If we need to make a board that use two voltage levels (3.3V and 5.0V) and the board can be powered both from the USB or from an external 5V power bus, then good practice is setting a diode on the USB power to prevent a reverse current to host. With this diode on the board, the "5.0V" power bus will have only 4.5-4.7V because of the diode voltage drop. If we need a true 5.0V, we should use a DC-DC step-up converter.

  1. Is a current reverse protecion diode so nessesary?
  2. Is it a lesser evil to remove a current reverse protecion diode if a true 5.0V is needed?
  3. If good practice requires a current reverse protecion diode, a 5.0-to-3.3 LDO and a DC-DC step-up converter, what is the best cascading of this devices:
  • 5.0USB->diode->4.7V->LDO->3.3 and 5.0USB->diode->4.7V->DCDC->5.0PCB
  • 5.0USB->diode->4.7V->LDO->3.3 and 3.3->DCDC->5.0PCB
  • 5.0USB->diode->4.7V->DCDC->5.0PCB and 5.0PCB->LDO->3.3
  1. If we need a very clean 5.0V power bus, do we need to step-up the voltage to 5.5V and then use a pecision LDO to cut 5.5V to 5.0V?
  • 1
    \$\begingroup\$ You could start with asking yourself, do you really need 5.0V? Sure, it's common. But many applications work just as well with 4.5V. \$\endgroup\$
    – Klas-Kenny
    Sep 13 at 13:04
  • 1
  • \$\begingroup\$ This is a common issue. Just pull the power handler off of some open-hardware platform. For example, Arduino's tend to do this in a fairly bullet-proof way. \$\endgroup\$ Sep 13 at 14:45

2 Answers 2


If the voltage drop across the reverse current protection diode is an issue, you could use an "ideal diode" circuit.


simulate this circuit – Schematic created using CircuitLab

Transistor M1 has a body diode which behaves as a normal diode. However, the body diode is forward biased, and Vin is greater than (the absolute value of) the threshold voltage of M1, (4 V at most for the IRF4905), then the MOSFET is turned on, and the voltage drop across the MOSFET becomes very small.

enter image description here

In this particular circuit, the voltage drop when Vin > 4 V, and Vin > Vout, then Vout = Vin - 0.06 V! This is for a load of only 5 \$\Omega\$!

  • \$\begingroup\$ Bipolar transistors for the "ideal diode" must have equal temperature and equal characteristics. This scheme will be unstable if descret bipolar transistors will used. There are ideal diode controllers (such as LTC4357) for this application. \$\endgroup\$
    – Arseniy
    Sep 14 at 9:04
  1. Only if you or a potential user of your board might ever connect USB and an external supply simultaneously. If you can somehow rule out that this ever happens you might get away without the protection.
  2. Only if it is more important to have "true 5V" than to have a non-destroyed computer. Although you don't specify what "true 5V" really means. And beware the fact, that the USB 5V supply might also be only 4.75V, depending on the supply capabilities, the quality of the USB cable and the load current. It might also be around 5.25V for low load currents...
  3. That depends on soooo many factors. I would not use your second suggestion (don't use an LDO to create the 3V3 and then increase that again with DC/DC, that makes no sense).
    // Edit: Actually in this case I would always opt to use variant 1: Get both your 3.3V AND your 5V directly from the input voltage.
  4. Yes, that might make sense, again depending on your definition of "very clear 5.0V". You can get down to a ripple of a few mV with a DC/DC, so something like "5.0 +- 0.01 V". If you need a cleaner 5V, then an LDO as filter is probably necessary. Make sure, that the LDO has a high enough PSRR in the switching frequency range of your DC/DC.
    For fixed PWM DC/DC converters this will be in the range of a few hundred kHz up to something like 2 MHz. If the converter switches to PDM for low load currents to run with higher efficiency the ripple will be more in the range of a few kHz (the exact frequency will change with load current).

Because the USB rail has such a "wide" tolerance range you should not use a simple boost (or step-up) converter, but you must use a buck-boost. That way it can either increase or decrease the voltage. Or you opt to always boost to a higher voltage like 5.5V and than use an LDO, than a simple boost converter will do.

  • \$\begingroup\$ "don't use an LDO to create the 3V3 and then increase that again with DC/DC, that makes no sense" Why does it make no sense? I think this action allow us to use a Step-up DC-DC instead of a Step-up/Step-down DC-DC. Is't it? \$\endgroup\$
    – Arseniy
    Sep 13 at 14:37
  • \$\begingroup\$ Fair point, that might actually be a plus for this configuration. I hadn't considered this. But be aware, that the DC/DC will induce some noise to the 3V rail in this config. \$\endgroup\$
    – jusaca
    Sep 13 at 14:47

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