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TLV62130 Efficiency

Charles Gervasi
Expert 2270 points
Other Parts Discussed in Thread: TLV62130

I have a circuit using the TLV62130 to charge a battery.  The input voltage is 5.0V and output is 4.0V.  The input current is 1.5A, and the output current is 1.5A.  I expect the efficiency to be a little higher.  As a result the chip rises 45C over ambient even though it's connected to a ground plane.  This is consisten with the thermal resistance in the datasheet: (5V - 4V) * 1.5A * 29.1 = 44C rise.  

My question is why I'm getting only 80% efficiency and how I can dig into the circuit and find what I'm doing wrong.  

CJ Gervasi (No relation to the CJ in Austin) :)

  • 0
    TI__Guru**** 150650 points
    How and where did you measure the voltages and currents? There are voltage drops across cables and meters that cause the voltage at the IC's pins to be different than what it says on the meter. The voltages need to be measured on or very near the IC to measure the IC's efficiency.
  • 0 in reply to Chris Glaser
    Expert 2270 points
    I measured it at the input and output caps. Tomorrow I could verify I'm as close as I can probe. I also may try and X7R output cap in place of the X5R cap in the circuit now.
  • 0 in reply to Charles Gervasi
    TI__Guru**** 150650 points
    The caps likely won't change anything, unless you are measuring at an elevated temp?

    How did you measure the currents? Is it possible to take a photograph of your setup?
  • 0 in reply to Chris Glaser
    Expert 2270 points

    Chris Glaser said:
    How did you measure the currents?

    I just measured both of them.  I'm measuring voltage at the input and output caps with a DMM.  I'm measuring output current with a cheap DMM and measuring input current with the bench supply.  

    1.59A * 4.88V = 7.8w
    1.78A * 3.93V = 7.0w
    Efficiency = 90.2%

    1.50A * 4.90V = 7.35W
    1.78A * 3.67V = 6.5326W
    Efficiency = 88.9%

    I'm seeing a 34C rise in temperature on the plane under the part.  This is more than I expect.  No matter how you slice it, though, rise will be (power in - power out) * thermal resistance.  Our ground board and gound plane are 3.0x0.7 inches.

    Does any of that sound unexpected to you?  

    Thanks!

  • 0 in reply to Charles Gervasi
    TI__Guru**** 150650 points
    Are the two readings from 2 different circuits? The output voltage is different for each, so this is confusing.

    I suggest measuring the input current with a meter as well. Most of the input supplies I have seen are not calibrated and can be off.

    But your measurements do show input current < output current. This is correct.

    You can compare your circuit and its configuration to the D/S efficiency curves or to Webench to see if your results are close to what you should expect. They look like it to me.

    There is likely other power lost on your PCB--any other power supplies and any loads. All of this power will heat up the PCB.
  • 0 in reply to Chris Glaser
    Expert 2270 points

    Chris Glaser said:
    Are the two readings from 2 different circuits? The output voltage is different for each, so this is confusing.

    We're using this part as a charger for a Li-Ion battery from a 4.75V source.  We're using the tracking pin to control the voltage.  The efficiency is about 90%.  The heat is raising the temperature of our thermistor and making it hard to know the temperature of the battery.  We are adding better coupling of the thermistor to the battery in the next spin.