Regulators without inductors

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engr_david_ee

Joined Mar 10, 2023
176
I was looking at the charge pump regulator CAT3200TDI-GT3 datasheet that used a voltage doubler inside. The efficiency is 80 %.

If I have input 4 V and I need output 5 V with load current 80 mA for example. Does it mean I need more then 80 mA on the input side ?

We can not have higher output voltage and higher output current with low input voltage and low input current, right ?
 

Ian0

Joined Aug 7, 2020
10,039
Calculate the output power.
Divide by 0.8 to give the input power (because the efficiency is 0.8)
Input current = power divided by input voltage.
 

Papabravo

Joined Feb 24, 2006
21,261
That is correct. The simple rule is that power out will always be less than power in. Working backwards:

\( 5\text{ V}\times80\text{ mA}\;=\;400\text{ mW, Output power} \)

\( \cfrac{400\text{ mW}}{80\%}\;=\;500\text{ mW, Input power} \)

\( \cfrac{500\text{ mW}}{4\text{ V}}\;=\;125\text{ mA, Minimum input current} \)
 
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