# 5V boost converter efficiency?

Discussion in 'General Electronics Discussion' started by Cirkit, May 16, 2016.

1. ### Cirkit

132
10
Oct 28, 2015
I purchased a couple of different 5V boost converter boards from eBay to power a 5V solenoid from 2xAAs. The solenoid normally draws about 120mA from a 5V power supply. If I use either of the boost converter boards powered by 2xAAs, the total current consumption (boost converter board and solenoid) is greater than 340mA.

The specifications of the boost converters state an efficiency of up to 96%.

Why would using the boost converter boards result in such a higher current consumption?

2. ### BobK

7,682
1,686
Jan 5, 2010
Are you measuring the current going into the boost converter and the current going through the solenoid and adding them? That would count most of the current twice. The input current is just the current coming from the battery. Hopefully this is closer to 200mA, the ideal amount.

Edit: Another possiblity, is your meter really reading the average current? The boost converter will use bursts of higher current alternating with lower current. Perhaps you are measuring the peak current.

Bob

3. ### Cirkit

132
10
Oct 28, 2015

I am measuring the current between the battery positive terminal and the boost converter positive terminal. The meter has a logging function and it measures around 340mA average and over 400mA (max) for one of the boost converter boards.

4. ### AnalogKid

2,425
690
Jun 10, 2015
96% efficiency is a fantasy, achievable only in one specific set of conditions (input voltage, output voltage, output current). And that spec might be the calculated peak efficiency of the switching regulator chip by itself, not including other circuit losses. Boost converters are the least efficient of all non-isolated converter circuits because of the large input current pulses that create losses in the inductor resistance and the switching transistor ON impedance. Also, a relatively low output voltage means increased losses (in terms of a percentage) in the output rectifier. If the output rectifier is a Shottkey with 0.3 V Vf at 120 mA, that is 6% of a 5 V output all by itself.

60% efficiency is not great, but not surprising for a generic circuit module optimized for cost.

ak

Cirkit likes this.
5. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,448
2,809
Jan 21, 2010
it's also quite likely that the voltage across the battery falls under load. Try measuring the voltage across the cells under load.