UCC28910: won't start

Hello Nguyen,

Please correct me if I am wrong, but I presume that the UCC27325 IC (in green) is a placeholder for the UCC28910 for pcb layout purposes. Let’s separate this issue into two independent problems: start-up and regulation.

Removing D8 allows the circuit to start. D8 is a clamp across the primary winding to limit the peak voltage on the drain of the switching MOSFET (internal to the UCC28910). This peak voltage is the result of leakage inductance energy that does not couple to the output winding(s). If the value of the clamp voltage is too low, D8 will act as a load during the flyback phase and absorb most of the energy that was intended to go to the outputs. The clamp voltage should be significantly higher than the reflected voltage during the flyback phase of switching. The schematic indicates SMBJ100A which may be too low of a clamp voltage for your application. Removing it allows the leakage energy to spike up the drain voltage unclamped, but since the UCC28910 MOSFET is rated for 700 V and since the input bulk voltage is only ~163Vdc, the leakage spike probably does not exceed the MOSFET rating. Now, all primary energy can be transferred to the output windings, including the Auxiliary winding which powers the IC. As a test of this, temporarily connect two D8s in series, and expect that the circuit should start up because now the clamp operates at twice the voltage and should not interfere with the reflected voltage and load down the primary winding.

For the regulation issue, this may be the result of unbalanced loading. Please refer to theuCC28910 datasheet, Section 10.2.3 (page 42), for information about working with multiple outputs. Although that section uses two positive outputs as an example, your configuration can be treated equivalently. In addition, at no-load on the outputs, the auxiliary winding and control IC can be considered to be yet another “load” on the transformer. Since primary-side regulation (PSR) regulates the output by indirect sensing of the reflected voltage on the Aux winding, an unloaded output may float a little higher and the IC will effectively regulate the auxiliary bias voltage. This is because the average Aux loading becomes a higher load than that of the output(s). You show a 2K pre-load on each output, which is ~11mW total. I suggest to increase the pre-loads to 1K each or more, to ensure that the output loading dominates the PSR waveform that the IC is trying to regulate. Adjust R26 as needed to obtain 3.3V at no (external) load. Then add your normal load in small increments to see how the output voltage varies as the load is increased from zero.

Regards,
Ulrich

Hi Ulrich,

Thank you for your help. I connected two D8s in series and the circuit couldn't start. I also tried three D8s in series but it still couldn't start.

-Nguyen

Hello Nguyen,

I'm sorry that idea didn't work.  I have another suggestion: try reducing the value of the VDD resistor R28(?) from 120ohm to 60 ohm or 30 ohm.  Also, leave the 3 D8's in series in place.  I do think that a single 100-V TVS clamp voltage might be too low, in any case, even if that alone is not the root cause of the no-start problem.   

If lower VDD resistance doesn't cure the start-up problem, I'll need to look at VDD waveforms for more clues.  Also, please provide the input and output line and load conditions applied during the start-up, the specification for the transformer, and a clearer schematic drawing.  The one in the posting is a little fuzzy.    

I hope the reduced VDD resistance does the trick.

Regards,
Ulrich

Thanks Ulrich,

I'll try your suggestion to see if it resolves the start-up issue.  I currently don't have the access to oscilloscope, but I will post more waveform once I have access to it.  I forgot to mention that the power supply design is from TIDC-CC3200SMARTPLUG (page 6).  I used the same transformer as the reference design and component values as well.

-Nguyen

Hello Nguyen,

Thank you for the information that your design is basically a copy of the TIDC-CC3200SMARTPLUG power supply section. That tells me that the design is already proven and we can concentrate on your physical implementation of it, instead of the schematic. 

Since you had mentioned earlier that removing the TVS-clamp D8 allows your circuit to start and that installing 3 D8’s in series still prevents it from starting, I wonder about the effect of D8. You had also mentioned that you had checked the soldering and component orientations, but I suggest double-checking the orientation of D8.  If it were installed backwards, that would prevent start-up because it would clamp the reflected winding voltage to about 2 diode drops and the VDD capacitor could be kept charged by the aux winding voltage.  It is simple enough to double-check.

If that proves fruitless, then we’ll need to look at waveforms to make any progress. Useful waveforms include the DRAIN voltage (U1-8), VDD voltage (U1-6), and AUX winding voltage (U13-4), all with respect to GND1 (U1-1).  Be careful of the high voltage on the DRAIN switching node and around the input section in general.  You will need to have an isolated AC or DC source so that the oscilloscope GND can connect to the directly primary-side controller GND1 net.  Failure to use an isolated source may result in high circulating currents and damaged circuitry or equipment.  Or you can use isolated high-voltage differential probes, but these are uncommon and cumbersome.

Regards,
Ulrich  

Hi Ulrich,

I reduced R28 to 60 ohms and 30 ohms and the circuit still couldn't start. I also double checked the orientation of D8. I rotated D8 around and the output voltage is still zero. Unfortunately, I didn't have an isolation transformer or high-voltage differential probes so I couldn't observe the waveform on the primary side.

I put D8 back and removed D2. The circuit starts up so I think removing D2 has the same effect as removing D8. So I decided to put a potentiometer in series with D2 (on the anode side) and vary the resistance of the potentiometer from 0 to 10 Kohms. Right around 3.3 kOhms, the circuit starts up. I measured the positive output voltage with different load (each LED draws about 24 mA) as below:

First LED: positive output voltage dropped from 3.24 V to 3.15 V
Second LED: voltage dropped from 3.15 V to 3.14 V
Third LED: voltage dropped from 3.14 V to 3.13 V
.........

Under light load, the new resistor in series with D2 seems to be fine, but around 200 mA, the resistor heats up. If the output is short-circuited, the new resistor seems to be fine and I think the power supply is in protection mode. The new resistor resolved the start-up issue but it starts to heat up under heavy load is my concern. I'm not entirely sure if placing a series resistor in series with D2 is a proper way to fix the issue.

-Nguyen

Hi Nguyen,

Adding the series resistor with D2 is not a proper solution if we do not know WHY it helps, however, it does provide another clue, and I have another idea for you.

Removing D8 helped it start. Removing D2 help it start. Adding high enough R in series with D2 helps it start. This is pointing to the clamp circuit as the location of the problem. The TIDC- design uses RS1G-13-F at D2, which is a medium-fast recovery diode with trr of about 150 ns. Your schematic shows this part number and an alternate 1N4004 diode. The 1N4004 is a standard recovery diode with trr maybe 2~3 us long.

If you are using the 1N4004 in your circuit, this extra reverse recovery time may be letting high peak current back through each switching cycle when the supply tries to start up. This heats up the diode and high junction temperature makes trr worse. Meanwhile, if the diode hasn’t fully recovered when the next switching pulse starts, the remaining (unwanted) recovery current could be tripping the current sense limit of the IC, which would cut short the pulse width and starve the outputs of power, including the auxiliary winding. Hence it could be attempting to start but keeps shutting down with continual restart attempts (you need an oscilloscope to see this). Adding the series resistance higher than ~3.3K is high enough to reduce the reverse current and heating of the diode and enough energy makes it through to the outputs to start up and stay running.

Please check to see if you have a 1N4004 diode at that location and if so, I suggest to replace it with a fast recovery diode. The RS1G seems to be fast enough, but if you can’t get one, there are many other diodes that can work there. It does have to be rated for at least 400V, 1A, and trr< 150ns. Faster ones are okay, higher V okay, higher I okay, (but size might become cumbersome even as a test case). Are you using a surface-mount part or axial part? If you have limited availability of parts in the university environment, you can tack-solder either package type (whichever you are able to get) in place with short jury-rigged connections. Other possibilities are: 1N4936, 1N4937, UF4004, US1G, MURS140, MURS160, MUR140, etc… Many of these are far faster than actually necessary for the application, but the point is (if my hypothesis is correct) to not use a diode that is too slow.

Regards,
Ulrich