Adding fully controlled full wave rectifier to Buck circuit

Discussion in 'Homework Help' started by PsySc0rpi0n, Feb 5, 2018.

  1. MrAl

    Distinguished Member

    Jun 17, 2014
    4,945
    1,062

    Hi,

    Just to recap a little, i can see how we might place a largish inductor in series with the input to the buck. That would make the input filter of the buck a "choke input filter" and that is an option used in high power systems because the peak currents can be outrageous. I could even see a swinging choke there too, but a large one would be good for study.

    So if we just take an SCR bridge like you have, and a standard buck circuit with large input caps withj low ESR, and connect the two grounds together and connect a large inductor from the output of the SCR bridge to the input of the buck, we have s system that meets the requirements of your problem statement and it is also practical.

    The addition of the inductor actually makes the problem a little easier because the inductor makes it easier to calculate the reaction switch times of the SCR's (or diodes) which is very very hard to calculate with a capacitor input filter alone as load. The reaction swtich times are the times when the SCR's or diodes switch off, but can also be the times when they switch on if the gate pulse does not turn them on right away.

    And yes, there are many times a difference between school work problems and real life problems until you get into higher technical education. The school problems are very good though, and as we see this one can be turned into a practical problem quite easily.
     
  2. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    1,369
    9
    Hello again.

    I'm still trying to understand some equations regarding Buck converter.

    I'm reading the article here from AAC hosted here.

    I'm having some issues understanding how Imax and Imin equations are evaluated right before Figure 7 - Capacitor Current!

    [Equation 1] and [Equation 3] are used to evaluate [Equation 4] and [Equation 5]. I don't understand how did they combined [Equation 1] and [Equation 3] to obtain [Equation 4] and [Equation 5].


    I also know from my notes book that:
    Imax = Iout + (ΔiL/2)
    Imin = Iout - (ΔiL/2)

    I need help here because I can't understand where both equations (the ones from my text book and the ones from AAC article for Imin and Imax) comes!
     
    Last edited: Feb 13, 2018
  3. MrAl

    Distinguished Member

    Jun 17, 2014
    4,945
    1,062
    Hi,

    I can offer a quick short explanation but we can always come back and go right into the equations to derive this stuff too, no problem.

    The short answer is that the current waveshape is assumed to be triangular because of the usual size of the inductor and the assumption that the output voltage does not change much. It's really a curve, but because of the short time periods it looks like a straight line up and straight line down, which is triangular. Also, the average DC is considered to be a constant.
    Now for 50 percent duty cycle, we see as much above the average DC output as below, so we will see the peak at 1/2 of the total delta and the valley at minus 1/2 of the total delta, so with a peak to peak of 1 amp we'd see 0.5 amp above the average DC and 0.5 amp below.

    So really it's all amount thinking in terms of short time periods where the time period is much less than the time for one cycle if the LC was allowed to run free. Since it's almost always like this, we think in terms of straight lines and constant DC so we see:
    i=C*dv/dt
    v=L*di/dt

    happening very clearly as straight line responses with small dt.
    In fact, with constant DC output dv is zero, so:
    iC=C*0/dt=0

    Since vL is not constant, we have:
    vL=L*di/dt

    and rearranging:
    di/dt=vL/L

    and we know vL is just Vin-Vout, we have:
    di/dt=(Vin-Vout)/L

    and when we assume a straight line, we get something like:
    di(t)=a*t=(Vin-Vout)/L

    because Vin-Vout is constant and L is constant we just have a ramp, a*t.
    When the switch goes off, we assume Vin=0 so we have:
    di(t)=(0-Vout)/L

    which is again a ramp, this time negative.

    So we get a ramp up, and ramp down.

    See if that helps, if not we'll go into more detail.
     
  4. ebp

    Well-Known Member

    Feb 8, 2018
    1,229
    396
    There is all sorts of basic info on buck converters available at TI, most of it from Unitrode, which TI acquired several years ago. Some of the good stuff was written by Lloyd Dixon.
    When I look at the linked article, I see a bunch of equations with all sorts of embedded formatting codes, so they are unreadable. When I read the words, I'm left with the impression of an article that explains the math but doesn't help a lot in terms of basic circuit operation.

    I always say all you need to know to understand switchers is di/dt = V/L (d's should be Greek deltas)
    - the differential of current with respect to time is equal to the voltage across the inductor divided by inductance
    V in volts, i in amperes, L in henries, t in seconds
    Well, that and "you can't instantaneously change the voltage on a capacitor" - which is described by a similar equation.

    The two equations
    Imax = Iout + (ΔiL/2)
    Imin = Iout - (ΔiL/2)

    simply describe the maximum and minimum inductor currents for a single switching cycle
    ΔiL is the peak-to-peak "ripple current" in the inductor.

    When the switch in the buck turns on, the current in the inductor begins to rise linearly (ignoring some details that are generally safe to ignore) according to our equation di = V*dt / L
    In this case V is the input voltage minus the output voltage minus any drop across the switch. It is usually quite acceptable to assume that both Vin and Vout are constant for a single switching cycle. In reality, all voltages and the inductance may change slightly, but not enough to make it worth fussing with in a practical circuit. It is often the case that the "equivalent series resistance" in capacitors is the biggest player in change in voltage during a single switching cycle.

    When the switch opens, the current will begin to fall linearly according to the same equation, but V is different. At one end of the inductor we have the output voltage, same as before, and which we'll again assume remains constant, held by the output capacitor. The other end is now at a small voltage above ground - the forward voltage drop of the "free-wheeling" diode, which is typically between about half a volt and 1 V, depending on the diode and the current. Don't forget the current in the inductor is still flowing in the same direction. Energy stored in the inductor is being transferred to the load and output capacitors.

    Now it gets a little more interesting. Two things can happen with the current. It may drop all the way to zero or it may still be at some positive value when the switch turns on the next time. If it doesn't drop to zero, the converter is said to be operating in "continuous mode" which really means "continuous inductor current mode." If the current drops to zero, even for a very short time, the converter is said to be operating in "discontinuous mode."

    When the switch turns on again, the inductor current will of course be at exactly the same level it was just before the switch turned on - some positive level in continuous mode, or zero in discontinuous mode. It will again ramp up and the cycle repeats. The current from the input will exactly match the current the current flowing in the inductor as long as the switch is on, and be zero when the switch is off.

    In continuous mode operating in equilibrium (input & output voltages constant, load constant), the average inductor current must be exactly equal to the output current. The peak to peak ripple current is centred on the average output current, which is what those two equations are saying. If you do the calculations, ignoring the voltage drop across the switch and that across the diode, you will find that the largest peak to peak inductor ripple current occurs at 50% duty cycle where Vout = Vin/2

    The basic buck converter equation
    output voltage = input voltage * duty cycle
    applies only to the continuous current case. It is more complex for the discontinuous case because each switching cycle essentially delivers a discrete amount of energy.

    Most buck converters are designed so that the peak ripple current is about 30% of the full rated output current, but there may be reasons to chose a different value. If the peak to peak ripple is 30% of rated output, then the discontinuous-continuous boundary is at 15% of full rated output power. The very first thing you do when designing a buck converter inductor is decide what peak to peak ripple current you want. That will lead you to the required inductance. You then move on to core selection and winding calculations.

    Brief summary:
    - during switch ON time current to the output (and filter capacitors) is delivered from the input supply through the switch and the inductor, and energy is also stored in the inductor
    - during switch OFF time, no current flows from the input, but all or some of the energy stored in the inductor is transferred to the load and the filter capacitors

    I don't like to think of the inductor and capacitors as a two pole low pass filter, though they are. I don't think it helps in understanding the basic circuit. I think thinking of the circuit in terms of two energy storage devices is more helpful. Considering the LC as a two pole filter is important in dealing with error amplifier frequency compensation.

    Not sure any of that helps you.
     
  5. ebp

    Well-Known Member

    Feb 8, 2018
    1,229
    396
    Ha. Yet again I repeated what someone else said because I didn't look before clicking post.
     
  6. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    1,369
    9
    @MrAl, I think I understood your explanation but still I can't see how [Equation 1] and [Equation 3] becomes [Equation 4] and [Equation 5] in terms of mathematical manipulations.
    I mean, did they made [Equation 1] equal to [Equation 3] and solved for one of the unknowns or what was done?

    Thanks

    If you can show me the steps used in that article to go from [Equation 1] and [Equation 3] to [Equation 4] and [Equation 5], I would appreciate!

    @ebp thank you for the practical explanation about the circuit working method. I think I understand the basics at least about how current flows, when and how! It's just the mathematical manipulation of equations to evaluate Imin and Imax generic equations.

    Thanks
    Psy
     
  7. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    1,369
    9
    No problem. I read it anyways!
     
  8. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    1,369
    9
    Ok, I think I got it...

    I just need confirmation that all steps are correct.

    Will post a screen with my math in a bit! I did a bit of "reverse engineering"! :)
    20180214_193715.jpg
     
    Last edited: Feb 14, 2018
  9. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    1,369
    9
    Well, anybody is going to check this, I guess...

    Anyway, I have one question yet!

    When we want to evaluate the resistance value for the border between CCM and DCM condition, we pick up the Imin equation. I don't want to ask if we could do it using the Imax equation because it's obvious result would be different, but why we evaluate this R using Imin?
     
  10. ebp

    Well-Known Member

    Feb 8, 2018
    1,229
    396
    Looks OK.

    The R you want to find is the load resistance at which the average output current is such that Imin is exactly zero.

    You can use either Imin or Imax. Right at the CCM-DCM boundary, Imin is zero and there is no "dwell time" during which the inductor current remains at zero (the current waveform would not go negative, if that were possible). So at the boundary, all of the following say exactly the same thing, and are just rearrangements of the equations

    Iout(average) = Imax/2 = ΔI/2 = Imin + ΔI/2 = Imax - ΔI/2
    These are all correct at the boundary, but the last two are also correct anywhere in the CCM region

    When you derive equations and have a choice of using different forms to describe the same thing in different ways, pick the way that you think shows what is happening in the most obvious way for the situation you are trying to describe with the equation.
    When you are are trying to describe the CCM-DCM boundary condition, you are trying to describe what happens when the minimum inductor current is zero, with no dwell time at zero. I think it makes more sense, then, to use an equation that uses Imin because Imin is always zero at the boundary. You don't really directly care about Imax, though you know exactly what it will be.
    If you were trying to come up with an equation for the peak magnetic flux in the inductor core, which is something you must consider for a real circuit, then you would most sensibly want an equation using Imax directly.

    Trying to come up with an equation that puts everything in one equation is an exercise in algebra, and in my opinion often does not help in understanding the circuit, which is the real objective in engineering. I think it is often easier to understand what is happening with a set of simple equations with very few terms instead of one equation that has everything. Often you want to know all of the results of the simple equations anyway. But some people like all-in-one equations.
     
  11. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    1,369
    9
    Ok, thanks. Sounds reasonable!

    I think I'm going to start a new thread for the Boost circuit and some questions also about the AAC article about it!
     
  12. MrAl

    Distinguished Member

    Jun 17, 2014
    4,945
    1,062
    Hello again,

    I didnt realize you were still after this. Yes, i think you got it now.

    It is easier to 'see' with a 50 percent duty cycle, but a triangle wave like this is always centered at the average, so that means a full excursion:
    di=(Vs-Vo)/L

    will be centered at:
    Iavg=(Vs-Vo)/(2*L)

    However, the writeup in that article could be just a tiny bit better, if it read that you need equations 1 THROUGH 3.
    I say this because Vo=Vs*D is equation 2 and that's not specified yet, only 1 and 3.

    So starting with di=D*T*(Vs-Vo)/(2*L)

    and replacing Vo with Vs*D (which is equation 2):
    (Vs*D*T)/L-(Vs*D^2*T)/L

    rearrange:
    (Vs*(1-D)*D*T)/(2*L)

    factor out Vs and D:
    di/2=((1-D)*T)/(2*L)

    we are left with the normalized upper excursion, which will be the same as the normalized lower excursion so we will see plus and minus Vs*D*di/2 from the mean:
    Imax=Iavg+Vs*D*di/2
    Imin=Iavg-Vs*D*di/2

    I think you got the same thing now.

    So the only thing really missing was that you should also use equation 2 with equations 1 and 3.
     
    PsySc0rpi0n likes this.
Loading...