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I'm using a buck-boost converter RT6150B with fixed 3V3 output. I followed their application notes, and here's my circuit setup: enter image description here Note: I used 22uF for C4 intead of the recommended 20uF for output capacitor. I couldn't find a good SM component for 20uF and I thought 22uF is close enough.

Testing condition: I made a small test PCB based on this schematic, and I used a external power supply to feed different input voltage levels into the IC. From Vin = 3.1V-4.2V, Vout was kept pretty stable around 3.9V (Iout~20mA), which is too high comparing to the expected 3V3 output. When I try lower Vin = 3V, Vout dropped significantly to around 1.7V (Iout~140mA).I can also hear very small buzzing sound from the IC when I was testing with it.

There are several issues here, but first of all I don't get why the output voltage is around 3.9V instead of 3.3V? I used the recommended inductor value from the datasheet for generating 3V3 at output. Other things are just capacitors to decrease the voltage ripples. I really don't know what is causing this issue.

Any help is appreciated!

[Update] Here's the screenshot for Vout: enter image description here (Sorry for the bad quality. Had to take the picture with my phone...)

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  • \$\begingroup\$ What is the load on your test PCB? \$\endgroup\$ – John Birckhead Apr 19 '18 at 18:13
  • \$\begingroup\$ Are you sure you have the fixed version and not the adjustable version? \$\endgroup\$ – EE_socal Apr 19 '18 at 18:19
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    \$\begingroup\$ Just curious. But are you absolutely sure you got the RT6150B-33GQW? They do make a variable output version, RT6150BGQW. Can you verify the complete part number? \$\endgroup\$ – jonk Apr 19 '18 at 20:16
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    \$\begingroup\$ @jonk Yes. I'm sure the one I have is the fixed output version. Double-checked the marking on the IC based on datasheet Page 2. \$\endgroup\$ – C.Sun Apr 19 '18 at 20:26
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    \$\begingroup\$ Because your inductor is turning into a simple wire when the core is saturated. The max coil current is 2 or 3 times higher than the load current in switched mode. \$\endgroup\$ – Dorian Apr 19 '18 at 20:53
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Your circuit will likely work with an inductor with an appropriate current rating. Remember that the load current is the average of the current in the inductor. The current has a triangle waveform; in boost mode, the inductor is switched to ground to "charge up" the inductor until its current is greater than that required by the load. Then the switch is opened and the inductor is current is dumped into the load, with the current decreasing as the inductor "discharges:"

enter image description here

In buck mode, The voltage input voltage is higher than the output and the inductor charges up while the supply is connected to the load through the inductor. The supply is then disconnected and the inductor current must flow through the diode. Since the inductor has a negative voltage, the current decreases:

enter image description here

The inductor B_H Curve looks like this:

enter image description here

The operating range of the inductor must be in the relatively linear area. Since the inductor only has current flowing in one direction, it must operate in the upper right quadrant only. The area marked "minor B-H loop" would be acceptable, with the BH loop sliding up and down the curve as the inductor charges and discharges. However, too much current will drive the operating region up into the non-linear part of the curve. The inductance is not constant in this area and it will drop and change value at different points in the boost or buck cycle.

In boost mode, the inductor must store enough energy on each charge cycle (1/2Li^2) to provide average power on the discharge cycle. You just don't have enough core to do this when L drops due to saturation, and the result is too low voltage.

In buck mode, the inductance drops during the charge cycle, causing a current surge into your load and capacitor and fully charging your inductor. The load must then discharge the inductor during the time before the next charge cycle.

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  • \$\begingroup\$ Thank you for your answer. I think now I understand that for boost mode, because my inductor can't store enough energy, my Vout results in very low voltage like the ~1.7V that I was getting. But for the buck mode, I still don't quite understand how the inductor results in the higher voltage level like 3.9V at Vout? Why would the inductance drops during charging? \$\endgroup\$ – C.Sun Apr 20 '18 at 21:22
  • \$\begingroup\$ Thank you for your answer. I think now I understand that for boost mode, because my inductor can't store enough energy, my Vout results in very low voltage like the ~1.7V that I was getting. But for the buck mode, I still don't quite understand how the inductor results in the higher voltage level like 3.9V at Vout? Why would the inductance drops during charging? \$\endgroup\$ – C.Sun Apr 20 '18 at 21:22
  • \$\begingroup\$ Even for buck mode the peak current might be twice as the average output current to avoid switching while the coil is still charged for efficiency resons. In the B-graph the slope is the coil inductance, over 30mA you are getting into the flat region , zero inductance and the current raises extremly fast so the switchoff is not fast enough to avoid a current peak getting to output. \$\endgroup\$ – Dorian Apr 21 '18 at 12:39
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    \$\begingroup\$ It is not really clear to me either - the buck mode reason is more dependent on your switching IC and is more complex. But in general, in buck mode, if you are saturating on each cycle, the inductor voltage will drop to near zero and the output will rise quickly. It would be interesting to see if your voltage decreases when you reduce the load. \$\endgroup\$ – John Birckhead Apr 21 '18 at 21:05
  • \$\begingroup\$ Finally got to test it with a new inductor! Used one with a 700mA current rating, and the circuit was able to give me ~3.3V voltage output. \$\endgroup\$ – C.Sun May 7 '18 at 14:31

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