I am working on a circuit connected to computer with FTDI FT232RL. The circuit have got some op-amps and one MCU. Total power consumption (including FTDI chip) is near by 0.4 Watts. My op-amps need well regulated 5V power (single supply). Here comes the time to design power supply. USB Vcc pin is connected to the FT232RL Vcc and MCU's Vcc. The question is this how i should use USB's Vcc pin?

First connecting USB Vcc to LDO regulator IC like AMS1117 and powering both FTDI IC and MCU + Op-amps with AMS1117's regulated output is good?

USB Vcc-->AMS1117-->FTDI chip+MCU+Op-amps

Or connecting USB Vcc to FTDI Vcc and a regulator IC on the MCU +Op-amps is better?

USB Vcc-->FTDI chip-->AMS1117-->MCU+Op-amps

  • \$\begingroup\$ Read this note: google.com/… ... If your current draw is only 80 mA it should be ok to use a USB 2.0 port as those can supply up to 500 mA. Just make sure your linear regulator is efficient. I don't know enough about your design to recommend one configuration over the other though. \$\endgroup\$ – FullmetalEngineer Jan 30 '15 at 14:33
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    \$\begingroup\$ mA/h is not a unit of power consumption. \$\endgroup\$ – Olin Lathrop Jan 30 '15 at 14:35
  • \$\begingroup\$ @OlinLathrop mA/h is current consumption right? But 5V-80mA can be said as power consumption maybe. \$\endgroup\$ – user30878 Jan 30 '15 at 15:10
  • \$\begingroup\$ @pikafu What is wrong with not efficient regulator? Is it dangerous or bad for another ICs? \$\endgroup\$ – user30878 Jan 30 '15 at 15:11
  • \$\begingroup\$ @user30878 - power is voltage times current. mAh is a unit of charge, which is current times time. To get watt-hours you times amp-hours by the voltage of the system. Amp-hours are used for batteries mainly, because the voltage of the battery changes as the battery charge changes. Your system draws 80 mA and is run at ~5V so it consumes 0.4 Watts. No need to use Watt-hours since a watt-hour merely means the system expends a watt constantly for an hour. \$\endgroup\$ – I. Wolfe Jan 30 '15 at 15:35

The XX1117 is not really an LDO regulator, more of a medium-drop-out.

You need to give it quite a bit more input voltage than you expect out of it. So if your circuitry is happy with 3V (say), you may be okay. You can only count on 4.35V at the USB connector, and the regulator can use as much as 1.3V depending on current etc. so that leaves you with 3.05V. At 80mA, 3.3V may be acceptable, but you'll have to sharpen your pencil to figure that out.

If you want to create a low-noise well-regulated 5V supply from a USB input, you would have to boost the input to something like 5.5V and use a true LDO or to more like 6.5V and use an XX1117 regulator. At 80% efficiency that means you'd be at about 150mA input current. That's acceptable, but you're supposed to negotiate for current above 100mA. Possibly the FTDI chip does that for you. If you're dealing with two supply voltages in your system you may have to have voltage translators and so on, and the details will vary a lot depending on exactly what you are trying to accomplish.

Maybe you can operate all the digital stuff from the USB Vcc and use regulated 3.0V or 3.3V for the analog stuff. That would be my first plan of attack.

  • \$\begingroup\$ Another question but small for being new: Does USB Vcc is stable enough to feed op-amps? \$\endgroup\$ – user30878 Apr 1 '15 at 13:57
  • \$\begingroup\$ Maybe- depends on the requirements. \$\endgroup\$ – Spehro Pefhany Apr 1 '15 at 16:08

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