First post here.

I'm making a DC/DC converter which will generate an output of 19.5V 18A. Its a synchronous buck converter with the LM5145 TI chip.

When I have light load on the converter the output is within +-20mV of the 19.5V which is perfectly well.

When the load is increased enough the immediate outputvoltage falls quite a bit, which makes the output sink below acceptable values. This in turn makes the converter not usable under higher load.

Is it possible to place a "big" cap on the output to soften the immediate voltagedrop?

Any other suggestion is happily taken.

enter image description here

enter image description here

  • \$\begingroup\$ What’s with the series connected capacitors? How’s you layout? \$\endgroup\$
    – winny
    Aug 20, 2019 at 14:07
  • 1
    \$\begingroup\$ Yeah the caps don't look right. That is 3.3uF of output capacitance. That is very low. \$\endgroup\$
    – Stiddily
    Aug 20, 2019 at 15:13
  • \$\begingroup\$ Since there a relation between increasing load, higher duty cycle, less time to recharge the boostrap cap: measure/check the BST voltage (pin 17 w.r.t. pin 19) and make sure it stays high enough to decently drive the upper mosfets. Maybe the boost capacitor is too small or an external bootstrap diode is needed. \$\endgroup\$
    – Huisman
    Aug 21, 2019 at 20:45

1 Answer 1


In most DCDC converters there will be a feedback loop that controls the output voltage. Your design has such a loop as well, R9 and R10 are a voltage divider and the output of this voltage divider feeds back into the FB pin which compares the DC voltage on FB to an internal reference voltage (see the block diagram on page 16 of the datasheet).

In a schematic, all wires have zero resistance right? In the schematic the top node of R9 (where it connects to L1) and the voltage at J3 (the output terminal) should have the same voltage.

However when we actually make this circuit in the real world we have to use wires and copper traces and these do have some resistance. These small resistances cause a voltage drop which is directly proportional to the load current.

To check if this is what is happening, do your measurement again but measure at the actual feedback point, that is the top node of R9 (where it connects to L1). Also, voltage drop can happen in the ground connections as well so choose the meter's - (minus) connection carefully as well. I would choose the AGND (pin 6) of the chip (a copper trace that connects directly to it is also OK).

So multimeter between R9 and AGND and repeat your measurement. Do you still see the same voltage drop?

If the voltage drop is now much less then you have too much resistance in the connections where the large currents flow. If possible make these connections better (more low-resistance). What might also help is connecting R9 not directly at the L1 but connecting it at the output terminal J3, connecting it to there with an extra wire is OK as almost no current flows through R9. You might also need to connect the chip's AGND directly at the output terminal J4.

Hint: if possible look at other high current power supplies to see how these are designed, you might notice "star ground" structures and such that are used to eliminate the effect that large currents have on circuits. The trick is to make the large current flow there where you want them and have separate wires for low "sensing" currents.

Adding a capacitor is usually not a good idea as that can also affect the (high frequency) loop-transfer of the feedback loop, that could cause instabilities and other issues. Regarding output capacitors it is usually best to follow the datasheet's recommendations.

You added a voltage over time plot, if you are switching on / connecting the load "instantly" then it might be that the voltage drop you see is to be expected. The DCDC converter has a limited speed , it has that by design. Maybe you need to make your load such that it starts loading the supply more gradually so that the supply has time to respond.

  • \$\begingroup\$ Next to the mentioned non-zero resistance wires on the PCB have, the non-zero inductance may play an even bigger (negative) role. \$\endgroup\$
    – Huisman
    Aug 21, 2019 at 20:48

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