I am using TPS61021 boost converter IC to increase input voltage from 1.2V to 3.3V. I have designed the boost converter from the application called TI WEBENCH.

Input Voltage: 1.2V

Output Voltage: 3.3V

Min current: 300mA

The PCB that I have designed, is measuring 3.3V, but the efficiency is pathetic. It is well below 25%. In the BOM, TI WEBENCH software recommends using 10UF 10V capacitor, but I am using 10uF 6.3V. I am not able to understand why efficiency is so low? Datasheet for Inductor.

Below in the tabular format I have written current values that I have measured from multimeter.

enter image description here

Fig.1 Measurement of current at different load resistance.

Fig.1 TI WEBENCH application output Fig.2 TI WEBENCH application output

Schematic in PCB designer software

Fig.3 Schematic design of Boost converter PCB picture

  • 1
    \$\begingroup\$ Datasheet link for that inductor? \$\endgroup\$
    – user16324
    Commented Jan 20, 2018 at 15:59
  • \$\begingroup\$ Also, the cap you are using, after derating, is less than 40% of 10 uF, if you add the tolerance on top of that you are probably below the minimum cap required on the datasheet. Can you poke around with a scope? \$\endgroup\$ Commented Jan 20, 2018 at 16:00
  • 1
    \$\begingroup\$ "...current values that I have measured from multimeter" - be aware that there will be a voltage drop across the multimeter. At 570mA on the 2A range my meter dropped 230mV! For accurate efficiency calculation you should read all voltages and currents (input and output) simultaneously. \$\endgroup\$ Commented Jan 20, 2018 at 17:33

2 Answers 2


Pin = 1.2V × 570mA = 684 mW

Pout = 3.3V × 105 mA = 346.5 mW

Efficiency = Pout/Pin = 346.5/684 = 50.7%

How on Earth did you come up with 18.42%?

Similarly, I get 64% and 70% for your other two loads. These numbers actually sound pretty good to me at these voltages. Did Webench give you any efficiency estimates for this specific configuration?


What you did looks OK to me.

I suggest that you re-create the exact conditions for which an efficiency is mentioned in the datasheet and see if you then do get a good efficiency.

You use Vin = 1.2V but the datasheet only mentions efficiency for Vin = 2.4 V as far as I can see.

It is possible that the resistance of the power switches on the chip is much higher at Vin = 1.2 V although there's a higher voltage available at Vout, I am not sure if the internal power MOSFETs would use this voltage for their Vgs in order to decrease their Rds_on resistance.

Another thing to check is the capacitors, make sure they are low ESR types.

Also the inductor, make sure it can handle the current, from looking at the model you used this inductor is very likely to be OK in that regard.

Use a scope to see how much ripple there is at input and output and also if the SW pin of the IC properly reaches GND voltage and output voltage.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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