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UPDATE

I've provided a full outcome report in one of the answers below with an updated schematic and description of the operating principles as I've come to understand it.


I'm studying switching converters to feed a strange craving to understand how they work. I'm just getting to the part about off-line AC-DC converters in the books, but being a practical sort, I thought I'd open up one I have handy and see what I could explain so far.

Here's what it looks like after opening:

enter image description here

And here's the schematic I reverse-engineered from it:

enter image description here [click to expand]

Here's what I think I understand so far. All component labels are as printed on the PCB:

  • C1 gets charged to roughly 170V DC by the line bridge rectifier and supplies the input current.

  • B1 is the transformer (no idea why it's not T1). B1P12 is the primary winding terminating at pins 1 and 2. I believe this is the main primary inductor/winding.

  • R3, C3, and D7 comprise a snubbing network for the main inductor. The "R1A" designator signifies a "rectifier-style diode, about 1A in size". I'm not able to see the markings without desoldering it, which I wanted to postpone for now. Also, given the provenance of the other parts, I'm not sure I'd discover much.

  • R6 provides base current for U2, the main switching transistor (a TO-220).

  • U1 is a base driver for the main switch, shunting base current when turned on. This is a TO-92.

  • Moving to the output, D10 (LED) and R11 provide indication when output voltage (nominally 12V) is present on the output.

  • C8 is the output capacitor.

  • B1S (secondary) is the only secondary winding and pulls current out of the negative end of C8 during the off stroke, providing the output energy. D9 blocks reverse current through the secondary.

Here's what I don't understand yet:

  • There's no clock/oscillator. How the heck does it switch periodically? The only thing I can think of is some resistor and capacitor make up an RC circuit or something.

  • What does B1P34, the second primary winding (on pins 3 and 4) do? I've heard of these being used to power a \$V_{CC}\$ rail, but there are no ICs in the circuit to power. Maybe it provides bias current for the opto and base driver or something?

  • I expect that D11 is a zener, maybe 11.5 V or something. I can't tell by inspection; it just looks like a signal diode package. But it makes sense to me in that location to turn on the opto when \$V_{out}+\$ goes above 12 V or so. I don't get what R10 does though.

  • I also don't get what C5 or C7 do, but I've probably asked enough.

Can a more experienced eye help me decode some of this?

share|improve this question
up vote 10 down vote accepted

Well done so far.

R6 is too big to provide all the base bias to U2 in normal oscillation, but it does 'tickle it into life' at start up.

There is no clock because it's self-oscillating. That's what the B1P34 winding is for, through components like D5,8 and R2. This network is disabled when the opto turns on.

When U2 starts to turn on, the feedback is such that it turns on harder. It stays on with the current growing steadily in B1's inductance. Eventually B1 becomes saturated, when two things happen. U2 collector current increases rapidly as the transformer inductance collapses, and the feedback voltage starts to drop for the same reason. U2 comes out of saturation, and the collector voltage rises rapidly. This is fed back and U2 starts to turn off. The feedback now turns it off harder. U1 takes part in this as well by shorting the BE junction to remove the base charge rapidly. This flyback phase ends eventually when the core has transferred its energy to the secondary. I haven't analysed it completely, but I suspect it's the R6 bias that restarts the whole conduction cycle.

R10 is to pre-bias the zener. Zeners don't have a sharp turn-on curve, they can draw quite a few uA at volts below their rated voltage. R10 keeps the zener well into conduction, so the turn on of the opto is better defined.

This doesn't answer all your questions, but may redirect your investigations. Try redrawing the components around B1P34 to emphasise their feedback role.

Bear in mind that some components' function may not be obvious, if they've been added to reduce EMI for instance.

share|improve this answer
    
Awesome! Very helpful user44635! :) – scanny Jan 29 at 7:18
1  
Aha! So your "self-oscillating" pointer was a key clue, I was having trouble on search, finding any circuits that looked anything like this one; but have now come upon the term 'ringing choke converter' from the Wikipedia page when I searched on 'self-oscillating converter'. Now I'm seeing circuits that look very much like this. Thanks very much user44635 :) – scanny Jan 29 at 7:26
    
Ok, I've made a lot of progress I think based on your guidance; I've added a full outcome report below with an updated schematic in case you want to see what I came up with :) – scanny Jan 30 at 8:29

OUTCOME REPORT

Based on the very helpful answer of @user44635 I was able to make substantial progress in understanding this circuit.

The critical link was the notion of "self-oscillating", which led to the search term "self-oscillating converter" and from there to "ringing choke converter" (RCC). This resource was especially helpful: http://mmcircuit.com/understand-rcc-smps/

I've redrawn the schematic below based on user44635's advice to emphasize the feedback role. I've changed some of the symbol names to more conventional designations, e.g. U1 -> Q1:

enter image description here (click schematic image to expand)

Here's my expanded understanding of the operation:

  • C1 gets charged to roughly 170V DC by the line bridge rectifier and supplies the input current.

  • T1 is the transformer, with a primary, secondary, and auxiliary winding.

  • Q2 is a power transistor in the role of main switch. R3, C3, and D7 form a snubbing network to protect the switch by dissipating the 'switch off' transient. Switching on is soft.

  • R6 provides "startup" base current for Q2 to begin the on stroke. As Q2 turns on, current flows though T1_PRI, inducing a voltage across T1_AUX (dot end positive). Current flows through D8, R7, and R2, rapidly turning Q2 on hard.

  • A positive voltage appears across T1_AUX during the on stroke. This causes a current to flow through R5 and charge C6. When that charge reaches the \$V_{BE}\$ of Q1, Q1 shunts base current from Q2. The resulting decrease in voltage across T1_PRI causes the voltage across T1_AUX to decrease as well. This lowers the Q1 base current further and that feedback loop rapidly turns Q2 off hard. This interaction of R5 with C6 determines the oscillation frequency, which is proportional to \$\frac{1}{R_5C_6}\$.

  • As in any flyback converter, the energy stored in the core is released in T1_SEC once \$\frac{d\phi}{dT}\$ changes direction at the transition to the off stroke. This charges the output capacitor C8 which stores that energy for output.

  • While the voltage across T1_AUX is reversed, C4 is charged through D5. I believe this provides a "turn-on pulse" to Q2's base at the end of the off stroke, kick-starting the on-stroke.

  • Control is provided by the ~12 V zener D11. When \$V_{out}\$ (via R10) rises enough to switch on the zener, current flows through R9 to energize the optocoupler. R9 limits current through the opto LED. When energized, the opto phototransistor provides base current to Q1 which then shunts Q2 base current. This ends any current on-stroke early and delays the start of the next one until the optocoupler is de-energized.

  • On the output side, D10 (LED) and R11 provide indication when output voltage (nominally 12V) is present on the output. D9 prevents reverse current flow through T1_SEC as is conventional for a flyback converter, allowing T1_PRI to accumulate flux in the core during the on-stroke and preventing discharge of the output capacitor C8.

  • I presume C5 performs an EMI suppression role, but don't understand the specifics of that yet.

  • I expect C7 bypasses noise in the secondary that might otherwise find its way to the output.

Special thanks to user44635 for setting me on the right track!

Let me know if I've got any of this wrong :)

share|improve this answer
    
Not wrong, just not quite the right emphasis. Q1 does not simply 'shut the base current off', but more actively pulls the stored base charge out of the BE junction, that accumulates when Q2 goes into saturation, that if not removed quickly would lead to a delay in Q2 turning off, with resultant higher dissipation in Q2. It is that stored charge that made saturating TTL logic slow, leading to schottky clamped logic to prevent transistor saturation, and the development of non-saturating logic like ECL. – Neil_UK Jan 30 at 8:59
    
@user44635 -- Ah, I see now I completely neglected the role of R5. I'm thinking now that R5 forms an RC circuit with C6 and this must be what determines the switching frequency by turning on Q1 when C6 charges up to Q1's \$V_{BE}\$. That makes total sense now. So the opto would just bypass that timing network to skip cycles or whatever, but Q1 would be active in each cycle in either case. And that switch-off would be hard and fast because of the depletion role you mention. Have I got that right? That was definitely a missing piece of the puzzle :) – scanny Jan 30 at 9:19
    
You're ahead of me now, <hoarse breath> the student is now the master! </hoarse breath> Like I said, I haven't analysed it completely, am just spotting the elements that are obvious to me, and will give you a leg up. Cycle skipping as you suggest sounds entirely plausible, I thought the opto turning everything off in the feedback sounded a bit crude. – Neil_UK Jan 30 at 10:20
    
I've updated the circuit operation description based on these comments. – scanny Jan 30 at 18:53

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