Solar capacitor charger no longer works with PNP instead of P-MOSFET - Page 1
Forum section: electronics / projects

author: Artlav
author_rank: Frequent Contributor
timestamp: July 13, 2014, 10:57:10 am
content: A solar panel have it's maximum power point at 16V and 0.58A.I wanted a circuit that would limit the current being drawn from it in such a way as to keep it's voltage above 16V, simultaneously stepping it down.I.e. output is 2A at 4V, 1A at 8V, 0.5A at 16V, etc.Think solar capacitor charger.First thing i tried was this:It worked nicely, but there were a few problems - the panel can go as high as 22V with no load, which nether FET gates, nor the driver can handle very well.And the output current wasn't limited in any way, easily overheating near 0V.So i tried to redesign it with a PNP transistor, which is what TL494 was supposed to be used with in the first place, and add an output current limit.The result is:And it does not work at all - it gives out no more than 0.6A at any voltage, while drawing no less than 0.5A at 16V.In other words, acting essentially as a linear regulator.What did i do wrong this time?Are there other (better) solutions for the problem (that would scale up to 30V or so)?

author: rs20
author_rank: Super Contributor
timestamp: July 13, 2014, 11:02:30 am
content: Just because I'm a n00b, what's R8 for?

author: Artlav
author_rank: Frequent Contributor
timestamp: July 13, 2014, 11:12:32 am
content: Quote from: rs20 on July 13, 2014, 11:02:30 amTo help the transistor close faster?

author: rs20
author_rank: Super Contributor
timestamp: July 13, 2014, 11:19:11 am
content: Oh, to prevent it from going into deep saturation? I'm naively coming from the point of view that BJTs are current-driven, so you don't need to "pull them high" to turn them off, just stop pulling them down.

author: Artlav
author_rank: Frequent Contributor
timestamp: July 13, 2014, 03:04:05 pm
content: To clarify the big idea (for the "Are there better solutions" question):It is something like s a solar phone charger with a supercapacitor buffer.The converter in question goes in between the panel and the capacitors.Without it everything works fine until a cloud covers the sun for long enough to discharge the capacitors below 10-12V.After that the panel can't recover - forced out of it's maximum power point, it can't supply enough power to charge both the capacitors and the phone, so the voltage keeps dropping until the charger shuts down.With the converter the charger remains functional, stepping down to 5V from whatever the capacitor's charge level, and the excess power is filling them back up.

author: dannyf
author_rank: Super Contributor
timestamp: July 13, 2014, 03:19:11 pm
content: TL494 is well understood so I would encourage you to take a look at the datasheet / schematic to see how a step-down converter is done using TL494.QuoteGoogle ride through.An issue universally true for wind and solar power (to a lesser extent). Both are big headaches for grid operators.

author: Artlav
author_rank: Frequent Contributor
timestamp: July 13, 2014, 03:24:09 pm
content: Quote from: dannyf on July 13, 2014, 03:19:11 pmExactly what i did.The only big difference from the datasheet's reference circuit is in the voltage feedback, and that feedback configuration worked fine in the FET version.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 13, 2014, 03:25:14 pm
content: BJTs are voltage driven, just like MOSFETs. In fact their transfer curve is even simpler: Ic = Is * exp(Vbe / Vth), where Is and Vbe are constants (which vary by semiconductor type and temperature, but not with voltage or current). It just so happens that BJTs do it with a junction, which draws current that clamps the base voltage just as collector current is going up; so it looks like hFE is a thing, but because it's not constant (it varies strongly with Ic and Vce), it's not a good design factor nor of theoretical significance. Anyway, the B-E junction stores charge, just like reverse recovery charge in a junction diode; this must be discharged, and the B-E resistor assists in this.Anyway, the operating frequency looks awfully high, 70kHz if I'm reading the graph right? That's a lot to ask of a TIP42C, especially with only 1k drive resistance (18mA?). To get up to 2A, you need more like 200mA base current. Which is obviously a pain for those times you only need 100mA output. A darlington connection would help, at the expense of even more switching speed and voltage drop.In short, your problem is exactly as stated in the subject. It works with a PMOS, because it works that way. It doesn't work with a PNP, because it doesn't work that way!Tim

author: Artlav
author_rank: Frequent Contributor
timestamp: July 13, 2014, 04:48:22 pm
content: Well, how to do it then?With a MOSFET, i can't see anything simpler than making an isolated supply for a few mA and using an optocoupled FET driver running off it (with an N-fet switch).With PNP, i can't quite figure out how to make it work, or why exactly does it not work, even at lower currents.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 13, 2014, 09:00:20 pm
content: Here are some examples. The darlington isn't very efficient, but not much worse than the TIP42 to begin with -- max Vce(sat) 1.5V at 6A I think the datasheet say, which leaves plenty of room for a darlington to do its work. The B-E resistor on the TIP42 could stand to be smaller still. Note I dropped the switching frequency to ~32kHz, and increased the inductance.Oh, and I don't think your compensation was correct -- the COMP pin can be overdriven, but I'm pretty sure it's intended as an op-amp output. I doubt it's a transconductance type output, which would be compensated as you had it. There should be an R+C to the other -in as well, I just changed the one.The other example shows a fairly simple discrete bootstrap MOSFET driver. The 33k and 2.2k may have to be smaller, simply because the TL494's output pins will take quite some time to charge, which will delay the driver turning off. Actual rise/fall time at the gate should be under 200ns, making this a reasonably efficient solution, if more complicated. The complexity can be entirely rolled up in a common driver chip.Tim

author: Artlav
author_rank: Frequent Contributor
timestamp: July 14, 2014, 12:34:19 pm
content: Quote from: T3sl4co1l on July 13, 2014, 03:25:14 pmShould be 55kHz, no?F=1.1/(R*C), according to the datasheet.Quote from: T3sl4co1l on July 13, 2014, 09:00:20 pmThe problem is that then the output won't exceed input minus 2-3V, since there is nowhere to drive the gate higher from.Neither does it solve 22V problem.Or else, i missed something obvious in your schematic.I'll try out the darlington version later today, thanks.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 14, 2014, 02:09:06 pm
content: Follow the 1N914 and 0.1 bootstrap cap

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: July 14, 2014, 02:21:37 pm
content: QuoteTypo?I assume you mean thermal voltage for Vth. In which case Vth would be the constant and Vbe would depend on how it is connected.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 14, 2014, 02:38:16 pm
content: Nice catchYes, Vth, which is k_B T / q_e, the average thermal voltage.Tim

author: Artlav
author_rank: Frequent Contributor
timestamp: July 14, 2014, 05:24:16 pm
content: Quote from: T3sl4co1l on July 14, 2014, 02:09:06 pmAh, sorry.Now i get it.Even worse, now it clicked - this is called a high side driver, and there are ICs, like IR2125, to do that job.I found the missing link.So, all the problems are to be solved in something like this, right?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 14, 2014, 11:42:39 pm
content: YupHmm, emitter output, so that needs to be a non-inverting type -- I don't recall offhand what the 2125 is.78L15 will be fairly marginal for most of the range (if the panel's much under 18V, it'll start dropping out). A 7812 would be fine as well. It doesn't really matter, as long as it's enough output for the transistor and driver, after all -- more important is limiting the maximum voltage, which it will do just fine.I would suggest a bigger CDRV and smaller CBS, and maybe a dash of resistance in series with D2, usually something like 10 ohms. Just to keep the charging spikes down.Tim

author: Artlav
author_rank: Frequent Contributor
timestamp: July 15, 2014, 06:55:52 pm
content: No, it still does not work.All this "realizing that it's a high side driver" made me forget what the problem was.Current schematic below.The problem with using a MOSFET was this:Consider what happens when the capacitor is nearly done charging.The output voltage is, say, 14V.The input is 16V.When the FET is closed, the bootstrap capacitor is between 16 and 14V, so it can only charge up to 2V, not nearly enough to open the FET.The high side driver solved the problem with driving the FET when the input is above 20V, but it does not help against the original issue of lack of headroom.I.e. now this works great with a pure load, but still does not work as a charger.So, back to square one...

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 16, 2014, 12:19:48 am
content: You need some way to limit maximum duty cycle, so it keeps switching, and a minimum load so the inductor continues to swing below 0V when the transistor switches off.Tim

author: Artlav
author_rank: Frequent Contributor
timestamp: July 16, 2014, 05:34:19 am
content: Sorry, i don't get it.How can the indductor ever swing below 0 here, and how would a load help when it's in parallel with a supercapacitor at the output?

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: July 16, 2014, 12:47:27 pm
content: Some FET drivers (high side) generate their own boost voltage. Maybe these would be more suitable.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 16, 2014, 10:00:17 pm
content: Load draws DC current through the choke, which lets it flyback and forward-bias the reaction diode: V = L * dI/dt. Can't have any dI/dt in this circuit without I > 0.Tim

author: mikerj
author_rank: Super Contributor
timestamp: July 17, 2014, 07:24:22 am
content: R8 is way too low a resistance in the OP's first schematic which is robbing the base of drive current.

author: oldway
author_rank: Super Contributor
timestamp: July 18, 2014, 08:31:37 am
content: Your first schematic seems allright and would works with the right values of components.Try with :R8 : 2k2 instead of 100RQ1: TIP135 or TIP137 instead of TIP42Oscillator frequency seems be too high for a darlington (more or less 60Khz) and would be lowered.Cosc: 2n2 instead of 1nL1: 200µ instead of 100µ

author: Wolfram
author_rank: Frequent Contributor
timestamp: July 18, 2014, 01:15:15 pm
content: I think the simplest solution will be to use the original circuit with the P-channel FET, and clamp the gate voltage to (VCC - 15V) using a zener diode.

author: Seekonk
author_rank: Super Contributor
timestamp: July 24, 2014, 01:04:45 am
content: I had a similar application to this with a solar panel operated at power point and used one of those LM2596 converter modules you can find on ebay for only about $1.50 shipped. These are good up to 35V and have a simple on/off control pin that is tied to common. I lift this pin and tie it to the collector of a PNP transistor with a 27K pullup resistor. The base is tied to the input voltage through a zener at close to the power point voltage. The small difference is adjusted through a pot. Another 27K resistor from C to E forma a voltage divider to prevent the on/off pin from seeing more than 15V given in the chips rating. This wham bam type of operation is really quite smoth and similar to the operation of many wall wart power packs. You should add about 1,000uF to the input. If anyone is interested I could dig up some more details.