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I have identified the LDL1117 (specifically the LDL1117S33R) to reduce a 4V DC voltage to 3.3V DC. The max current draw is 500mA. I had initially selected this part because I read the dropout voltage is 350 mV @ 1.2A.

However on reading the datasheet in detail the electrical characteristic section says that: VIN = VOUT + 1 V or 2.6 V. Since my output voltage requirement is 3.3V, does this mean my minimum input voltage must be 4.3V? If yes then I am confused as to what the dropout voltage of 350mV means.

Can I use this part?

The datasheet can be found here.

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2 Answers 2

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You read the datasheet correctly, the drop-out voltage is what matters regarding your 4V input voltage requirement. The statement that "VIN = VOUT + 1 V or 2.6 V" is just stating the conditions under which the data in the table following was taken. For example, the noise rejection numbers vary strongly with the voltage drop and are much worse when you are close to the drop-out voltage.

Be careful that 350mV is only the typical value and could be a lot higher according to the 'max' value. But as your current requirement is a less than half the maximum, this should not be a problem and the actual drop-out voltage should be around 200 mV typical (although this is not stated in the datasheet).

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  • \$\begingroup\$ Dropout condition is mentioned in datasheet. It is when output has dropped 100mV from nominal output. \$\endgroup\$
    – Justme
    Aug 28, 2021 at 11:36
  • \$\begingroup\$ @Justme Yes, but this particular datasheet doesn't mention anything about the relation of drop-out voltage and output current. \$\endgroup\$
    – asdfex
    Aug 28, 2021 at 11:48
  • \$\begingroup\$ Drop-out voltage is measured at 1.2A \$\endgroup\$
    – Justme
    Aug 28, 2021 at 12:01
  • \$\begingroup\$ @Justme And this is the reason I wrote the last sentence. The OP doesn't want to operate at this current. \$\endgroup\$
    – asdfex
    Aug 28, 2021 at 12:17
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The test conditions for accuracy, tempco etc are taken with Vin = Vout+1 V.

Note that the 350 mV dropout voltage is a typical, not a guarranteed, specification.

That still leaves the guarranteed 600 mV dropout specification as an oddity since it seems to be measured under the same test conditions of Vin = Vout+1 V.

Note that the dropout voltage is defined as the voltage across the device which results in the output voltage dropping 100 mV from the programmed voltage. Is that still 'working' as far as you are concerned?

However, with you planning for 700 mV across the part, at less than half the maximum current, you will almost certainly be able to use it successfully.

Do bear in mind the tolerance of the 4 V input, and the tolerance on the 3.3 V output voltage, both of which will eat into your 700 mV device drop.

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