5VSB TOP244PN principle of operation explanation - Page 1
Forum section: electronics / repair

author: tony359
author_rank: Frequent Contributor
timestamp: September 24, 2020, 08:36:08 pm
content: Hi all,I've recently repaired an SMPS from a server. It went bang and I discovered that the 5V Stand By section had failed. A Zener diode on the circuit had gone open and the offline switcher TOP244PN had basically exploded.I do not have a schematic of the PSU but I feel the typical schematic found on the TOP244PN datasheet can apply.I would like your help to better understand how the circuit works - for my own education.If I am not mistaken the two diodes (standard and zener) connected to the "D" pin (Drain of the integrated Mosfet) act as a snubber network as, when the Mosfet opens the circuit, the magnetic field in the transformer is still there and, while it collapses, it generates very high voltages which would be otherwise reaching the Drain of the mosfet, potentially damaging it.The Zener diode in question is a 150V one.So out of the bridge I have roughly 390VDC. When the Mosfet closes the circuit, the current flows through the transformer, creating a magnetic field. Then the Mosfet opens the circuit and here the snubber network should prevent dangerous voltages to happen.Can someone try to explain me how that zener diode prevents this high voltage from happening and why a shorted zener would cause havoc? I hope this is not too much of a silly question!ThanksTony

author: drvtech
author_rank: Regular Contributor
timestamp: September 25, 2020, 12:09:20 am
content: You've just about got it - when the switch is on, the regular diode at the bottom is reverse biassed so all the current flows through the inductor, storing energy in the magnetic field. When it switches off the field collapses and ideally all that energy would be transferred to the output, but in the process the voltage on the drain of the switch rises. Because the bottom end of the coil is now generating a positive spike the bottom diode is forward biassed so you can forget it's there. There is, therefore, a positive spike across the zener which is reverse biassed so it doesn't conduct until the spike reaches the zener voltage. The zener clamps the voltage across the coil to (in your case) 150V. Now note that this voltage is positive with respect to the supply rail so the total voltage across the switch is the supply voltage plus the 150V. This means the switch has to be able to withstand approx 550V. If we didn't put a zener there and just had the bottom diode a lot of energy would be lost in shorting out the spike, making the supply very inefficient. By the way, that is exactly why we put a protection diode across a relay for example - it absorbs the spike when the relay is switched off.As for the devastation you observed, it's difficult to know what failed first but if the zener initially went open circuit then the voltage spike would have killed the switch. Or perhaps if the zener goes short the core saturates too soon on the next cycle and causes overcurrent in the switch. Not really sure about the chain of events. Hope that helps anyway.

author: tony359
author_rank: Frequent Contributor
timestamp: September 25, 2020, 06:25:51 pm
content: Thank you for taking the time to explain me, appreciated.I read your message yesterday and I could not understand the "supply + 150" concept. I was writing a reply now and I suddenly got it! It is very clear now thanks!I see what you mean when you say that a shorted Zener doesn't cause anything obviously wrong at first sight. It must be as you say. Well, it's working nowCheers again!