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12v to 9.5 converter — Parallax Forums

12v to 9.5 converter

koentonkoenton Posts: 8
edited 2013-02-22 19:07 in General Discussion
Hi all
I have a converter that reduces a rechargable 12v lead acid battery to 9.5v dc, I need the output to drop volts as the 12v drops. Do they make anything
that will do this? What i want to do is use the 12v battery to replace the 6 c cell batteries 1.5v each that goes inside of game camera. I have
the converter that drops the right volts for camera but the output stays at 9.5v all the time. The 12v battery will drop below the rechargeable state before the camera shows any drop.

Can anyone help!

Comments

  • Mike GreenMike Green Posts: 23,101
    edited 2013-02-18 20:25
    I don't quite understand what you need. You say you already have a converter (voltage regulator) that will accept the output of a 12V lead-acid batter and produce the 9.5V that your camera needs. What's wrong with the output staying at 9.5V? Do you want to monitor the output voltage of the battery to determine when it needs recharging? Why do you need the output voltage to drop as the battery voltage drops? You can use an LM317 voltage regulator to produce a fixed voltage drop. A 12V lead-acid battery actually produces up to 13.8V when fully charged. You'd need a voltage drop of around 4.3V. Is that what you want? How much current are you talking about?
  • Duane DegnDuane Degn Posts: 10,588
    edited 2013-02-18 20:36
    @Mike, My guess is koeton wants to use the camera's low battery detector to shut off the camera before the battery is drain too low.

    @koeton, I can't think of a regulator that will do what you want. IMO, you'll need to find some other way to monitor the 12V's voltage than using your camera's low battery detector.

    There are lots of ways to drop voltages. My new favorite are these regulators on ebay. They work great IMO. I used one in my cheap bot.
  • GadgetmanGadgetman Posts: 2,436
    edited 2013-02-19 00:08
    He could just stick something like this in the box, between battery and converter?

    http://www.amazon.com/PAC-BG-12-BATTERY-GUARD-VOLT/dp/B001EAQTQO
  • LoopyBytelooseLoopyByteloose Posts: 12,537
    edited 2013-02-19 03:00
    A few facts..
    Batteries vary greatly from their labled voltage.
    You may not need to provided regulated input.
    It might be cheaper to find the right battery pack than to get involved with converters.

    To do so, get some realistic parameters of the operating voltage of the device. Never exceed the labled high side limit, but you may be surprised by how far you can go below.

    Of course a 12v lead acid battery is never really 12volts, it is nominally 12 volts but actually higher.. maybe 13.2 volts. In automotive context when charging the voltage might go as high as 14.2 volts.

    The whole project might a lot easier, less wasteful with 9.6 volt battery pack.. Lithium ion (at 3.2 volt cells) or NiMH (?) in series.

    If the 9.6 volt is too high, it can be dropped with a resistor or a variety of diodes. Rectifier diodes drop about 1.0 volt, and Schottky diodes can drop at little as 0.3 volts. I do suspect that 0.1 over-voltage is not serious.

    Completely eliminate the converter as it just wastes energy and cost more money. Select a cell that is easy to buy with a charger, such as the 18650 Lithium Ion or the NiMH.

    I think if you test a fresh set of 1.5 volt cells, you are going to see they are significantly over 1.5 volts and 6 in series is delivering more than 9.5 volts.. the camera likely has an internal regulator and accepts variable voltage within a range. Hopefully the external input goes through that protection and regulation as well.

    You can buy an 8 Volt Lead Acid Gel cell, charge it with a 500ma 9 volt wall wart and likely use that directly. The six 1.5 cells are also an optimal estimate. Do a test to see if the camera runs fine as low as 8 volts DC.
  • GadgetmanGadgetman Posts: 2,436
    edited 2013-02-19 04:05
    Adding my bits...

    If I found 14.2V in my car when charging, I'd assume that the alternator is dying.
    (14.4V is a requirement when living in a country covered in ice for so many months every winter.)

    Converters...
    If you're lucky, the converter hits 85% efficiency. In reality, a good DC-DC converter will get you around 80%.
    A bad one (containing a 780x chip or similar) will waste any voltage over the specified output voltage.
    So, at least 1/5 of the battery pack will be wasted...
  • LoopyBytelooseLoopyByteloose Posts: 12,537
    edited 2013-02-19 05:10
    Usually we think of electrical damage being limited by the high voltage, and that is often the case.

    But even a 3.2 Lithium battery may provide 3.9 volts when fully charged, and 3.7 lithium may go to 4.2 volts. So every battery chemistry requires either the insertion of a voltage regulator or fully knowing what the batteries voltage range is going to be in use. Testing brand new fully charged batteries will tell a lot about the highest voltage an appliance will accept.

    Reading consumer electronics information is likely to just leave you confused.

    For the lowest operating voltage, one needs to provide power via a good variable voltage supply and see when the device stops operating.

    That should solve the problem for a device that suffers damage only from excessive voltage, but the world has grown more complex. LED actually can handle very high voltages, if the current is limited. So for those we tend to use a combination of current and voltage limiting to avoid damage. It might help to consider the this situation an excess of watts in or watts out.

    And to make matters more awkward, newer batteries these days operate very differently. Not only excessive charging can cause damage; the newer batteries have both a high volt limit when charging and a low volt limit when discharging. Lithium batteries will short circuit if you discharge below a certain voltage. So you have to be sure the device shuts down before the cell reaches the low voltage boundary. And this gets more tricky if you are driving high power drains on a lithium cell.

    The good news is that if you understand all that, you can eliminate regulators and get a lot more power from your batteries.
  • koentonkoenton Posts: 8
    edited 2013-02-19 06:44
    Duane Degn wrote: »
    @Mike, My guess is koeton wants to use the camera's low battery detector to shut off the camera before the battery is drain too low.

    @koeton, I can't think of a regulator that will do what you want. IMO, you'll need to find some other way to monitor the 12V's voltage than using your camera's low battery detector.

    There are lots of ways to drop voltages. My new favorite are these regulators on ebay. They work great IMO. I used one in my cheap bot.


    Thanks for all the replyes
    What I am trying to do is make the camera work as the 6 c cells batteries does, the camera shows % as the c batteries drop from taking pictures. That is why I would like
    to make the output drop as the input drop. The regulators you are talking about looks like they will only take 3 amps in, the 12v battery has 7amps. What I really want to do
    is make the battery last longer then the c cells do, but need to know when the battery is getting low.

    Thanks for your help

    Koenton
  • koentonkoenton Posts: 8
    edited 2013-02-19 07:11
    Mike Green wrote: »
    I don't quite understand what you need. You say you already have a converter (voltage regulator) that will accept the output of a 12V lead-acid batter and produce the 9.5V that your camera needs. What's wrong with the output staying at 9.5V? Do you want to monitor the output voltage of the battery to determine when it needs recharging? Why do you need the output voltage to drop as the battery voltage drops? You can use an LM317 voltage regulator to produce a fixed voltage drop. A 12V lead-acid battery actually produces up to 13.8V when fully charged. You'd need a voltage drop of around 4.3V. Is that what you want? How much current are you talking about?

    Thanks for your reply
    The input 12v 7amp rechargable battery will be 13.8 v after I charge it, but the converter drops it to 9.5v. What I want to do is make the camera work the same as the c cells battery does.
    The camery shows the drop in % as the c cells battery drops. I want to use the 12v battery to make the battery last longer than c cells do, but don't have any way to monitor the 12v battery.
    I thought if the output would drop as the input drops it would make the camera show the drop so i could tell when i needed to recharge battery. Or what do you think would work?

    Thanks for your input
  • kwinnkwinn Posts: 8,697
    edited 2013-02-19 07:37
    Much simpler to monitor the 12V battery directly. A volt meter would do that.
  • Duane DegnDuane Degn Posts: 10,588
    edited 2013-02-19 11:24
    koenton wrote: »
    The regulators you are talking about looks like they will only take 3 amps in, the 12v battery has 7amps.

    Batteries have several ratings. One is the capacity often measured in milliamp hours (mAh) or amp hours (Ah). The regulator doesn't care at all about the capacity of the input voltage battery just the current (and voltage) passing through the regulator.

    If you mean your camera needs 7 amps, then there's a problem using a 3 amp regulator but I'd be very surprised to learn a camera running off of 6 C cells draws anywhere near 7 amps of current. Now it it were a radio controlled airplane, I wouldn't be as surprised.

    How long does the charge of 6 C cells last? If you don't know the current draw of the camera we might be able to make an educated guess based on how long a charge lasts.
  • CircuitsoftCircuitsoft Posts: 1,166
    edited 2013-02-19 16:56
    Is your converter switching or linear? You could do this with an LM317 and a trio of resistors, but there'll be a fair bit of math and/or experimentation to make it work.
  • LoopyBytelooseLoopyByteloose Posts: 12,537
    edited 2013-02-19 22:55
    koenton wrote: »
    Thanks for all the replyes
    What I am trying to do is make the camera work as the 6 c cells batteries does, the camera shows % as the c batteries drop from taking pictures. That is why I would like
    to make the output drop as the input drop. The regulators you are talking about looks like they will only take 3 amps in, the 12v battery has 7amps. What I really want to do
    is make the battery last longer then the c cells do, but need to know when the battery is getting low.

    Thanks for your help

    Koenton

    Ummm. There are some added problems with using the camera's battery guage. If it only measures the change in voltage it is not useful for new battery chemistries. Newer batteries have an extremely flat middle section of their discharge curve. Most of the time a 3.2 volt Lithium just hangs at 3.2 volt with a sudden fall as it nears empty.

    So newer batteries require monitoring time in use, change in voltage, change in current, status as charging or discharging. This is quite a bit of data to accumulate and interpolate. Personally, I'd use an 8 volt lead acid cell and hope that it would reflect on the existing guage. The 8 volt lead acid will provide 9.2 volts if fully charged and there is a good possibilty that the camera will operate well at 8 volts. You would just have to adapt to what the gauge tells you to know when a shut down is extremely likely.
  • koentonkoenton Posts: 8
    edited 2013-02-20 16:52
    Hello
    Where can I buy a 8v sealed lead acid battery, will try one if I can find out where to buy

    Thanks for info.
  • Mike GreenMike Green Posts: 23,101
    edited 2013-02-20 17:03
    A simple websearch comes up with this.
  • koentonkoenton Posts: 8
    edited 2013-02-20 17:23
    No the camera does not need 7amps, it is set up for 6 c cell batteries. The 12v battery that i am trying to use is 7 ah amps, but looks like I will have to use
    something else. What I wanted to do is use the 12v battery to make it last longer than the c batterys, wanted to know what volts were left in battery when I
    go to check it. I only check it about 1 time a week. Was told I could get a 8v sealed lead acid battery that might work. But only found 2 ea on ebay and they
    were 3.2 ah also were very high.

    Thanks for your input.
  • kwinnkwinn Posts: 8,697
    edited 2013-02-20 18:17
    The difficulty here is that the camera electronics is basing the remaining life % in the battery on one of two things, either the measured voltage of the battery, or the estimated capacity of the battery. By changing the chemistry, voltage discharge curve, or capacity of the battery you will be invalidating that assessment. To get a valid assessment of the remaining battery capacity you need to have a circuit that bases that capacity on the characteristics of the battery you are using. Using a zener diode to decrease the nominal 12V to 9.5V (with a 2.5V zener) might give you a better estimate if the camera uses the battery voltage to determine remaining capacity.
  • koentonkoenton Posts: 8
    edited 2013-02-21 07:20
    kwinn wrote: »
    The difficulty here is that the camera electronics is basing the remaining life % in the battery on one of two things, either the measured voltage of the battery, or the estimated capacity of the battery. By changing the chemistry, voltage discharge curve, or capacity of the battery you will be invalidating that assessment. To get a valid assessment of the remaining battery capacity you need to have a circuit that bases that capacity on the characteristics of the battery you are using. Using a zener diode to decrease the nominal 12V to 9.5V (with a 2.5V zener) might give you a better estimate if the camera uses the battery voltage to determine remaining capacity.

    Kwinn


    Where could i get the zener diode and would i put it in + or - line, between battery and converter if i try that? Also would that read the drop in the battery?
    Would the zener diode drop as battery drops?
    Thanks for info.
  • kwinnkwinn Posts: 8,697
    edited 2013-02-21 20:00
    Any of the electronics suppliers have zener diodes. Make sure you get one with a high enough wattage to handle the current your camera needs, and a voltage rating that is the difference between the nominal voltage of the 12V battery and what the camera needs.

    The diode would be used instead of the regulator. By placing it between the 12V battery and the camera it will reduce the voltage the camera sees by the voltage rating of the zener diode. For example, if the 12V battery put out 12.6V and you had a 3V zener between the battery and the camera the camera would see 9.6V. When the battery voltage drops to 12.0V the camera would see 9.0V.
  • CuriousOneCuriousOne Posts: 931
    edited 2013-02-22 12:18
    There are "battery protection pcb" -s available. When battery voltage drops below 10.5v, they disconnect the load. Or you can do same by yourself, just TL431, couple of resistors, a mosfet or relay.
  • koentonkoenton Posts: 8
    edited 2013-02-22 12:33
    Thanks for all the replies, will see if i can find some zena diodes and try that first. Will let everyone know what i find out.
    Thanks again to everyone
    Will in Al
  • LoopyBytelooseLoopyByteloose Posts: 12,537
    edited 2013-02-22 19:07
    Make sure the zener diodes can actually handle enough watts for what you want, or they will just burn up.
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