whats going wrong here?

Discussion in 'The Projects Forum' started by bisctboy, Jun 26, 2009.

  1. bisctboy

    Thread Starter Member

    Nov 3, 2008
    26
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    I have 3v dc going into a dc/dc converter which I verified is outputting 5.04v. The 5.04v is being fed into a li ion battery charger IC that outputs 4.21v to charge a battery. Again, I have verified the output. My li ion battery is obviously connected to this output to charge it. This is where things go wrong... The 4.21v output is not only connected to the positive terminal on the battery, but it also connected to another dc/dc converter so I can boost the 4.21v to 5v so I can charge my ipod and Blackberry. For some reason, the output is roughly the volatage of my li ion battery (3.7v). I have switched out the dc/dc converters, inductors, resisitors, and capacitors in an attempt to find the issue. The results are the same 3.7v instead of the expected 5v? Is there something with combing the 2 dc/dc converters that puts a wrinkle into this? Help...
     
  2. R!f@@

    AAC Fanatic!

    Apr 2, 2009
    8,754
    760
    kekeke. you are one funny guy.
    The thing about li-on chargers is that they will charge at different rates, cause this is how the battery likes it.
    Please do not use the charger for any other purpose when charging cause then the smart charger will tend to operate at different charge levels.
    What I mean is the charger will charge approx 0.5C to 0.7C or may be 1C, I donno what's the type you have. After a few charge cycles, the battery voltage will rise to tha max and after that the charger will switch to float charge mode.
    So the out put voltage will be at 3.7V when started and ends up at 4.2V at full charge.
    This is for one li-on cell only. may that be any current capacity.
    And during charging it's not advisable to load the battery.

    Rifaa
     
  3. bisctboy

    Thread Starter Member

    Nov 3, 2008
    26
    0
    Thanks for the reply. However, I did not mention in my previous post that the 3.7v output I got was not under battery load. In other words, the battery was not hooked up when I got these unexpected results. Also, the people at Sparkfun claim that the battery charger IC I am using (MAX1555) can be charged even while under load. I'm glad you think I'm a funny guy though....
     
  4. RolfRomeo

    Member

    Apr 27, 2009
    17
    0
    Why don't you just use the 5 Volts from the first DC/DC-converter?
    I think alot of gadgets have internal charging circuitry and only needs a regulated supply.
    I agree with Rifaa in that stepping up the output from the charger chip seems like an odd way of doing things.
     
  5. bisctboy

    Thread Starter Member

    Nov 3, 2008
    26
    0
    I could use the 5v from the 1st converter, but that would defeat the purpose of having the battery. The original 3v input is from a solar panel that is fed into the 1st converter to boost the voltage so I can charge the battery through charger IC. The recharged battery's voltage then needs to be boosted through another converter so I can charge my ipod and BB at 5v again. I am a newbie and I am assuming that I can't take the volts from the battery and input them through the 1st converter again? To me, it would seem that I am just running the volts around-and-around through the converter, charger, and battery. Can I do that? I would think that this would put an early end to the battery's life.
     
  6. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    I'm not sure if I understood you. You mentioned that the battery was not hooked up when you got these unexpected results. Are you saying that it works with the battery, but not without the battery? Have you tried the design with the battery in the circuit?

    The second DC/DC converter will need to operate from a voltage source (such as the battery) if you drive it with the output from the charger without the battery in the circuit, it is not likely to work. Li-ion batteries are typically charged with a current source initially, and then a voltage source later. This direct output is not a good source for a DC/DC input voltage.

    It's possible that it could work with the battery in circuit. I'm not sure if the MAX chip is designed for that, but if you try it and it doesn't work, then you have your answer. You either need to modify the circuit, or go to a different charging IC.
     
  7. steveb

    Senior Member

    Jul 3, 2008
    2,433
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    You could get more productive feedback if you post the entire circuit diagram. There are a number of issues that come up with what you are trying to do. For example; what happens when there is no light hitting the solar cell? Do you disable the first DC/DC converter then? Also, the issue you are seeing might be more clear if we see the schematic.
     
  8. bisctboy

    Thread Starter Member

    Nov 3, 2008
    26
    0
    Yep, I realize that my original post in this thread was clear as mud. Sorry. I have attached my schematic. JP1 is attached to my solar panel and JP2 is attached to my li ion battery. I am assuming that when there is no sun, the charged battery will still power the 2nd converter. I guess I wouldn't be writing this if my assumption was correct. I know the schematic is showing the USB GND and VBUS are backwards, but when I make a pcb it is the correct way (that took me a couple of pcb's to figure out).

    A previous reply to this thread got me thinking about the possibility of eliminating the 2nd converter. I'm pretty sure I could take out the 2nd converter and add a switch/relay or two to get the same result. Instead of the battery output going to the 2nd converter, I could have it going back to the input of the 1st converter. I could put a switch/relay in between the battery output and converter input that would be open if there is sun and closed if there is no sun. Also, if there is no sun, I would have to put another switch/relay between the converter and charger IC that would be open so it is not trying to charge the battery with its own supply when its under load. To this newbie, its sounds like it would work?

    Which would be better to use...a switch or relay?

    Thanks,
     
  9. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    I can make a few comments:

    1. You didn't answer whether or not your circuit works with the battery in place. This is important information.

    2. The idea of eliminating the second converter using switches should only be considered when you've got all the bugs worked out with the basic operation of the circuitry.

    3. Try powering the second converter off of the battery without any other circuitry. That is, no charging chip or solar cells etc.

    4. If #3 works, then try adding the charging chip powered from a benchtop power supply to make sure the charging chip can work while you are loading the battery with the dc/dc converter.

    5. If #4 works, then add the first dc/dc converter but power it from a benchtop power supply.

    6. Well you get the idea now. Go step by step and if problems are encountered report back if you can't fix them.

    7. Once everything works with the sun shining, you need to see what happens when you have insufficient light power. At that point, do you need to disable the dc/dc converter and charging chip? This is important to know. How do those devices respond to a brown out condition (i.e. too low a voltage). This is a tricky aspect to any design and is often overlooked. It's important to consider brown-out even when power comes from the mains. Protecting against brown out is just as important as protecting against overvoltage. Speaking of which, are you sure that the solar cell won't deliver too much voltage in the brightest sunlight? Some protection (zener diode for example) might be needed.
     
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