MPS MP1584EN Buck Regulator

[RBeaubien]

I've built a schematic using the MPS MP1584EN regulator ( http://www.monolithicpower.com/Products/Product-Detail/101/DC-DCPowerConversion/SwitchingRegulators/Step-down(Buck)/MP1584 ), pretty much using their design tool.  Problem is the low voltage cutoff. It is about 2.6v.  I'm regulating to 10v at up to 2amps and the input battery power can drop down below 12v.  How do I change the design to switch to direct battery instead of letting the regulator simply shut down?  Any guidance would be appreciated.

Thanx.

[TrickyNekro]

Hello,

it could be that you´ll have to look at a Zeta converter implementation, theoretically you only need a High side switch,
which I think this IC has. You will be trading an inductor for a transformer, but a lot of manufacturers (like coilcraft for example) do
provide coupled inductors for something like this (they would specify them for example as SEPIC converter loosely coupled inductors).

It could be a bit more difficult to compensate for the feedback loop, but hey that the price you pay for choosing specifically this IC.

Hope this helps,

Cheers,
Lefteris

[RBeaubien]

Doesn't sound easy.  I'm certainly open to suggestions on a different regulator that will gracefully/gradually lower its voltage instead of turn off.  Input voltage from 11-28v battery, output 10v 2amps.  I'm sorta lost at this point as to what to use.

Thanx.

[bktemp]

Problem is the low voltage cutoff. It is about 2.6v. 

It don't see why this should be a problem. It only says the regulator should work as long as the input voltage is above 3V.

But your problem is the high duty cycle of >90% needed for generating 10V output voltage from 11V input voltage (page 8-9):

As long as VIN is sufficiently higher than SW,
the bootstrap capacitor can be charged. When
the power MOSFET is ON, VIN is about equal
to SW so the bootstrap capacitor cannot be
charged. When the external diode is on, the
difference between VIN and SW is largest, thus
making it the best period to charge. When there
is no current in the inductor, SW equals the
output voltage VOUT so the difference between
VIN and VOUT can be used to charge the
bootstrap capacitor.
At higher duty cycle operation condition, the
time period available to the bootstrap charging
is less so the bootstrap capacitor may not be
sufficiently charged.
In case the internal circuit does not have
sufficient voltage and the bootstrap capacitor is
not charged, extra external circuitry can be
used to ensure the bootstrap voltage is in the
normal operational region. Refer to External
Bootstrap Diode in Application section.

Did you add the external diode?

[Ammar]

Could you use a step up/step down converter or something like a cuk converter to ensure 10V on the output?

[RBeaubien]

Did you add the external diode?

That is the part of the datasheet I didn't understand.  If it is referring to D1, then yes, it is installed.  If not, then it must be the diode referred to on page 14.  Since BST is regulated to 5V, I am unsure how to connect this for my needs.  I am not operating this at the full 3amps and instead will be keeping the current demand under 1.5amps so I am unsure if the duty cycle will be above 90%.  What I didn't do was add the resistors to EN due to earlier instructions that stated I could leave it floating to enable the chip.  Per the section on page 14, I need to add the divider resistors to program the UVLO voltage to Vout+3V.

[bktemp]

Did you add the external diode?

That is the part of the datasheet I didn't understand.  If it is referring to D1, then yes, it is installed.  If not, then it must be the diode referred to on page 14.  Since BST is regulated to 5V, I am unsure how to connect this for my needs.

That's the diode on page 14. Try adding the diode to see if it helps.
I can't find any max duty cycle specification in the datasheet, so I'm not sure if the diode is enough to make the stepdown running at >90% duty cycle.

I am not operating this at the full 3amps and instead will be keeping the current demand under 1.5amps so I am unsure if the duty cycle will be above 90%.

The duty cycle depends on output voltage/input voltage ratio. As long as the converter operates in continous conduction mode, the duty cycle shouldn't change much when changing the load.

[RBeaubien]

That's the diode on page 14. Try adding the diode to see if it helps.

Where does it connect to?  VIN?  The diagram shows 5V.

Thanx for help on this.  The COMP and BST sections of this datasheet really confused me.  :-)

[bktemp]

It needs 5V. The 5V can be generated from either the input or the output voltage using a 78L05 regulator, or a simple zener diode shunt regulator.