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The following is the schematic of the SMPS stage of an old (mid 90s) CRT TV set that I often use to show my students how a real SMPS design might look like.

enter image description here

More ore less I've already understood the details of its workings, but a couple of components placed here and there (highlighted in yellow below) defy my full understanding.

enter image description here

Could someone please shed some light (if it is possible without further design notes) on why the designer put those components where they are?

The following is the internal block diagram of the controller chip (an obsolete TEA2261):

enter image description here

Here is what I think so far:

  1. C12 (270 pF): put there for stabilization purposes of the switching transistor T1 against oscillations/ringing at high frequencies.

  2. D4: it's in parallel with R13, which should be part of the snubber for the primary transformer winding together with C13. Could it be a freewheeling diode for when the switch turns off? In this case why is it not in parallel with the primary winding, but in that unusual (to me) position?

  3. D1: it's a Zener across the electrolytic cap C9. I would guess it's there as a protection against over-voltage across the cap, but it's rating is funny: it clamps at 3.3V whereas the cap can withstand 35V. Is it just a huge safety margin or could it have other functions?

  4. L1: really no clue at all, it just bypasses R3 (which is part of the AC-coupled voltage divider used to drive the base of the BJT switch). It lets DC pass, but DC is then blocked by C9. I'm puzzled. Could it be intended to interact with D1 in some way?

  5. C17: No clue. Why would a designer put a small cap (470pF) in parallel with a rectifier diode? My only guess is it somehow bypasses the junction capacitance, but why?


EDIT (n.2 solved!)

Thanks to a comment of G36, n.2 was sorted out!

R13,C13 and D4 together form what is called an RCD snubber network, also known as polarized turn-off snubber network. See, for example, this document (fig.5 at page 4). As in most documents I've seen, the D+R sits on top of the cap, but they are in series, so it doesn't change the overall network behavior.

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  • \$\begingroup\$ Did you ever heard about "Snubber" network ? obrazki.elektroda.pl/2881388100_1472413327.jpg also C9 and D1 helps turn-off BJT faster (the negative voltage at base). \$\endgroup\$ – G36 Aug 28 '16 at 19:45
  • \$\begingroup\$ @G36 This is spot on! I've never seen a snubber network with a diode in that position. Have you got a reference which is not a simple image? BTW, if you format your information as a question I'll be happy to upvote you. \$\endgroup\$ – Lorenzo Donati Aug 28 '16 at 19:59
  • \$\begingroup\$ As for C17 his job is to prevent the spikes caused by the rectifier reverse recovery characteristics to lower RFI emissions generated by diodes. Or yet another "snubber" network fairchildsemi.com/application-notes/AN/AN-6093.pdf (page 11 [step 10]) \$\endgroup\$ – G36 Aug 29 '16 at 16:07
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I think we can all do with partial answers:

Pretty sure D1 has the function of ensuring that if T1 is quick and clean to turn off as soon as the voltage drops. Also, notice that the C9 is polar and wouldn't like being subject to larger voltages "in the wrong direction".

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    \$\begingroup\$ Yes it's definetevely to speed up switch-off. D1/C9 parallel can be thought as a 3.3V DC battery series connected to the base. This shifts base driver output voltage swing from something like 0V-12V to -3V to 9V or so no matter dutycycle and hence average base drive. Negative off voltage speeds up base depletion. \$\endgroup\$ – carloc Aug 28 '16 at 18:09
  • \$\begingroup\$ @carloc Isn't reversing the BE junction by more than 3V dangerously close to BE breakdown? From the BU508AF datasheet I see max reverse BE voltage is just 5V. \$\endgroup\$ – Lorenzo Donati Aug 28 '16 at 20:51
  • \$\begingroup\$ @carloc Scrap that! It's 9V! I've got an older datasheet which says 5V! \$\endgroup\$ – Lorenzo Donati Aug 28 '16 at 20:53
  • \$\begingroup\$ @carloc Actually the Fairchild part has a 5V BE max voltage. Oh well, probably it is underspecced. I think reversing by about 3V could be safe after all. \$\endgroup\$ – Lorenzo Donati Aug 28 '16 at 20:57
  • \$\begingroup\$ Well from 3V to 5V is just 60% of rated voltage, plenty of headroom given no extra ringing on top of base voltage. But anyway that was just some ballpark estimate. Just consider around 3.3V below base driver low voltage ....nothing on IC datasheet, just reckoning, may be 0.5V to 1.5V so all in all we get somewhere between -2.8V to -1.8V \$\endgroup\$ – carloc Aug 28 '16 at 21:06
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C13/D4 might be a kind of non-linear high pass for over-current detection: If the voltage at the top of T1 rises very quickly, voltage drop across C13 should be low, and current through D4 should be high, quickly pulling up the Emitter voltage, hence reducing \$U_{BE}\$, hence reducing current flowing through T1, saving that from a potential burn out.

That is a bit speculative; one would need to test that hypothesis.

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  • \$\begingroup\$ Uhm I believe this is nothing more than a switch-off snubber. At switch-off it slows down vce rate of rise as \$\text{d}v_\text{CE}/\text{dt}=I_\text{C,peak}/C_{13}\$ since C13 is now charging through D4. This reduces switch-off energy or if you prefer you may say it shapes \$v_\text{CE}/i_\text{C}\$ switching locus to stay away from SOA boundary. Then at switch-on C13 discharges via CE and R13 to get ready for next cycle. Discharge takes about \$t_\text{dis}\approx 5\,R_{13}C_{13}=5 \times 220\,\Omega\times 1\,\text{nF}\approx 1\,\mu\text{s}\$ \$\endgroup\$ – carloc Aug 28 '16 at 20:02
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    \$\begingroup\$ @carloc Yes you are right! After G36 posted a comment with a link to a picture on my question, I did a bit of research. C13+R13+D4 form what is known as a RCD snubber, or also polarized turn-off snubber network. \$\endgroup\$ – Lorenzo Donati Aug 28 '16 at 20:35
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C12 and C17 are there "to take the edge off fast signals", to short any high frequency signals that occur when switching elements, like diodes, switch on/off very rapidly. Shorting out these high frequency signals prevents them from having to take a long path through other components (working as an antenna !) and causing too high electro-magnetic emissions.

L1 I think is used also to "to take the edge off" of the signal going to T1 so that it switches on in a slightly less abrupt way also to reduce EMI emissions/ cause less RF problems.

Note that a TV also has a sensitive RF frontend (the tuner) so the other circuit must keep the amount of high frequency disturbances low enough as to not disturb the signal reception.

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  • \$\begingroup\$ re L1: Hm, if I mentally turn the transistor circuit around, L1/C9 form an oscillating circuit with \$\omega_0=\frac1{\sqrt{LC}}=(5\mu\text H \cdot 100\mu \text F)^{-\frac12}= (0.5\text{ms}^2)^{-\frac12}\$... \$\endgroup\$ – Marcus Müller Aug 28 '16 at 18:53
  • \$\begingroup\$ This is a switching circuit, it operates in a non-linear mode meaning it's not meant to oscillate. This is more about shaping the signal (going into T1) than creating an oscillation. \$\endgroup\$ – Bimpelrekkie Aug 28 '16 at 19:30
  • \$\begingroup\$ Yes: C12 resonates with inductors primary stray inductance while C17 with its secondary ones. In both case they add to stray capacitances (transistor CE and D5) and take down resonator characteristic impedance \$Z_o=\sqrt{L_\sigma/C}\$ till stray series resistances and other losses effectively damp the system. \$\endgroup\$ – carloc Aug 28 '16 at 20:31
  • \$\begingroup\$ @Marcus Müller I believe it would be more interesting to find L1/R3 time constant as \$\tau=L/R=5\,\mu\text{H}/27\,\Omega\approx 200\,\text{ns}\$ and consider that upon a current driven transient they will then behave pretty much as a resistance for times much shorter then this and as an inductor for times much longer. So given relatively slow transients we can expect in such an old (slow) switcher I'd mainly consider L1/R1 parallel as an inductor. Its overall effect will indeed be to slow down transient, but getting closer is hard given the absolutely undocumented base driver. \$\endgroup\$ – carloc Aug 29 '16 at 6:55

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