Input current at the dc-dc converters

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engr_joni_ee

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I was looking at the datasheet of LTC3200/LTC3200-5 which is a charge pump dc-dc converter. The input can be from 2.7 V to 4.5 V. The output voltage is fixed 5 V. The output load current is up to 100 mA for the input voltage more than 2.1. The efficiency is 80 % of that device. The link to the datasheet is given below.


My question is regarding the input current requirement. I know that if the efficiency of the dc-dc converter is 80 % then we can get maximum 80 % of the input power at the output. In other words the output power can not be more then 80 % of the input power.

Let's suppose I have 3.8 V as an input and I need 80 mA load current at the output with fixed 5 V output. Does it means that I need more then 80 mA current at the input ? In order to get higher voltage at the output then the input voltage, the input current has to be higher then the output load current, right ?
 

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betwixt said:
Think of it in terms of power W = V*I.
Click to expand...
Unfortunately efficiency of SC voltage is also involving voltage ratio dependant losses. An ideal SC doubler performs 1:2 conversion, for 3.8 V input, the doubled voltage is reduced by generating additional losses (imagine a voltage doubler followed by linear regulator as first order model).

According to datasheet efficiency with 3.8V input is about 63 %. Expected input current is 127 mA for the given operation point.

 
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engr_joni_ee said:
Does it means that I need more then 80 mA current at the input ?
Click to expand...
Surely. Just think about it...

The output voltage is higher than the input voltage. On equal current this means the output power is higher than the input power.
So if you can build such a device -- you are a rich man. Solved the world power problem.

Not only would it be a perpetual motion machine, it would actually generate energy from nothing

engr_joni_ee said:
for the input voltage more than 2.1. The efficiency is 80 % of that device.
Click to expand...
No.
Neither the one nor the other.
* I can´t see where the datasheet says it works down to 2.1V
* and 80% are only for the given conditions. 2.1V is out of specification. and the "Efficiency vs Load Current" stops at 50mA on 2.7V input (so no 80mA at all) while the efficiency on 3.8V input surely is below 65%.

Klaus
--- Updated ---

FvM said:
According to datasheet efficiency with 3.8V input is about 63 %. Expected input current is 127 mA for the given operation point.
Click to expand...
A typo?

Output power: 80mA @ 5V = 40mW
input power = output power / efficiency = 400mW / 0.63 = 635mW
input current = input power / input voltage = 635mW / 3.8V = 167mA

Klaus
 
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