2.7v Voltage Clamp - Page 1
Forum section: electronics / beginners

author: wilheldp
author_rank: Regular Contributor
timestamp: December 30, 2013, 07:47:14 pm
content: I want to use a solar cell to charge a supercapacitor. The problem is that the open circuit voltage of the solar cell is 6V, and the capacitor is rated for 2.7V. All of the voltage clamp circuits that I have found online use a zener diode with a Vz of the desired clamp voltage. Is there any other way to do it with other components? I was planning on using a voltage divider and suddenly realized the obvious today...the divider is going to lower the voltage even when the solar cell is outputting a voltage lower than 2.7V.

author: Jon86
author_rank: Frequent Contributor
timestamp: December 30, 2013, 07:48:38 pm
content: A voltage clamp is surely going to be the least efficient way to charge your capacitor in this application. I'd go for a LDO regulator instead, if possible.

author: mariush
author_rank: Super Contributor
timestamp: December 30, 2013, 08:00:02 pm
content: Either a LDO or a switching regulator in SEPIC mode ... for example LT1308 ( you have buck-boost example on page 16 of LT1307 datasheet, lt1308 is just the higher current version of lt1308 :http://cds.linear.com/docs/en/datasheet/1307fa.pdf)

author: tszaboo
author_rank: Super Contributor
timestamp: December 30, 2013, 08:11:12 pm
content: The correct way to do this would consist from two "blocks"A solar energy harvesting IC to produce a load which is optimal to get the maximum energy from the solar panel.And a supercapacitor charger, which would limit the currents to safe limits, and wouldn't allow you to overcharge.But of course that is not easy nor simple.

author: SeanB
author_rank: Super Contributor
timestamp: December 30, 2013, 10:19:51 pm
content: Simplest is a shunt regulator and a blocking diode, or use 2 supercaps in series and a LDO 5V regulator to feed them via a low forward drop diode to disconnect the LDO when not charging.

author: mrkev
author_rank: Regular Contributor
timestamp: December 30, 2013, 10:54:46 pm
content: Quote from: SeanB on December 30, 2013, 10:19:51 pmJust be carefull with the two capacitors in series as the voltage distribution may not be equal for them... Something like 1meg resistor in parallel with each cap is usually needed.

author: sleemanj
author_rank: Super Contributor
timestamp: December 30, 2013, 11:15:48 pm
content: LM2577 SEPIC module from ebay. Panel in one end, stick a multimeter on the outputs and twiddle the trimpot to get your 2.7v output. Job done.ebay search:http://www.ebay.com/sch/i.html?_odkw=LM2577&_osacat=0&_from=R40&_trksid=p2045573.m570.l1313.TR0.TRC0.XLM2577+auto&_nkw=LM2577+auto&_sacat=0If it's LM2577 and has 2 inductors, it's the SEPIC variety (can boost and buck to maintain output voltage regardless of input [within limits] -- actually, with 2.7v set as output it can't boost anything to that, but it can buck from 3v+ down to the 2.7v, anything under 3v of your panel is probably wishful thinking for useful power output anyway).

author: wilheldp
author_rank: Regular Contributor
timestamp: December 31, 2013, 02:25:49 am
content: I like the efficiency of those buck converters, but they are literally twice the size of the rest of my circuit combined. Space is limited since I'm trying to pack this system in an existing landscape light enclosure, so I don't think the converter will fit.How is the Zener inefficient? Will it allow 2.7v to pass into the cap even when it produces voltages in excess of 2.7v? I'm really confused as to how a Zener works.

author: sleemanj
author_rank: Super Contributor
timestamp: December 31, 2013, 02:58:14 am
content: Quote from: wilheldp on December 31, 2013, 02:25:49 amIf the voltage is above 2.7v, then yes, from the top of the (2.7v) zener to the bottom (call it ground), is 2.7v (give or take, tolerances etc etc).But the extra volts don't just vanish, the power has to go somewhere.... heat.Let us imagine that your load is going to be fed 100mA, and that your panel is producing that at 6v.The resistor must by definition drop 6-2.7 = 3.3v, if 100mA is passing through that resistor, then how much power is being burned off as heat: 3.3*.1 = 330mW of power, that's a lot of power loss. Gets worse, because the zener also must pass current through it, lets say 5mA, so your resistor passes 105mA (100 to the load, 5 to the zener), increasing the power dissipation further.All that power is thrown away, which for solar applications, is generally the opposite of what you want :-)You also have the problem that if your load stops being a load, that is the amount of current it is drawing drops, because you have a fixed resistor, all that current it was using must by definition now go through your zener, which in the best case get's hotter (now the zener has to dissipate .105*2.7 Watts of power), or in the worst case goes pop.Zener based regulation (and linear regulators in general for that matter) is fine if you don't care much about chucking out useful power, and if the load is fairly constant.

author: gxti
author_rank: Frequent Contributor
timestamp: December 31, 2013, 03:01:15 am
content: A simple way to think about zener regulators is that you pick a resistor R that will drop just the right number of volts at your maximum output current, and then the zener diode simply "soaks up" the current your load is not using. For example. if your input is 5V and your output is 3V and your max current is 1A then you need to drop 2V at max current which means a 2 ohm resistor. No matter how much current the load is drawing, the zener will draw the remainder so that the voltage remains at exactly 3V.This is actually not a terrible way to charge a solar cell but a LDO would be cheap and takes up very little space. You can also use a second LDO to limit the current going into the cap but I bet you won't need it since solar cells have a very high supply impedance. The big advantage over the zener is that it would not be putting out heat once the cap is fully charged.

author: wilheldp
author_rank: Regular Contributor
timestamp: December 31, 2013, 03:10:04 am
content: I'm curious about capacitor voltage ratings too. The cap isn't supposed to be charged to more than 2.7v, but it can handle more than that during charging, correct? I've I put that Zener or buck converter in there before the cap, does that mean that the cap will charge slower? I guess the real question is will the capacitor charge faster at a higher voltage?

author: mariush
author_rank: Super Contributor
timestamp: December 31, 2013, 03:33:38 am
content: No, the capacitor is rated for a maximum voltage of 2.7 volts. It won't tolerate more during charging.Well, some series of supercapacitors specify that the capacitor will tolerate spikes of up to let's say 2.9-3v or something like that, but that doesn't mean you can subject the capacitor to voltages higher than 2.69v really.

author: Jon86
author_rank: Frequent Contributor
timestamp: December 31, 2013, 11:34:30 am
content: Quote from: wilheldp on December 31, 2013, 03:10:04 amNo, definitely do not try and put anything over 2.7v into that cap.If you really want to keep the footprint of this circuit small, you could always try a DC-DC module, some of them only need a resistor to set the voltage and they can be as small as you want them to be

author: amyk
author_rank: Super Contributor
timestamp: December 31, 2013, 12:20:38 pm
content: A zener or LDO will both dissipate the extra voltage as heat, whereas a buck converter won't (neglecting switching losses etc.)

author: wilheldp
author_rank: Regular Contributor
timestamp: December 31, 2013, 10:01:45 pm
content: We'll, i will continue to look for a small buck converter, but I will use a Zener in the meantime. By my calcs, it should only dissipate 66mW (3.3v * 20mA) with the current solar cell, or 165mW with a cell rated for 50mA (thinking about ordering one). That shouldn't generate too much heat, and is well within the limits of the 400mW Zener I ordered.New question...is the resistor required for the Zener to work if I don't need to drop the power dissipated in the Zener?

author: sleemanj
author_rank: Super Contributor
timestamp: December 31, 2013, 10:56:33 pm
content: Quote from: wilheldp on December 31, 2013, 10:01:45 pmIf you start with 6v, and your zener drops 2.7v, you are left with 3.3v. That has to be dropped somewhere to get you to zero volts.

author: wilheldp
author_rank: Regular Contributor
timestamp: January 01, 2014, 03:13:51 am
content: I whacked this circuit in iCircuit on my iPhone while I was sitting around the living room with my family earlier. With just a solar cell (6v, 50mA), a 2.7v Zener and 50F 2.7v cap in parallel with the cell, and two white LEDs on the other side of an open SPST switch, the Zener successfully limited the voltage on the cap to 2.755v. The clamp seemed to work fine (other than the 55mV overshoot). Whenever I tried to put a resistor in series with the supply, the clamp either didn't work, or it significantly retarded the charging of the capacitor.

author: mrkev
author_rank: Regular Contributor
timestamp: January 01, 2014, 03:31:40 am
content: Quote from: wilheldp on January 01, 2014, 03:13:51 amWell, what value of the resistor did you use? For 3V and max. 50mA the value should be about 60ohms. But the thing is that solar cells of this kind have already quitte a big Ri...

author: wilheldp
author_rank: Regular Contributor
timestamp: January 01, 2014, 03:35:42 am
content: I varied the resistor from 1 ohm to 1k. The 47R range seemed to work the best, but still slowed the charging quite a bit. The model for a solar cell includes a series resistance, so I'm wondering if that might be enough to allow the Zener to work.

author: mrkev
author_rank: Regular Contributor
timestamp: January 01, 2014, 04:25:55 am
content: Quote from: wilheldp on January 01, 2014, 03:35:42 amOf course it will slow the charging process, but i would use at least few (f.e. 15) ohms... Solar panels are not that efficient in the current mode, but it seems to me that you wouldn't mind... If 40% is enough for you, it's easiest sollution...

author: wilheldp
author_rank: Regular Contributor
timestamp: January 01, 2014, 04:57:37 am
content: I have both Zeners and a buck converter on order right now. I will try both, but the Zener definitely wins in terms of simplicity even though it is far from the most efficient.

author: wilheldp
author_rank: Regular Contributor
timestamp: January 08, 2014, 10:11:51 pm
content: I ordered one of these...http://www.ebay.com/itm/DC-DC-Auto-Boost-Buck-Step-Down-Converter-Module-Solar-Voltage-LM2577-A858-HM-/400629032075?pt=LH_DefaultDomain_0&hash=item5d4759e48bI've tried powering it with up to 7.5V from both the solar panel and the bench power supply (DP832), and no matter what I power it with, I get no voltage on the output. I have tried adjusting the trim pot several turns in each direction, and it has no affect on the output.It is interesting to note that there are quite a few components shown in photos of the board on eBay that do not appear on the board that I received (most obviously the two caps, inductor, and diode right next to the Out+ terminal). Also, there is an unpopulated space for an SMD component next to the trim pot on my board. There is approximately 1k ohms across the unpopulated pads (also doesn't change with movement of the pot).Any thoughts?

author: sleemanj
author_rank: Super Contributor
timestamp: January 08, 2014, 11:44:26 pm
content: Quote from: wilheldp on January 08, 2014, 10:11:51 pmSounds like they sent you an unpopulated board!Take a picture and attach it.

author: wilheldp
author_rank: Regular Contributor
timestamp: January 09, 2014, 12:16:37 am
content: The first image shows the whole board and the second shows the unpopulated pads next to the trim pot. Compare that to the image from the eBay listing that shows the exact same angle as my second photo (looking at the corner containing the Out+ pad).

author: Unknown
author_rank: Unknown
timestamp: January 09, 2014, 05:18:33 am
content: Quote from: wilheldp on January 09, 2014, 12:16:37 amMight be the same angle but it is a completely different board.

author: sleemanj
author_rank: Super Contributor
timestamp: January 09, 2014, 07:05:25 am
content: Quote from: wilheldp on January 09, 2014, 12:16:37 amWhat you have there is a LM2596 based buck converter (you can read the markings on the chip to confirm), it looks complete at least (erp, maybe not, see next post), but if you bought an LM2577 auto step-up/down converter then that's not what you got.What you have is a step-down (buck) board only, input must be higher than output.You may have to give the pot a lot of turns to get it in range before it starts stepping down, these are 20 turn potentiometers, just dial it all the way in one direction until it clicks, and if still nothing, go in the other direction.These are the typical specs:Input: 4 to 35vOutput: 1.25 to 30vMaximum output power: 12w without heatsink, 20w if you slap a nice heatsink on the bottom of the board (not included)Output current: 2 Amp but remember the power limit above, if using a heatsink you can probably push this to 2.5A or 3A peakRipple: 30mV

author: sleemanj
author_rank: Super Contributor
timestamp: January 09, 2014, 07:10:47 am
content: Oh that missing component I didn't notice, looks like it may have been an SMD resistor which has been knocked off (have a look in the mailing bag it came in). If you want to try and get it working, youmayfind it's actually a 0 ohm resistor (aka, jumper), I have a bunch of buck boards which use a 0 Ohm in that position (but the board design is slightly different).

author: mariush
author_rank: Super Contributor
timestamp: January 09, 2014, 07:37:58 pm
content: See if it's in series with potentiometer. It might be a resistors chosen in such a way so that along with the trimpot the lowest you could go would be 1.25v.

author: wilheldp
author_rank: Regular Contributor
timestamp: January 11, 2014, 03:14:04 am
content: Thanks for the suggestions, guys. I finally got around to tinkering with this again tonight. It is indeed a LM2596 controller. That isn't necessarily a bad thing, because I don't really need to boost (although it would be nice in low-sunlight conditions). Regardless, I'm going to complain to the eBay seller and see if they'll send me what I actually ordered.I also tried turning the hell out of the pot in both directions. I swear, I turned it at least 100 times in both directions, and it never once clicked or produced a voltage on the output. I'll try to add a low value resistor to the unpopulated pads tomorrow and see if that changes anything. BTW, throughout testing, I had roughly 2.1v on the input to the board.EDIT: I just realized that the minimum input voltage is 3v on the board. But still, there should be some sort of indicator that the trim pot has reached it's upper or lower limit after a certain number of turns, no?

author: sleemanj
author_rank: Super Contributor
timestamp: January 11, 2014, 04:48:43 am
content: Quote from: wilheldp on January 11, 2014, 03:14:04 amThey usually sort of tick per revolution if you turn past the limit. But you got a multimeter right, measure it, turn it some, measure it again,turn it the other way, measure it againAssume you paid by paypal, if the seller doesn't come to the party, dispute the transaction.