How can output voltage and current of a DC-DC converter both be higher than input?

I'm working on a TFT LCD Supply and using a step up DC- DC converter from Maxim, the MAX17062. I want to find the input current. The datasheet lists that as 2.5mA. I have a doubt: if the input current is 2.5mA and the input voltage is 5V, how can the output be of higher voltage and higher current? I'd say that that would violate the conservation of energy.

• Where exactly do you see these numbers? I can't find the output power.
– user17592
Commented Jun 4, 2013 at 5:55

The 2.5 mA referred to in the datasheet is the quiescent current, i.e. the current drawn by the device when there is no power being drawn from the output.

The operating input power will be found by computing the output power drawn, taking into account efficiency, and with the quiescent current as an overhead.

I'd say that that would violate the conservation of energy.

Rightly so! The efficiency for this part is listed at ~90%. So your input and output current voltage are going to be liked according to:

$$I_{in} \cdot V_{in} = \frac{I_{out} \cdot V_{out}}{0.9}$$

Note that efficience typically depends on your implementation and the load -- see page 4 of the datasheet.

The 2.5mA rating related to the quiescent current, which is the minimal current that is consumed by the part, when no load is applied:

$$I_{out} = 0 \rightarrow I_{in} = 2.5 mA$$

I think you are looking at the wrong spot in the data sheet. The maximum input current for the MOSFET inside the converter would be 5.3 A.

If you want to calculate the input currents, output currents and peak current for the whole setup, you can have a look at page 9 in the datasheet. There is a short description of how to do that. There is also an example calculation for the default setup. The values in the example are:

Output: 15.0 V, 0.6 A
Input: 4.5 V, 2.35 A with a ripple of 0.97 A. The peak current would be 2.85 A.