No, I don’t have a resistor at the 12V input of the DC converter. The filtering happens at the +/-15V output of the converter. There is a 0.2V drop across each of the resistors over (theoretically) 20mA, so (theoretically) about 0.004W heat across the resistors. If I had anything like 14W of power across the resistors, I think I would have noticed!
Oh, sorry, I was thinking it was 12 V at the converter output. But 15 V at the output is worse.
When you have 20 mA that’s I^2R = 0.004 W, yes. But when you turn on power you don’t have 20 mA. You have 15 V at the converter output and 0 V at the capacitor, a 15 V drop across 10 Ω, or 1.5 A. (In fact you may not actually have 15 V in the first moments, if the supply and converter can’t provide that much current. In that case you have the voltage corresponding to the maximum current they can supply.) Only after the capacitor charges up to near 15 V do you drop down to 20 mA. And so you do have I^R = 22.5 W power dissipated by the resistor, at the instant of power on, and it drops down to 0.004 W over the next few fifteenths of a second. That’s what inrush current means: The current required to charge up the capacitor. It doesn’t fry the resistor because it isn’t high current long enough to do so, but the resistor and the converter and upstream power supply are under some strain.
If you need dual rail voltage, as long as it doesn’t draw too much current, you could always use a 555, a MAX1044, or a ICL7660S, to do it.
I don’t know about the XL6007, but the XL6009 I’ve even seen being used to power small tube amps on a japanese blog, after some modifications, of course.
I get that but I still don’t think that the inrush current is 1.5A. I think I would have noticed something going bad, like the converter module which cannot deliver that much current (at +/-15V, it is rated at 0.6A/0.12A respectively), or the switching supply (rated at 1A).
I think the answer lies at the XL6007 which has a built in soft start circuit that limits the inrush current. So I assume that inrush current after the XL6007 will be less of (or none of) an issue.
Since this is a learning opportunity for me, I will also try to see empirically how much of an inrush current I have due to the capacitors.
The MAX1044 and ICL7660S have a maximum operative voltage of 10V so they cannot be used here. I have used the 555 before, but the regulation is very poor, nothing compared to what you get from a dedicated DC/DC converter.
Certainly the PSU and converter have upper limits to what current they can deliver, though that may be higher instantaneously than DC. 1.5 A is what the initial current would be if they can deliver it. Otherwise they’re maxed out at however much they can deliver. That’s the concern, that you’re (briefly) running them at or above their rated maximum. And even at 0.6 A, the initial power is 3.6 W, so a higher power resistor than 1/4 W is probably advisable.
Back with a few empirical measurements:
I connected an 1A analog ammeter at the XL6007 input, to see how much current the converter draws with the capacitors connected at the other end. Previously, I used this method to see how much current the cold tube filament draws. For a split second I could just see the needle jump, and this was roughly consistent with the theoretical expectation when I used different—yes, beefy 2W ceramic—current limiting resistors).
The needle jumped to about 0.5A, nowhere near 1A (the maximum).
I then tried the same approach using 100Ω resistors instead. Theretically, we should expect a lowpass filter at 1.5Hz but a voltage drop of 2V (which would still be acceptable for the circuit in question).
Interestingly, with 100Ω, I got a negligible (<0.1V) drop at the rails, nowhere near the theoretical 2V, and the inrush current draw at the input was much lower, about 0.2A.
I then tried measuring the inrush current at the XL6007 output, i.e. the 100Ω resistor. Unsurpisingly, it did not register at all in the ammeter, as the voltage drop was negligible.
To ensure that the rough ammeter measurements were not way off (e.g. the needle did not have enough time to jump higher), I used a normal (i.e. not slow blow) 0.5A fuse at the XL6007 input. The fuse did not blow.
Given the above, I can only assume that:
- The XL6007 has excellent voltage regulation at the output, which is not affected by the series resistor at its output as much as we would expect.
- The XL6007 internal slow start circuit is efficient in limiting the inrush current to the presence of capacitors at its output. Besides, the datasheet implementation of XL6007 as a double SEPIC (which these Chinese modules are) has 220μF capacitors at the output, but it seems that its ability goes far beyond that.