I have a 12V/3.2A switching PSU powering an LTC3780 buck/boost converter rated at max 2-30V/0-10A.
Let's assume 100% efficiency, can the LTC3780 achieve more than 3.2A if set for example at 5V? I guess it is a general electronic question. Do we have to 1) think in watts here or 2) assume the amps out can't be higher than the amps in?
If we have to think in watts, would it mean that…
PSU: 12 * 3.2 = 38.4 Watts
LTC3780 @ 5V: 38.4 / 5 = 7.68 A
The LTC3780 would be able to push 7.68 amps @ 5 volts? (again, let's assume 100% efficiency)
Thank you!
Best Answer
Assuming 100% efficiency, yes.
Your calculations are correct and makes sense.
\$P_{efficiency}=\frac{P_{out}}{P_{in}}\$, you are saying that this ratio is equal to 1.
If we plug in some numbers... we get this:
$$ \begin{align} 1 &= \frac{P_{out}}{38.4W}\\\\ 38.4W &= P_{out}\\\\ 38.4W &= V_{out}×I_{out}\\\\ I_{out}&=\frac{38.4W}{ V_{out}} \end{align} $$
So this will be your maximum current draw at any given voltage.
For P to be the same and impedance is constant, then with lower voltage the current increases, or higher voltage and lower current.
But... in the real world there's no such thing as 100% efficiency... Also, what you will notice is that it will be hard for you to get like 1000 A and 0.0384 V. But it will be relatively easier to get 1000 V and 0.0384 A. This is because there's always parasitic resistances everywhere. Except for in superconducting materials (as far as I know).
There's not much else to say really.