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My DC/DC setup is going to power a GSM modem. The behavior of these modems is that they typically consume only a few tens of milliamps, but when transmitting they are drawing 2A for microseconds. On average these pulses can add up for an average of ~300mA.

Of course I'll need low-ESR caps and all, but can I choose smaller current rated inductor and Schottky diode if the average current is ~300mA? My circuit needs to be very small and the 3-4A inductors and diodes that are rated for the max power output of 2-2,5A are really big. I am curious whether you think I can get away with smaller components rated for like 500mA or 1A max.

I know it is not officially according to the specifications, my question is really practical and looking for practical answers.

toolic
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Szundi
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    If you know your current draw, duration, and duty cycle, and your inductor current limit you can calculate how much output capacitance you need to meet a certain voltage dip during transmit. So add that amount of capacitance and you can use a smaller inductor. – John D Jun 29 '22 at 21:23
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    Diodes in buck regulators is very 1980's. There are 2A synchronous buck chips in tiy packages. If the inductor is too big, you need a buck chip with higher frequency. – bobflux Jun 29 '22 at 21:35
  • Semiconductors yes, inductor no chance. – winny Jun 29 '22 at 21:36
  • Often, systems with embedded GSM modems incorporate a small lithium ion battery to help supply the peak currents. The peak currents needed during transmit are a major issue and are well known by people who have had to do what you are doing. The pain is real. As others have said, the inductor saturation current needs to be greater than the peak output current + current ripple. It is very doubtful that a capacitor can hold up the voltage long enough to support the transmit pulse. One thing that can help is going to a very high frequency DC-DC. this will allow a smaller inductor to be used. – user57037 Jun 30 '22 at 04:51
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    Szundi, you aren't writing any sense to me. You write "2 A" for "microseconds" (call this **x**) but that the "average is 300 mA". This is easy to set up as solving \$\frac{x\,2\:\text{A} +\left(1-x\right)\,30\:\text{mA}}{1\:\text{s}}\$ for **x**. (I used \$30\:\text{mA}\$ to replace your "few tens of milliamps.") For which I find \$x=137 \:\text{ms}\$. Which is a ***lot*** more than "microseconds" to me. What's up? The numbers you are throwing out don't jibe well. – jonk Jun 30 '22 at 04:58
  • @jonk of course the point is lots of pulses in a second, some microseconds each. average is what i see on my power supply, I hope this makes sense now. – Szundi Jun 30 '22 at 09:16
  • @JohnD thanks, this should have been an answer, makes sense! – Szundi Jun 30 '22 at 09:17
  • @Szundi: That sounds like **real** GSM (2G), which is a TDMA system with 8 timeslots. IIRC, those timeslots are significantly longer than microseconds. They have to be, the speed of light is only 300 meters per microsecond and GSM cells can be >30km. – MSalters Jun 30 '22 at 11:52
  • @Szundi Ah. So I needed an **N** in the equation, with **N** unknown. Got it. So to get **x** down into 'microseconds' I must at least have N > 200, something like that. (And maybe a lot more than that.) Would that be correct? – jonk Jun 30 '22 at 17:34

2 Answers2

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The inductor should be sized to handle the peak current loads without saturating.

Most parts will spec both saturation (usually at 20-30% drop in inductance) and DC/thermal limits (typ. 30 C rise). Pretty much any practical inductor that can support 2A without saturating should be fine for a 300 mA average, but always check the datasheet.

As current through an inductor increases, eventually the magnetic flux density becomes higher than what the surrounding material can support. So it doesn't, and the effective inductance drops. That is called saturation.

Diodes (esp. Schottky diodes intended for power supply applications) also typically specify both an average forward current and a pulse current. The average is based on DC thermal behavior, and the pulse current is based on what happens within the packaged die at small timescales.

For example, there might be a part rated at 500 mA DC, but it can support 2A peaks as long as the pulses are <1 ms and the duty cycle is such that the average current is within the 500 mA DC limit.

Not directly asked, but your regulator will basically transmit the output load directly to the input. So your input will also see pretty large transients whenever your modem does a TX burst. Adding capacitance to the output may not help much since the regulator will try to fill that capacitance as soon as charge leaves and the voltage begins to drop.

matt
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The inductor current limit is thermal only. RMS current is what matters, and averaged over the thermal time constant, so, some seconds. The regulator however needs nominal inductance, so saturation must be avoided.

The main thing is, there aren't many parts available, with a low current rating and high saturation rating, so you're not going to save much.

Similarly, the diode will have a thermal time constant of some fraction of a second, probably fine; but it needs a low enough Vf at that peak current for the regulator not to malfunction.

The comments suggesting a larger (including synchronous) regulator, or extra capacitance, may prove worthwhile. Note that the regulator's compensation depends on load capacitance, so this must be done cautiously.

Tim Williams
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