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question about switch mode supply

Discussion in 'Electronic Basics' started by Louie, Aug 12, 2003.

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  1. Louie

    Louie Guest

    can someone explain to me something about how switch mode power supplies
    work? i read an article that states:

    "The duty cycle of chopped DC will effect the AC voltage level generated
    on the transformer's secondary. A long duty cycle means a larger output
    voltage (for heavy loads) and a short duty cycle means lower output
    voltage (for light loads) (for heavy loads) and a short duty cycle means
    lower output voltage (for light loads)"

    what i don't get is the "long duty cycle means a larger output voltage"
    part. i figure that the secondary voltage is determined by the primary
    voltage and the ratio of windings. how does this work?

    thanx, Louie
     
  2. John G

    John G Guest

    I would say the article was written by someone who does not know much.
    If it is anys smart the output voltage will be controlled by the circuitry
    which controls the duty cycle of the transformer so that the rectified
    filtered voltage is constant. And as you think this means wider pulses not
    higher pulses from the transformer.
     
  3. This description is so simplified that it no longer makes sense.
    There are lots of different topologies with different waveforms and
    concepts so I am guessing about which one this is trying to describe.

    I think the above description is about a version that produces pulses
    in alternating directions from some DC source into a transformer, and
    then rectifies the pulses out of the secondary, before passing those
    pulses through some low pass filter. Duty cycle refers to the percent
    on time of the pulses. Since the output, here, is rectified AC, when
    the primary sees a 50% of cycle pulse in each direction, the rectified
    output is on 100% of the time, and the DC out is at maximum. If the
    primary pulses are shorter than 50% of a cycle, and the rest of the
    half cycles consist of zero volts, the rectified output will be a
    pulse train (the pulses being the same peak voltage as the first case)
    with some zero volt sections between them. The low pass filter after
    the rectifier will average this pulse waveform as a lower value of DC
    than the full duty cycle version.

    The aim of most switching supplies is to keep the DC output voltage
    constant, in spite of both source voltage variations (height of pulses
    varying) and load current variations (changes in the resistive voltage
    drops of the switches, transformer windings, rectifiers and filter
    caused by changes in current).

    Here is a good general introduction to switching regulator concepts
    and circuits:

    http://www.national.com/appinfo/power/files/f5.pdf

    found on this list of papers:

    http://www.national.com/appinfo/power/0,1768,95,00.html
     
  4. Bill Bowden

    Bill Bowden Guest

    I think it works more like an inductor than a transformer.
    Say the primary inductance is 1mH and the applied voltage
    is 12 and the pulse width (time on for applied voltage) is
    100uS. This means the current will ramp from 0 to about 1 amp
    in 100uS and the average current is 1/2 amp. If the duty cycle
    is 50%, the average current will be 1/4 amp and the average
    power is 12*.25 or 3 watts. Assuming 100% efficiency, and
    a load resistance of 100 ohms, the output should be around
    17 volts DC. A longer duty cycle will increase input current
    and output voltage on the load.

    But this doesn't account for the transformer turns ratio.
    I can't quite see how the turns ratio effects the operation except
    maybe to reduce losses. If you put 3 watts in, and 3 watts goes
    into a 100 load, the voltage must be 17, regardless of turns ratio.

    What am I missing?

    -Bill
     
  5. Louie

    Louie Guest

    thanks, John, that's what i thought.
     
  6. Louie

    Louie Guest

    yeah, i sorta understand now how they work. but i have to test this PS
    unit. how can i trick it into thinking its in circuit so i can measure
    its output?

    louie
     
  7. Louie

    Louie Guest

    thanks, John, that helps alot. it wasn't easy, but i read through the
    reference you cited, so now i have some understanding of how they work.
    now what i have to do is troubleshoot this PS that came out of a
    projection TV.
    if i could figure out how to load the output so it thinks its in
    circuit, then i could just measure the voltages and see if its working
    or not.
    problem is, i have no documentation telling me what the voltages are
    supposed to be.
    however, there are several zeners in the control circuit, which is on
    the output side of the board, seperated from the input side by the
    multi-tap trans and the opto coupler. maybe the zeners can tell me what
    the levels are supposed to be.
    or maybe i just don't have the slightest clue what i'm doing.
    and, yes, i do have a resistor to drain the caps.

    Louie
     
  8. Louie

    Louie Guest

    i believe the reason for the high frequency of switching is so that you
    can have a smaller transformer. with the unit i'm working on the caps
    and switching regulator are almost as big as the transformer.

    as for the ratio, i think you can be pretty loose with it as long as the
    secondary has comparable inductance with the primary. as far as i can
    tell, that is.

    louie
     
  9. uuuuu

    uuuuu Guest

    Regarding the first link, why does the current in Fig 30 (the current
    through the L) rise and fall linearly ?

    Anoher question:
    There is a law that current change through an inductor cannot happen
    instantly, so what happen in the switch is opened in one go ?
     
  10. The answer is the same for both questions. The rate of change of
    current through an inductance is proportional to the voltage across
    it. This property defines inductance.

    In figure 30, the ramp up rate of change is proportional to the input
    voltage minus the output voltage, because this is the voltage across
    the inductor when the switch is on. The ramp down slope is
    proportional to - the output voltage and a diode forward drop, because
    the inductor can produce no more reverse voltage than this without
    forming a conduction loop to the output through the diode, which then
    carries the inductor current as it winds down its stored energy.
    Since the voltage in both directions changes little during that part
    of the cycle, the current rate of change is similarly constant for
    those two ramps.

    There is a discontinuity in the inductor voltage as the switch opens,
    and the inductor voltage changes polarity. This voltage rate of
    change is limited only by the speed of the switch and the stray
    capacitance of the inductance and the other circuit components tied to
    it. But the magnitude of total voltage change is limited by the turn
    on of the diode.
     
  11. Sure. For example, if the inductor has 7 volts switched into it, and
    5 volts out, there is only 2 volts causing current ot increase when
    the switch is on, but when the switch turns off, there is -(5 +.6) or
    so across it as the inductor produces enough induced voltage to
    forward bias the diode.
    Yes. The induced voltage is limited only by the speed that the switch
    turns off, and the capacitive load across the inductance and the
    parallel resistance across the inductance (that represents the current
    induced in the core). If the inductor and switch were perfect and the
    stray capacitance approached zero, the induced voltage would approach
    infinity (but would last for a time approaching zero) and the inductor
    current would quench to zero almost instantaneously.
    This equation just describes the averaging effect of a low pass filter
    with no losses. The average voltage over the time of one pulse cycle
    (Tp) of a pulse that spends Ton time at Vpk, and (Tp-Ton) at zero
    volts is:

    Vout = ((Ton*Vpk + 0*(Tp-Ton))/Tp = (Ton*Vpk)/Tp

    Of course, it a real circuit, the off state voltage is one diode drop
    negative, and the inductor has some resistive voltage drop, so this
    formula overestimates the output voltage.
    The voltage across L is assumed to have one of two fixed values during
    a cycle. when the switch is on, it is assumed to be the unregulated
    voltage supply, and when the switch is off, it is assumed to be the
    negative of the output voltage plus a diode drop.
     
  12. uuuuu

    uuuuu Guest

    OK, amlost done :) now how about the current,
    I meant to ask why the current is rising and falling in a triangle
    wavefrom (figure 30). Is it derived from integrating di/dt in the
    V=Ldi/dt formula ?
    Thanks!!
     
  13. Precisely. The voltage and the rate of change of current (the slope
    or di/dt or the current waveform) are proportional to each other. The
    proportionality constant that relates them is the inductance, L.
     
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