Capacitor charging at Converter Output

Discussion in 'The Projects Forum' started by jpborunda, Jun 1, 2014.

  1. jpborunda

    Thread Starter Member

    Apr 9, 2014
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    Hello everyone.
    I'm analyzing this circuit but I'm having a difficult time understanding how a part of it works.

    I have basically a 2 stage circuit, A and B. Stage A (Attachment 1) is a 5 to 200V converter and stage B is a constant current sink (Attachment 2). Stage B will use the 200V as the voltage compliance, so it can apply a current of 20 to 100mA(max) to a load resistor of up to 1kO. The ON time is small, with pulse width of 200-500µs max, and at a frequency of 20-100Hz.

    My question is with the 5-200V converter. (Here is the datasheet http://www.datasheets360.com/pdf/5050517326450757206). A 200V converter was used. The datasheet says a 0.1µF-1000V capacitor is needed at the output to prevent surge. From the paper Im reading: "R10 ensures a minimum load for the 200 V converter. The converter has a maximum output of 6.25 mA, and R11 ensures the current is always below this amount."

    If the converter sources only 6.25mA max, how do they manage to source 200V @ up to 100mA?

    Im pretty sure it has to do with capacitors C5 and C6 from attachment 1, which are actually pretty big (100µF, 350V) , so they must be able to store a lot of energy. I think that they would have to charge the capacitors first, then turn of the converter, act as a Voltage source for a couple of pulses, then charge them again with the converter and so on.
    If this is the case then how can I determine the amount of pulses that the capacitors provide before they need to be charged again?


    If my question isn't clear, please let me know.
    Any help appreciated! Thanks.
     
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    Last edited: Jun 2, 2014
  2. #12

    Expert

    Nov 30, 2010
    16,284
    6,797
    According to the drawings,
    There are no capacitors in attachment 2.
    C5 and C6 are in attachment 1 and are a pair of 100 uf size.

    You don't turn off the 200 volt converter between pulses. It merely stops charging the capacitors when they arrive at 200 volts because both voltages are the same.

    The 200 uf worth of capacitors hold 200 micro amp seconds per volt.
    The load is 500 microseconds times .1 amp = 50 micro amp seconds.
    The capacitors will only lose 1/4 volt per pulse.

    Is that enough information?
     
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  3. jpborunda

    Thread Starter Member

    Apr 9, 2014
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    Sorry, Im afraid its still not cristal clear just yet.. hehe elaborate maybe? :|


    Ok, so I know that capacitance is 1 F= 1 A seg/volt.
    Therefore, 200µF = 200µA seg /volt "available from the capacitors right?".
    The highest pulse required times the longest time is (.1 A)(500µs) = 50µA seg.
    200µA seg => 1 Volt
    50µA seg => .25 Volts

    Ok and from the basic formula, C = Q/V,
    V = Q/C = (50µA seg)/200µF = 0.25V

    But I just dont see it. I mean what happens if the converter is always ON, and the current sink demands a higher current that the converter can source. Does it come from the capacitors?
    Thank you, and sorry for the typos from the original post.
     
    Last edited: Jun 2, 2014
  4. #12

    Expert

    Nov 30, 2010
    16,284
    6,797
    I described the short term operation. At this point, you are asking about the long term operation. (At least you followed the math well.) In this case, "long term" is one second.

    In one second, the maximum load will be 500 usec x 100 pulses = .05 seconds of, "on" time.
    .05 sec x .1 amp = .005 amp seconds per second.
    That's 5000 ua seconds per second.
    The capacitors hold 200 ua sec/volt.
    5000/200 = a 25 volt loss if there was no replenishment from the power supply.
    If the capacitors ran continuously at 175 volts instead of 200 volts, the replenishment current through R11 (47k) would be 25V/47k = 532 ua seconds per second.
    This is obviously not enough to keep the capacitors fully charged because full load is 5000 ua seconds/second.
    What voltage would the capacitors drop to under full load? That's the point where 5000 ua flows through 47k and that is 235 volts. Obviously a, "fail"...if I did the math correctly.

    Will you double check me, please?
     
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  5. jpborunda

    Thread Starter Member

    Apr 9, 2014
    55
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    That is, if I charged the capacitors up to 200V, they would show a voltage drop of 25V per second right?

    Here, I´m not sure how are you calculating that current. Under which conditions I mean. Are you saying that the 47kΩ resistor is having a voltage difference of 200V(from converter)-175V(the actual capacitors voltage after 1 sec) = 25V? Wouldnt that be the case of leaving the source ON all the time?

    And finally here, this would mean that either the frequency, pulse width, or amplitude would have to decrease, in order to maintain the voltage compliance of 200V right?

    Im afraid Im still kind of confused when it comes to the capacitors charge, and with the example you provided of the 175 V.
    :confused::confused::confused:

    Thank you very much for all your help!
     
  6. #12

    Expert

    Nov 30, 2010
    16,284
    6,797
    yes, yes, and yes. You seem to be getting along just fine.

    I think it is time for a different person to, pretty much, say the same things I said. A different way of phrasing the answer might connect better with you. (That's why we have a lot of different helpers.)
     
  7. jpborunda

    Thread Starter Member

    Apr 9, 2014
    55
    0
    Fair enough.... I get what you're saying, that's what I like about this forum everyone shares their opinion and it helps to 'see' things from another perspective.


    Getting back on the subject... The reason I'm studying this topology is because I'll be implementing it eventually( or some variation of it). I've managed to obtain the 5v200 converter, so I'll be using it as my voltage source. Do you think I should start a new thread asking for assistance in the design of the 200V circuit using the 5AV200 converter?
     
  8. #12

    Expert

    Nov 30, 2010
    16,284
    6,797
    Yeah, I think so. You are changing the question from, "How does this work?" to, "How shall I design a different load circuit?" Very different point of view.
     
  9. jpborunda

    Thread Starter Member

    Apr 9, 2014
    55
    0
    Ok thank you for your answers! Really appreciated!
     
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