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Buck Regulator VIN under required VOUT
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tomshirvo:
Hi All,
So I am working on a little project and I am interested in knowing if the regulator I have chosen (or any buck regulator) will work if you end up feeding in less voltage than you have configured the regulator to output.
I have a Buck regulator outputting 24V. (TPS54560DDAR) (http://www.ti.com/lit/ds/symlink/tps54560.pdf)
That 24V powers a 19V LDO linear and a 3.3V Buck
There could be a case where I would need to power the device with only a 24V power source that could drop down to 21V. ( Normally 36VIN )
The regulator (TPS54560DDAR) states after doing the calculations that the Minimum voltage required at my current would be about 25.3V.
What happens when I go under that... does the device stop working and not output any voltage at all?
Or does it just start outputting whatever voltage is coming in?
What if the VIN goes under the configured VOUT of 24V.
With my 19V LDO I can go down to 22V so if my VIN is 22V will the buck regulator just output that 22V?
Thanks in advance and I hope I have given enough detail :)
Zero999:
It depends on the topology of the buck regulator. The one you've linked to has a bootstrapped N-MOSFET output stage, which means it can't work at 100% duty cycle, which will probably limit the drop-out voltage more than the MOSFET's on resistance.
The data sheet says if the output voltage is close to the input voltage, the MOSFET will turn on, until the voltage bootstrap capacitor is too low, causing it to turn off for the time necessary to recharge it. The duty cycle will be near 100% but not quite. The calculation for the drop-out voltage is given at the bottom of page 13: 7.3.4 Low Dropout Operation and Bootstrap Voltage (BOOT)
http://www.ti.com/lit/ds/symlink/tps54560.pdf
In short, if the input voltage drops below the set output voltage, the output transistor will turn on for the maximum duty cycle possible. There will be some voltage drop, but not much.
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