tps40170 start up problem

Hi Jae,

Could you provide the schematic and layout?

Regards,

Na

Hi,

Here I attached a  schemtic and layout.

The component which is not a value is not used.

When input power is on by switch ,R12 and  TPS40170 pin3 is burn out.

Regards,

7180.tps40017.xls

Hi Jae,

I think the major reason for the blow-up of R12 and M/S pin is the large AC ripple on the input capacitance (C2-C8). The input capacitance is not big enough.

For a rough calculation, when Vin = 36V, Vout = 12V, Iout = 20A, Fsw = 250kHz, the peak-peak ripple on C2-C8 can be estimated as

Iout *(Vout/Vin) / (Fsw*Cin) = 20A * (12/36) / (250kHz * 2.2uF * 7) = 1.73V

Assume R12 and C18 are enough to filter out the AC ripple on VIN pin of the IC, the peak-peak AC current in R12 will be

1.73V / 2.2ohm = 786mA

Then, the power on R12 is

(786mA/2)^2 * 2.2ohm/2 = 0.17W

The power stress on R12 is high. Similarly, the large AC voltage ripple applied to M/S pin will induce large AC current on the internal capacitance and burn the traces. Hence, more input capacitance will definitely help.

Regards,

Na

Hi, Na

First, thanks to  your tip.

But I have another queation.

That problem happens in no load condition.

I think ac ripple current is not large at no load.

what do you think about this?

Regards,

JAE

Hi Jae,

There are two things I can think of to contribute to the input ac ripple.

1. At start-up, the input source needs to provide not only the load current, but the charge for output capacitance. But with the parameter provided in the schematic, i.e. Vout = 22uFx10,Css = 330nF, the average charging current is only 12V*22uF*10/0.09*330*1e-3) = 0.089A.

2. The inductor current ripple is not considered in my previous reply for simplicity. If considering that, the inductor ripple is about (Vin-Vout)*Vout/(Vout*L*Fsw) = 6.8A @ Vin = 36V, Vout=12V, L=4.7uH, Fsw = 250kHz. The input voltage ripple is only Vout*6.8A/(8*Vin*Cin*Fsw) = 0.07V.

These cannot explain the failure on R12 and M/S pin. The layout also looks good. I would suggest you measure the input ripple and the current in R12 on your board to get some clue.

Regards,

Na

Hi Na,

Here I attached a ripple voltage and current.7725.tps40017_wave.xls

Hi Jae,

From the waveforms, the power stress on R12 at full load is apparently high. In my calculation, I didn't take the derating of input capacitance at high DC bias voltage into consideration. Due to the derating, the effective input capacitance is even smaller, which can cause higher ripple voltage and more power loss in R12 eventually.

At no load, the waveforms are also something as expected and I didn't see why R12 can be burned out. 

Anyway, to have more input capacitors will help.

Regards,

Na

Hi, NA

Your evaluation module PMP6680 has almost same capacitor value(from C2 to C9 : 2.2uF X 8).

I think the difference is a  large capacitor of  C1(220uF).

Is threre any possibility due to input surge when switch on.

Hi Jae,

Does the failure happen frequently, or just on one part? Could you check whether there is abnormal positive or negative voltage spike on the M/S pin, input capacitors before and after L1 as the part starts to switch? The wires from the input supply to the input socket on the board should be as short as possible.

Regards,

Na

Hi NA,

The failure happen just on time.

Once the module start ok without failure, it does not happen during operation.

I agree with your opinion that more capacitor is needed but I think tps40170 is very weak to spike.

Hi NA,

The failure just happen on time.

Once the module start ok without failure, it does not happen during operation.

I agree with your opinion that more capacitor is needed but I think tps40170 is very weak to spike.