LM64460-Q1: 5V fixed output model (CPP), no-load testing resulting in IC overheat and destruction

Hi David,

The links you have provided do not work. Can you try again to upload the images?

Ben

Thanks for the heads up, I fixed the links

Hi David,

The links are now blocked by TI's firewall. Can you upload the pics by attaching them to this thread?

Ben

Hi David,

Thanks for the pictures. Do you have any scope shots of the SW node, VOUT, inductor current and VIN? I see that pin 7 is tied to GND so the part has spread spectrum on with AUTO mode. With no load, the part should be in PFM mode and a scope shot of the SW node should confirm that.

Also, I recommend checking your design with the LM64460-Q1 quickstart calculator: https://www.ti.com/tool/LM63460-LM64460DESIGN-CALC

From the layout, are the pads (that I have circled) shorted or not shorted when the inductor is soldered to the board?

Ben

Hi Ben,

The circled pad is from the SW1/2/3 node, and this is not shorted to GND when a fresh LM64460 is soldered to the board. Both times I've tried removing current limits on my bench supply the IC has overheated and then a short forms from SW1/2/3 to GND until the IC is replaced again.

Here are the measurements I took last week of operation at Vin=~3.8V (the highest Vin I could achieve in CV mode without entering CC when my bench supply is current limited to 100mA). 

Vin:

SW1/2/3:

Vout:

SW4:

Cboot:

At shorter timescales all I was seeing was HF noise, no clear switching pattern. E.g Vin:

For inductor current, what is the suggested method to measure this? I don't have access to a suitable current probe but I could possibly bodge in a sense resistor. Any suggested value? The measurements above I think suggest a high inductor DC current, higher than seems possible given the inductor stays cool to the touch - I'm not sure how to explain that.

Hi David,

The pads to the left and right of the center pad are GND. The center pad is SW. Are you sure when you are soldering down the inductor to the board that the pads on the left and right of SW pad are not shorted?

Other than what you have told me, I the setup looks correct. It is difficult to measure inductor current without a current probe.

Ben

Hi Ben, they're not shorted. The GND is not actually a pad because there's no exposed copper. Only SW pad is exposed and Vout on the opposite side of the inductor. Sorry, it's not clear in the image.

So nothing looks out of the ordinary with the scope measurements I posted?

Hi David,

Do you have the EVM so you can confirm your setup conditions on a different board? 

VIN scope shot: The ripple is very large. I would try adding more capacitance on VIN. Try adding a 100uF bulk capacitor.  Are the ceramic capacitors close to the IC? A pic of the board will be helpful.

SW1/2/3: Can change x-scale and zoom in on a single switching event? This scope shot looks very different than SW4. 

Vout: Looks like SW1/2/3.

This might be a silly question but before the part was placed on the PCB, were you able to ohm out the pins and traces on the PCB to ensure the board is laid out correctly? Is the PCB board a 2-layer board?

Ben 

Hi Ben,

I assembled another copy of the board from scratch and this time added a bulk capacitor per your suggestion. I also tried to run with a 7 ohm load, and no current limit this time on my bench supply in case the limit was preventing a proper startup.

The result was the IC smoking as soon as 12V input was applied. 

Unpopulated board:

The assembled board before power up:

Destroyed after powerup

Bottom of burned pcb

This is indeed a 2 layer pcb, I expected reduced performance as a result of not using the recommended 4 layers, but not complete failure to operate.

I did ohm the board beforehand and didn't see anything unexpected. I don't own the EVM but it has different design (e.g.  inductance), so I'm not sure how I'd use it to determine how to fix this design. 

Hi David,

David Grant said:

I did ohm the board beforehand and didn't see anything unexpected.

Before adding the IC onto the PCB, you were able to ohm every pin and make sure there were no shorts from VIN to GND or any other pins? Did you repeat the same task after the part was soldered down onto the board?  From the pic that shows the bottom of the PCB, it seems that there was a short from VIN to GND around the exposed pad area. Is the exposed pad on the PCB connected to GND through the three vias?

Ben

Hi David,

Just a thought, try to use very little solder paste on the exposed pad. I am thinking there might be too much solder around the exposed pad and the extra solder paste is shorting VIN/SW to GND under the package.

Ben

Hi Ben,

The exposed pad is connected to GND through the 3 vias. Vin runs fairly close to these 3 vias on a trace on the bottom layer. I wonder if there is ringing on Vin during startup if this this could form a short. 

I skipped using solder paste on the exposed pad completely for this test. I think it's mainly needed for thermal reasons and I'm not testing a large load so I thought better to skip it to avoid any shorts. 

Assembly process was to first solder the IC, then ohm the board for shorts, then solder other components. A short may have formed when reflowing.

I'll assemble another copy, adding IC last, and check everything. 

Hi David, 

Ohm every pin again after adding the IC. I am curious what is causing your error. Keep me posted.  

Thanks,

Ben

Hi Ben,

I ohm'd the burned board and it looks like there's a short from Vin to SW1/2/3. I didn't test this resistance before applying power so I don't know for sure if it failed to a short or if the short pre-existed. 

I think it's unlikely this short is caused by solder under the IC based on pin positions and neither node being shorted to GND as well. 

As a reminder, on previous failures, SW/1/2/3 was found shorted to GND, so this is a new failure mode.

(The bolded lines are higher risk to short due to pin proximity)

I will assemble another board and run this kind of meticulous measurement prior to applying power.

Regards,

David

Hi David, 

This is good information. Also, I noticed the burnt via (exposed GND via) on the back side of the board. Do you know how that happened? If you were to  disabled the IC (EN = 0V) does the IC still experience a catastrophic failure if VIN is applied to the PCB?

Ben

Hi Ben, 

I desoldered the failed IC and confirmed the short from SW1/2/3 to Vin is inside the IC.

I'm not really sure the cause of the burnt via, my guess is this is proximate to where overheating occurred within the IC and the heat conducted into the exposed pad and down this via.  

It's interesting that the bottom of the IC is scorched closer to pin20 and 16, but the top of the IC has a protrusion closer to the right side of the chip (viewed from above)

I'll see if EN can be easily grounded for testing

Hi David,

Is the exposed pad on the PCB the correct size for the exposed pad on the IC? Just trying to figure out what is causing your issue.

Ben  

I used the ECAD file from TI's website.

...which matches the datasheet.

Hi David, 

Thanks for the confirmation. By the way, what inductor are you using? Is it an Eaton 5.2uH inductor? Another option is to not solder down the inductor when applying VIN to the new board.

Here are a few options to try (in order):

1. check PCB board first
2. solder down IC and check again for shorts
3. Do not solder down inductor
4. EN=0V, with a current limited power supply
5. if not short is found with EN=0V, enable the device
6. Try soldering a resistor in place of inductor (10kohm). Check current from power supply. 

Ben

Thanks, I tried to scrape at soldermask to to access the EN trace but it is too small, I may need to print another board design to try EN=0V. 

While I wait I may try your idea of running without inductor to see if the problem only comes when it is in place.

Regarding the inductor it is 5.8uH 17mohm DCR from the Webench BOM. Part # is Eaton DR1040-5R2-R.

I checked the Excel-based Quickstart converter design tool and it shows that 5.8uH is below inductance for 35% ripple at Vin.

Is this a problem? I used 412kHz F_SW because this is what Webench says it will be given these operating specs, I don't see in the datasheet where PFM mode frequency scaling calculation is given.

HI David, 

I thought you were using the LM64460CPP part, correct? If so, the frequency is set internally at 2.1MHz. See below:

Also, you have pin 7 shorted to GND so you are setup for AUTO mode which means the part goes into PFM mode. In PFM mode, the current limit is:

Ben

Hi Ben, I am using the LM64460CPP, yes. However in the datasheet it indicates

and since I am mostly operating at 0.1A, with occasional peaks to 0.7A, I expect I will always be in PFM mode. This seems to suggest the switching frequency will be reduced, but I don't know what to. Webench says the frequency will be around 400kHz for this part.

If this is incorrect and the switching frequency does NOT reduce below 2.1Mhz, then the efficiency predicted by the excel tool for light loads like mine is very poor (less than 65%) and this is not the right part for my application even though Webench reccomended it to me.

Please confirm. 

I assembled a board with no inductor and checked all resistances (see below). I powered with 12V and 100mA current limit. I am observing <1mA current draw and seeing 0V on SW1/2/3 with no switching activity. VCC is only 0.5V. EN is still tied to Vin.

David Grant said:

If this is incorrect and the switching frequency does NOT reduce below 2.1Mhz, then the efficiency predicted by the excel tool for light loads like mine is very poor (less than 65%) and this is not the right part for my application even though Webench reccomended it to me.

Yes, you are correct about the quickstart calculator not outputting light load efficiency information. Regarding the switching frequency, I see from your layout and schematic that the MODE/SYNC pin is tied to GND. This sets the IC into AUTO mode which will allow the part to decrease its switching frequency at light loads. The switching frequency will be very low as the load decreases to 0A. At light load, the part will operate like a COT (Constant On-Time) and the high-side FET will only turn on if the divided down output voltage falls below the reference voltage to the error amplifier.

Ben  

David Grant said:

I assembled a board with no inductor and checked all resistances (see below). I powered with 12V and 100mA current limit. I am observing <1mA current draw and seeing 0V on SW1/2/3 with no switching activity. VCC is only 0.5V. EN is still tied to Vin.

VCC is only 0.5V? Something is not right about VCC. Can you remove the capacitor C5 from the board and measure VCC again?

Ben

Ben C said:

Yes, you are correct about the quickstart calculator not outputting light load efficiency information. Regarding the switching frequency, I see from your layout and schematic that the MODE/SYNC pin is tied to GND. This sets the IC into AUTO mode which will allow the part to decrease its switching frequency at light loads. The switching frequency will be very low as the load decreases to 0A. At light load, the part will operate like a COT (Constant On-Time) and the high-side FET will only turn on if the divided down output voltage falls below the reference voltage to the error amplifier.

So if we've established switching frequency will be below 2.1Mhz, how can I determine what the switching frequency will be for the range of light loads I expect, and how can I use this in the quickstart excel calc to determine inductor requirements?

Ben C said:

VCC is only 0.5V? Something is not right about VCC. Can you remove the capacitor C5 from the board and measure VCC again?

I removed the IC and resoldered it, now I have 3.3V at VCC with C5 in place.

This looks normal: I have Vin (12V) present at SW1/2/3, with a switch event happening every ~10us, dropping to 0v for about 0.2us then back to 12V. Tomorrow I'll try connecting a 10k resistor in place of L1.

Ben C said:

Yes, you are correct about the quickstart calculator not outputting light load efficiency information. Regarding the switching frequency, I see from your layout and schematic that the MODE/SYNC pin is tied to GND. This sets the IC into AUTO mode which will allow the part to decrease its switching frequency at light loads. The switching frequency will be very low as the load decreases to 0A. At light load, the part will operate like a COT (Constant On-Time) and the high-side FET will only turn on if the divided down output voltage falls below the reference voltage to the error amplifier.

So if we've established switching frequency will be below 2.1Mhz, how can I determine what the switching frequency will be for the range of light loads I expect, and how can I use this in the quickstart excel calc to determine inductor requirements? I'm not sure if I should trust the 5.8uH recommendation from Webench.

Ben C said:

VCC is only 0.5V? Something is not right about VCC. Can you remove the capacitor C5 from the board and measure VCC again?

I removed the IC and resoldered it, now I have 3.3V at VCC with C5 in place.

This looks normal: I have Vin (12V) present at SW1/2/3, with a switch event happening every ~10us, dropping to 0v for about 0.2us then back to 12V. Tomorrow I'll try connecting a 10k resistor in place of L1.

David Grant said:

So if we've established switching frequency will be below 2.1Mhz, how can I determine what the switching frequency will be for the range of light loads I expect, and how can I use this in the quickstart excel calc to determine inductor requirements? I'm not sure if I should trust the 5.8uH recommendation from Webench.

At light loads, the part is not at FPWM because the MODE/SYNC pin is tied low. In order to keep efficiency high, the switching frequency drops and the IC will switch over to a COT like control scheme where the part will only need to switch on if the output voltage droops. If the inductor has a higher inductance, more of the inductor energy (0.5*L*I^2) energy will get delivered to the output capacitors and basically raising Vout voltage. A higher output voltage will cause a longer "sleep interval" which will cause the switching frequency to slow down more. 

Regarding WEBENCH, it would be good to check their recommendation inductor value against your own calculation.

David Grant said:

I removed the IC and resoldered it, now I have 3.3V at VCC with C5 in place.

This looks normal: I have Vin (12V) present at SW1/2/3, with a switch event happening every ~10us, dropping to 0v for about 0.2us then back to 12V. Tomorrow I'll try connecting a 10k resistor in place of L1.

This is great! We are finally taking a step forward.:-)

I would check to make sure the SW pins are not shorted to GND or VIN. What is the current consumption at the moment? I hope it is less than a 50uA. 

Ben

I took a chance and went straight to returning the inductor to the board. The good news is that the current consumption did not increase to several amps as before, and the IC did not burn. 

The other good news is that I now have a steady voltage on Vout

The bad news is that Vout is not 5V, it is equal to Vin (it's not shorted to Vin). As I increase Vin from 8 to 12V Vout follows. 

I can still see the switched output on SW1/2/3 and a smoothed version of this on Vout. The duty cycle on SW1/2/3 is very high... 99% or more. It is not adjusting to regulate output to 5V.

I confirmed that Vout is present on pin 5 (FB). Any ideas?

Ben C said:

What is the current consumption at the moment? I hope it is less than a 50uA. 

It is about 50mA.i hope this will decrease when the regulation is working properly. 

Ben C said:

At light loads, the part is not at FPWM because the MODE/SYNC pin is tied low. In order to keep efficiency high, the switching frequency drops and the IC will switch over to a COT like control scheme where the part will only need to switch on if the output voltage droops. If the inductor has a higher inductance, more of the inductor energy (0.5*L*I^2) energy will get delivered to the output capacitors and basically raising Vout voltage. A higher output voltage will cause a longer "sleep interval" which will cause the switching frequency to slow down more. 

Regarding WEBENCH, it would be good to check their recommendation inductor value against your own calculation.

I understand but I'm unsure what switching frequency to use in my calculation. As you say it will be reduced from 2.1MHz due to my very light load, but to what? Fsw is critical to inductor selection, right? 

David Grant said:

I confirmed that Vout is present on pin 5 (FB). Any ideas?

Is there a short on somewhere? What is the Fsw?

David Grant said:

It is about 50mA.i hope this will decrease when the regulation is working properly. 

I still think there is a short somewhere. Can you measure CBOOT's voltage?

David Grant said:

I understand but I'm unsure what switching frequency to use in my calculation. As you say it will be reduced from 2.1MHz due to my very light load, but to what? Fsw is critical to inductor selection, right? 

When you calculate the inductor, assume full load with a Fsw of 2.1MHz. I would not worry about the light load condition.

Ben

Ben C said:

David Grant said:

I understand but I'm unsure what switching frequency to use in my calculation. As you say it will be reduced from 2.1MHz due to my very light load, but to what? Fsw is critical to inductor selection, right? 

When you calculate the inductor, assume full load with a Fsw of 2.1MHz. I would not worry about the light load condition.

I suspect my highest load (0.7A) will be a light load condition. Webench suggests this because it shows reduced Fsw in this operating scenario. It would be good if you can confirm. 

David Grant said:

I suspect my highest load (0.7A) will be a light load condition. Webench suggests this because it shows reduced Fsw in this operating scenario. It would be good if you can confirm. 

Yes, that is correct. Looking at the efficiency curves in the datasheet, the part will be in light load mode at 700mA. Figure 9-2 is the figure I am referencing too:

Ben

I think I'm missing something. Figure 9-2 suggests fsw doesn't reduce at light load but from our discussion earlier, it does.

Hi David,

If you set the part to FPWM by connecting the MODE/SYNC pin high, the efficiency at light loads will be a lot lower. Here is an example of a comparison between AUTO mode and FPWM mode.

To achieve higher efficiency at lower output load currents, the switching frequency has to be reduced.

Ben

HI David,

Wow, this has been a great learning experience for the both of us. Glad to be of help.

Regarding the switching frequency in light load, I would suggest reading up on COT control schemes. Our ICs exhibits COT behavior when it is in light load mode. 

I am glad that your setup is finally working.:-)

Ben