Improve this solar controller?

[Nigel Goodwin]

The current which the panel provides at 14.2V

Unless you leave the diode out!

 

[MikeMl]

I always assumed that the solar panel has an in-built reverse leakage diode such as D1 in the schematic below. Here we are talking about placing a diode a the big X, or not. I re-simulated the shunt regulator in the case shown. It is dark. The panel voltage is a few mV due to starlight. The SLA battery is fully-charged from the day before, and has sagged to 13.6V resting voltage, which is typical of a 12V SLA at room temperature with surface charge...

A .DC simulation of the circuit yields the branch currents shown in blue. Since V(plus) = 13.6V (because of the battery), the TL431 is off, and sinking the minimum amount of current of 431uA. The total drain out of the battery is 543uA.

If you added a Schottky diode at the big X, so you would reduce the nighttime battery leakage to from 543uA to 113uA. Now if the sun wasn't going to come up the next day, it might be worth adding that extra diode.

Notice that I added R4 to the circuit. I'll discuss that in my next post.

 

[Nigel Goodwin]

I always assumed that the solar panel has an in-built reverse leakage diode such as D1 in the schematic below. Here we are talking about placing a diode a the big X, or not.

Nothing to do with a diode in the panel or not - but without the missing diode the shunt regulator will drain power from the battery as well as from the panel.

With the diode the regulator will prevent overcharging, and leave the battery fully charged.

 

[MikeMl]

Nothing to do with a diode in the panel or not - but without the missing diode the shunt regulator will drain power from the battery as well as from the panel.

With the diode the regulator will prevent overcharging, and leave the battery fully charged.

You obviously never read my conclusion in post #42. Some current will drain from the battery at night with or without the added diode. Without the diode, the overnight current drain is not worth worrying about...

 

[MikeMl]

Here is another simulation of the shunt regulator. Here, I disconnect the battery, so all of the solar panel output is dissipated as heat by the regulator (nothing goes to the battery, this is a worst case for the regulator). I'm showing what happens if you add a load resistor R4 to move some of the dissipation out of Q1 into R4. You could use a power resistor for R4, or as previously suggested, R4 could be an automotive 12V taillight lamp, or headlamp.

For the simulation, I plot the regulated voltage V(plus) as green. The voltage at the collector of Q1 V(c) as violet. The power dissipation (Watts) in R4 as red, and the power (Watts) in Q1 as blue. These are plotted vs the simulated current out of the panel. I am using a log sweep to better show details at low panel currents. The simulation shows what happens as the sun comes up and the panel current goes from 100uA to 1A (this simulates a "15W" panel).

Note that the regulator begins regulating with a panel current of ~700uA, and holds V(plus) steady at 14.2V until the panel current reaches 500mA. This is because Q1 fully turns on, and the regulator cannot sink any more current, allowing V(plus) to rise above 14.2V. The actual current at which this happens is dependent on R4. Lowering R4 would let the regulator work for higher max currents.

R4 was added to dissipate some of the heat that would otherwise go to Q1. Note the blue trace. With R4 there, Q1 now has to dissipate a maximum of ~2W which occurs at a panel current of ~250mA. Note that R4 is dissipating 2.5W at a panel current of 300mA, and it is dissipating over 7W at a panel current of 500mA.

For a <= 300mA panel, a value of R4 = 47Ω rated for 5W would keep the dissipation in Q1 under 1W.

 

[EccentricDyslexic]

My solar panel has no diode, I have fitted one downstream of the solar panel and shunt regulator. I have a small clip on heat sink on the tip30 but will get a 47Ohm 5w resistor to take the pressure off it. This is only every likely to do anything unless it has been a very sunny day/period.

Using abulb is an idea, but for reliability (the bat and charge controller will be hiden in a box) I prefer a resistor.

I do have a 20 w panel with built in diode, I am not sure if I can tap in to the panel before the diode, but if I can them that is ideal. I just need a 10ohm 20w resistor.

Thanks chaps!

Steve

 

[Nigel Goodwin]

You obviously never read my conclusion in post #42. Some current will drain from the battery at night with or without the added diode. Without the diode, the overnight current drain is not worth worrying about...

Your conclusion didn't make any sense, and still doesn't - without the diode the shunt regulator is draining high current from the battery while it's in operation - with the diode it prevents overcharging while leaving the battery fully charged.

 

[MikeMl]

Your conclusion didn't make any sense, and still doesn't - without the diode the shunt regulator is draining high current from the battery while it's in operation...

No, prior to the battery reaching 14.20 V, all of the panel current (except the ~700uA that goes into the TL431 and the voltage divider) goes into the battery. The battery voltage is slowly increasing as it charges.

After the battery reaches 14.20+Δ V, there is zero current that flows backwards out of the battery towards the shunt regulator. At that moment almost all of the panel current flows into Q1 (and its optional resistor R4). There is still a small current (few tens of mA) that continues to flow into the battery. This is because at a float voltage of 14.2V, an SLA is converting some of its electrolyte to O2 and H.


Actually, 14.2V is too high for long-term floating of a sealed battery where electrolyte cannot be added. If electrolyte can be added (as in a flooded-cell battery), 14.2V is ok.

With no sun, and no panel output, the shunt regulator never turns on, so the only current that flows out of the battery is the few hundred uA I already showed you in post #42.

with the diode it prevents overcharging while leaving the battery fully charged.

You just keep showing us that you do not understand how the circuit works! Can you demonstrate the origin of this mythical current that the diode is supposed to block???

 

[Nigel Goodwin]

After the battery reaches 14.20+Δ V, there is zero current that flows backwards out of the battery towards the shunt regulator. At that moment almost all of the panel current flows into Q1 (and its optional resistor R4).

So you've got a switched ON power transistor directly across a 7AH battery, yet no current flows from the battery through this low impedance load?. Sorry, I don't see it, and don't believe it.

I'm much happier by the way with a series resistor rather than just the transistor.

 

[MikeMl]

So you've got a switched ON power transistor directly across a 7AH battery, yet no current flows from the battery through this low impedance load?...

Up
Remember, the Solar panel current gradually drops below a few mA as the sun goes down. The battery voltage only has to decrease from 14.201V to 14.198V (3mV) after the last rays of sun, and the TL431 turns off Q1.

How long does it take to turn off the TL431 and PNP transistor? Maybe 10us? Do you think that the battery loses any significant charge in 10us? And Q1 is only conducting a few mA or so as the sun sets. I have showed (several times) that after Q1 turns off from a few mA to zero, the current flow out of the battery is ~543uA, which, in the 12 hours or so before the sun shines again would deplete 0.000543A*12h = 0.0065Ah out of a 7Ah battery overnight.

 

[EccentricDyslexic]

If the whole circuit is the solar panel side of the diode, then there is no battery drain?

 

[Nigel Goodwin]

Up
Remember, the Solar panel current gradually drops below a few mA as the sun goes down. The battery voltage only has to decrease from 14.201V to 14.198V (3mV) after the last rays of sun, and the TL431 turns off Q1.

How long does it take to turn off the TL431 and PNP transistor? Maybe 10us? Do you think that the battery loses any significant charge in 10us? And Q1 is only conducting a few mA or so as the sun sets. I have showed (several times) that after Q1 turns off from a few mA to zero, the current flow out of the battery is ~543uA, which, in the 12 hours or so before the sun shines again would deplete 0.000543A*12h = 0.0065Ah out of a 7Ah battery overnight.

The hysteresis will mean it drops more than that, why not simply fit the diode and leave the battery fully charged instead of 'nearly' fully charged?.

 

[MikeMl]

If the whole circuit is the solar panel side of the diode, then there is no battery drain?

Are you absolutely sure that your panel doesn't have an intrinsic diode? To find out, cover your panel, hook your bench supply across the panel (+ to +, - to -), and crank the voltage up to ~14V. If your panel has an intrinsic diode, then no measurable current will flow from the supply into the panel. If your panel does not have an intrinsic diode, then a few mA will flow from the supply.

Only if the panel has no intrinsic diode would you have to add one (somewhere). You have a choice of four places where to put it, three of which I will discuss:
1. Between Panel and everything else. Keeps the panel from discharging the battery overnight. 560uA of TL431 current comes out of the battery overnight. Preserves excellent voltage regulation.
2. Between everything else and the battery positive pole. Keeps either the panel or the TL431 from discharging the battery. Screws up the voltage regulation as described below.
3. Place it inside the regulator at the big X. Prevents discharge by the panel, and reduces the overnight discharge by the regulator itself from 560uA to about 120uA (the voltage divider). This preserves the excellent voltage regulation.

In Case 2, the TL431 is sensing a voltage that is higher than the actual battery voltage by the forward drop of the diode. If you carefully set up the shunt regulator to switch on/off at 14.20V like I did in the simulation, at room temperature, the actual battery voltage would wind up about 0.25V lower (13.95V) if the diode is Schottky, or about 0.65V lower (13.55V) if the diode is Silicon. The forward drop of the diode and its internal resistance makes for a much more gradual turn on of Q1 as the battery voltage climbs. The forward drop of the diode is effected by temperature, so adding the diode makes for a less precise turn-on voltage where the regulator begins shunting current. If you put the diode there, you will have to use trial and error to get the correct float voltage for the battery.

If I was doing this, and started with a panel that had an intrinsic diode (every panel I have ever bought has one), then I would not use a secondary diode....

 

[EccentricDyslexic]

Yes I am sure there is no diode, it was stated on the eBay listing.

So connect straigh to the battery and the diode on the solar panel?

And with my 20w panel with internal diode do the same?

Steve

 

[EccentricDyslexic]

Hi again,

I am trying to ge this up in ltspice but have an error 'unknown sub circuit called in .... Tl431' I used the above posted files and put a copy of each in the lib folders. Is there a simple way to add extra components in to ltspice? IE an expansion .exe?

Thanks,

Steve

 

[MikeMl]

Google "adding a new library component to LTSpice"

 

[EccentricDyslexic]

I did and added the files to the apropriate places. I have asked on the ltspice yahoo group if anyone has the files, see if different one help.

Ta

Steve

 

[killivolt]

Hey, Mike. Do you have a better model or are the ones I provided correct.

kv

 

[MikeMl]

Unpack this into a clean subdirectory. It should have everything you need to run the sim.

 

[EccentricDyslexic]

Hi Mike, the TL431 gets quite hot under load, granted when dumping a fair amount. is it better to just fit a little heat sink to this to92 to keep temps as low as poss?

Ta

Steve