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Original question:

LP5912: LP5912 Vdrop

LP5912-Q1: LP5912 Vdrop

Kami Huang60
Genius 3830 points
Part Number: LP5912-Q1

Hello Sir/Ms.

The datasheet has a description of the following paragraph.

VIN = VOUT(NOM) + 0.5 V or 1.6 V, whichever is greater; VEN = 1.3 V, CIN = 1 μF, COUT = 1 μF, IOUT = 1 mA (unless otherwise

stated).(1) (2) (3)

What's the VIN = VOUT(NOM) + 0.5 V or 1.6 V mean? Vin must over 1.6V?

My client's application environment is Vin 1.8V, Vout 1.5V and Vdrop observed. Vdrop showing following fig.

Yellow Output voltage. Blue System Load (Iout). Green  SoC trig signal.

They observed that Vout would drop 0.2V when Iout was from 0A to 0.3A. 

Is this voltage normal?

Does TI have software to simulate a Voltage drop? How to get it?

Best Regards,

Kami Huang

  • 0
    TI__Genius 10200 points

    Hi Kami,

    Unless otherwise indicated, the data collected for the Electrical Characteristics table was taken at the listed operating conditions. For example, if Vout(nom) = 1.5 V, the data was collected with Vin = 2.0 V, since Vout(nom) + 0.5 V = 2.0 V > 1.6 V. You are correct in that the minimum recommended input voltage is 1.6 V for this device:

    The output is likely dropping during the load transient because the regulator does not have sufficient headroom (Vin - Vout) to regulate the 1.5 V output when the load current is 300 mA. In other words, the dropout voltage for this device at Vout = 1.5 V and Iout = 300 mA is larger than 300 mV. Note that the dropout voltage was not characterized for output voltages lower than 1.6 V due to the minimum input voltage requirement of 1.6 V. This is because for small output currents, the dropout voltage approaches 0 V. In order to comply with the minimum recommended input voltage of 1.6 V, the output voltage when measuring dropout voltage should also be at least 1.6 V.

    Transients, dropout voltage, and other characteristics can be simulated with spice models found in the "Tools & software" tab of the product page:

    http://www.ti.com/product/LP5912-Q1/toolssoftware

    The dropout voltage when Vout = 1.5 V can be simulated by applying Vin = Vout + 0.5 V = 2 V with a 300 mA load and then decreasing the input voltage until the output voltage falls to 1.35 V (150 mV below the nominal 1.5 V output). The dropout voltage is then calculated at this operating point as Vdo = Vin - Vout. 

    Thanks,

    Gerard