12V to 6V DC converter help needed

Discussion in 'The Projects Forum' started by Shapath, Jul 24, 2010.

  1. Shapath

    Thread Starter New Member

    Jul 10, 2010
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    Recently to power my Panasonic cordless phone via a 12V 55Ah lead acid battery.

    Using multimeter i measure the supply voltage = 12.69V;
    output Voltage = 4.70 V.

    Now my adapter ratings are: input: 220-240V~; 50/60Hz; 30mA
    output: 6V DC; 500mA.

    & at the base of the phone its written: Power Source: 9V DC; 500mA

    The battery is a rechargeable Ni-MH 3.6V 650mAh battery.

    Initially the battery didnot charged when connected with my converter but now it works fine. But my question is "one of my co-worker is saying to me that eventually i will destroy the device" but he cant explain why...!!...so I basically wanna know if he is correct or not and what might be the reason behind that.
    I have attached the schematic of my circuit and please suggest me for any improvements in the circuit.

    Thank you:)ψ
     
  2. tom66

    Senior Member

    May 9, 2009
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    The picture is far too blurry.

    Your coworker is talking crap.

    No, it will not damage your device. You should preferably have it outputting the nominal 6 volts (the 9 volts printed on the phone is probably a mistake), as it will not charge as efficiently. A phone (and many modern battery powered devices) is using a switch mode converter to charge the battery, so has a fairly wide operating range.

    For extra protection, put a few 6.2V zeners back-to-back on the output. When the voltage goes over 6.9 volts (6.2 volts + 0.7 volts dropped across a diode), the zeners will short out, shutting down the power supply.
     
  3. Shapath

    Thread Starter New Member

    Jul 10, 2010
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    hmm...i have used two diodes...one in the input and one on the output line. N i will put a better picture soon.
    And also thanks.:D
     
  4. tom66

    Senior Member

    May 9, 2009
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    Since you appear (the picture is far too low resolution to be sure) to be using a 7805, why not substitute a 7806, it has a +6V output. If you drop half a volt across the output diode, you've still got +5.5V to play with, much better to charge off.
     
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  5. Shapath

    Thread Starter New Member

    Jul 10, 2010
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    hey,
    I have tried to edit it but then the time limit allowed for editing exceeded so cant put a new schematic.
    Anyways , the IC i have used is yeah 7805 with 2 diodes 5404.

    Hey please tell me if having voltage less than rating is bad for the device?
    n thanks for your help.
     
  6. tom66

    Senior Member

    May 9, 2009
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    Generally, no, undervoltage is not harmful to a device. However, it may not properly charge the device, or the device may not run properly.

    When you reply, use the full reply editor (not just quick reply), and post a schematic there.

    1N5404 = schottky diode?
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    You will not get good performance using a 7805 or 7806 regulator, as the power dissipation in the regulator will be more than in the load. At the least, you will need a fairly large heat sink. You could also use a 9.1 Ohm 10W power resistor between the battery + terminal and the IN terminal of the regulator to take most of the heat off the regulator itself. It won't be any more efficient, but at least the regulator won't get so hot.

    Stepping 12.7v down to 6v at 500mA means (12.7-6v)*0.5A = 6.7*.5 = 3.35 Watts power dissipation in the regulator, and 6v x 0.5A = 3 Watts power dissipation in the load. If you are depending on battery power, you need to use it efficiently. If your lead-acid battery goes below around 85% fully charged, it will age much more rapidly than if it is kept above 85% charged. Take a look at this post: http://forum.allaboutcircuits.com/showpost.php?p=262143&postcount=38

    I have no idea as to what kind of parts you have access to in Bangladesh. If a 7805 regulator is all you have at the moment, then you can increase it's output voltage by adding resistance between the GND terminal and ground.

    If you can order parts from a website, then let us know where the website is. Hopefully it is available in English, as I can only read other languages with great difficulty.

    Every 100 Ohms of resistance you add between the GND terminal and GND will increase the output voltage of the regulator by approximately 0.5V. Therefore, a 200 Ohm resistor should give you right about 6v out.

    Note that with 78xx regulators, you must use a 0.33uF capacitor from IN to GND, and a 0.1uF capacitor from OUT to GND, and they must be located as close as possible to the regulator. If you don't have these capacitors in the circuit, the regulator may oscillate at high frequency (in the MHz range) and you will not get the voltage output that you expect; it will be closer to 1.2v to 1.8v.

    1N5404 are standard 3A rectifier diodes. They will likely have a Vf of around 0.6v with 500mA flowing through them.
     
    Last edited: Jul 24, 2010
  8. Shapath

    Thread Starter New Member

    Jul 10, 2010
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    I have used a very large heat sink with these circuit. I have uploaded the schematic of the circuit (drawn in MsPaint!). I didn't use any breadboard, veroboard or pcb. I just joined the IC with the heat sink and then shouldered the two diodes, one at the input end and other one at the output end...thats it!!!

    I have studied that 2 capacitors are needed when using 7805..but no one was able to explain me properly why its needed. Thanks a lot SgtWookie for your information. Since I was confused about the use of capacitors so i didnt add those in the circuit, but now I will modify it and definitely use that.

    Here parts are available but I dont know why many times I found that these parts does not perform as it should.

    To design this converter at first I wanted to use IC 2N3055, Silicon Power Transistor. I have also attached its picture as attachment. This circuit is not designed by me rather provided by one of my friends. Please tell me if you feel that this one is better than the 1 i have been designed.

    N thanks to Tom and SgtWookie a lot..:D
     
  9. Shapath

    Thread Starter New Member

    Jul 10, 2010
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    this is the original one that i have designed.
     
  10. tom66

    Senior Member

    May 9, 2009
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    That circuit will work, but it will have a lot of variation; that is, even though it is a constant voltage supply its voltage will change over a few hundred millivolts depending on load, due to its poor "line regulation".

    Instead, consider an LM317. How much current are you expecting? I wouldn't think any more than about 500mA for a phone. You might need a big heatsink on it.
     
  11. Shapath

    Thread Starter New Member

    Jul 10, 2010
    11
    0
    in the second circuit using Zener Diode and IC 2N3055, i was afraid that if the zener diode shorts then my device will burn. So i didnt use that circuit.

    now you are suggesting me to use LM317...hmmm....but it will also require heat sink..but yeah i will get better regulations.

    Anyways thank you again for your help...:D!!!

    I will let you know what modifications i will do to this circuit...!!!
     
  12. SgtWookie

    Expert

    Jul 17, 2007
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    The LM317 is capable of more output current (1.5A vs 1A), but will require the use of two resistors to control the output voltage. In this case, the 7805 with a 200 Ohm resistor from the GND terminal to GND will work just about as well as an LM317 with appropriate resistors from OUT to ADJ and ADJ to ground.

    In either case, both the IN and OUT terminals should have capacitors on them. Metal poly or ceramic are fine to use. If you wish, you can add 10uF electrolytic capacitors on the IN and OUT to ground, which will improve transient response (sudden current demands). However, the 0.33uF and 0.1uF capacitors are still required.

    Wiring is inductive (roughly 15nH for a 10mm piece of 18ga straight wire), and there are also small amounts of stray capacitance, in the pF range. This creates a resonant LC circuit, whos' frequency will be difficult to predict without performing a number of calculations. The regulators have a fairly low frequency response range. Adding the 0.1uF cap on the output and the 0.33uF cap on the input ensures that the resonant frequencies are low enough that the regulator won't oscillate.
     
  13. tom66

    Senior Member

    May 9, 2009
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    0.33µF (330nF) ceramic capacitors are sometimes tough to find. In my case, I substituted them with a 1µF tantalum bead capacitor, which was easier to find. Be careful in getting the polarity correct, tantalum capacitors are well know for lighting up like matchsticks when abused. The low-ESR of tantalum capacitors helps.

    The 7805 circuit will work, however it may have a few problems. Since there is a resistor on the ground pin does this affect the transient response? The regulator requires a few mA

    I recommend you use a circuit similar to the one I have attached. It gives a 5.5V output approximately. But I'm not sure what output you want. If you want 6 volts, just replace the diode with wire.
     
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  14. SgtWookie

    Expert

    Jul 17, 2007
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    Have a look at the attached schematic. The circuit is based on Roman Black's 2-transistor Black regulator; Roman is also a member here.

    Roman's website: http://www.romanblack.com/

    His 2-transistor regulator page:
    http://www.romanblack.com/smps/smps.htm

    It's a 2-transistor switching buck regulator. There is no provision for overload detection, but it's certainly more efficient than the linear regulator route.
     
  15. SgtWookie

    Expert

    Jul 17, 2007
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    Metal poly film caps are preferred over ceramic. You could also parallel three 0.1uF caps if you can't find the 0.33uF caps, or simply wanted to reduce the variety of caps you needed to keep on hand. EVERYONE should have a good-sized stock of 0.1uF caps on hand; they're required for bypass caps on just about any IC that you can think of.

    Yes, tantalum caps can be used. However, if not used for a long period of time, their dielectric will weaken, and they'll act like firecrackers when power is suddenly applied. For reliability, use metal poly film or ceramic instead.

    There is roughly 5mA current via the GND pin. 5mA/200 Ohms = 1v.
    Note that the GND pin and the tab are electrically connected. Transient response really shouldn't be affected any more than an LM317 configuration with R1/R2. However, if you wish to avoid overshoot on start-up, add a 10uF cap between the GND terminal and GND.

    Adding a diode in the output path will cause considerable variation in the output voltage, as the Vf of a diode depends upon it's temperature and current flow through it.

    A 7805 regulator is generally more useful than any other 78xx series regulator, as you can get whatever voltage out that you wish, simply by adding a resistor from the GND terminal to ground.
     
  16. tom66

    Senior Member

    May 9, 2009
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    The 5.5V output is nominal.

    The OP is charging a phone. There is more variation on your typical USB port or adapter (usually ±5%) than this regulator has, so I think it's fine.
     
  17. SgtWookie

    Expert

    Jul 17, 2007
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    You're adding another 5% (VERY roughly) deviation on top of the 10% (factory specified) deviation of the 7805.

    If one wanted to be nit-picky, a 500 Ohm 10- or 21- turn pot from the GND terminal to ground could correct much of the deficiencies of the 7805 regulator (10% variation, adjustable), and a 1.2k resistor (5mA load) from OUT to GND would satisfy the minimum 5mA current draw requirement to provide guaranteed regulation.

    I'm not arguing with you; I'm just trying to help you with some factual information.
     
  18. tom66

    Senior Member

    May 9, 2009
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    If we're correcting each other: The deviation of a 78xx regulator is ±4% under ideal conditions or ±5% under non-ideal conditions. Not ±10%.

    As far as I understood, the OP was just looking for a circuit to charge his 6V or 5.5V phone from a 12V SLA. Well, the circuit I posted was perhaps the simplest and guaranteed to work. Sure, it's less than 50% efficient at full load, and it might not have the greatest load regulation, but it works. (I'm invoking the principle of good enough here.)

    I feel that interfering with the ground path of the regulator could cause more problems. I'm not certain, but it might interfere with overcurrent/short circuit/overtemperature protection.
     
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