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I've been trying to educate myself on general power supply operation and stumbled across several schematics for ATX power supplies. In general I understand how a SMPS operates, and an ATX PS is just one example. However, in reference to the following schematic, I am totally stumped on several fronts:

enter image description here

And enlarged:

zoomed image

  1. Why is the primary of T1 fed off the un-rectified input 'AC2'? Why should it not be connected to the + output from the bridge rectifier? From what I've seen on other SMPS, the primary side on the main switching transformer is fed by a DC source and switched on/off at several kHz usually by a low side power NFET.

  2. I also don't understand the point of coupling the un-rectified AC line signal (through ZNR1) into the center point of the filter network formed by C1/C2/R2/R3

  3. How can any current flow in T1's primary at all? Cap C9 blocks all DC on T1's primary.

  4. I understand T2 is an isolation transformer; Q6 and Q5 run off the PWM from the controller, and by way of T2 provide current to turn driver transistors Q3 and Q4 on and off. What I don't understand is what's happening as a whole on T2's secondary: Is the pin #10 a center tap? An off-center tap? Why not just have one NPN turning the current through T1's primary on and off? Why is pin 5 connected to the lower half of T1's primary?

I understand I am lacking some basic knowledge here about how this works, so if you may be so kind please direct me to books or other tutorials on what is happening here. Thanks!

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  • \$\begingroup\$ AC2? which component is this? \$\endgroup\$ – Andy aka May 17 '13 at 22:55
  • \$\begingroup\$ @Andyaka: The bridge rectifier "BD1" is near the top center of the schematic. Its top leg is labeled "AC2". AC2 is more-or-less directly connected to the (unrectified) mains AC power connector labeled "L" (for "Line") at the top. \$\endgroup\$ – davidcary May 18 '13 at 3:33
  • \$\begingroup\$ ZNR1 is a "metal oxide varistor", for surge protection. It doesn't couple until the AC voltage exceeds normal levels. Why it's coupling over there rather than to ground I'm not sure. \$\endgroup\$ – pjc50 May 18 '13 at 9:49
  • \$\begingroup\$ Sorry about the size of the schematic; S.E. won't allow me to post a picture until I have rep=10. \$\endgroup\$ – smoothVTer May 20 '13 at 2:27
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    \$\begingroup\$ A few guesses ... If you start at the output side and look at how T1's secondaries are connected, each output looks like a center-tapped 2-diode full-wave rectifier setup. With this in mind, you'd want to drive T1's primary with a symmetrical AC waveform (hence C9 ensuring no DC). Q3 & Q4, while both being driven through T2, also provide the drive to T1 through T2's pin 5. But that's about as far as my brain can stretch on a Friday afternoon ... \$\endgroup\$ – brhans Oct 17 '14 at 20:54
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The schematic SmoothVTer included is based on an Asymmetric PWM Half Bridge Converter design. There are papers (search for "asymmetric half bridge converter") and documents (e.g. Fairchild's application notes) explaining the design details. Here are the specific answers to your questions:

  1. Actually T1 is driven by the halved rectified DC voltage after the bridge rectifier BD1. You can see that if you remove the ZNR1 line altogether (the voltage select switch ZNR1 will be opened anyway when the input supply is 240V. I will explain the use of that below.) Therefore it is indeed the + and - DC output from BD1, halved by the pair of resistors R2 and R3 that is driving the power transformer.

  2. The line with the ZNR1 switch going between R2/R3 C1/C2 allows the circuit to optionally operates as a simple rectifier (ZNR1 open) and/or Voltage Doubler (ZNR1 short.) See wiki's Voltage Doubler and/or Voltage Multiplier articles for more details. The varistor ZNR1 is often a physical switch for selecting between 240V (switch opened) and 110V (switch closed) in many ATX PSUs. (I am not sure if ZNR1 is really a varistor because that would make the switch open at 110V but short at 240V...) The two big caps (C1,C2) after the bridge rectifier BD1 is operating as a [Voltage Doubler][3] in the 110V (i.e. ZNR1 open) case. So the DC voltage across R2+R3 is always ~340V (240V*sqrt(2)) whether the input is 110V or 240V.

  3. How can any current flow in T1's primary at all? Cap C9 blocks all DC on T1's primary.
    

    Of course current can still flow in T1 with a capacitor, the voltage is being switch on and off all the time by PWM so it is effectively AC! Transformers relies on current/flux changes to operate. So cutting out the DC component is fine.

  4. T2 is just operating as an isolation transformer (to keep the controller chip side away from dangerous high voltages) for the switching signals for operating the two NPN power transistors Q3 and Q4. The entire block between T2 and T1 (and including T1 actually) is the source side of the Asymmetric PWM Half Bridge Converter. The coil section between 10 and 5 is just used as the additional inductor connector in series needed for that circuit. The number of turns between 9 and 10 is the same as that between 6 and 7 but 5-10 is really a different inductor so 10 is generally not a centre tap.

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i will answer for the first question only: enter image description here This is the basic smps circuit, t3 and t4 is the driver PWM t1 and t2 is power amplifier of the input signal generated by the pwm as push pull amplifier, the ladder R1, r2 is to get the half dc voltage -at no signal T1, t2 is tuned off because thair base connected to ground, - when we applied a positive signal transistor T2 turned on and t1 turned off c=1uf will charge, the current will pass trough R1,T2 because vce ot T2 is approximatly =0 enter image description here

  • when we applied a negative signal transistor T1 turned on and and t2 turne off c=1uf will discharge, the current will pass trough R2,T1 because vce ot T2 is approximatly =0

the out put transformer will driven by ac current and rectified by a diode to get dc voltage

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