Need Help Understanding SMPS - Page 1
Forum section: electronics / beginners

author: Adam60
author_rank: Regular Contributor
timestamp: December 11, 2016, 05:30:00 am
content: Good Evening. I have a few SMPS units kicking around with the same problem but I wanted to first understand a few things about these units. I am a newb when it comes to power supplies and I sort of half assed understand them, but still some things make no sense to me. I have attached a diagram of part of an SMPS unit and would like a clear explanation of what is going on in this portion of the circuit. What I am puzzled by is the voltage on one side of D4 being -165 VDC and on the opposite side of the diode being -33 VDC. I would expect them to be relatively the same as the voltage drop across the diode should be 0.4-0.7 V, correct? So I am not understanding something. Perhaps someone could explain in a voltage sense how this circuit works. I realize frequency comes into play somewhere, still bending my mind around that one. One thing at a time. Thank you for the help.These voltages are from a working unit. The dead ones are blowing Q1 or Q2, can't remember which one or both.

author: oldway
author_rank: Super Contributor
timestamp: December 11, 2016, 06:37:46 am
content: As long as you measure a continuous voltage without ripple or a 50 or 60 Hz sinusoidal voltage, you can trust what you read on your multimeter.But we must be careful in other cases.Your multimeter also seems to have a problem as it should indicate 0V and not -0.010V.As for your measurement of -33Vdc, it is a false reading for one simple reason: the voltage at this point is not a DC voltage, but it is a PWM waveform that ranges from -165V to + 165V at a frequency of about 25Khz and your multimeter is unable to make a correct measurement of this waveform.You need an oscilloscope with differential probe to make such a measurement.

author: Adam60
author_rank: Regular Contributor
timestamp: December 11, 2016, 06:50:58 am
content: Thank you Oldway for your reply. I can live with that. Can you explain how it becomes a PWM signal? Like I said, I am a newb and I work on voltages making transistor turn on and off. Now I have to learn about frequencies.

author: salbayeng
author_rank: Frequent Contributor
timestamp: December 11, 2016, 06:54:27 am
content: Hi Adam.Your circuit looks a bit like a self oscillating compact fluorescent lamp half bridge inverter.Some more circuits and explanations here:http://www.pavouk.org/hw/lamp/en_index.html.Basically the bottom half makes +/- 166v DC as you have already determined.The node you measure as -33VDC is actually swinging up and down (50kHz) between +166 and -166v, it spends slightly more time at -166v , hence the average voltage is ~ -33v.So for the moment, assume D1,R8,C12 are just a short circuit, and assume R6 and R7, D3 are non existent, ditto D2,R12,C13 are shorted, R11,R10,D4 are open circuit.....----first half cycle----If the top of T1 swings up say 1v, current will flow into the base of Q1, and Q1 will turn on, in the process it will pull "-33" up to +166, and draw current through the middle winding of T1, this causes even more voltage at the top of T1, ensuring Q1 stays hard on. After about 10uS T1 saturates, and the voltage drops to zero, this forces Q1 off, and without any current through the middle winding, the top winding of T1 goes negative (turning off Q1 even more), at the same time the bottom winding goes positive (it's drawn wrong here, there should be a dot symbol that is anti-phase to the top winding).---next half cycle---So when the bottom winding is positive, Q2 turns on , and drags -33VDC to -165v, this time the current flowing through the middle winding pulls the bottom winding more positive turning Q2 on harder, after another 10us the flux collapses, Q2 turns off. and the cycle repeats.--------------So now you have a +/- 166v square wave at "-33VDC" , this voltage is then applied to the primary of T2 , and the load is connected to the primary====================So that's the important part, the rest of the components are "garnish"(a)C10 is used to ensure no DC flows through T2 (else it would saturate),(aa) additionally C10 can be used to make a resonant circuit with the inductance of T2, this will reduce switching losses of Q1 and Q2(aaa) if driving a compact fluorescent, C10 can cause resonance when the lamp is cold , creating a very high voltage that causes lamp to strike.D3, D4 are free wheeling diodes, they conduct should the voltage at t2 exceed +/-167vC11 and R4 are a snubber, basically the leakage inductance and interwinding capacitance of T2 ring at several MHz, so to suck up this energy we use a snubberD1 R8 C12 are used for waveshaping, to provide a fairly constant pulsating base drive to Q1, with a max positive voltage of say ~ +1v , and going -1v negative with low mains voltage to -3v with large mains voltage, so R8/R7 gets warm with high mains voltage, R7 is there to discharge C12 on the negative swings.R6 and R10 are for startup, a small trickle folows through each, cause both transistors to partially conduct, and they have a tug o war, the winning transistor gets to work the first half cycle.You can control the power output of the convertor by periodically shorting out the left winding of T1, crude, but effective.Hope that helps , Bob.

author: salbayeng
author_rank: Frequent Contributor
timestamp: December 11, 2016, 07:20:13 am
content: If you are having troubles with these PSU's , the first thing to try would be replacing C12 and C13 , preferably with bipolar units.Under certain load and line combinations they will be reverse biassed and depolarise.The Apple][ power supply had this problem when you filled all the card slots.

author: oldway
author_rank: Super Contributor
timestamp: December 11, 2016, 08:16:03 am
content: I think it is rather a circuit of a switching power supply of a computer.This circuit is controlled by a TL494.To understand how it works, see data sheet and application notes of TL494

author: salbayeng
author_rank: Frequent Contributor
timestamp: December 11, 2016, 01:20:47 pm
content: Yep I checked the transistors later, seem to be designed for ~3A operation , so ~ 200w power supply.The extra winding on the drive transformer threw me off course!Here's a similar schematichttp://www.eleccircuit.com/switching-supply-computer-5v-15a-12v/And one from Pavel , showing the extra winding on the drive TFR.http://www.pavouk.org/hw/atxps.gifThis guy made his own , but easier to see how it workshttp://www.qsl.net/xq2fod/Electron/PS40/Figure1.gif

author: Adam60
author_rank: Regular Contributor
timestamp: December 11, 2016, 03:04:40 pm
content: Bob, thank you for that great description of operation. I wish more people had that type of patience when asked that question. I understand it much better with your explanation. Now I must figure out how to look at it with a scope. A little intimidating as scope skills are limited and only have a couple of older analog scopes. Thanks guys for your help.

author: salbayeng
author_rank: Frequent Contributor
timestamp: December 11, 2016, 11:21:48 pm
content: Hi Adam,As Q1 and Q2 do all the heavy lifting, they will fail for most types of faults. With bipolars , a lack of drive will cause them to overheat and fail promptly.You will need an isolating transformer for the AC before you can scope anything out on the convertor side, however all the waveforms will be shimmying because of the ripple on the filter caps, and because the TL494 is regulating, you won't be able to make much sense of it. You should also solder short lengths of insulated wires to all the points you want to measure, having a probe slip can have spectacular results!The finger is still pointing at c12, c13. Remove some of these from failed units and check their capacitance (and check date code , usually first two digits are year) (check whether they have 85c or 105c rating) If your multimeter can't measure capaciance then you should buy one that can, an LCR meter is even better, Failing that, if you rig up a small transformer to make 2- 5VAC at 60Hz, then connect the capacitor and a 2.7kohm resistor in series across that, the AC voltages across 1uF and 2.7k will be the same at 60Hz.A better way to test these SMPS is (obviously remove any shorted transistors first, check C10 isn't shorted, and make a note if the failure is shorted or open (the latter means the emitter lead has blown open and R5 R7 and c12 are suspect) you could just cut the collector lead if it's too awkward to extract the part )Procure power supplies to make 50-60v (two 30v supplies stacked up)(could be unregulated) but a bench supply with current limiting is useful (you will see something even as low as 12v here, but you won't be able to load up and check regulation until about 60v,) a 12.0v current limited supply, some dummy loads (car taillight bulbs, and some car headlight bulbs, with wires soldered on.)(a) connect some longish blue and orange wires across the main filter caps, then connect these up to a 60VDC power supply . So basically +/-165v is now +/-30v . Connect red and black wires to the filter cap on the auxillary supply this is usually a 12v supply, and may actually be the +12v PSU output, apply 12.0v here from another supply, also solder another red wire to the pilot supply capacitor (e.g. C23 on the ATXPS.gif) and put 12v in there too. You might need to strap the power button if the supply has one.(b) withjust the 12v supplyconnected, the TL494 should be running at close to 50% duty cycle, you should see square waves on T1, Check for the voltages around the base drive circuit (connect scope ground clip to the emitter) R5 should have ~ 1.0v on the left end and 0.6v on the base end for the positive half of cycle (this is ~ 200mA into the base of Q1) and maybe negative 2 or 3 volts for the negative half cycle. The waveform will decay away a bit particularly on the negative half cycle (maybe a volt of droop) If you see a rapid decay then the 1uF capacitors are dry.(c) now wind up the big supply slowly and with a scope (ground to middle of supply caps) check the waveform on Q1 emitter, you should see a squarish wave here swinging up and down to the full extent of your supply. With no load on the 5v output, the supply should start regulating as the "5v" lead reaches 5.0v , if you can't get to 5.0v , try back driving 5v into this lead from a fourth power supply (or maybe a resistive divider from your 12v supply to make ~ 6v)(d) If you have enough volts on the big supply to get regulation, then try loading up the 5v with some light bulbs.Note with power on the "big supply" you only connect the scope ground safely to either side of C9, Top side to measure most waveforms, and the bottom side to measure around Q2, (you can't connect it to "-33v", unless the big supply is off)At this point you can safely and effectively probe around the circuit.Most common faults will be dried up caps, and dry joints particularly on the big parts (transformers, anything on heatsinks, connectors), eyeball with a strong magnifying glass, Best just to smear flux all over the bottom of PCB and resolder everything.