How to get the same high efficiency as it on datasheet?

Hi，

The x-axis should be current and the unit is mA.

You may turn off the REG_OUT and measure the efficiency of the boost as the first step. Please take the SYS and BAT power together.

Then, you can enable the REG_OUT and combine power of the boost and REG_OUT together as the output power.

Thanks!

Thank you for help!

We noticed that ADI provided EVAL piece of ADP5091. And we revised some place on our PCB to follow the EVAL one. For example, we connected REG_D0 and REG_D1 to GND. Other resistors, capacitors and inductors are the same as those on EVAL borad.

However, the efficiency is still unable to beyond 50%.:(

To give more details:

We uses voltage source which connect to VIN, and set at 0.5V in the test. MPPT is working correctly, so the real input voltage is 0.6V*80%=0.48V. And load is a potentiometer conncted to SYS, changing from 10Ω to 10kΩ.We use a capacitor on pin BAT, voltage between two pins  is  2.5V.

Input current is very low(we didn't set any limit), typical value is about 10mA or lower. Output voltage is steady around 1.86V, when load is 2kΩ. The efficiency is poor 36%. Far from 80% in datasheet.

Could you give us some advice to sovle it? Maybe we shouldn't use resistor as load to measure output power? Or any mistake we made on this PCB?

Thanks again.

Yes, it's not easy to use resistor to measure the efficiency unless the BAT capacitance is very large because you may need make the BAT and SYS voltage between SETSD and TERM threshold. If the BAT voltage is lower than SETSD, the boost is working in asynchronous mode, the efficiency would be low. If the voltage is larger than TERM, the boost would stop switching. The better way ot measure the efficiency, you can use a source meter to tie to SYS because it can sink current from ADP5091.

Hi, we solved the efficiency problem with your help. But another problem rises, which is about the power.

We use voltage source as VIN, which was set at 0.5V. We adjust the load resistor at REG_OUT pin from 5Ω to 10kΩ and observe output voltage and current. The question is, output current at REG_OUT pin never reached 150mA, which is the typical value on the datasheet. Output voltage has been set at 3.3V by resistor at VID, so a 22Ω resistor could be suitable. However, output voltage doesn’t rise beyond 2V until load was turned up to 1kΩ in practice. In short, we can’t recurrent it on our PCB. We need higher output current on the chip ,so we care abort this.

Or the 150mA current on datasheet is under some condition, please tell us and we will do whatever is required.

Thanks!

Hi, we solved the efficiency problem with your help. But another problem rises, which is about the power.

We use voltage source as VIN, which was set at 0.5V. We adjust the load resistor at REG_OUT pin from 5Ω to 10kΩ and observe output voltage and current. The question is, output current at REG_OUT pin never reached 150mA, which is the typical value on the datasheet. Output voltage has been set at 3.3V by resistor at VID, so a 22Ω resistor could be suitable. However, output voltage doesn’t rise beyond 2V until load was turned up to 1kΩ in practice. In short, we can’t recurrent it on our PCB. We need higher output current on the chip ,so we care abort this.

Or the 150mA current on datasheet is under some condition, please tell us and we will do whatever is required.

Thanks!

Hi,

The high current from REG_OUT should be relied on the storage element at BAT pin. The boost from VIN cannot deliver so high current because of the peak current limit is only 200mA and the average is only 100mA. The equivalent current at output should be much lower as it's a boost from 0.5V.