Going from Bench Power to a fixed-DC power supply

[LoveLaika]

I have a pair of buck switching converters in a circuit, the LM2673 and the LM2679. However, the LM2679 is being used in a buck-boost configuration as shown in this app note below, AN-1157, and I'm converting a positive voltage to -10 volts. The LM2673 is being used to convert a DC voltage down to 5 volts. You can see my circuit below.

So far, I've tested it with a bench power supply supplying 3 amps, outputting at 12 volts and 15 volts. However, when a load is connected to the output, the power board and load seems to be drawing more current than 3 amps, bringing the bench supply to current-limiting mode. The work around I've found is to change out the current-adjust resistors on the regulators (increase them to limit the current to lower values, 1 amp and 3 amps). Changing out the resistors seem to work that way.

I think the problem is that the bench supply can't supply enough current on start-up. I want to try and use a 20 V (4.5 amps) supply, but do the regulators have a limitation on the output-input voltage difference? The ICs are rated for 40 volts input max, but with a 20 volt supply, the voltage difference would be 15 volts and 30 volts respectively (for positive and negative voltages). Would this be within safe operation?

https://www.ti.com/lit/ds/symlink/lm2673.pdf

https://www.ti.com/lit/ds/symlink/lm2679.pdf

https://www.ti.com/lit/an/snva022e/snva022e.pdf?ts=1598635634497&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FLM2679

EDIT: Using a shunt resistor (1-ohm power resistor), it appears that the power supply board is drawing 4 amps upon startup. I found a 12 volt 5 amp supply, but is it safe to use it in place of the bench power supply, or would it be better to limit via the current limiting resistors for the ICs?

[David Hess]

I think the problem is that the bench supply can't supply enough current on start-up.

That may be exactly the problem.  Switching regulators have a negative input resistance where at lower input voltages, they draw higher current.  This can cause the current limiting of the supply to hold its output voltage low during startup.  Further, foldback current limiting makes this worse because the current limit is lower at lower output voltages.

The above is why integrated regulators are designed to "hard start" and provide full output current at low voltages when recovering from thermal overload.

[LoveLaika]

Thanks for replying. I made an edit to my post about the tests that I ran. In this case, is the solution to run it at a higher voltage?  The IC can have an input voltage of 40 volts but my input caps are rated for 25 while I start at 12.  If i use higher voltage rated caps, assuming same capacitance, that would reduce the in rush surge current?

[Psi]

It's pretty common to get odd readings and sometimes lockups with switch mode controllers if the psu voltage drops out right when they try to start.

[LoveLaika]

Well, it locked out when my psu was tested with a single supply and max current is 3 amps. When i mean locked out,  psu reached current limit, so psu showed 1 volt 3 amps. In parallel tracking, with max 6 amps, it worked fine. That was how I saw the 4 amp start up current draw.   In any case, would starting at a higher input voltage help in any way?

[Berni]

This is normal for switching supplies.

They need to draw more current at low voltages to get the same amount of power. Also all the capacitors are empty on startup so the switching regulator will be pulling all the power it can get to get those capacitors filled up to the correct output voltage. All this combined makes them draw an abnormally large current on startup, but most power sources have no current limiting so they tolerate the short spike of startup current.

The only issue is that a lab power supply tends to have very fast current limiting (in order to help protect circuits from damage in case something goes wrong). So this current limit intervenes during startup, reducing the power and causing the voltage to drop. If the voltage drops too low for the switching regulator then it might give up and stop the startup sequence or get into some weird faulty mode.

Only way around it is to have a lab power supply that is powerful enough to supply the startup spike, or if there is no way around it then you can place a large capacitor across the power supplies output and add a switch to your circuit. This lets the capacitor charge up and help supply the startup current once you flip the switch. However this is not good if your circuit actually does have a fault since the capacitor will happily supply 100A into a short circuit for a brief moment (potentially welding the switch, blowing a chip into pieces etc..)

[LoveLaika]

Thanks for replying.  Looks like there's no way about it, aside from finding a high current supply. I found that the start up current depends on the load, and decreasing the switch current reduces the start up current but that would reduce overall current to the load.