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I'm prototyping with a development board which has a 3.3 V rail and I have an IC that needs to be partially powered at 1.9 V. From my research, LDO linear voltage regulators are ideal for this application. I am unsure as to how to select the current rating for the regulator.

Background

The IC has one power input at 3.3 V, one power input at 1.9 V, and a third power input close to but smaller than 1.9 V. Ignoring the 3.3 V input (because I'll be taking that straight from the 3.3 V rail) the 1.9 V power input pin VDD has a typical operating current at 1.7 mA and the other power input pin LED_P is for an integrated LED which requires 50 mA. The datasheet also specifies that a 5.6 Ω ± 1 % resistor should be used between the 1.9 V source and the LED_P pin to obtain the 50 mA current.

Additionally, the datasheet specifies the maximum (0 to 1.9 V) current for the VDD pin is 70 mA.

Armed with this information, I'm trying to select a voltage regulator to create a common 1.9 V rail for both VDD and LED_P pins from the 3.3 V rail that is available.

Questions

Based on the information above, I have the following questions below.

Do the following factors fully describe the current-specific decisions that need to be made for a regulator, or are there more factors to consider?

Design current

Since VDD runs at 1.7 mA and LED_P requires 50 mA, my selection should be based on a nominal current of 50 mA + 1.7 mA ≈ 52 mA, is that correct?

Considerations for maximum current of VDD pin.

Should the maximum startup current of VDD (70 mA) factor into my selection?

If so, what total value should affect this? Would it be max(70 mA (VDD), 50 mA (LED_P)) = 70 mA or 70 mA (VDD) + 50 mA (LED_P) = 120 mA?

Should this number correspond to the regulator's nominal or maximum current rating? For example, just picking a regulator that looks promising, the LP2951 has a typical current limit of 160 mA and maximum current limit of 260 mA. Would this not be too high?

Considerations for dropout voltage

I think this is straightforward but just to double-check, I should aim to select a regulator whose dropout is more pronounced outside of my target operating current of 52 mA, is that right? Taking LP2951 for example again, its dropout is 380 mV @ 100 mA:

Dropout graph for LP2951

This doesn't look like the best but also wouldn't necessarily be the worst, is that right?

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  • \$\begingroup\$ Perhaps you would get more specific answers if you told what chip you are using. \$\endgroup\$
    – Justme
    Commented Feb 10, 2023 at 7:34

1 Answer 1

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You need to design for the maximum current. (You do not want the LDO to shut down during startup.)

If the datasheet does not clearly tell you whether the maximum is 70 mA or 120 mA, then you have to ask the manufacturer, or measure, or use another device, or just assume the worst case of 120 mA.

The dropout voltage is the minimum different between input and output (and a bit more voltage typically allows better regulation). Your circuit has 3.3 V − 1.9 V = 1.4 V, which is more than enough.

The LP2951 has a rated output current of 100 mA. The current limit is a last-ditch protection feature to prevent it from burning up.

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