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Emergency Power Supply Schematic Diagram

Discussion in 'Power Electronics' started by guitarnoob123, Feb 25, 2014.

  1. guitarnoob123

    guitarnoob123

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    Feb 25, 2014
    I am designing an emergency power supply.
    GOALS
    -The regulated source should provide power to the load when there is power in the mains.
    -When a power outage occurs, a backup battery should take over.
    -When the mains power is available again, the regulated source should power up the load again and charge the battery at the same time.
    -When the battery is full, the charging should stop

    This is my current circuit:

    [Moderator's note: I've embedded the schematic into this post. It was originally at http://i.snag.gy/du6Wf.jpg. -- KrisBlueNZ]

    [​IMG]

    V1: Regulated Buck supply
    V2: Battery
    R3: LOAD

    Switch States: With Mains power, battery is charging
    1) Q2 is switched on which in turn switches Q3 on. Battery starts charging.
    2) Q4 is switched off since the Regulated supply voltage is greater than the battery voltage.
    3) Q1 is switched off since the battery is not yet full.
    4) Substate: Battery is full. Zener diode D1 lets current flow to the Gate of Q1 which switches it on.
    This switches off Q2 and Q3 since the Gate voltage is insufficient.

    Switch States: Without Mains power
    1) Insufficient voltage switches Q2 and Q3 off.
    2) Q4 switches on since the battery voltage is now greater than the Regulated supply voltage.
    3) Battery supplies power to the load.

    Any wrong assumptions and mistakes in my design? Any ideas? I am not yet sure if this will work.
    All suggestions for improvements are welcome
     

    Attached Files:

    Last edited by a moderator: Feb 26, 2014
  2. Arouse1973

    Arouse1973 Adam

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    Hello
    What battery are you using? I don't know of any 10V batteries I might be wrong. Q4 is around the wrong way I think, you will be supplying current through the body diode of the FET. The FET will always be conducting. I doubt also you will have enough voltage difference between the gate and source to fully turn on the FET.
    Adam
     
  3. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    There are many problems with your schematic.

    D2 is a 1N4148, and D3 and D4 are BAT54s. These are small signal diodes rated for 200 mA maximum!

    Q1 and Q2 don't need to be power MOSFETs. The ones you've specified are rated for 14A or more!

    If you used those parts because of a limitation with your simulator, you should at least mark the actual component type numbers next to the placeholder component types.

    I'm not sure what Q2 is supposed to do. Connected like that, it will act as a shunt voltage regulator with an unpredictable and temperature-dependent voltage.

    In the charger, R1 will limit the charge current to a trickle. Q3 will not turn on properly because of its gate-source voltage requirement. I think you've used a transistor circuit and changed the transistor to a MOSFET. That won't work. Also, Q1's gate threshold voltage is also not well controlled so the charger's voltage regulation will not be stable, and in any case, Q1 would need a gate-source pulldown resistor.

    If you delete R2, change Q3 to a Darlington NPN transistor, change D2 to a Schottky diode with very low voltage drop, and replace Q2 with a shunt regulator such as a TL431 set for nearly 12V (reduce R1 to ensure at least 1 mA through the TL431), you could delete Q1, R3 and D1, and it MIGHT be workable, but I doubt it.

    I think you'll need to use a PNP or a P-channel MOSFET for Q3 because there's not enough voltage difference between the input supply and the battery voltage.

    It might be better to use a 12V battery with a boost converter to charge it.

    Q4 will work, I think. The body diode will pull the source up to a positive voltage, and when the input voltage disappears, Q4's gate will drop to 0V and it will conduct. I would add a series resistor to the gate, and a zener (e.g. 10V, 1/4W) from the gate to the source, to protect Q4.

    Have a look at some MOSFET data sheets to see how the gate-source voltage controls the drain current. For a standard N-channel MOSFET, the gate needs to be at least a few volts higher than the source before it starts to conduct, and several volts higher than that if you want significant current to flow through it.
     
  4. guitarnoob123

    guitarnoob123

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    Feb 25, 2014
    Hello. This is my newer schematic

    [Moderator's note: I've embedded the schematic into the post again. It was at http://i.snag.gy/hTTeA.jpg. -- KrisBlueNZ]

    [​IMG]

    I removed Q2.
    I removed R2. What will happen if I change Q3 to a Darlington NPN? Increased Beta = increase in charging current? Will it also serve the switching application I wanted to implement?
    I replaced Q1 from a FET to a BJT. So that it is now current controlled instead of voltage controlled.

    What will cause the current flow through the body diode? I thought the body diode's orientation is reverse that of the conduction path of the FET

    Please comment on my newer schematic :D Thanks for the knowledge.
     

    Attached Files:

    Last edited by a moderator: Feb 26, 2014
  5. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    You haven't replaced the diodes with properly rated ones.

    You don't need R5. The gate-source pulldown resistor I mentioned in my last post was on Q1 not Q3.

    Changing Q1 from a MOSFET to a junction transistor doesn't make it current-controlled instead of voltage-controlled. The circuit still regulates the output voltage and has no explicit current limiting.

    The charging circuit won't work with an N-channel MOSFET for Q3 because of the gate-source voltage requirement. Have you looked at some data sheets or tutorials and learnt about this?

    Changing Q3 to a Darlington NPN would reduce the gate-source (base-emitter) voltage needed, but not enough to charge a 10V battery from a 12V supply, especially with D2 in the circuit.

    Neither the MOSFET version nor the Darlington NPN version has any current limiting. This could be added by inserting a resistor in the source or emitter lead, with some kind of feedback from the bottom of the resistor to the base, but only at the cost of adding extra voltage drop.

    You could use a low-value current sense resistor in series with the negative lead to the battery to sense the charge current. Feed it into a circuit using a single-supply op-amp to reduce the gate bias to limit the charge current. If you Google those keywords you may find circuit examples.

    Using an N-channel or NPN device to feed the 12V supply to a 10V battery is not workable because of voltage drops, unless you geneate a higher-voltage rail from the 12V rail that can provide at least 8V more, in which case you could use an N-channel MOSFET.

    This rail doesn't need to supply much current, so you could generate it pretty easily using a charge pump driven by an oscillator. A CD4584/CD40106/74C14 is a good fit here; one gate can be the oscillator and the other five can be buffers to drive the charge pump.

    I don't know what you mean by the "switching application I wanted to implement". That circuit will automatically change over between input power and battery.

    Edit: Do you really want to use a 10V battery? If I was designing this, I would probably use a 12V battery, charged with a boost converter. What is the part number of the battery you have chosen?
     
    Last edited: Feb 26, 2014
  6. Arouse1973

    Arouse1973 Adam

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    I think Q2 on the original diagram is used to track gate voltage changes due to temperature as both devices are the same the turn on voltage generated at the drain is used to turn on the second FET. Q3 would have probably worked for the low current in this circuit. But Q4 is still wrong.
    Thanks
    Adam
     
  7. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nope. Q3's source voltage is about 10.6V higher than Q2's source voltage! And Q3's source cannot supply more than about 7~8V (the 12V input voltage minus its gate-source voltage) even if Q2 wasn't pulling its gate voltage down.
     
  8. Arouse1973

    Arouse1973 Adam

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    Yeah Duh what was I thinking? go easy on me Kris, only been doing this a few months :)
    Adam
     
  9. guitarnoob123

    guitarnoob123

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    Feb 25, 2014
    I haven't found schottkey diodes w/ sufficient current rating in Simetrix. I will edit the schematic that I will post later.

    I see. I was doubting if a voltage drop will occur on the FET gate if there is no current flowing. So either way a BJT or a FET for Q1 would serve the same purpose?

    I have overlooked this. I always assumed that I can always increase the supply voltage so that it is sufficiently higher than the terminal voltage of the battery which I realized was not good.


    More on my progress about this later. So basically I have to regulate BOTH the battery terminal voltage and (especially) the charging current?
    The battery specs would come later. I can always rate its required current and voltage based on the finished design and vice versa (or am I wrong?).
    What would be the advantages if I am to use a boost converter? I don't know how to integrate it into the design and I have no experience in creating one (I'm still studying how to control switching converters using UC3843)

    Also what are the changes that I will make if I use relays instead of FETs for switching? I realized that I have to avoid the switching loss of the FET because there may be a point wherein it is held ON for a long time.

    (BTW thanks for the comments. Its always fun learning)
     
  10. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Functionally, yes, but a BJT has a better-defined input voltage threshold. You would be best to use a shunt regulator such as a TL431. These have a very accurately controlled input threshold voltage. Feed it from the battery voltage via an adjustable voltage divider (e.g. a multi-turn trimpot) instead of using a zener diode. This will improve the voltage accuracy.
    Yes, if the battery is a lead-acid type (SLA or VRLA). If it's NiCd, NiMH, Li-ion, Li-pol or something similar, you're supposed to use a proper charge algorithm to recharge them, and trickle-charging may or may not be safe. Do some research!
    Well, you need to specify a battery that exists in reality, so it's a good idea to do some research beforehand and pick some possibilities. What voltage range does the load accept? Is it important that the output voltage doesn't drop too far when switching to backup mode?
    The UC3843 might be a good choice for a boost converter. It's designed as a flyback converter, but a boost converter is just a flyback converter bootstrapped from the input voltage.
    I just realised that a boost converter is probably not appropriate because the bottom end of the output voltage range will be too high if the battery is significantly discharged. But a flyback converter would work.
    The idea is to match the input voltage fairly closely to the battery's nominal terminal voltage, so the output voltage to the load doesn't change much when the circuit switches to battery backup. If that's not a problem, then a linear or buck charger is fine.
    If that is a problem, another approach is to use an input voltage that's higher than both the battery voltage and the output voltage, and add a regulator to drop it down to the desired output voltage when it's running from the input supply.
    You can get MOSFETs with pretty low losses!
    0.0041 ohms for USD 1.82: http://www.digikey.com/product-detail/en/IPD90P03P4L-04/IPD90P03P4L-04INCT-ND/2269941
    0.0062 ohms for USD 1.14: http://www.digikey.com/product-detail/en/AO4423/785-1288-1-ND/3050792
    0.0075 ohms for USD 0.94: http://www.digikey.com/product-detail/en/DMG4413LSS-13/DMG4413LSS-13DICT-ND/2279225
    0.008 ohms for USD 0.68: http://www.digikey.com/product-detail/en/DMP3010LK3-13/DMP3010LK3-13DICT-ND/3076574
    If the load resistance marked on your schematic (8 ohms) was right, at 12V supply the current will be 1.5A. Even the cheapest of those MOSFETs will drop only 12 mV (0.1% of the total supply voltage) and dissipate only 18 mW. MOSFETs are a lot smaller and cheaper than relays.
     
  11. Arouse1973

    Arouse1973 Adam

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    Kris can you explain what you mean, a boost converter is just a flyback converter bootstrapped from the input voltage.
    All the flyback converters I have seen have transformers and boost convertors tend to have just a single inductor.
    Cheers
    Adam
     
  12. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Exactly. In both the boost and flyback converters, energy is delivered to the load when the switch turns off and the polarity of the voltage(s) across the winding(s) reverses. In the flyback converter, the output voltage is supplied entirely from the flyback voltage, whereas in the boost converter, it's supplied from the flyback voltage added onto the input voltage (bootstrapped).

    This is just an observation I have made; I don't think I've ever seen it explained elsewhere. If anyone disagrees, please explain why.
     
  13. Arouse1973

    Arouse1973 Adam

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    I see what you mean now.
     
  14. guitarnoob123

    guitarnoob123

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    Feb 25, 2014
    Hello guys. This is my newer schematic diagram
    http://i.snag.gy/cTENZ.jpg

    [Moderator's note: I've attached the schematic to this post again. You should do this yourself. Click the Go Advanced button, click the paper clip icon, and use the dialogue box to upload the image. Files are limited to about 100 kB, or about 150 kB for JPG images. If you want the image to appear inside the post, as well as in a thumbnail at the end, after you've uploaded it, on the upload dialogue box, right-click the filename that appears in the "Current Attachments" area, select "Copy link location" or similar, close the dialogue box, click the Insert Image icon below and right of the paper clip, and paste the URL into the field. -- KrisBlueNZ]

    [​IMG]

    I decided to use LM338 (LM317 in the figure) to regulate voltage. It is rated 5 amperes
    I replaced the FETs with BJTs so that less voltage is needed to switch them on.

    R2 and R1 sets the regulator output to
    Vreg = 1.25*(1+(R2/R1)) = 14.27 volts

    Charging current is limited by R3, R4, and Q1.
    Ichg = Vbe,sat/R3 = 0.8/0.25 = 3.2 amperes**
    **asuming base current is very small. I got the circuit from http://www.ti.com.cn/cn/lit/ds/symlink/lm138.pdf page 16. I recalculated but I am not sure of the values because Vbe,sat is in the range 0.7 volts to 2 volts depending on the collector current

    Charging stops when the battery voltage reaches Vzener + Vbe,sat of D6 and Q2 again assuming that base current is very small. Vzener + Vbe,sat = 13 + 0.8 = 13.8. Q2 shorts R2 which decreases the LM338 output which stops the charging.

    Battery supplies power to the load through Q3, when V2 (POWER SOURCE) is open/short. V2 provides power to the load otherwise (it is not shown in the schematic but basically V2 is fed to another LM338).

    The extra capacitors and diodes are for improving response time and protection of the LM338 regulator.

    Please comment again. Thanks guys!
    It is always fun learning!
     

    Attached Files:

    Last edited by a moderator: Mar 5, 2014
  15. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    That's quite an improvement. Here are some random comments.

    1. Q3 can't be driven like that. You'll need at least a resistor in series with the base. But to saturate it, you'd need some significant base current. And you need to insert a diode between where Q3's base connects to the input supply, and the input to the regulator, so that the base drive voltage will drop when the input power source disappears. If you don't have that diode, the battery will back-feed through D1 and prevent that voltage from going low and turning on Q3. Also Q3's base-emitter junction should be protected against reverse voltage. You would be much better to stick with the MOSFET from your earlier design. Adam (Arouse) has queried it, but I think it will work fine, if you add the protection components I recommended in post #3. Simulate it using LTSpice to be sure.

    2. Also, Q3 is connected in the wrong place - there's no separate path from the regulator to the load. When input power is present and Q3 turns OFF, the load will lose power!

    3. Q2, R9 and D6 should be replaced with a shunt regulator such as a TL431 as I mentioned in post #3. Use a multi-turn trimpot to provide the input voltage to the TL431 and connect it where Q2 is now. This will give you a much more stable voltage limit. This will probably make the circuit oscillate though; you may need to slow down the response using a small capacitor (e.g. 10 nF) from the cathode to the control pin of the TL431.

    4. You're monitoring the battery current using a 0.2 ohm resistor. It will drop significant voltage when the battery is supplying the load. You would be better to use a much lower value with an op-amp (or at least a PNP long-tailed pair) to monitor the voltage and/or amplify it.

    5. C1 should be marked 1000 µF. Millifarad (mF) is a valid unit but it's almost never used - for regular electrolytics, at least.

    6. You've connected R2 to the negative terminal of the battery, so the regulator will actually regulate the voltage across the battery, ignoring the voltage dropped across R3. Clever!

    7. The relevant Vbe voltage of Q1 is around 0.65V, not Vbe(sat). Q1 will not saturate unless the battery is shorted or nearly; it will normally operate in its linear region where Vbe is around 0.65V. You can use that figure for your calculations.

    8. Personally I would use separate regulators for the load circuit and the charging circuit. This allows you to set the charge voltage and the output voltage independently, and have a clean, stable output voltage while the input supply is present. It also avoids the problem I mentioned in 2, above. You may need to move the series Schottky diodes around to be more like the schematic in post #4. You won't need D3 to isolate the voltage regulator from the output if you put the reverse-connected diode across it (D1 in post #14) and put a series diode on the input so the input supply voltage will fall to zero in battery backup and turn on the backup MOSFET.

    9. You're now using an input voltage of 18V and a 12V battery. What are the restrictions on the input voltage? Can you make it whatever you want?

    10. I designed a power supply with these features (and a lot more) that used switching regulators for the main regulator and the charger. They are much more efficient than linear regulators and don't need much heatsinking. They perform better with higher input voltages - 24V for example.
     
    Last edited: Mar 5, 2014
  16. guitarnoob123

    guitarnoob123

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    Feb 25, 2014
    I just assumed that I can get any voltage I want from a step down transformer and a full bridge rectifier at the moment.

    I'm still studying about switching regulators. I have no confidence yet in building one and I think that it might take some time considering that I have no experience yet on the controller feedback stuff (AS3843 IC) and winding my own inductors. Maybe 2 or 3 months from now.

    Would I be better off if I monitored the current (plus set the maximum and minimum voltage) like the circuit I posted here,https://www.electronicspoint.com/12v-battery-charger-schematic-t267632.html#post1601558

    then implement the "full charge detect" function using the shunt voltage reference you suggested?

    By the way thanks for your answer in that thread, that cleared things up.
    I am loving power electronics more :D
     
  17. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Yes, I think you can abandon this design and go with something like the circuit in the other thread.

    As I said over in that thread, that circuit will over-charge the battery because its maximum output voltage is too high - 16.3V. But changing the resistor values as I described will fix that. Then the circuit can be used as-is because it includes voltage and current limiting. There's no need to add a shunt reference.

    It's a lead-acid battery you want to use, right?
     
  18. guitarnoob123

    guitarnoob123

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    Feb 25, 2014
    But I want to disconnect the charger to the battery when it is fully charged. I think that the regulator might still provide/consume power when the battery is full.

    Yes. But its ratings were not specified. That is why I am assuming things like 3 amperes maximum charge current and over-rating the components
     
  19. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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