Understanding TOP258PN current limit design

Øystein, please do not feel bad to ask what you do not know. We can not know everything. So it is good we ask and would be my pleasure if you can learn new things from me

I am not sure if I understand what you are measuring when you said

I have measured it to be:

207V AC input -> 12V DC current limit = 0.86A

230V AC input -> 12V DC current limit = 0.88A

253V AC input -> 12V DC current limit = 0.92A

The current you are measuring,  is it the DRAIN current (ID) or it is the Output current (IOUT) of the power supply? I think it is IOUT. Am I right

The DRAIN ILIMIT you set with the resistor connected to the M pin is not intended to set up the IOUT-LIMIT

I need you reply and tell me if I am right before I keep going. I think what you are trying to set is the maximum IOUT the power supply can deliver, am I RIGHT?

Alos would be nice if you can attach an schematic  with more resolution. I can not see the details

Hi,

The current I am measuring is the 12VDC (schematic net name V12VB) output current (Iout) of the power supply which varies with the AC input voltage. See uploaded complete schemtic with higher resolution.

My chinese colleague who desinged this tells me that you set the output current limit with resistor R44 connected to M pin, but if I understand correctly it sets the input current limit, not the output?

Our product is already in the market so we cannot do a redesign, but I really need to know how our output current limit will vary with AC input voltage (measured as described), temperature, component tolerances, other parameters? etc, so I know what to tell our customers , how much power/current out power design can deliver.

BR

Øystein

I have a better understanding now. 

The M pin IDRAIN_LIMIT It is not intended to be used to set the current limit of the output voltage.

IOUT chances as VIN changes

There is a  time delay for the controller to turn off itself when it detects ID reaches IDLIMIT. At high line voltage, the power supply delivers more power because the slope of IDRAIN is larger. A current with larger slope will reach higher IDRAIN_PEAK because the slop is larger and the time delay of the controller is fixed. Larger ID_PEAK means more power delivered to the secondary side of the power supply and more power to the load.

It surprised me how your designer did not implemented the circuit as shown in figure 36 of the data sheet.  With this simple resistor connected from M pin to high voltage DC bus you can have a much narrow window in the variations of the MAX PO of the power supply. 

To calculate what is your MAX PO your supply will deliver at high input voltage is a time consuming task.  Variations of MAX PO depend more on input voltage capacitance (tolerances) and on transformer primary inductance (tolerances)

One approximation I would use if I were you is to measure your maximum POUT at Maximum input voltage. Then whatever results you get add 10 %.  This number is a very educated approximation.   

Hi

Thanks. I need to have an idea about minimum and maximum variations in Power out. Can I use the same approxiamtion for minimum AC input voltage (207V AC), but instead of adding 10%, I substract 10%?

As an educated guess of course.

BR

Øystein

For your minimu POUT at MIN VIN you can use the same aproximation. Also  you can reduce your Vin 10% and then measure your POUT (at Vin=176VAC). Whatever POUT you are measuring. you can use it as your minimum POUT.

Hi

Just to clarify:

Our product needs to be guaranteed to customer that it will work with AC input voltage 230V AC +/- 10%. The AC/DC part will also source power to many different unit over a cable from the 12V DC output-. I need to figure out at least the minimum soucring capability for this design which we can communicate to over customer. So it is the minimum guaranteed Pout I need to find.

If I use 230V AC - 10% input = 207V AC then I measure maximum:

V=11.9V

I=0.76A (max average)

P=9W (max average)

-10% from this gives minimum guaranteed output power=9W * 0.9 =   8.1W

If I use 176V AC (which is much lower than I product is sold as, but the design will of course work at this volatage)

V=11.9V

I=0.72A (max average)

P=8.6W (max average)

Which one should I use as a qualified guess, 8.6W or 8.1W?

BR

Øystein

If your customer needs to deliver POUTMIN at 230 -10% = 207. Then you specify at this voltage. But also you can specify at the lower voltage.  In other words, You can specify POUTMIN at two or three or as many input voltages as you like to do. There is not a problem about it and is perfectly acceptable.   But also you can specify the absolute minimum POUT.  

Just make sure that when you are measuring your POUT , your power supply is thermally stable.  In other words , when you measure POUT, your power supply should not go into thermal shutdown or being overheated.  You can put a note when you specify your POUT that it is at ambient temperature

But my question was

If I measure Pout at 207AC input voltage to be 9W in the lab at 25 degrres celcius ambient, I should susbtract 10% from this (as an educated guess), and incustomer documentation promise that we can source at least 8.1Watt output power?

As I understood you I could use the edcuated guess and measure in lab at a specific aAV input voltage and then add or substract 10% to Pout to get maximu and minimu level.?

I reckon this is worst case scenarioes whichj will probably never happen, for all practical purposes the max, min Pout level window will be narrower?

Yes, it is a very educated guest if the ambient temperature does not changes much. If you power supply is rated for a wide ambient temperature, then I suggest you repeat the measurements at the spec ambient temperature window. Then you can specify absolutely minimum POUT at the worst case temperature.

Yes, it is a very educated guest if the ambient temperature does not changes much. If you power supply is rated for a wide ambient temperature, then I suggest you repeat the measurements at the spec ambient temperature window. Then you can specify absolutely minimum POUT at the worst case temperature.