1
\$\begingroup\$

I am making a circuit which has buck converter LM3671 inside: http://www.ti.com/lit/ds/symlink/lm3671.pdf Schematics are attached at the bottom of this post. In real application, power source is 3xAA/AAA batteries which should buck-convert voltage to 3.05V.

Input capacitor: 4.7uF

Output capacitor: 10uF

Inductor: 2.2uH (CDRH2D14NP-2R2NC)

Set output voltage(with resistor feedback network): 3.05V

The problem I am having is that output is always 0.2V lower than the input. It starts at very low voltage(below 2V) and continues up to 5V. At larger voltages, the output is 0.1V lower than the input. That is at the output current of 20mA.

At smaller output currents, the difference between input and output is even smaller.

I tried a lot of things, can not find the source of the problem. I am using suggested schematics from the datasheet and suggested components from it. What needs change so IC stops following voltage?

schematics

\$\endgroup\$
2
  • 1
    \$\begingroup\$ The schematic looks OK to me. Did you use the recommended capacitor types for C4 and C5? See datasheet page 17. If those are OK, did you follow the PCB layout example on page 22? \$\endgroup\$
    – Bimpelrekkie
    Jan 18 '18 at 9:21
  • 1
    \$\begingroup\$ Try bypassing your voltage adjustment circuit (connect FB directly to VCC) and see if it outputs the specified FB voltage. If it doesn't you may have a defective part. \$\endgroup\$
    – Dmitry Grigoryev
    Jan 18 '18 at 9:24
3
\$\begingroup\$

I believe you have the fixed 5 volt version of the chip (where feedback is directly fed from Vout to Vfb) but you think you have the adjustable version that requires an external potential divider.

If I am correct, the device will always be trying to make the output as high as it can in order to produce 5 volts at Vfb and this is, what you appear to be seeing.

enter image description here

See the red box I've added to the bottom of the above diagram.

\$\endgroup\$
2
  • \$\begingroup\$ Perfect. I ordered wrong chip version by mistake - I got 3.3V version. Now, when I connect Vout(Vcc at my sch) to FB pin and remove feedback resistors, everything works as a charm. Thank you for the tip! \$\endgroup\$
    – davaradijator
    Jan 19 '18 at 13:17
  • 1
    \$\begingroup\$ @davaradijator it had to be that or you'd blown the chip up! \$\endgroup\$
    – Andy aka
    Jan 19 '18 at 13:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.