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I have a source of 24 V and I cannot draw more than 2.4 W from the same. This means I cannot draw a current more than 100 mA from this supply. Alongside the LM5001 boost converter, it also has some other circuits to supply. Therefore I would want to charge the output capacitor slowly. This means I want my input current to be limited to at around 40 mA.

From the design, you can see that I have got big output capacitance of 2.5 mF. Because of which, I have high inrush current while charging these capacitor.

What methods can I try to slow down the charging time so that I can limit input current from the 24V power supply?

P.S.: I cannot reduce the output capacitance as I would want to store the energy and keep in that for dedicated operation.

enter image description here

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    \$\begingroup\$ My suggestion would be to add current limiting. The LM5001 doesn't have build-in support for this. I would therefore search for a boost converter IC which does support current limiting. Current limiting can be added by manipulating the voltage at the FB pin using an opamp and diode (so that the opamp can only pull up the FB pin) and a current sense resistor. However, this is complicated and you would need some experience to get that working. \$\endgroup\$ May 19, 2021 at 10:24
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    \$\begingroup\$ I also have doubts wether such a large capacitor directly at the output of the boost converter will not disturb the operation of the boost converter. The capacitor(s) at the output are part of the regulation loop and need to have a certain ESR. Your circuit might become unstable. I would consider a current limiting circuit between boost converter and 2.5 mF capacitor. \$\endgroup\$ May 19, 2021 at 10:29
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    \$\begingroup\$ You can try taking a look at "eFuses" and/or "hot-swap controllers". The main difference is the eFuse has the current limiting MOSFET internal to the IC, and the controller uses an external one for when you need something more efficient. ti.com/power-management/power-switches/… \$\endgroup\$
    – Ste Kulov
    May 20, 2021 at 5:29

4 Answers 4

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So, you need a boost switching current source. Most chips of this style, like LED drivers, will regulate the output current. But you need regulation on the input current instead.

The boost chip you used will not control its input current, so even after your inrush problem is solved, it will still try to charge the cap at its maximum current, which will crash your power supply.

So, to solve both problems at once, why not use a SEPIC hysteretic converter?

enter image description here

I didn't search for parts or spice models, so I used function sources, but that should work as an example.

L1 and the FET are the usual input stage of a boost converter. However C2 and L2 make it a SEPIC. There is no inrush current because the input is connected to the output through a small capacitor only. This also means shorting the output will not draw high current from the input either. These features make this circuit a good match for your requirements, I think. I won't explain how the SEPIC works, there's plenty of online docs for that.

To control it, the FET has to be switched. This can be done in many ways, for example using fixed frequency PWM, etc. I'll use hysteretic because it is super simple and there are no compensation headaches.

Shunt resistor "Rsense" monitors current in L1, which is also the current drawn from the power supply. Function source E3 models a current sense amp, for example MAX4173 (if you don't like this one, pick one with wide bandwidth). This amplifies the voltage on the sense resistor and references it back to GND. Then E1 models a fast comparator, for example MCP6561. The resistors around it set the hysteresis.

So, here's how it works:

When the current in the inductor is below the set value, it turns the FET ON. When it is above the set value, it turns the FET OFF. Si inductor current zig-zags between the two values, making a sawtooth. Therefore, input current is, on average, constant. You can set the value with the voltage "Iset".

Frequency is determined by how much hysteresis you put in, the inductor value, and delay introduced by the circuits.

There should be a FET driver, and a suitable FET.

You should pick a comparator or a FET driver with an ENABLE input, to turn the circuit on or off. Especially to turn it off once the output capacitor is charged to the desired voltage.

For only 40mA average it's difficult to keep the frequency below 1MHz without ending up with a large inductor value, so if you prefer a small inductor, you could use a much higher current like 1 Amp, and PWM the thing at high enough frequency that the input current averages to the value you want.

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  • \$\begingroup\$ Your answer is really readable and presents the key ideas well. I think it would be educational for anyone not already comfortable with the arrangement. Nicely targeted. +1! I hope the OP takes heed because I think this is a far better approach than "stick a resistor here." (Though what's "better" is obviously up to the OP.) \$\endgroup\$
    – jonk
    May 19, 2021 at 17:48
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    \$\begingroup\$ Thanks ! I hope that works out for OP \$\endgroup\$
    – bobflux
    May 19, 2021 at 18:27
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Hopefully this current hogging problem occurs only at the startup.

If that was guessed right have 2 output capacitors. One small and the big one in parallel, but not directly in parallel. Let the big one be charged through a resistor and have in parallel with that resistor a diode which allows the hopefully occasional load current peaks.

enter image description here

Another approach is to replace D1 with an active switch which cannot be auto-conductive when the output voltage is lower than what's available as input. Implementing that needs some design.

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  • \$\begingroup\$ Yes I do have a 100nF in parallel to these big capacitor. \$\endgroup\$
    – IamDp
    May 19, 2021 at 11:18
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How about an inrush current limiter? It's basically a negative temperature coefficient thermistor with a certain resistance at room temperature and negligible resistance at higher temperatures (like when current is flowing through it.) You choose the room temperature resistance based on your maximum current needs. Digi-key has a wide selection.

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You could place a resistor in series with the pos input .Such a resistor would waste too much power or cause bad chip operation .Dont worry because a mosfet can bypass this when the cap has enough charge .Mosfets at these power levels are cheap ,easy to find and have low on resistance .

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