Suppose I am trying to boost the voltage of my 3.7V Lipo battery to 8V using a boost converter (MT3608 module) and lighting up some LEDs.
Does the voltage of the battery reduce just because of the boosting circuit, if no load is connected at the output(which is 8V) of boost converter?
Does the boosting module play a role in reducing the battery voltage when no load connected, or does the battery drain as usual in the same way with or without a boosting circuit when no load is connected?
I'm doing it practically now. My lipo battery voltage is 3.7V, I boosted the voltage to 8V (without any load) and left it to check whether boosting circuit reduces the batery voltage. In 1 hour battery voltage has reduced to 3.67
What is the reason for this and how can I avoid this problem, since suppose say if we are trying to make a PCB, Lipo battery will always be connected to boosting circuit and it can drain the battery overnight.
Best Answer
What you're asking as called the quiescent current consumption of the boost converter. This is the current that the boost converter takes when it outputs the boosted voltage ( 8 V) but no current is pulled (by the load) from that 8 V.
When I want to know this \$I_q\$ I look at the datasheet of the chip, on page 3 there's a table with the information:
The Shutdown current is the lowest but then the converter is shut-down so there will not be 8 V at the output!
There are two Quiescent Current figures, one for PFM and one for PWM.
Usually PFM is used when the load uses a small current, the converter then operates in small bursts and shuts down in between bursts. This saves supply current as we can see in the numbers: \$I_{q,PFM}\$ = 100 uA typically
PWM is used when the load is higher. The converter switches continuously. The \$I_{q,PWM}\$ has a higher value (1.6 mA typically) compared to the PFM mode but that usually is not an issue as the load will draw a lot of current as well. So the 1.6 mA that the converter needs is usually small compared to the load current.
So yes, the converter will take some current, it will take around 100 uA continuously. If that will drain your battery overnight depends on the capacity of your battery. With a 1000 mAh battery, at 100 uA it takes 10 thousand hours to discharge, that's 400 days. My guess is that you will be using a battery with more capacity than 1000 mAh so I do not think the converter can drain your battery overnight.
The change from 3.7 V to 3.67 V doesn't mean much, that can be due to temperature changes as well. I would measure the actual quiescent current, and if it is more than 200 uA (the specified maximum) then I would investigate what's going on.