Hello everyone,

Would this circuit charge a six volt battery. I was bored and came up with the idea of making an inverter and charge controller for a 4.5 AH 6 volt battery I have and some 5 watt solar panels. I made one circuit earlier which performed well in a simulator, but was made up of too many components and an IC, so I thought I would keep things simple. This is the new design I came up with. I just need to know if charging would stop to a trickle when the voltage across the battery and diode hit 7.7v (0.7 for the blocking diode).
Would shorting the solar panel when the battery is full be of any good; it seems to me like shorting it would lower the voltage to a point where it wouldn't charge the battery.
Appreciate your time.



Isaac

 

Who is going to open a zip file? (not me)

 

Sorry mike, the file was too big (beyond upload limits) and I therefore had to compress it.

 

Please k' just download and open it huh?

I was bored see,,,'' huh?

seriously screen shot it and upload the gif

 

There it is. Sorry about before. I need some answers ASAP, please.

 

on what action are you basing the 7 volts claim?

to my rather primitive eyes I see the unit barely maintaining a decent state of discharge on the 6 volt battery.

 

Hello imbaine13.

Well, what I see is that you have a series diode between the panel and the battery, so the panel itself is the only icurrent limiting device.

The transistor circuit is not really accomplishing anything but turning on the LED, and running the LED at a bit over 20 ma's.
(the max current for some standard LED's). It might have a shortened life span.

Why not post your other circuit.

 

Would this circuit charge a six volt battery.

Yes, and then it will continue to overcharge it.

I just need to know if charging would stop to a trickle when the voltage across the battery and diode hit 7.7v (0.7 for the blocking diode).

There's nothing in the circuit to make this happen, although your arrangement could be adapted to dump more current.

Would shorting the solar panel when the battery is full be of any good; it seems to me like shorting it would lower the voltage to a point where it wouldn't charge the battery.

Yes, that's one approach. The shunt circuit needs to be able to take the full current of the panel without overheating. Rather than just hard switching a shunt on or off, a voltage-regulated diversion such as you've drawn is probably a better approach. It only needs to shunt just enough current to hold the voltage down.

 

What you should do is to shunt the excess solar panel current around the battery after the battery terminal voltage reaches a "fully-charged" voltage.

Is this for a SLA battery?

What is the open-circuit panel voltage in bright sun?

 

on what action are you basing the 7 volts claim?

to my rather primitive eyes I see the unit barely maintaining a decent state of discharge on the 6 volt battery.

Actually it's a solar charge controller, meant to charge a 6 volt battery, not a load circuit.
I'm basing my 7 volt claim on the fact that the battery is connected in parallel with the solar array. When the voltage reaches 7.7 volts (plus diode drop), the transistor would turn on shunting the current away.

 

Hello imbaine13.

Well, what I see is that you have a series diode between the panel and the battery, so the panel itself is the only icurrent limiting device.

The transistor circuit is not really accomplishing anything but turning on the LED, and running the LED at a bit over 20 ma's.
(the max current for some standard LED's). It might have a shortened life span.

Why not post your other circuit.

Hello tubeguy,

I do appreciate your response. I have seen a lot of charging circuits, and for the love of me, can seem to figure out why a current limiting resistor is placed in the circuit to burn away precious power. I honestly don't understand why. Please help me understand. I'd want all the current from the array to be dumped into the battery, saving time, after all, the panels are only rated at about 5 watts!
I have some 0.5cm (5mm) LED's and if memory serves, can handle up to about 28-30ma, that's why that is the current here. I'll adjust that down to about 12.5ma, hoping that works (good brightness in the daylight is what I'm hoping to achieve from this). Would shorting the solar panel work better? Thanks for your input otherwise, it was very helpful.
Here is my other circuit.

Isaac

 

Yea, about the 5 volt zener, I know that doesn't actually exist, but I can make adjustments to that.

 

It only needs to shunt just enough current to hold the voltage down.

Thanks wayneh, could please give ideas on how to do this?

Isaac

 

Is this for a SLA battery?

What is the open-circuit panel voltage in bright sun?

Yes Mike, its an SLA battery. The short ciruit current of the panels in series is about 800mA.

 

Yes Mike, its an SLA battery. The short ciruit current of the panels in series is about 800mA.

I think you just answered your question of why you need to burn more current than 20mA led.

Its actually a problem with technologies such as wind and solar because when your not using the energy it is wasted.

and the second image is well, I don't exactly see a current path between the solar array and the battery....maybe im missing something.

 

Thanks wayneh, could please give ideas on how to do this?

Take a look at the right-hand side of the circuit shown here. Everything to the right of C1 are used to shunt current around the battery when the solar panel voltage rises above the value set by the zener. Note the correction I made in the post following that one.

This is what I meant by a soft switch, as opposed to an all or nothing. This only shunts enough current to pull the voltage down.

 

Take a look at the right-hand side of the circuit shown here. Everything to the right of C1 are used to shunt current around the battery when the solar panel voltage rises above the value set by the zener. Note the correction I made in the post following that one.

This is what I meant by a soft switch, as opposed to an all or nothing. This only shunts enough current to pull the voltage down.

I suppose it depends on what the batter manufacture has listed as 'float charge' current and voltage.

i have a 12V sla. that indicates a cycle use voltage of 14.4V-14.7V and a standby charge voltage of 13.5V-13.8V.

I'm pretty sure the float current is a function of the self discharge of the battery.

depending on the open circuit voltage of the solar array a simple lm317 could deliver the constant float voltage to the SLA

 

Here is a simple circuit that will work. If you have a 4.5 AH battery you should limit the current to about .45 amps.
All of your shunt current has to flow thru the led. With the resistor to limit the current it won't shunt enough to keep the voltage from going to high.
There are other circuits that will charge the battery a little faster, but alas more complex.
The 317 will need a small heat sink.

 

I'm curious how you determined 7.15v? Is that the standard float charge of a 6 volt SLA?

I always thought the rule of them for the SLA's was 2.2volts per a cell. 3*2.2=6.6volts

 

Hello tubeguy,

I do appreciate your response. I have seen a lot of charging circuits, and for the love of me, can seem to figure out why a current limiting resistor is placed in the circuit to burn away precious power. I honestly don't understand why. Please help me understand. I'd want all the current from the array to be dumped into the battery, saving time, after all, the panels are only rated at about 5 watts!
...

Isaac
View attachment 63161

Thank You.
Batteries have recommended charging currents. So, if the available current exceeds this, then current limiting is needed. But that also means you can use as much of the available current while charging as the battery allows.
Curious, what are you using the battery for?

EDIT I missed ronv's post above. Your 5 watt 12 volt panel can supply quite close to the current recommended in his circuit.