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I want a switching regulator (eg LM2596) set to 5v output voltage to work with 5V - 36V input voltage.

I know regulators have min dropout voltage. So how does it behave in such a case (Vin=5V when Vout=5V)?

What is the best practice of this application I want?

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3 Answers 3

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It depends on the specific regulator.

Some buck regulators have 100% duty cycle mode. In that case, the top FET turns on, and it's just like having a resistor between Vin and Vout with a value equal to the RDSon of the FET.

Some buck regulators have a maximum duty cycle, (often so they can charge the bootstrap cap). In that case, the part would not be able to regulate or even keep the input close to 5V.

Using a 4 switch buck-boost regulator would allow you to regulate the output voltage when Vin is above and below Vout, so that could be a solution for your application.

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Depends on type. LM2596 for example, I believe can run up to 100% duty cycle, but has a Sziklai type output stage, so drops about a diode drop minimum, or about 5.8V minimum input (unless the control needs more; and certainly more at higher load current, hence the 7V recommended minimum input).

Most bootstrap (NPN/NMOS) buck converters will not run up to 100%, as the bootstrap supply needs less than 100% to remain charged and active. The minimum pulse width (high/low) or maximum duty is usually given, which will in turn tell you the dropout.

Complementary types (PMOS switch) can run up to 100% with low dropout (Rds(on) limited), though they might not for other reasons; they're also less common, and seem to have less desirable controls (like voltage-mode or hysteretic control). YMMV.

You can certainly design one yourself, to behave that way; even using NMOS switch, using a charge pump or isolated supply to power it statically. This is usually more work than it's worth, though (or there are controllers specific to the application, like wired-OR or hot-swap controllers).

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Most DCDCs configured in buck mode will try to operate at 100 % duty cycle -- i.e. the high-side switch will be 100 % on.

In the LM2596, since the output is a BJT device, this means that the output can't rise to closer than ~ 0.8 V of the supply. So with 5 V in, you would get 1bout 4.2 V out maximum.

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