Automatically selection of voltage sources?

Question

I'm trying to design a microcontroller board. There will be a main power connector and also an ISP programming port on my board. The microcontroller is running on 3.3V using a linear voltage regulator. I want to use the following circuit. Is it OK? The voltage from the ISP port would be 5V. The Diode D2 will decrease the input voltage of regulator would it be ok to have less than 5 volts on input?

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Here's several solutions, [1](https://electronics.stackexchange.com/questions/347826/switching-between-two-power-supplies-without-reverse-current), [2](https://electronics.stackexchange.com/questions/266312/passive-power-supply-or-ing), [3](https://electronics.stackexchange.com/questions/96398/or-ing-power-supplies-diode-or-mosfet), [4](https://electronics.stackexchange.com/questions/303250/prioritized-power-path-controller-for-battery-backup). — Harry Svensson, Apr 10 '18 at 22:00
@Finbarr I'm using the AMS1117 and for 3.3V it needs 4.8V on input. — Masoud, Apr 10 '18 at 22:52
@HarrySvensson Thanks that links helped a lot! I wasn't aware of the keywords. — Masoud, Apr 10 '18 at 22:53

score 4 · Accepted Answer · answered Apr 10 '18 at 21:02

Providing a link to the data sheet usually helps muchly. As, often, does reading said data sheet. IF in spec then a capacitor at regulator input may be a good idea.

Datasheet here

Page 12 says:

Vd Dropout voltage typical / maximum
IO = 100 mA ... 1.00 / 1.1
IO = 500 mA ... 1.05 / 1.15
IO = 800 mA ... 1.10 / 1.2

With a 1n4007 at nominally 0.6V drop you have 5-0.6 = 4.4V IF the ISP nominal 5V is actually >= 5V.
Acceptable dropout voltage is then 5-0.6-3.3 = 1.1V.
This is exceeded in above table only at 500 mA 800 mA worst case.
Using a Schottky diode for D2 puts Vin_regulator in spec for all currents up to 800 mA.

If the ISP 5V is sourced fron a "5V" USB port and there is potentially connector and wiring drop as well worst-case-V_ISP may be say 4.8V and maybe lower. If so, deduct this from the allowed regulator 'headroom'.

Read the data sheet to be sure all other requirements are met.

score 1 · Answer 2 · answered Apr 10 '18 at 20:50

If you're not worried about current flowing back into the 9V node, you might not need a diode from 9V to auctioneer the 9V and 5V lines, if you're sure the 9V line will always be higher.

You're using a low drop-out regulator, so the diode drops from D1 and D2 shouldn't prevent you from getting 3.3 V on the output pin.

Automatically selection of voltage sources?

2 Answers2