Boost-Converter: What does "no-load 18mA" input current spec mean?

Amazon: XL6009 DC-DC Adjustable Boost Converter

This is my first boost converter purchase. I have a question about this specification:

Input Current: 4A (max), no-load 18mA (5V input, 8V output, no-load is less than 18mA. Higher the voltage, the greater the load current.)

Does this mean I have to know in advance that my circuit will draw no more than 4 amps?

Also what does the "no-load 18mA" mean?

Does this mean I have to know in advance that my circuit will draw no more than 4 amps?

The maximum current into the boost converter is 4A. And, since you're boosting the voltage you'll get less current out and into your load.

Power (Watts) is calculated as Voltage X Current and since there is some power loss in the converter you'll get less power-out than power-in depending on it's efficiency.

Also what does the "no-load 18mA" mean?

With nothing connected to the converter's output (no load) the converter takes 18mA to run.

In the description, they give the formula: Vin * Iin * Efficiency = Vout * Iout (or Vin * Iin = (Vout * Iout )/ Efficiency ) The "4A" refers to Iin max.

Assuming you select Vin and Vout, you can calculate the maximum Iout available if you know the efficiency.

Unfortunatly, they did not give much information about it...

But by googling " XL6009 efficiency ", if you search for images, you may be able to find some curves or tables for efficiency.

DVDdoug:
The maximum current into the boost converter is 4A. And, since you’re boosting the voltage you’ll get less current out and into your load.

Power (Watts) is calculated as Voltage X Current and since there is some power loss in the converter you’ll get less power-out than power-in depending on it’s efficiency.

Understood. As the product description noted, efficiency is just Pout/Pin. Since Pout < Pin or VoutIout < VinIin with Vout > Vin, we have Iout < (Vin/Vout)*Iin which says Iout is a fraction of Iin. As oignon pointed out, this is approximate.

DVDdoug:
With nothing connected to the converter’s output (no load) the converter takes 18mA to run.

I see. So in other words, the boost converter draws a minimum of 18 mA.

Yes how else are you going to size it. If you have more amps that will be OK, if less you will have problems. Look at your load and convert it to watts (volts * amps), then multiply that by 1/ by the efficiency then add 10%. That is the wattage you use for the rest of your calculations. Roughly if you have a boost to 10 Volts from 5 Volts every amp at 10 Volts will require (2 amps + losses) at 5 volts. Supply watts = Load watts + loses. No-load is a term that indicates there is no load on the device or another term would be quiescent current. This response is to help you get started in solving your problem, not solve it for you. Good Luck & Have Fun! Gil