voltage-regulator
Could the output current of step-down switching regulator be larger than its input current?
Assuming that the efficiency of your switching regulator is 90% and you are converting 5 V to 1.8 V. If you measure that input current is 100 mA, do you expect that the output current would be 250 mA?
(5 V * 100 mA * 0.9 / 1.8 V = 250 mA)
If so, where do the charges come from?
I believe I may need a buck-boost however, because the components will need at least 1.8 V and 3.3 V, and my 3.6 V battery could drain below that
First hit on Linear Technology might be useful: -
Try going to LT to get an idea and use this parameter search engine.
for the 1.8 V rail it's 1.154 mA
Use a linear regulator - it's just not worth the effort I believe to use a switcher from 3V3 to 1V8. However, if current was over 50mA I might start considering a switcher.
Generally, you can assume a buck regulator will give you about 90% power efficiency near full load and a buck-boost will be about 85%. Ball-park figures.
The input must not affect the output, it's a closed loop system, which means that in wide range of input voltage, it will regulate the output to what you configured.
And since i haven't seen such unregulated DC/DC for a long time, i suggest to first understand how do you connect it and what are you measuring. Specifically, you have to measure on the module itself to ensure regulation. The wires must be thick enough to hold charge current. Of course, in the end the measurement should be taken near the phone. If the cable is poor, you may need to raise the DC/DC output voltage.
Best Answer
If you're using a 90% efficient DC/DC system (IC, inductor, diode, etc), then output will be the 277mA you calculated *0.9 (250mA).
The "charges" come from the input. Nothing is "gained". You exchange voltage potential for current (and heat, maybe some noise, etc).
1 - Power = Current * Voltage
2 - Power in = P out + P losses
Pin = 5V * 100mA = 0.5W
Plosses = Pin * inneficiency = 0.5W*0.1 = 0.05W
Pout = Pin - Plosses (or Pin * efficiency)= 0.5W - 0.05W (0.5W*0.9)= 0.45W
So, if Vout = 1.8V, Iout = 0.45W/1.8V = 0.25A
p.s.: In step-up systems the reverse is true. The output will be capable of less current throughput. The "original" current does not go anywhere, you just exchange it for more voltage potential (and heat, etc).