I am trying to analyze a circuit where AOZ1014 buck converter is used to generate 7.5V and negative voltage(-5V) is generated with following arrangement.


As shown above Lx pin of AOZ1014 ( left sided pin of inductor L2 )is assigned to 7905 regulator with C33,D16 and D18 arrangement. Here, as per my consideration, D18 is for suppressing +ve peak to ground since we need -ve input to 7905 IC.

What is the purpose of D16, R57 and C33 here? How changing C33 will change my output voltage?

I used same concept with TP5430 Buck regulator as shown.


But I am getting random results in my case here. Sometimes it is regulating as per my requirement but randomly misbehaves with loading 7.5V to 1.5V and -5V to -2v. Am I correct in assigning C33 pin to in between of C31 and L2? Can someone guide me to make it work.

Thanks in advance.


1 Answer 1


D18 and D16 will rectify the AC voltage at pin 8 of U14. It is similar to a voltage doubler rectifier.

This can work usefully however it relies upon U14 continually producing a square wave at that pin. It may not do so if the load on the main output (7.5V) is at a low current level.

Many switching regulators go into various low power nodes when the load is reduced - pulse skipping, PFM (pulse frequency Modulation) or similar. These will result in the signal to D18 not being continuous and could give the symptoms you see.

Make sure you have a load on the main output - use a resistor to simulate the final product load if necessary.

  • \$\begingroup\$ Thanks, Kevin.. But, are u saying load at 7.5V node..how to choose value..even I connect 500E load, it's loading sometimes. Why C33 used and how it's value is calculated? \$\endgroup\$ Commented Sep 11, 2015 at 17:36

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