2012-03-21 1V 3A Solar Panel to 7V Converter

I received the following comment from Mayank and this is the reply to his comment.

Hi Watson,

I have a 1V, 3A solar panel that I want to use to charge 6V SLA battery. I checked some other circuits on the web and they use big torriod cores with 2N3055 transistors. Can I use a mostet like IRF44N to charge the battery? What circuit do you suggest I use? I can hook up the multimeters and check charging current etc.

Regards
Mayank

First off, I went to International Rectifier’s website to find a datasheet for this IRF44N MOSFET, but the search came up null.  For our purposes, I’ll assume that this is a low voltage, very high current, logic level MOSFET that needs a gate voltage of 2.4 to 5V to turn fully on.

I received another comment with more info:

thanks for the info. I meant IRFZ44N which has RDS(on) of 17mohm (though will be 1-2 ohm in our operating range). Vgs (2-4V).

Can I not use your MOSFET thief (one using 0702?). If I use fatter wires for primary and 3.7V Li-Ion battery to provide startup voltage, the circuit can provide higher voltage – right?

I am not limited by 1V on solar panels, I just happen to have 2-3 spare solar cells of 3×6″ that I picked from ebay. I can cut them and resolder them in series to get higher voltage at lower current. So I can cut 2 cells to get 2V and 1.5A. Would that help?

Also, how do I decide how many turns I need? Smaller number provides higher frequency so more current but what is the downside of less turns?

Can I keep different number of turns in the secondary? What would be the impact of that?

I am sorry to bombard you with real basic stuff but I do belive your understanding of the JT circuits is rock solid.

The added information is very helpful.

But let’s look at what we have to convert.  The solar panel is 1V at 3A, which means it needs a load of 1/3 ohm to supply its rated current.  That’s 333 milliohms, which is a very low resistance.  If your converter circuit has a tenth of an ohm DC resistance, it will waste almost a third of the power.  The circuit will have to have very heavy wire, very heavy copper traces, and very low resistance for the coil.

(As suggested in the comment, it would be very helpful to change the solar cell to 2V at 1.5 amps, because the demands would be much less on the MOSFET and the rest of the circuit.)

Another question we must answer is how much power does the SLA (sealed lead acid) battery need to charge.  If we have 3 watts from the solar panel and the converter is 75% efficient, then it will deliver 2.25 watts to the SLA.  The SLA will need 7 point some fraction of a volt to charge, depending on the temperature.  2.25 watts divided by 7.2 volts is a little over 300 milliamps.  That is going to be a lot less when the sun isn’t very bright.

I have found that it is difficult to get a Joule Thief to draw a supply current of 1 amp at 1.5 volts, so it will be even more difficult to get 3 amps at 1 volt.  I’ll discuss how I would do things.  Here is one reference where such a low voltage at high current is converted to 5V to run a small computer.

To minimize wiring losses, I would locate this converter at the solar panel. That will allow the  higher voltage, lower current to go from the converter to the SLA battery over reasonably sized wire.  I won’t need an isolation diode between the solar panel and the converter, because the output of the converter will block any current from the SLA from going backwards when the sun isn’t shining.

Further design decisions.  I would start out by making a regular Joule Thief, with a NPN transistor such as the PN2222A or BC337.  Instead of the LED, I would use the JT to switch the MOSFET gate on and off.  I’ll draw a preliminary schematic of what I would start with to get the DC to DC conversion.

Update Mar 23 – Mayank had a few more questions:

Questions (again):

Transformer: Shall I just wind same number of turns on all 3 coils?

1N5240 Zeners: These are 10V zeners, I understand the logic for 10V so the Vgs doesnt exceed 10V, but why the zeners in both directions?

Output Zeners: You say 5V 1W zeners, 2 of them in series. Is there an advantage of using 2 in series instead of one 10V zener? I suppose that is only to make sure that voltage to the battery doesnt exceed 10V, but wouldnt we be safer with something like a 7or 8V zener?

Thanks again in advance.

Mayank

I would start with a high permeability toroid such as the FT87-75 (20mm outside diameter) (easy to wind) or a Fair-Rite 2673002402 (10mm O.D.), and wind 10 turns for the primary and feedback windings.  The secondary to the FET will have to have enough turns to give the +- 10V peak to peak or whatever the voltage is for that particular MOSFET.  Also, the number of turns will influence the frequency of the JT, and that may be too low or high for the 22 uH inductor of the MOSFET.  This inductor will have to handle a lot of current, and should have heavy wire and a core that will handle the power.

The reason for back-to-back zeners in series is that if you use a single zener, it will limit the peak positive voltage to 10 volts, but the peak negative voltage will be only 0.6 or so volts.  There will be a net DC current flow through the secondary winding that could cause the core to saturate.  However the primary has a DC current flow, so if the secondary’s net DC current was opposite, it would counteract the primary’s.   I think that two zeners are better than 1.  Whatever you do, don’t leave them out, the gate voltage could rise to very high levels and damage the FET.

The  solar panel will put out 3 watts, and I figure that a bit more than 2 watts will be output, so I used two 1 watt zeners, such as the 1N4733 series.   I could have used a single 2, 3, or 5 watt zener.  No zener is not an option, because the charger could be powered on without a SLA battery, and that would probably damage the FET.

One rule is you can’t put two zeners in parallel because one will be slightly lower voltage than the other and it will hog the current and overheat.  You could use two 1 watt LEDs in series, but their voltage would add up to only 6.6 volts, and the SLA is going to need 7 or more volts to fully charge.  You could put two or three 1 amp diodes (1N4002) in series with the LEDs to add a volt or so.  Just use a load that doesn’t conduct until the voltage rises well above the battery voltage – you don’t want the load to waste power when the SLA is getting nearly fully charged and the voltage rises to 8 volts.  A 9V, 1/2W zener between the base and collector of a TO-220 power transistor would make a 5 or more watt zener.  The collector voltage rises to 9.6V, and the zener conducts and turns on the transistor, where most of the power is dissipated.

Just remember that what I’m saying is from what I know and my experiences with the JT.  I haven’t put this together yet, so I can’t say how well it will work.  You will have to do some experimenting with the values to see what will give you the best results.  Also, remember that the solar cell will put out 1V when it’s in full bright sunlight, but when it’s cloudy or at the beginning and end of the day, the voltage will be much lower and so will its output.  That’s why I say it will be difficult to get a converter that can boost watts at 1 volt or less.  If you can do it, get two volts input and it will help a whole lot.

Update Mar 24 – I have looked at a number of circuits that drive a MOSFET and many if not most of them use a two transistor circuit to drive the gate (see Fig.2 in the latest attached schematic).  The problem is that the gate of high power and especially high current MOSFETs is relatively large in area, so it has a high gate to source capacitance, thousands of picofarads.  In order to switch the MOSFET on and off quickly, it is necessary to charge and discharge this gate capacitance very quickly. If the MOSFET takes too long to turn on or off, power is wasted in the MOSFET and it gets hot and the circuit is inefficient.

The gate of a IRL3302 is 1300 pF typical according to the datasheet, but it might be more than 2000, even though the datasheet doesn’t give a maximum value.  The IRL3302 needs only 5 volts to turn its gate fully on.

I added a 1N4148 diode, 10uF cap and 1N5231A 5.1V zener to the JT to rectify, filter and regulate a 5V supply to turn the gate on and off. The third winding might not be necessary if the output from the JT collector is coupled through a capacitor to the gate driver transistor.  I added a fuse to the output to protect against battery polarity reversal.  It happens!  Without the fuse, the zeners and capacitor would be damaged.

Update Mar 25 – I soldered together the schematic as seen in Fig.2.  I posted a pic of the jungle of wiring and parts.  I used PN2222A transistors, and the coil with heavy red wire is L1, and is probably more than 10 uH.  The output load is two 1 watt white LEDs (half of one can be seen in the far right of the pic). I added a 1/2 ohm resistor in series with the LEDs to measure the output current.  Right now with no adjustments, the input current at 1 volt is 1 amp, and the output is 6 volts at 100 milliamps, so it’s about 60 percent efficient.  I think that’s pretty good for the first try.  The MOSFET is mounted on the heavy heatsink, as is the 10 amp dual Schottky rectifier, and neither one gets warm enough to feel.   They just came that way, pulled out of an old PC power supply.

Update Mar 25 evening – I made a few changes to the circuit and got another 20 mA, total 120 mA at 6 volts.  Input current was 1.24 amps at 1 volt.  I noticed that the voltage across the 5.1V zener was only 4.65 volts, which meant that the JT wasn’t putting out enough current to feed the hungry gate driver transistors with a few mA to spare for the zener.  I thought that it might help if I cut back the current from the coil to the bottom gate driver by increasing the resistor from 1k to 2k.  I checked and found that the output current increased a small amount.  I changed it to 3.3k, and left it at that value.  I think that lessened the load on the secondary or driver winding of the coil.

I thought I could get some more current by reducing the number of turns on the big L1 choke seen in the foreground of the picture.  I reduced it a turn or two, but I didn’t notice any change in the output current so I put the turns back.,  Then I decided to take a turn off the JT coil, from 10 turns to 9 turns.  That added more milliamps to the output current. I might take another turn or two off later.  At the end, the frequency was 76 kHz.

The circuit looks like less than a 1 ohm load to the 1 volt power supply.  I haven’t made really strong attempts to minimize the milliohms of resistance in the circuit.  I used heavy wire between components and some high current parts, but there are still many other places where the IR losses could be minimized.  I still have to squeeze some more power out of the JT to bring the 5 volt supply up to a full 5 volts.  Another question is what is the optimum frequency to drive the MOSFET.

Update Apr 2, 2012 – I worked for awhile on trying to get the circuit to put out more current to the LEDs.  I was using a PN2222A for the JT transistor, so I changed it to a BC337-40.  Strangely, the LED current went down a few mA – it originally was a tad over 100, it went down to 95 millliamps.  The reason I changed the transistor was to get more power out of the JT so

I tried swapping the two leads of the winding that drove the gate driver transistors.  In the schematic it showed it was connected to a 1k resistor but that resistor is now 3.3k.  When I swapped the leads, the LED went dark and the power supply showed it was overcurrent at 3 amps, so I shut it off and switched the leads back to their original positions, and the LED lit back up again.  This leads me to conclude that that winding is very polarity sensitive.

I then removed a turn from the primary winding of the JT coil.  The LED current went up and the supply current went up.  So I removed another turn from the same winding, and the current increased again.  The frequency measured 72 kHz and the voltage across the zener was still only 4.7 volts.  Right now I measured 110 mA LED current, and 1.3 amps from the supply.  It looks like the small increase in LED current is accompanied by a much larger increase in supply current, which means the efficiency is decreasing – not a good sign.  One other thing I noticed was when I grabbed the base lead of the gate driver transistor with my long nose pliers, the supply and LED current increased.  I’ll have to investigate that further.

More minor tweaks on the evening of Apr 3rd.  Only thing notable was I took off another turn from the JT primary winding.  The supply current at 1V was 1.45 A, and the LED current was 112 mA.

Update Apr 5 – I checked the supply voltage across the filter capacitor and found that i was only 0.93V, nearly a tenth of a volt was being lost in the resistance of the wiring from the power supply.  After I adjusted to compensate, the supply current was 1.6 amps, the LED current was 136 mA with a voltage of 6.1V across the LEDs, and the frequency was 91 kHz.

I thought that the waveform at the MOSFET gate might be rounded a bit by the low pass filter formed by the gate capacitance and the 100 ohm resistor in series with the gate lead.  To see if it would help, I put a 47 ohm resistor in parallel with the 100 ohm resistor, and to my surprise the LEDs dimmed noticeably.  I removed the resistor, and soldered in a 200 ohm trimpot in place of the 100 ohm resistor.  I set it to halfway or about 100 ohms and powered up the circuit.  The LED current was about the same, somewhere around 130 mA.  I adjusted the trimpot and found that the LEDs were noticeably brighter when I increased the resistance a bit, then got dimmer as I continued to increase the resistance.

This peak was not broad as I had thought it would be, it was a relatively narrow peak, and I was puzzled as to why such a minor resistance change could make such a large difference in the LED current.  I powered it off and measured the trimpot at 118 ohms.  I removed the trimpot and soldered a 18 ohm resistor in series with the 100 ohm, for a total of 118 ohms.  I powered it back up and was pleasantly surprised to find that the supply current had risen to 2.25 amps, the LED current was 144 mA, and the frequency was 103 kHz.

I calculated the efficiency as being about 40 percent, which does not look like it’s all that good, but considering the challenge of having such a low supply voltage, I think it’s about par for the course, so to speak.  Think about it:  The 1 volt at 2.25 amps is a load resistance on the power supply of less than half an ohm, 0.45 ohms to be more precise.  Those small fractions of an ohm in the circuit wiring are probably causing substantial losses, which if eliminated could bring the efficiency up considerably.  The circuit should be mounted on a PC board with heavy copper traces and a ground plane to minimize the resistance in the ground.  I think that having the components closer together would help minimize those stray resistances, and minimize losses.

Conclusion (?)   The circuit design is solid and the circuit is very capable of doing the original objective of converting the 1 volt output of a solar cell to 6 to 8 volts to charge a SLA battery.  I think “the devil is in the details,” the small but important resistances in the circuit have to be minimized, and that should help increase output and efficiency.

I’m still pondering why the 100 ohm (changed to 118 ohm) gate parasitic suppression resistor’s value is so sensitive to change.  One theory might be that the high current pulses are causing voltage changes in the supply line which is fed back into the Joule Thief and interacting with its operation.  I could put a choke in series with the supply line to filter out any feedback.  Maybe I’ll try that soon.

Back to experimenting…