TPS62864: Output regulation drops to 0.5V under load.

Hi,

Thanks for sharing your schematic and issue.

Can you share a photo of your setup, including the meters where you measure the voltages and currents?

Your input impedance may be too high or the current limit on your input supply set too low, so that Vin is dropping with the load.  Can you look at Vin and Vout with a scope, with and without the load?

Thanks,

Chris

Hi Chris,

Please see attached picture of our setup.

Here is the measurement tabulated:

Plane input impedance around the switching frequency is very low:

Thanks,

Arnav

Hi Arnav,

Thank you for sharing your board and the data.

Some of the trace appear thin.  Can you share the actual top copper around U10?

What is the programmed voltage and current measured in your table?  I don't know what these refer to in your design.

How are you calculating the efficiency?  Can you share the raw data: Vin, Iin, Vout, and Iout?

Do you have an oscilloscope to measure some waveforms?

Thanks,

Chris

Hi Chris,

We have done an IR drop simulation and there should be sufficient copper. I will dig up the artwork and simulation for you.

Programmed voltage is what we set via I2C interface. 

The current measured is from the Agilent supply, we can see how much current is being drawn from 3.3V. 

To calculate the efficiency we theoretically know how much current is being drawn from U10 by the resistors. Then simply calculate the power drawn from U10 and divide by the power drawn by the 3.3V supply. 

Vin = "input measured V"

Iin = "Current measured" we can only measure 3.3V current

Vout = "Output measured" 

I out = We have no way to measure this

Also no way to prove this tiny board on a scope. I have some headers for communication IO, but not for power.

Thanks,

Arnav

Hi Arnav,

Thank you for explaining.

This is not a good way to measure efficiency.  If you are just attaching a 286 mOhm resistor and assuming some amount of current is drawn, you don't actually know how much current is drawn.  Especially with such a small resistance, the solder, jumpers, traces, etc. can easily add additional resistance which would decrease the load current.

In addition, there are probably other devices on the 3.3V bus and they are drawing current.  This should not be included in the device's efficiency calculation, since the current goes to other devices.

If your power supply is set to 3.3V, then why is the voltage you measured 3V or lower?  Please try to touch the tip of the scope probe to the exposed terminals on the inductor and/or caps to see what the voltage is actually doing.

Thanks,

Chris

Hi Chris,

I understand it is not a great way to measure the current (FYI we checked and subtract out other devices that draw current from 3.3V by pulling all the regulator enable down), but we have to make do. The simulation shows us that the plane resistance is 13mOhm (yes quite high) from the regulator to the resistors. Do you know the symptoms if the regulator reaches close to maximum current output? We could be close to 4A output, for example 1/(.286+.013) = 3.3A. 

I don't understand why the voltage is 3V or lower at the input. The resistance of that 3.3V plane is only 2.5mOhms from simulation. It should not drop voltage even if the current drawn is excessively high. We will try to solder some coaxial wire to look at the 3.3V input on the scope. 

Thanks,

Arnav

Hi Arnav,

I don't believe you are reaching the current limit.

Yes, we need to find out why the input voltage is lower than expected.  I imagine is has to do with the connection from your power supply to the chip.  Can you post a photo of your whole setup?

Chris