I am a student doing a project and I have an interesting dilemma and I was wondering if anyone had experience solving this kind of issue.

I am designing the input protection / line filtering for a power supply module with aerospace applications.

I want the power supply to remain functional despite a 50ms power interrupt (input power goes to 0V for 50ms), so I am trying to determine a topology that can achieve this without massive capacitor banks.

Here are a few specs. for my system: Input Voltage: 18-32V Current Draw: 200mA Temp. : -40C to 125C ambient

So far I have looked at the following options:

  1. Capacitor bank on the output of the filter circuitry
  2. Capacitor bank + boost driver on the output of the filter circuitry
  3. Integrated Chip Solutions

Options 1 and 2 are quite expensive and require lots of space, so I would like to determine if there are good integrated chip solutions. I found the following in a design magazine but it was intended for a 12V input power supply.

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Does anyone have experience with this type of problem? What was your solution? Most of the IC solutions I have found are intended for lower voltage applications, I need an IC output of 24V (ideal) or 18V (minimum).

Any advice / ideas / guidance would be greatly appreciated! Not necessarily looking for answers; general wise advice would be appreciated.

  • 2
    \$\begingroup\$ You need 0.24 J of energy to span a gap of 50 ms with a 4.8W load -- not much at all, really. A capacitor of just 2000 uF charged to 24 V would only droop by 5 V (down to 19 V) in that amount of time. \$\endgroup\$
    – Dave Tweed
    Jun 8, 2017 at 19:22
  • \$\begingroup\$ LTC3350 gives similar functionality to the LTC3643 in your question, but its good for up to 35V. \$\endgroup\$
    – brhans
    Jun 8, 2017 at 19:22
  • 1
    \$\begingroup\$ There is no topology choice to get this for free or ease the problem. You need to store the energy somewhere and supply it to the output. \$\endgroup\$
    – winny
    Jun 8, 2017 at 19:23
  • \$\begingroup\$ Another solution would be a rechargable battery, but still the capacitor is: cheaper, safer, simpler, has longer life than a battery. \$\endgroup\$ Jun 8, 2017 at 19:41
  • \$\begingroup\$ What are your output requirement? 24 V at what load? \$\endgroup\$ Jun 8, 2017 at 20:25

3 Answers 3


Your input power can be as high as (32 V)(200 mA) = 6.4 W. You want to ride out a 50 ms outage. One way or another, something has to come up with (6.4 W)(50 ms) = 320 mJ. That's basic physics.

So now you look around at what can store 320 mJ and be able to cough it up again over 50 ms. The only realistic choices are capacitors or batteries.

Your spec of 18-32 V is a bit unclear, but let's say you arrange to keep a cap charged to 32 V when input power is present. When input power goes away, the cap can't go below 18 V (drop by more than 14 V) within 50 ms. (200 mA)(50 ms)/(14 V) = 720 µF.

That's the simple system, but uses only 68% of the cap's capacity due to it still containing some energy when charged to 18 V. Draining a cap more fully is possible, but requires more sophisticated electronics.


You need 5x the energy storage that a typical device operating from the mains (at 50Hz) needs. Not difficult at all. Easier because the mains device causes a ripple current in the capacitor and your failure will be infrequent so there will be less heating of the capacitor to contend with.

Since your desired output voltage lies within your input range you likely need something like a buck-boost regulator or an isolated (eg. transformer coupled) supply. At 200mA/24VDC out (4.8W) a flyback converter with a coupled transformer-like inductor could be used.

Since you are going to design it, you'll not be using off the shelf modules, but consider a 24V output DC-DC converter such as a DJ06S2424A. It will accept 9V~36V in and supply 24V at up to 250mA. So a Schottky diode (or a regular diode, doesn't matter much at your input voltages) and input capacitor plus the module will do the trick. The capacitor will be charged to a minimum of (say) 17V and can be allowed to discharge to 9V before the DC-DC fails. Let's assume average 75% efficiency during the discharge- check this assumption carefully (usually efficiency is worst at minimum input voltage and efficiency is often quoted under the best possible conditions).

So we need to supply 0.05s * 0.2A * 24V = 0.24J at the output, or about 1/3 J at the input to the DC-DC.

The energy in the capacitor is C*V^2/2 so the delta in energy is:

\$\Delta E = (C/2) (V_i^2-V_f^2)\$ where Vi = 17, Vf = 9, so C = 3.2mF.

So you could meet the spec with an off-the-shelf module, a diode and a 3300uF/35V capacitor. Total cost around $12-13 in 100 quantity.

  • \$\begingroup\$ Why do you need the diode? \$\endgroup\$ Jun 8, 2017 at 21:22
  • \$\begingroup\$ @AliChen Good question. The diode isolates the capacitor from the input voltage so if "interruption" means pulled down to 0V by other loads, the cap will not be affected. If "interruption" always means "open the connection" then the diode is not required. \$\endgroup\$ Jun 8, 2017 at 23:21
  • \$\begingroup\$ Good answer, thanks. :-) I didn't explore the meaning of "interruption" to that depth. I didn't come to me that in aerospace a dog can chew on supply wires :-). The whole "power goes down to 0 for 50 ms" still remains a mystery then. I could understand cranking a car engine, but aerospace... \$\endgroup\$ Jun 9, 2017 at 0:00
  • \$\begingroup\$ @AliChen It could be something like a surge from a power supply starting up (diode required) or a slip ring opening up (no diode). Or something else altogether. \$\endgroup\$ Jun 9, 2017 at 1:08

As I understand, you need to make 24 V output from a fairly variable input, from 32 V down to 18 V, and make it tolerant to 50 ms interruption. The normal solution is to design a SEPIC-type converter for a range from 32V to well below the input minimum (18 V). The minimum value of input capacitor must be selected to sustain the voltage from 18 V down to SEPIC lowest acceptable input.

Something like LT8494 will do the job. In this case the Vmin = 3 V, so the cap must keep voltage from 18 V to 3 V for the converter to remain fully functional. You didn't specify the output current, so it is difficult to come up with exact value of input cap, but I think a small 1000 uF capacitor will do the job.


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