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Pulse Skipping - Calculating Frequency Limits

  1. Dec 7, 2011 #1
    I'm trying to design my first switching regulator circuit. I require three voltage rails, 5V, 3.3V and 1.8V. But to get started, I'm just focusing on 1.8V. I've decided to use the LT3507 as it seems to be commonly available here. It also has 3 buck regulator outputs, so it's a one-chip solution for me.

    http://www.linear.com/docs/Datasheet/3507fa.pdf [Broken]

    However, as I'm new to this, I struggling a bit with calculating the upper frequency limits fmax1 and fmax2. Please see page 10 for the formulas used to derive fmax1 and fmax2. For them, I need to determine Vin(ps), Vout (=1.8V), Vf(=0.4V), Vsw(=0.3V) and I need to know ton(min) which is 130ns, as given in the datasheet.

    OK, to determine Vin(ps) seems easy enough. I just need DC(min) which is

    [itex]t_{ON(MIN)} \times f_{sw} = 130ns \times 450kHz = 0.0585[/itex]

    I assume fsw is the switching frequency I'll work at. Let's take 450kHz as an example. If that's the case, then DC(min) is just 0.0585. The expression for Vin(ps) is:

    [itex]\frac{V_{out} + Vf}{DC_{min}} - V_{f} + V_{SW} = \frac{1.8 + 0.4}{0.0585} - 0.4 + 0.3 = 37.506V[/itex]

    If I understand Pulse-Skipping correctly, if I exceed the above voltage then pulse-skipping will kick in to make sure the output voltage doesn't exceed 1.8V.

    To compute fmax1, I subs. the above into:

    [itex]\frac{V_{out} + V_{f}}{V_{IN(PS)} - V_{SW} + V_{F}} \times \frac{1}{t_{on(min)}}= \frac{1.8 + 0.4}{37.5 - 0.3 + 0.4} \times \frac{1}{130ns} = 450kHz[/itex]

    No matter what I do, my fmax1 always computes to be equal to fsw that I chose above. Why is this? I've used Excel and I've done it on paper and I get the same result.

    I feel like I'm missing something basic and I would really appreciate some help with this. I don't really have a co-worker that I could ask and so the web is the only place I can ask. Googling this doesn't seem to return much.

    If the above result is correct, then what does it imply? The datasheet states

    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Dec 8, 2011 #2


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    Staff: Mentor

    Regulating 37V down to 1.8V in one step is a bit much. That's why you are getting such a small duty cycle (5%).

    What is the input voltage to your circuit? What will the current drain be for each of the rails? It would probably be better to do a single buck regulator for the 5V rail, and then use a linear LDO for the 3.3V. Depending on the current required for the 1.8V, you could either do a linear off of 5V or 3.3V, or do a buck regulator off of the 5V rail.
  4. Dec 11, 2011 #3
    Thanks! Sorry about the late reply - I didn't notice anyone replied.

    The input voltage is just 7V. Actually, I can make it anything I want as it's in my hands, so I'm going with 7V. The 5V and 3.3V rail have the highest current requirement (0.5A and 1A respectively). My initial estimates suggest this would make linear reg. too hot without a heat-sink, but with one it may be manageable.

    The other solution I came up with is that I ought to instead use 3 buck regulators instead of a single IC. With enough speed, the inductor and caps ought to be small enough.
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