# Two input voltages to buck voltage regulator, using diodes to protect the sources

I have a buck regulator that takes 4.4-36V input voltage, and gives 3,3V 1A output. It will have two voltage sources, one potentially 25V and one 5V. Do I need a mosfet circuit to make only one source active at a time? OR Can I use two diodes in this way to protect the sources? The buck regulator should be able to handle the total voltage of the two sources put together so I don't think that is a problem.

simulate this circuit – Schematic created using CircuitLab

You have the right idea but the wrong implementation. Yes, the concept of one diode in series with each source is good.

However, 1N4148 is totally inappropriate here. That's a small signal diode, not a power diode. Let's say for example that the buck regulator is 85% efficient. With 1 A at 3.3 V out, it needs 3.88 W in. When only the 5 V source is active, then the input voltage is about 4.3 V. That means the input current will be about 900 mA. That's way too much for a 1N4148.

Since your voltages are low, you should use Schottky diodes. Schottky diodes that can withstand 30 to 40 V and handle 1 A are cheap and easy to find. The lower forward voltage drop will also increase overall efficiency a bit, especially from the 5 V source to the output.

• Hey, yes that I am aware of. I forgot to edit the component name. this is my diode: mouser.com/ProductDetail/AVX/… Commented Apr 25, 2018 at 11:00

Since you're using a buck regulator @ 1A, the input current is very low. You can get away with using diodes in this case.

Assuming ideal buck converter and assuming a Vf of 0.3V for a Schottkey, you have 3.3V / 4V * 1A = 0.825A on the input side, at the worst case (when the input voltage is low). Multiply that by 0.3V and you get just under 0.25W of loss. Not a lot for a diode to take, and about 7.5% of loss, which I think should be acceptable.

This is in fact what the Arduino Uno board does on their power input, IIRC.

A logical extension to this problem is--what if your buck regulator draws a lot more than 1A? You should use FETs to control the sources, and a nice way to do it is to use chips like this: http://www.analog.com/media/en/technical-documentation/product-selector-card/2pb_linear_idealdiodesfc.pdf?domain=www.linear.com

More answers in this question: Multiple Input supply switching

• That was my thought aswell, but just to double check. Would this diode: mouser.com/ProductDetail/AVX/… Be a nice fit in this application if this is the buck regulator? the buck is based of the typical operating circuit: mouser.com/datasheet/2/389/l5970d-973965.pdf Commented Apr 25, 2018 at 10:57
• You can re-do the calculations with the Vf values from the diode, and you can double check the max. power dissipation from the diode, and also the max current for the diode. You also wanna ask yourself if you're ok with losing half a watt, which is not some insignificant amount. Commented Apr 25, 2018 at 11:05
• On a closer look in the datasheet I found out that the buck supports 4.4-36v, not 4-36v as previously stated. My diodes have a Vf of 0.5V. This is a little close on the 5V source, should I worry or do you think it is fine? Commented Apr 25, 2018 at 11:23
• I'd worry about that TBH. Perhaps you should use the IC. Also, the "5V source" can have voltage droops which would definitely push it below the 4.4V minimum, so yeah. Commented Apr 28, 2018 at 0:40