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Power Regulation Strategy

Discussion in 'General Electronics Discussion' started by NiGHTS, Feb 23, 2015.

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  1. NiGHTS

    NiGHTS

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    Nov 19, 2014
    I have an electronics related problem I have been trying to solve for at least a month and I'm hoping your guys could help me with a functional strategy to solve it.

    I am basically trying to power my project which requires 12 Volts DC and up to 5 Amps (typically runs at 3.7 Amps). I need to support AC and DC power from a range of 17V to 30V. The issue here is heat management. I am using two voltage regulators each capable of handling up to 3 Amps using a shared current configuration with 3 Amp diodes acting as load balancers. I am also using a 6 Amp bridge rectifier active only when AC power is applied.

    The Bridge rectifier and Voltage regulators get quite hot and require a very large heat sink. The heat sink does the job fine, but it idles at up to 70C which i'd prefer not to happen in this particular application.

    Is there a way I can lower the heat output of this circuit without adding too much extra cost? Could I somehow use a transformer to improve my heat problem? I can't use a fan in this application so its either use a very large aluminium heat sink or find some strategy that tempers the heat with a high amp load. I am also open to use 120VAC if this somehow improves my heat issue as long as it does not add too much cost.

    Thank you in advance for your help.

    Power.png
     
  2. davenn

    davenn Moderator

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    it is going to get very hot, you cannot get past that if you use linear regulators, you must implement ways to dissipate the heat

    consider if you are putting in 25V and outputting 12V @ 4Amps, that's a 13V drop across the regulator x 4A = 52 Watts of heat being generated


    you would be better to move to a buck DC-DC converter that is much more efficient and thereby producing much less heat
    and you would only need one regulator chip

    cheers
    Dave
     
  3. davenn

    davenn Moderator

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    Ohhhh BTW
    when drawing circuit diagrams, its conventional to have inputs on the left side and outputs on the right side :)
     
  4. Ratch

    Ratch

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    Mar 10, 2013
    I would use a variac (variable voltage transformer) or something else to lower the AC. Adjust the AC variac voltage until it is sufficient to power the load. The idea is to let the load use most of the power so the regulators don't have to dissipate such a heavy load. A heavy current load with a small DC load voltage together with a high AC input voltage puts the maximum stress on the regulator. A DC-DC converter is noisier than a linear, so you have to decide. Be careful. The variac is an autotransformer and has no isolation.

    Ratch
     
  5. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    I agree with Dave's suggestion - use a switching supply.

    If you always have mains voltage available, just get an "off-line" (mains-powered) supply rated for 12V and 5A (60W) output.

    If you need the option of feeding in a lower DC voltage, you could use that type of power supply to bring the mains voltage down to, say, 30V DC, "diode OR" that voltage with your external DC supply, and feed the ORed voltage into a buck regulator that converts it to 12V. In that case the off-line supply needs to be rated for more than 60W to cover the power lost in the buck regulator (and the diode).

    Suitable power supplies are available cheap on eBay.
     
  6. NiGHTS

    NiGHTS

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    Nov 19, 2014
    I am happy to say that I've learned something significantly important for my current and my future projects. It's a great concept to use a passive method to control voltage using a fractionally saturated inductor than an active drain method with the voltage regulator. I can see how both have advantages and disadvantages, though in my application the switching version seems to be the most correct strategy for the job.

    Assuming I continued to use a voltage regulator I would still need two because I can only find up to 3 Amp voltage regulators with 12V output, and my load averages on 3.6 Amps so I would need two regardless.

    Though for my application voltage variance is acceptable and so is some noise, so I don't really need the voltage regulators if I use the switching supply.

    Noted. Thank you for your advice.

    Well I think a Variac is overkill for my application, but I appreciate your comment.

    I can visualize what you are suggesting. Not quite sure yet if I will pursue it for this project. I will see if the switching supply strategy will work first.

    ---------------------

    Attached is a schematic of a power configuration I will be attempting very soon based on your feedback. I'm not sure how well this will work yet but it seems to be worth a try. The central component of this new configuration will be a Buck PWM Stepdown Voltage Regulator which controls a MOSFET to charge a coil. There are a few questions that this new schematic raises.
    1. I'm not sure what value of coil to use. In my experiment I will be using a 50uH 5A coil based on a few educated guesses, but I'm not completely certain about this. I believe this coil needs to be rated in relation to the switching frequency but I don't know quite what formula to use to make that determination.
    2. I don't expect this to improve the heat produced by the bridge rectifier component. How much would the heat generated by that component increase between 17VAC and 30VAC? Is it trivial or significant? I haven't tried this experimentally yet.
    3. I am aware of PWM configurations that use more than one MOSFET and coil to improve heat among other things. Should I be using more coils to drive my load? How much heat should I expect to dissipate from the coil? Is there a formula I can use to predict this heat output?
    4. The datasheet for the IC I am using (UC3573N) implies that I need an RC timer on the EAOUT pin. I can't seem to find any information on how to determine the values of the resistor and capacitor attached to this pin. The values I chose are based on recommended values from a datasheet of a similar part. Would anyone know whether I need these components at all and if so how to determine these values?

    Thank you again for all your help! You guys have been wonderful! Power2.png
     
  7. davenn

    davenn Moderator

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    Sep 5, 2009

    No, you misunderstand how switching regulators work
    The reg chip passes very little current as it is all bypassed through the external components

    Where did you get that circuit from ?
    Those 2 x 3A diodes are incorrect they are short circuiting the output to ground, you will see lots of smoke
    and not sure why you have 2 of them ?

    here's the sort of thing you should be considering
    so cheap you couldn't build it for that !! :)
    http://www.ebay.com.au/itm/4-38V-to...ter-DC-5A-Car-Voltage-Regulator-/281233821216

    Dave
     
    Last edited: Feb 25, 2015
  8. davenn

    davenn Moderator

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    Sep 5, 2009
    But you didn't take it ;)


    when convention is followed ... it makes it easier for everyone to follow what is going on :)
     
  9. NiGHTS

    NiGHTS

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    Nov 19, 2014
    That's correct. I put the diodes the wrong way. Thank you for pointing this bug out. Its supposed to be facing the other direction. I have updated the schematic to correct this.

    I have two of them because I want to use 3 Amp diodes for size and cost reasons and my load is higher than 3 Amps, so I am rating everything in the path of the load current to up to 5 Amps.

    This was on purpose. I was updating my original schematic in such a way as to keep the updates consistent for the thread's sanity. If I made a new thread this convention would be followed.
    Power2.png
     
    Last edited: Feb 25, 2015
  10. NiGHTS

    NiGHTS

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    Nov 19, 2014
    I'm sorry I misunderstood you. When you said "voltage regulator chip" you were referring to the PWM controller and I thought you meant the linear voltage regulator. They are both built for voltage regulation but have a different approach to tackle that job, so I was confused by your terminology.

    So in conclusion, ignore my comment about using two vreg chips as it was a misunderstanding.
     
    Last edited: Feb 25, 2015
    davenn likes this.
  11. davenn

    davenn Moderator

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    Sep 5, 2009
    Yes but the diodes are not conducting any of the output current are they .... ;)
     
  12. davenn

    davenn Moderator

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    Sep 5, 2009
    so, again, where did you get this circuit from, it varies a bit from the norm for the use of this chip
    ( not that I could find too many examples LOL)

    Did you look at the link to the one on eBay ? ... thoughts ?
     
    Last edited: Feb 25, 2015
  13. NiGHTS

    NiGHTS

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    Nov 19, 2014
    They are part of the time, though if the math were done it would probably show that a lower amp diode would be fine depending on the duty cycle. I base this on the conceptual understanding of the function of the diode which is to complete the circuit on the load during the time when the MOSFET is turned off and the inductor is discharging.

    Buck_operating.svg.png
     
  14. NiGHTS

    NiGHTS

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    Nov 19, 2014
    The schematic was inspired by the one in the Datasheet for UC3573N, page 4.

    I did not include the sleep function as its not applicable to my project. Some of the gaps in their design were filled by this Datasheet for a similar part of the same brand.

    Unless I can successfully reverse engineer it so I can incorporate it into my design it is useless to me. It looks like a multi-layer board so it might be difficult to trace a schematic out of it, and I very much doubt they will be supplying a schematic along with that part on ebay. Still I may buy it just to study what I can from it. Either way since my time is at a premium right now I need to see if this schematic I presented to you has the potential to get the job done.

    My 4 questions on the schematic still stand.

    Thank you again for all your prompt help and time given to this problem of mine. I know I am taking on a lot all at once very quickly but i'm sure with a little guidance I can achieve my goals.
     
  15. NiGHTS

    NiGHTS

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    Nov 19, 2014
    Here's a quick update. I've built a breadboard prototype of the schematic I presented in this thread and it seems to regulate voltage correctly. The main problem seems to be with the diodes. They get extremely hot with voltages greater than 17V, and that in turn drops the voltage and current output significantly. I am going to try using an 8 Amp TO-220 package diode with a basic heat sink attached and see how it fares. By my estimation the wattage through the MOSFET should be the reciprocal of the wattage going through the diode. At 12V input I'd expect the MOSFET to work at 100% power and the diode to work 0%.

    I would have loved to have the option to pass 12V power through the inputs and get 12V on the output, but it seems it will only do this starting at 13V. Would anyone know what component is likely to be causing this voltage drop? I would think its either the coil or the MOSFET.
     
  16. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Breadboard construction isn't suitable for a switching supply. Use stripboard or padboard, or ideally PCB. Layout is important and should be described in the data sheet or application note for the controller IC.

    What's the part number for the Schottky diodes you're using? What reverse voltage are they rated for?

    Schottky diodes have a negative temperature coefficient of forward voltage so they won't share current evenly unless they're on the same die. You should use a single diode with a higher current rating.

    What type number are you using for the MOSFET? You should also include this information on your schematic so we know what you're doing. Check your MOSFET's maximum gate-source voltage rating. If the input voltage is too high you may exceed that rating and the MOSFET will fail immediately.

    What is the maximum duty cycle of the controller? That will limit the minimum dropout voltage.
     
  17. NiGHTS

    NiGHTS

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    Nov 19, 2014
    I needed quick proof of concept. I am aware that breadboards are not designed for the load I am passing, but its far easier to experimentally reconfigure layout and part values with the breadboards and I try not to have it run for long periods of time.

    I am also aware of the importance of layout in electronic design. The schematic I constructed for this project, while not specifying net routing and part proximities, was at least arranged with that in mind. In my prototype I made sure short and direct routes were taken for all important parts.

    Also the datasheet for this controller IC is poorly documented. It leaves a lot to be desired.

    I have abandoned using axial diodes and instead am using a single TO-220 package for the diode. This particular part I picked up today is an NTE581 which according to the package has a forward maximum current of 8A and reverse voltage of 400V max. Can you explain to me what the significance of the reverse voltage is to this circuit?

    I am currently using FQP7P06 for my P-Channel MOSFET. It has a Gate-Source voltage of +/- 25V. Does this mean that if the voltage difference between the gate and source is say 30V that the gate would break and create an internal short? My testing up to this point must be pushing its limits as I've been using a 25V source.

    And I apologize about not supplying this part information in the schematic but at the time I had no idea what components I would use. I literally used whatever I had available at the time of assembly and this is the part I happened to chose out of a slim selection.

    The datasheet for the IC describes how to calculate this duty cycle on page 3. It recommends between 10kHz to 200kHz. From what I understand of the datasheet the capacitor on the "RAMP" pin programs the frequency with the following equation:

    F = 1 / (15000 * C)

    So based on that I would guess that the duty cycle of my current schematic would be about 98kHz. I suppose I should use an oscilloscope to confirm that. How does duty cycle play a role in dropout voltage? What are the advantages / disadvantages to higher or lower frequencies?

    And would you happen to have any idea what the R/C timing gate on the EAOUT pin is all about?

    I am very grateful for your detailed assistance with my problem! I am thankful and humbled by all of your efforts!
     
    Last edited: Feb 26, 2015
  18. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    High currents aren't the only problem. How about capacitance and inductance?
    Well good for you! Breadboard isn't a suitable way to construct a switching power supply regardless of the current, and regardless of how carefully you lay it out.
    That's true. The original data sheet from Unitrode might be better.
    Yes. When the MOSFET is ON, the full input voltage appears across the catch diode, so the catch diode must be rated for at least that voltage, with a safety margin. This isn't a problem for the diode you're using, but usually, Schottky diodes are used in the catch diode position because of their low forward voltage, which improves the circuit's efficiency. Schottky diodes have relatively low reverse voltage ratings, which is why I asked you about it.
    Yes, that's right. Why did you choose that IC? There are much better alternatives around - ones that will drive an N-channel MOSFET (these generally have better performance figures than P-channel MOSFETs) and provide a nice stable gate drive voltage. Check out the LM3477 (http://www.ti.com/lit/ds/symlink/lm3477.pdf).
    OK, fair enough.
    That's the switching frequency, not the duty cycle. The duty cycle is the percentage of each cycle that the MOSFET spends in the ON state. The LM3477 which I recommended has a maximum guaranteed duty cycle of 88%. So if your output voltage is 12.0V your input voltage needs to be at least (12.0 / 0.88) = 13.6V. Actually a bit more because of voltage drop in the catch diode.
    If you decide to use the LM3477 that will be a moot point. The LM3477's data sheet is very long and detailed. It's not the only suitable device, but it's the cheapest suitable one I could find on Digi-Key. You could also look into the LT3724 and LT3844 (they look great but the maximum duty cycle isn't specified) and the LTC1624, which looks to be ideal for your application and has a maximum duty cycle of 95%.
    No problem.

    Edit: When you redraw your circuit, you have another chance to put the input on the left!

    Edit2: If you do a search for other buck converter controllers, be aware that the modern ones generally use synchronous rectification, where a second MOSFET is used instead of (or as well as) the catch diode, to improve efficiency. This is a great idea but most of these modern regulators tend to be suitable for low-voltage applications only - for use in portable electronics and computer motherboards for creating sub-3V supplies for processor cores.
     
    Last edited: Feb 26, 2015
    davenn likes this.
  19. NiGHTS

    NiGHTS

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    Nov 19, 2014
    I would normally worry about this for very high frequency applications, though since I'm running this gate at almost 100kHz I suppose this should be more of a concern to me. At what frequencies would you be worried about capacitance and inductance using a breadboard?

    Duly Noted.

    So the takeaway here is that a lower voltage drop on the diode means higher efficiency which means less heat?

    Do higher voltages through the diode, comparing between 12V to 30V, produce substantially more heat (exponentially?) or is it trivial?

    How would you recommend that I calculate the heat output of a diode? Is it simply ohms law with Voltage Drop multiplied by maximum expected current to get me Wattage, or is it a little more complicated than that?

    What are the advantages / disadvantages to simply using a MOSFET configured as a diode? Seems they would have naturally very little resistance acting as a diode.

    I chose this part because it was through hole and matched my search criteria for a buck switching power supply controller. I am by no means married to it in any way, so I am happy to see all your alternative suggestions! I will be sure to experiment with one or more of these you have listed.

    And yes, I had noticed how the power P-channel MOSFET didn't quite have as nice ratings as the power N-channel MOSFET I'm accustomed to using. So that's an advantage right there.

    Hmm, I wasn't aware that these chips wouldn't naturally just idle to a solid "on" output at a low enough voltage. Why would they posses such a non-mathematical internal design limitation?

    These parts look fantastic. I will try to get my hands on one of these right away. Thank you for these suggestions.

    My schematic must look quite appalling to the conventional engineer to have generated such a chorus of criticism. I can see now that any efforts of mine to keep the schematic consistent for the threads sake was a poor decision. I will be sure to follow this convention when my schematic is touched up for another review if applicable.
     
  20. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    It depends on many factors. Although the frequency of the switching cycle may be only 100 kHz, there are some very rapid voltage and current changes which have large frequency components well into the tens or even hundreds of MHz. These are the one that cause the problems when inductance and capacitance are present, and when current loops are not properly controlled and kept as compact as possible. That's all I know about it.
    Yes, as long as the diode is fast enough - something which is also true of Schottky diodes.
    "Voltage through" isn't meaningful. Current flows through; voltage appears across. Voltage across a diode doesn't cause it to dissipate heat, apart from rapid changes which can produce current flow through the diode's capacitance, but that's not much of a factor. Current through a diode, in the forward direction, multiplied by the diode's forward voltage, is what causes most of the power dissipation.
    It's the forward voltage multiplied by the mean inductor current multiplied by the duty cycle of the diode, which is 100% minus the duty cycle of the MOSFET. You can estimate the MOSFET duty cycle under load as VOUT / VIN × 100%.
    You can't "configure" a MOSFET as a diode. If you're talking about the body diode that appears between the source and drain, this diode is just parasitic and doesn't have any impressive characteristics at all. When a MOSFET is used as a synchronous rectifier, its gate is driven by the controller IC and it is not used as a diode; it is just used to provide the path to 0V that the diode would otherwise provide. It's a lot more efficient than a diode because a MOSFET with gate bias has no equivalent of forward voltage in its drain-source path and its ON-resistance (RDS(on)) can be very low.
    Good!
    I guess there are different reasons for the different types of circuit arrangements. The ones I suggest here have a bootstrap circuit to provide operating voltage to the gate drive circuit for the N-channel MOSFET, and this circuit can only work when the MOSFET is switching; it can't sustain a constant gate voltage indefinitely. Have a look at the description of the bootstrap circuit in the data sheets and you'll understand why. As for the ones that use P-channel MOSFETs, I don't know offhand why they're limited in that way. Someone with more knowledge of switching supplies would be able to tell you.
    No problem.
    That's OK. "Input should be on the left" is the main criticism. Personally I like to show the high-current ground path explicitly in my schematics. Here's an example from a product I designed that shows what I mean. Ignore the fact that the power input is on the right! There were reasons for this; I'm not sure whether I still think they justify drawing it backwards or not!
    Demopsu.gif
     
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