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DC-DC Converters & KCL?

  1. Oct 6, 2015 #1
    I'm confused, how is it that DC-converters work with respect to KCL in terms of conservation of charge?
    When a boost/buck converter would output low/high current output than input?

    I_in =/ I_out

    I'm teaching myself the basics of circuits, understood KCL and KVL, but this point is confusing me when relating power converters(of all classes).
     
  2. jcsd
  3. Oct 6, 2015 #2

    berkeman

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    No, most DC-DC switching converters are constant power converters, so Iin does not equal Iout (unless Vin = Vout).

    The input current in a Buck converter only flows when the high-side switch transistor is conducting. When it snaps off, the flywheel diode at the output keeps the current flowing. Since the buck circuit is driving into an inductor, the inductor current (Iout) ramps up while the switch is on, and ramps down when the switch is off and the flywheel diode is conducting.

    Does that make sense? Use Google Images to find some current and voltage waveforms for Buck DC-DC converters... :smile:
     
  4. Oct 6, 2015 #3

    berkeman

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    Last edited by a moderator: Apr 14, 2017
  5. Oct 6, 2015 #4
  6. Oct 6, 2015 #5
    One thing though, is it valid to have a "varying" power converters? Where feedback loops would exist to change the Vout(and Iout) with respect to the change of the input.
    Acting similar to the behavior of constant voltage/current sources.
     
  7. Oct 6, 2015 #6

    berkeman

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    I'm not sure I understand the question. Normally you will use voltage feedback from the output in order to adjust the PWM circuit to maintain the output voltage at the desired value, independent of the input voltage.

    Alternately (as is used in some LED driver circuits), you can use feedback from the output current to control the PWM circuit to maintain a constant output current at some set value, independent of the input voltage.

    Does that help?
     
  8. Oct 6, 2015 #7
    Yes it does thanks, I confused myself.
    It should be a change input(V or I) to maintain a constant (V or I) depending on the setup.
     
  9. Oct 6, 2015 #8

    berkeman

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    BTW, one way to use a DC-DC buck converter as a constant current source is to have a low-side sensing resistor (small value) to give the converter a small voltage that represents the current flowing through the load. The DC-DC converter uses PWM to maintain that constant average current value.

    Like this Maxim circuit for driving LEDs with a constant current:

    https://www.maximintegrated.com/en/images/appnotes/3668/3668Fig01.gif
    3668Fig01.gif
     
  10. Oct 6, 2015 #9

    meBigGuy

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    Assuming a 100% efficient DC to DC converter, the power in ALWAYS equals the power out. Exactly. If you reduce the load, the converter draws less. If you reduce the input voltage, the converter draws more current (assuming constant output voltage and load).
     
  11. Oct 7, 2015 #10

    Baluncore

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    One simple way to view a buck converter is as an LC low-pass filter. The inductor input is switched rapidly between zero and the input voltage with a duty cycle that determines the output voltage. Since there are two distinct phases in each cycle, you can apply KL to only one phase at the time.

    Fundamentally there are two independent circuits in a buck converter. The first circuit is from the supply, through the switch, the inductor, to the load, then back to the supply via the ground return. The second is from the ground, through the diode, inductor and load.
    The polarity of the voltage across the inductor is reversed during each phase of the cycle. Since VL = L * di/dt, the inductor current alternatively rises and falls during the two phases of each cycle. di/dt = VL / L.
     
  12. Oct 10, 2015 #11
    I'm struggling with one aspect related to this topic, applying Ohm's law with the respect to the output. Let me give a simple example:

    Assume a constant DC power supply, that is 30W connected to a boost-converter to output higher voltage(with lower current) to the load, since Pin = Pout(assume 100% efficiency). The wire's resistance is 0.5ohms(total), the load's resistance is 5 ohms(so net resistance is 5.5), I think that the applied voltage from the PS would be 12.84V and the current is 2.33A. Now the input range of this boost-converter is 5-20VDC,0.5-5A(example numbers,their all made up), and I set the output voltage to 30V, in my head I know it has to be 1A current since P = 30W, but if I apply ohms law... I always get higher current than 1A which can't be true, so what am I doing wrong here?
     
  13. Oct 10, 2015 #12

    berkeman

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    I'm not following your example at all. First you set the output voltage at 12.84V, then to 30V. What do you set your boosted output voltage to? What is your load resistance? That gives you your output power. What is the input voltage set to? That determines the input current drawn to provide the output power.
     
  14. Oct 10, 2015 #13
    That would be the input from 30W PS, since resistance is 5.5ohms, that 12.84VDC will be inputted to the boost-converter.

    Output would be set to 30V(from 12.84 input), load resistance is 5ohms(with 0.5ohms from the wire so I totaled the resistance to 5.5ohms).
    Input voltage should be set to 12.84V, I assumed the input current would be 2.33A(not sure though).
     
  15. Oct 10, 2015 #14

    berkeman

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    Now you've got 5.5 Ohms at both the input and output?

    Can you just draw a sketch? The input current is determined by the output power and the input voltage.
     
  16. Oct 10, 2015 #15

    Baluncore

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    You have an input power of 30W from your supply, then the output power from a 100% efficient converter will be 30W.
    If the output load is 5.5 ohm total, then Vo/Io = 5.5R and Vo*Io = 30W. Then Io = 2.3355 amp and Vo = 12.845 volt.

    You are assuming too many fixed parameters.
    You cannot change the output voltage of the 30W power supply to 30 volt without increasing the load resistance from 5.5 ohm to 30 ohm. That is because that would require more than the 30 watt power available at the input.

    You should ignore power as an input parameter in your computational games.
    Energy is conserved in a 100% efficient converter. Power is the rate of flow of energy.
    Power is the only “computational bridge” between input and output.

    Follow this computational process.
    Specify the input voltage and the output voltage. Vi, Vo.
    Specify the output load resistance. Ro.
    Compute the output current. Io = Vo / Ro.
    Compute the output power. Wo = Io * Vo.
    For a 100% efficient converter. Wi = Wo
    Compute the input current. Ii = Wi / Vi.

    Or just note that since Wi = Wo then Vi*Ii = Vo*Io and Ro = Vo / Io.
     
  17. Oct 10, 2015 #16
    @berkeman , sorry for the confusion, but the load is 5ohm at output, the 0.5 is the wire's resistance.

    @Baluncore you made me realize my flaw here, thank you, but I'm stuck on a few things(bear with me):
    For the computational process what is the resistance prior to the converter and after? It should be 30ohms all around? Or 0.5ohms(for wires) before for input calculations, and 30ohms for output calculations?

    Ri = 0.5ohms(wires) ##\therefore ## Ro = 29.5ohms(load) + 0.5ohms(wires).
     
  18. Oct 10, 2015 #17

    meBigGuy

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    Where do you come up with this stuff? You say 30W power source, but do you even understand what that means?

    First off, the concept of a dynamic 30W power supply is not realistic. That in its self is a complex system guaranteed to confuse. It is neither constant voltage nor constant current. Is it an instantaneous 30W supply? Or does the supply average 30W. (for example does it limit the charge cycles of the DC to DC converter then go to infinite voltage when a switch opens). Supplying "30W" to a dynamically changing load makes no practical sense.

    When we speak of power-in to power-out in a dynamic converter, we are speaking average power averaged over the energy storage times of the converters energy storage components.

    So, first define your converters output characteristics. Is is constant voltage? Is it constant current? Is is current limited? Is the current limit "fold-back"?
    '
    Maybe you want to define some sort of non-linear input to output relationship? Well, define it and write the equations. But, playing with that can get tricky when you start talking dynamic systems with response times, feedback, damping factor, etc. That's a whole new subject.

    Then define your load. Is it constant, or changing? If it is changing, write the equations.

    Now, if you want to apply some weird power source to all that, then again, define the power source mathematically.

    Start with a constant voltage supply, define a converter, define a load, then look at what is happening. Then change 1 thing at a time and look at the effects back at the power source.
     
  19. Oct 10, 2015 #18
    lol, well I was in the process of doing that... and I think I've defined a lot about the circuit from the previous posts relative to the example(hopefully)?
    But I'll give that approach a go, it's perfect way of analyzing the circuit(and other things).

    I need a clarification on point #16 though because my calculations are based off that.
    I have a constant voltage DC power supply(forgot to mention "voltage" on post #11), that's supplying 30W.
    Ri (which is just the wire connecting the PS to other components) = 0.5ohms
    Vi = 3.8V
    Ii = 7.75A
    Pi = Vi x Ii = 30W

    That is the input to the boost-converter that is meant to output 30V to the 29.5ohm load:
    Vo = 30V
    Iout = 1A
    Ro = 30ohm( 29.5ohm + 0.5ohm)

    Pout = 30W

    Pi= Po ##\checkmark##

    Am I right?
     
  20. Oct 10, 2015 #19

    Baluncore

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    No. Your 0.5 ohm must be modelled on the output side with the load if you add it to the load. Wire resistance on the input side will lower the supply voltage to the converter. It is unimportant to the computations.

    A converter is the DC equivalent of an AC transformer. It transforms the V/I ratio from one side to the other.
    Do not cross the converter with anything other than the power. Treat input and output as totally separate circuits.
     
  21. Oct 10, 2015 #20

    Baluncore

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    By introducing a poorly specified series resistance, one that can jump around the circuit and across the transformer without being transformed, you are making it both unreal and more complex than it needs to be.

    The output load is 30 ohms. Now forget the 0.5 ohm wire. All wires are now perfect conductors.
     
  22. Oct 10, 2015 #21
    That 0.5 ohm of the wire is what made me determine the input current & voltage, If I neglect it, I don't know what the input values would be from the PS...
     
  23. Oct 10, 2015 #22

    Baluncore

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    The output voltage is regulated to 30 volt. The output load is fixed at 30 ohm.
    The output current must be 1 amp. The output power is then 30 watt.

    For converter efficiency = 100%, the input power must be 30 watt / 100% = 30 watt.
    The power supply has a regulated output voltage of Vs =3.8 volt.
    So the input current must be 30 watt / 3.8 volt = 7.895 amp.

    The load appears to be 30 ohms when viewed from the converter.
    But when seen from the power supply, the converter transforms the load to look like a different value resistance.
    That converted load resistance then appears to be 3.8 volt / 7.895 amp = 0.4813 ohms.
    So the converter is transforming the load of 30 ohms to look like 0.4813 ohms to the lower voltage power supply.


    Now, the idea that there might be Rs = 0.5 ohm of supply output resistance before the converter is clearly impossible.
    3.8 volt / 0.5 ohm = 7.6 amp. There is clearly insufficient voltage to supply the 7.895 amp current needed by the converter.

    The maximum Rs that could function would be when series resistance = apparent converter and load resistance.
    In that situation the input voltage to the converter will be reduced to half of the 3.8 volt = 1.9 volt.

    Because converter input voltage is halved, the converter input current would have to rise to 30 watt / 1.9 volt = 15.79 amp.
    1.9 volt / 15.79 amp = 0.1203 ohm. Which is the maximum possible supply series resistance for the circuit to still function.

    Overall efficiency would then have fallen to 50% simply due to resistance of the supply and cable to the converter.
    The total power supplied would then need to be 60 watt, 30W to the supply series resistance and 30 watt to the load.

    That is why you should use short thick cable between the power supply and the boost converter.
     
  24. Oct 24, 2015 #23
    I had something that I'm not sure about at all, here is a diagram of a buck-converter(a simple one):
    2V8EK4S.png
    The out voltage(##V_o##) is it in series with the load? Any component from point a-b are considered in series with the output current and voltage or parallel?
    I find it parallel, because of the diode and the capacitor are parallel, but not sure...
     
  25. Oct 24, 2015 #24

    meBigGuy

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    That question tells me you have no concept of voltage, current, and circuits.
     
  26. Oct 24, 2015 #25

    Baluncore

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    The voltage appears across the load. Vo is measured at (a) relative to (b).
    The current flows through the load.

    Series and parallel refer to the connection relationship between components.
     
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