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Alaeddine , 12-13-2024, 12:46 PM

Hi everyone, I'm a newbie in electronics and working on a project where I want to power a custom FPGA board via USB-C. The USB source provides 500mA and 5V.

I’m planning to use a buck converter to step down the 5V to 3.3V, and I need around 1 or 1.5A at 3.3V for my board. My question is:

Is it possible to increase the current at the lower voltage using a buck converter? and how much should be the load current of the converter

QDrives , 12-13-2024, 05:14 PM

Yes and no.
Yes you can increase current when you go down in voltage.
No you cannot get more power out of it.

First things first.
P = U * I
Pin >= Pout
P = 5V * 0.5A = 2.5W
2.5V / 3,3V = ~0.75A
This considers a 100% efficient transfer.
But more importantly, this considers the use of a switch mode converter. Is does not work with linear regulators (often called LDO, sometime incorrectly).

As for needing more current from USB -- most USB ports, especially the USB-C ports, allow more than 500mA current draw.
https://www.ti.com/lit/wp/slyy109b/slyy109b.pdf?ts=1734053255889
https://www.usb.org/sites/default/files/D2T2-1%20-%20USB%20Power%20Delivery.pdf

QDrives , 12-13-2024, 05:19 PM

Also note that higher current may cause a voltage drop. So taking 2A may cause the voltage to drop to 4.5V (depending on the source).
Forgot to mention that the average efficiency of a SMPS is 80...90%.

Alaeddine , 12-13-2024, 05:26 PM

thanks for the reply. I saw the last videos of robert's DIY keyboard and he used an LDO with an output current of 1A although he was using USB-C with 500mA. I'm a bit confused

QDrives , 12-13-2024, 09:48 PM

When you select an output current of a voltage regulator, it is the **maximum** current that the regulator can do.
With a linear regulator, the input current is the output current (+ quiescent current). This all regardless of the input voltage. All 'excess' voltage on the input is turned into heat Ploss = (Vin - Vout) * Iload

**It is not the regulator that determines the current drawn, but the load**.

I do not know why @Robert Feranec selected the LM3940 over the TLV1117 in this video. TLV1117 was much lower cost, better accuracy, better PSSR and lower quiescent current. Sorry Robert, I think I will always criticize your regulator choice, but I am not able to do such designs in a few hours.

Robert Feranec , 12-14-2024, 11:29 AM

Just would like to add, USB-C in many cases can deliver more than 5V/0.5A, but it has to be supported by the power adapter / host / downward facing port (DFP). Read for example this, it may help: https://hackaday.com/2023/01/04/all-about-usb-c-resistors-and-emarkers/