V Step-Down:80seconds3. Shunt regulator:60secondsThe power source was a 6.2 V battery pack through a 50 Ohm resistor.Amorphous thin film panel (OCV 9 V, s/c 130 mA):1.0 V2.0 V2.4 V2. Pololu 2.5 V Step-Down:30 s75 s100 s3. Shunt regulator:55 s145 s190 sPolycrystalline panel (OCV 6 V, s/c 65 mA):1.0 V2.0 V2.4 V2. Pololu 2.5 V Step-Down:30 s150 s240 s3. Shunt regulator:65 s140 s190 sThe only light source was a 2000 Lumen flashlight.The buck converter circuit always appears to be faster than the shunt regulator circuit when the solar panels receive direct sunlight.

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: January 31, 2017, 10:57:39 am
content: Quote from: BrianHG on January 27, 2017, 07:38:35 pmI have attached the complete schematics of both circuits.Any suggestions for improvement are welcome.

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: February 08, 2017, 04:27:07 pm
content: I have attached a schematic of the new circuit.Features:The super capacitor is charged up to 2.5 VThe 3.3 V rail is established once the super capacitor is charged up to 0.9 VThe 3.3 V rail collapses when the super capacitor is discharged down to 0.2 V (the LED starts losing brightness at 0.6 V)The LED turns on when the microphone detects a sound impulse (me opening the door to the room)The LED turns off when it is silent in the room for around 10 secondsThe LED only turns on when it is darkThings I will try to do next:Reduce the idle energy consumptionReduce the number of different resistance valuesCompactify the circuitAdd a switch that overwrites the sensor inputAdd a potentiometer that controls the follow-up timeAdd a potentiometer that controls the brightness thresholdMove the circuit onto a stripboardAnd maybe replace the microphone with a PIR sensor module.Edit 1:Is it possible to get the same functionality with a less complex circuit?Edit 2:What is the recommended way of transforming my breadboard circuit onto a stripboard? Pen and paper and some trial and error?Edit 3:I think about skipping the stripboard and going directly to a PCB.

author: 0-8-15 User
author_rank: Regular Contributor
timestamp: February 14, 2017, 06:50:06 pm
content: A new iteration of the solar light circuit (schematic down below).I found a way to "disable" the 3.3 V rail while the solar cell receives at least some light.Which allowed me to get rid of the comparator shown in the schematic above.And reduce the idle current (3.3 V rail enabled / LED off) from 3.1 mA to 2.0 mA.It would help a lot if someone with more experience than me could take a look at my schematic.And please tell me if I do something completely wrong or in a way that is not recommended.