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### Topic: How to modify the voltage of solar charger? (Read 3046 times)previous topic - next topic

#### falexandru

#15
##### May 24, 2018, 08:57 pm
Demo only. So I think time is OK. I guess the real current will be around 200 mA and my unit are 3 x 800 mAh = 2400 mAh. Then charging time (real conditions) might be 12 hrs. Assuming fully discharged of course.

If I use a Zener regulator that will take some current as well (maybe 20 mA? - there is a quite fantastic willy-billy method to label the Zeners so I am reluctant to go for a Data Sheet).

I also assume the variable current will destroy the NiMh within some 200 charging cycles (read 200 days) but this is very good for the kids demo project).

#### ted

#16
##### May 24, 2018, 09:08 pm
I also assume the variable current will destroy
What you mean.

When batteries are discharged on beginning of charging current will be max 200 mA - that how much can deliver your charger ? - then will drop to 0 mA when batteries voltage reach 5V.

#### falexandru

#17
##### May 24, 2018, 09:27 pm
I am talking about voltage depression and smaller "slices" when the panel only partially charges the NiMH battery.

However, at this cost and for demo only, that is not going to be  problem. I assume the units will last for at least for few months.

I am almost decided to go for NiMH instead of LiIon.

It may be a nightmare to find a Zener on line because of labeling, so I will go to "mortar-and-brick" store.

+++

Next step is to see whether my Savonius turbine can go the same way to charge the NiMH. The turbine's output  will be perhaps a 2V average, which means that I have to push up the voltage and then to regulate it via the same Zener regulator.

#### wvmarle

#18
##### May 25, 2018, 04:43 am
Demo only. So I think time is OK. I guess the real current will be around 200 mA and my unit are 3 x 800 mAh = 2400 mAh. Then charging time (real conditions) might be 12 hrs. Assuming fully discharged of course.
that would be if you place your batteries in parallel, but you'll have them in series so total capacity is still 800 mAh and your charging time at 200 mA about 4 hours.
Quality of answers is related to the quality of questions. Good questions will get good answers. Useless answers are a sign of a poor question.

#### falexandru

#19
##### May 25, 2018, 05:48 am
Right, my mistake. But is very counter-intuitive, I must admit. I have to find a simple way to explain this (first to myself :-)).

Question is whether units for a total of 4.5 V (or a 3 x1.2V = 3.6 V) can be charged by a panel which is rated 6 V, but can easily drop to 4V when in shadow. Maybe the prototype will show how (and whether) it works.

#### wvmarle

#20
##### May 25, 2018, 06:01 am
The current runs through all batteries. That's why.

PV cells are current sources rather than voltage sources, making that part more complex. I also still don't fully understand how that works.
Quality of answers is related to the quality of questions. Good questions will get good answers. Useless answers are a sign of a poor question.

#### falexandru

#21
##### May 25, 2018, 09:14 pm
Today I want to the field and I measured my PV panels. No load, only multi-meter. It is a cheap multi meter (some 4 USD per unit), so I assume not very accurate.

However, I got 7V in half shadow, out of panels rated 6V. So, no voltage drop as a I (wrongly) expected.

For practical purpose - still OK. But why voltage went up?! OK, maybe low internal resistance of my multi meter has something to do with this, but even so - I am taking about 1-3 V above the mark.

I am thinking to make a voltage divider and feed two LEDs out of this voltage, although it seems bizarre and I never came across such a solution for a PV panel.

#### 1steve

#22
##### May 26, 2018, 04:55 am
As someone else said solar cell are current devices. You gave the panel as rated 3.5 watt and .384 max current. (3.5w/.384a=9.1v ) That is going to give you an open voltage (no load) of  9.1 volts in full sunlight. Cheap solar controller short out the solar panel when the voltage get to high. and when voltage drops it unshorts. So there would be a diode,  the cathode connected to the battery, an the other end connect to the panel and something (circuit ) to short it out when voltage is to high. The battery gets all the current, when voltage is not to high. Now here the fun part, when the battery is pulling more than .384 amp (full sunlight) the voltage will drop to or near the battery voltage. The panel will only supply so much current, so the voltage must go down when more current is wanted than the  cell can supply.
I do not think I have done a very good job explaining this but hope it helps.

#### falexandru

#23
##### May 26, 2018, 06:16 am
Then what is the meaning of "6V" label on the PV panel?

#### Wawa

#24
##### May 26, 2018, 06:37 am
384mA is probably short circuit current, and 6volt open circuit voltage.
At MPP (maximum power point), voltage and current are both less than the above values.
Thus giving a lower power rating than 0.384*6.
Leo..

#### falexandru

#25
##### May 26, 2018, 06:50 am
It is more fun than I expected.

Attempting to get the picture:

a) in full sunlight, no load = maximum current 384mA(label) only if UxI < 3.5W (label), voltage as high as it can go (9, 10 whatever)

b) shadow, no load - the similar to above, except available power decreases as a result of current decreasing

c) sunlight, load = maximum current PV can provide to a specific load, voltage decreases to match the maximum current in these circumstance, but the available power is still 3.5W (this point I cant understand quite well)

d) shadow, load = similar as above, except maximum current decreases as a result of low light

#### Wawa

#26
##### May 26, 2018, 07:21 amLast Edit: May 26, 2018, 07:22 am by Wawa
Panel open voltage doesn't change much with light intensity (apart from maybe a low light threshold),
but panel current does (so panel is mainly a current source).

Short circuit current (panel@0volt) is not very useful (no power generated).
Open circuit voltage (no current) isn't useful either.
The trick is to find a voltage under load where "voltage times current" (power) is max.
That's about 0.5volt per cell, or about 2/3 of open circuit voltage.
Leo..

#### falexandru

#27
##### May 26, 2018, 07:37 am
Then 6V label refers to 2/3 of the open circuit voltage the cells are aligned to (open circuit voltage is assumed at 9 V then)?

If I use a Zener to pick 5V to charge my 3 x 1.2 V (NiMH, 800 mAh /unit), that is going to be low efficiency method - much lower current available than the 384mA?

This is a demo project only, thus optimisation is not an objective. But a clear picture of what happens is a must, for me in the first place :-).

#### Wawa

#28
##### May 26, 2018, 08:35 am
Counting the cells (0.5volt each) might give you a better idea about MPP voltage.

You must use a diode to stop current flowing from the batteries back into the solar panel when it's dark.
That could be a schottky (with a ~0.4volt drop).

You can't use voltage regulation (load dump) on the battery side, but you can on the solar panel side.
That could be a "super zener", made with a zener diode and a TO-220 transistor.
Leo..

#### falexandru

#29
##### May 26, 2018, 08:59 am
I found 1N581 Shotky - which looks as a general purpose one - 1A. Looks that it fits my circuit. I never ever used  a Shotky. From the net it looks like it is about mounting it as it is (no resistor, nothing but  "mount and forget").

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