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Buck dc-dc Converter w/Control System Analysis

  1. Feb 6, 2014 #1
    1. The problem statement, all variables and given/known data

    I'm given the following circuit
    http://imageshack.com/a/img850/8576/b3ny.png [Broken]
    and given [itex]L, V_{in}, D, P_{o}, f_{s}[/itex] and [itex]q(t)[/itex]. How do I go about finding [itex]T_{s}, V_{a}, i_{L}, i_{L} ripple, and I_{o}[/itex].

    2. Relevant equations



    3. The attempt at a solution

    I'm honestly not sure how to even begin to analyze the circuit. I'm not exactly sure what kind of transistor that is in the diagram. I've never seen a transistor symbol like this and am not exactly sure which one it is. I know that for most diodes the voltage across them is almost always [itex]0.7 V[/itex]. I know that [itex]V_{L}(t) = L \frac{d}{dt}i_{L}(t), i_{c}(t) = C \frac{d}{dt}V_{C}(t), V_{O}(t) = I_{o}(t)R[/itex] It's been a while since I've taken the class that studies transistors but I believe that the the voltage from the collector to base is [itex]0.7 V[/itex]. I'm not exactly sure how charge into the base as a function of time really helps here.

    Thanks for any help.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 6, 2014 #2

    berkeman

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    The transistor is drawn kind of funny, but it is probably just a vanilla NPN bipolar transistor. Vbe = 0.7V and Vce(sat) ~ 0.2V. You can start with those numbers.

    There are lots of good tutorial descriptions on the web about analyzing buck DC-DC converters. Trr a google search on Buck DC-DC Converter Analysis.

    There is a simplifying trick that you can use for an initial relationship between duty cycle and Vo/Vin. Can you find it in the tutorials, or deduce it by looking at the circuit? Assume 0V drop across the transistor Vce and the flywheel diode Vd for the simplification. Then after you use that for an initial guess at Vo/Vi, you can add back in the voltage drops Vce and Vd to refine the Vo/Vi relationship...
     
    Last edited by a moderator: May 6, 2017
  4. Feb 6, 2014 #3

    berkeman

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  5. Feb 6, 2014 #4

    donpacino

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    the switching device is an IGBT (isolated gate bipolar transistor). think of it almost like a bjt that has the isolated gate of a mosfet.

    have you ever analyzed an ideal buck converter? or is this your first ever?
    The diode and transistor together act as a switch. When the switch is on, the transistor conducts and the diode does not, so for all intents and purposes the node to the left of the inductor is at vin. when the switch is off, the diode conducts and the transistor does not, so the node to the left of the inductor is at ground.
    note that is ignoring the forward drop of the igbt and diode.

    that should be good to get you started
     
  6. Feb 6, 2014 #5

    berkeman

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    Ah, an IBGT! I've never used one of those. Thanks Don. What are their advantages?
     
  7. Feb 6, 2014 #6

    donpacino

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    well high power MOSFETs typically use the gate voltage to set a resistance from the drain to the source. So when the mosfet is conducting in high current situations they dissipate a lot of power.

    A high powed bjt always has 2 diode drops of voltage from the collector to the emiter, regardless of current. However they are usually slower to switch and will pull steady state current from the base driver, which is not ideal.

    IGBTs combine the fast switching speed of the high power FETs with the diode like quality of a bjt. So there is always 2 diode drops from the collector to the emitter, as opposed to the resistance quality of the FET. this way is high current situations, you get low power loss. And there is only gate current pulled during switch time.

    In basic terms, they are a high power/high current switching transistor.

    disclaimer, my knowledge of them is theoretical. I've only just started using them practically during the past month. I do know they are expensive and require a lot of supporting circuitry compared to regular transistors.
     
  8. Feb 7, 2014 #7

    rude man

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    Not so. Saturation voltages for a BJT are usually well below 2 diode drops, in fact they're almost always below 1 diode drop.

    "The IGBT combines the simple gate-drive characteristics of the MOSFETs with the high-current and low-saturation-voltage capability of bipolar transistors."

    http://en.wikipedia.org/wiki/Insulated-gate_bipolar_transistor
     
  9. Feb 7, 2014 #8

    rude man

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    I would not make any complicating assumptions regarding the exact nature of the switch. I would assume a perfect on-off switch.

    What concerns me is the lack of explanation of all those parameters cited in the OP's post. What are D, Ts, fs etc.? What is the duty cycle of q? I would assume q(t) is simply an on-off swiching function. (If the switch is an IGBT then q approaches zero anyway if q is charge). If this approach is not followed the problem is both undefined and impossible to sove without circuit simulating software and specific component definition.
     
  10. Feb 7, 2014 #9

    donpacino

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  11. Feb 7, 2014 #10
    Well you've stumped me to stay the least, and I'm a bit concerned by suggesting that the problem can't be solved

    I did not realize that I could think of the transistor and the diode as a switch. This may lead me in the right direction. I'm not exactly sure why it could be considered a switch though. I took a course last semester that studied transistors but have already forgot some of the material so I may need to refresh myself. At one point I had a understand of BJT MOSFET FET etc. and have never seen a transistor like the in one in this problem.

    I know from class and my text that for a buck converter [itex]V_{o} = DV_{i}[/itex] I was just lost when I saw this problem because I was confused by the diode and transistor, which I now know that I can consider to be a switch, in which case I have seen this circuit before. The only issue is the transistor in series with the capacitor. I haven't seen a buck converter in class or text with a resistor like this.

    This is the first time I've seen a IGBT. I'm not exactly sure what you mean by an isolated gate of a MOSFET. If I remember correctly there are two equations for BJT and MOSFET, and you have to solve them. You end up getting some kind of quadratic if I remember correctly. The only problem is that in the class in which I studied transistors the base of the transistor was always connected to the circuit some how or ground, and in this case I have a charge function going into it so I'm not exactly sure how to handle that. If I remember right, don't I need some kind of given conditions to determine the equations for the transistors, which I don't think I'm given?

    Oh. So when q(t) is a high, current passes through the transistor and not the diode? When q(t) is low current flows through the diode and the left node of the inductor is connected to ground? I guess that makes some sense as to me as why this configuration it's basically a switch.

    I believe D is the duty cycle, fs is the switching frequency and Ts is period. These values are undefined.

    [itex]V_{L}(t) = -V_{o}(t)[/itex]
    [itex]\frac{d}{dt}{i_{L}(t)} = -\frac{V}{L}[/itex]

    I believe I can sketch [itex]V_{L}(t)[/itex] and [itex]i_{l}(t)[/itex]. Is V_{A}(t) just 0.7 when the switch is off and Vin when it's on?
     
    Last edited by a moderator: Feb 7, 2014
  12. Feb 7, 2014 #11

    rude man

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    You have a good picture of the situation. Actually, V_A(t) is -0.7V when the diode is conducting. But for all practical purposes it's what you said: ground.

    When the transistor is ON (q=1), current flows thru the inductor to the load per di/dt = (V_in - V_o)/L. When it's OFF (q= 0), the diode takes over and continues the current thru the inductor and the load. But while the diode is conducing, the inductor current di/dt = (0 - Vout)/L < 0 so depending on your parameter values the L current might decay to zero until the next time the transistor turns ON again. (The current thru L never goes in the opposite direction.)

    To draw waveforms you really need numerical values for your parameters.
     
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