USB Solar Charger Current Question

Presumably you have checked the obvious, and tested how much current the unit you are trying to charge will draw if connected directly to a power supply?

 

Looking at the datasheet for the V7805-1000 it shows a specified input range of 6.5~32 Vdc and output current of 1000 mA. However, that 1000 mA is not a hard limit since the performance charts are shown for output current over 1000 mA. The more likely source of your problem is the minimum input voltage of 6.5 Vdc. What is the output voltage of your 6 volt solar array, especially under full current load?

 

Another possibility is related to the MBR150 diode. USB power voltage is specified as between 4.75 and 5.25 volts. However, the forward voltage drop of the diode would be about 0.5 volt @ 500 mA. So the solar charger may only be providing 4.5 volts while the AC adapter may be providing 5.25 volts. That difference in output voltage could account for the difference in charging current. Something to check.

 

Surely that proves the issue is on the output side. Given there's only two components after output, it's got to be something simple (assuming the DC/DC itself is ok).

Check voltages at each point. I'm betting Laplace is right - my MP3 player charges VERY slowly at <5V input, and at <4.5V input it stops and refuses to charge altogether (presumably to protect the Li-Ion cell).

 

The 7805 is a linear series regulator. Have you thought about using a shunt regulator instead? If the open circuit voltage of the solar array is 6.0 volts, try using a silicon power diode in series to drop 0.7 volt followed by a 5.1 volt zener power diode to shunt the current to ground. Then the load will be able to pull off all the current at a 5.0 volt load voltage. Sure, you will be "wasting" the current if it is not going to the load, but what were you going to do with that unused current anyway? But you will need a good I-V curve for the solar array in order to make good decisions.

 

Using Ajgiampa's situation, i have a question (directed to anyone) regarding his equipment.
He has two solar arrays that, in parallel, should output aprox. 6 volts and around 800 milliamps. That equals about 4.8 Watts on the input side. Lets figure about 50% efficiency to the output. That should be 2.4 watts to the output.
Output 5 volts, at 2.4 Watts gives 480 milliamps on the output.

Assuming efficiency is truly 50%, Is my math and expectations for his circuit correct?
I started another thread with different questions but still concern a solar power supplying power to a DC-DC converter.
FYI, i have been doing some research and, as far as switching regulators go, i saw a 7805 drop in replacement on a Y0utube video with the name "usb Solar Power". The author posted the following link
Mouser Part #:580-OKI78SR5/1.5W36C - $3.93
Datasheet for OKI-78SR
http://www.murata-ps.com/data/power/oki-78sr.pdf


Or

maybe the following as a solution.
http://www.linear.com/product/LT1302#overview
I saw a video online demonstrating the low 2volts dropout with a continous 5 volt output. The thing that concerns me is that with a Solar Power source, the voltage may be there, but the power available might be a limiting factor.
thanks

 

Your maths and expectations would be correct, but I do not understand where you get the 50% from.

In good strong sunlight, the output power of the solar panels is (pretty close to) 6V at 450mA.

The output will drop as the light level decreases.

Your 50% figure is probably accurate for a day on which there's a bit of cloud cover, or the unit isn't in direct sunlight, or it's mid-evening and the light level has dropped.

 

) mostly i made it up.....
However, to be honest, i did a quick google of "lm7805 effeciency", for a quick %.
Also, my % effeciency is focused on the regulator not the solar panel/cells.

I bought two 1.5 watt solar panels from Harbor Freight, which has a male car-power plug on each one. They each produce about 12 volts at 125 ma. So I went to Dollar General and purchased (2) Male to two Female Car-Power Y combination cables so i could daisy chain and use a standard car-charge plug. I havent had time to do my own board yet, but needless to say, in full sunlight yesterday, my Iphone 4 indicated "this is not a suitable charging device".
I realize that the Car charger is set up for much higher current output and it has a MC34063 converter in it. It also has the Data line resistors that sets the amp draw on the power supply.
I dont know all i need to know to make it work ,but I was looking for a solution as a step down converter at this point where i could use one or both to charge an iphone.
Thats why i was asking about my math.
1 panel produces 1.5 watts. 1.5 = 12 volts x 125 ma
so if i can get a 85% efficient DC-Dc convertor to work, then
1 panel + converter produce 1.275 watts = 5 volts x 255 ma

Then if I paralleled the inputs, I could use a dongle with the two inputs to charge an Iphone at 5vdc & 500 ma.

If my logic proves incorrect, dont hesitate to offer advice. I dont have an oscope, although I am going to try and find a cheap one at christmas or build a usb 'scope.
so, im operating blind other than my meter and using shunts to measure current.

I used to work in embedded hardware/software, but I do system type testing now and i have only my own tools to rely on for my projects.