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This is a very beginner's question - I apologize if this is something completely trivial but I am stuck trying to understand the exact functioning of the circuit below.

The circuit below is supposed to be a simple square-wave power inverter circuit and I am puzzled by: (1) the purpose of several elements, (2) the absence of some (what I think are) important elements.

More specifically:

  1. What is the purpose of D1 and D2 diodes? This looks like some kind of a flyback diode application but I don't see how it helps with flyback for the coil: what would happen when Q6 is suddenly shutdown? Wouldn't the induced EMF from T1 just fry transistor Q6?

  2. What is the purpose of Q2 and Q5 transistors? Couldn't I just connect the multivibrator outputs directly to Q1 and Q6? Is it done to increase the base-emitter currents in Q1 and Q6 respectively because of some kind of output power requirements? If it is only about power, why not just use MOSFETs for Q1 and Q6 instead of BJT?

  3. It seems that there is no resistors in the path from +12V to coil to ground. Wouldn't this path become a short-circuit if there is no load connected to the 220V side of the transformer (and thus, no back EMF)?

enter image description here

The circuit is taken from here: https://archive.is/0yRJc

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    \$\begingroup\$ Why don't you get a free sim like LTSpice and simulate it and make modifications (like bypass Q2 and Q5) and see how it performs. You should be able to see if D1 and D2 are needed to (I can't see why they are there BTW). Bear in mind that to most folk reading this, the circuit appears to be totally unproven and created by an potential idiot so, to try and justify answers to your questions YOU might as well do the work yourself and learn how to use a sim. \$\endgroup\$
    – Andy aka
    Commented Jun 19, 2017 at 7:41
  • \$\begingroup\$ Regarding 3, you might want to learn about inductance \$\endgroup\$
    – PlasmaHH
    Commented Jun 19, 2017 at 7:46
  • \$\begingroup\$ @Andyaka thanks! I see your point -- I may be too trusting with these circuits. \$\endgroup\$
    – akhmed
    Commented Jun 19, 2017 at 7:51
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    \$\begingroup\$ @Andyaka I always wondered whether we should have a rule that prohibits asking about circuits copied from anywhere without in-depth understanding of said circuit. I think I've seen about enough terrible inverters copied from copycat textbooks (looking at you here, Indian university system!), 1950's germanium semiconductor amplifier circuits copied from copied articles from copycat amateur radio mags, and BC54x-based "yeah I got this from some copycat Arduino article" MCU motor drivers... \$\endgroup\$
    – Marcus Müller
    Commented Jun 19, 2017 at 7:55

1 Answer 1

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1) What is the purpose of D1 and D2 diodes?

When everything's working properly, they shouldn't conduct. If for some reason Q6 fails to come on when Q1 turns off, then D2 will save it from reverse bias.

2) What is the purpose of Q2 and Q5 transistors?

2n3055s have very low gain, and need a lot of base drive to work well. Q2/5 are used as current amplifiers, rather than increasing the currents in the Q3/4 oscillator stage to drive them directly.

3a) It seems that there is no resistors in the path from +12V to coil to ground.

and indeed there should not be

3b) Wouldn't this path become a short-circuit if there is no load connected to the 220V side of the transformer (and thus, no back EMF)?

You will have back EMF from T1 due to the changing current, regardless of the load on the transformer. A load on the transformer will increase the currents flowing. The path will become short circuit if the oscillator is too slow, or stops oscillating for any reason, as either could cause the transformer to saturate. A fuse in the DC line is the minimum protection required to guard against this.

WARNING While the circuit looks plausible, it may need more development to be a good inverter. The point where conduction switches from Q1 to Q6 is critical. There is no active turnoff for either transistor, and these things tend to turn off slower than they turn on, resulting in a transient short circuit when both are conducting. Often we see a series inductor in the transformer supply to handle this, or more complicated transistor driving.

If T1 is a standard power transformer, it may have poor inter-secondary coupling compared to one that has been designed for inverter duty, which could produce damaging voltage spikes on Q1 and Q6.

Unfortunately, a good power inverter is anything but trivial. The above points are just a couple of things that can trip you up, there may be more. FETs generally make for a better design these days, they are faster, and most are avalanche rated to handle the commutation spikes. 2N3055 is a very old transistor, it was about when I was young, I'm now retired, and so should it be.

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  • \$\begingroup\$ Thanks a lot! Great point about slower turn off than turn on (I never even considered it!) I am not sure I understood one point: is it true that if inter-secondary coupling was really good then Q1 and Q6 would not have voltage spikes when they are shutdown? why? there seem to be no fly back diodes anywhere -- shouldn't Q1 and Q6 get a voltage spike in any case? \$\endgroup\$
    – akhmed
    Commented Jun 19, 2017 at 8:22
  • \$\begingroup\$ Voltage spikes are inevitable as coupling can never be perfect, and turn on/off timing never perfect, it's just a question of their magnitude. When you say shutdown, do you mean turn-off? A subtle possibly translation difference, they mean quite different things in English. \$\endgroup\$
    – Neil_UK
    Commented Jun 19, 2017 at 8:56
  • \$\begingroup\$ yes, you are right -- I meant "transistor turns off". Basically, my understanding is: once Q6 is turned off by multivibrator, there will be significant induced EMF present in the coil and now it has no escape path so it must build up the voltage at Q6 collector even if inter-secondary coupling was good. Would the above be an accurate statement? \$\endgroup\$
    – akhmed
    Commented Jun 19, 2017 at 9:00
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    \$\begingroup\$ This has been useful for me. Although I know all this stuff, having to explain it in detail has refreshed my understanding. Think of it this way. Were the two windings to be in exactly the same place, it doesn't matter which winding carries the current, the magnetic effect is completely equivalent. With a core there, even physically separated windings are in 'more or less' the same magnetic place, you get coupling ~99%, and only the magnetic field in the air round the wires themselves is not coupled to the other winding. It's the magnetic energy that creates the voltage, drives the current. \$\endgroup\$
    – Neil_UK
    Commented Jun 19, 2017 at 10:14
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    \$\begingroup\$ I think I finally got it! Because the two windings are tightly coupled, the collapsing magnetic field can "escape" through either one, it doesn't need to go through both equally. Therefore, instead of frying Q6, the easier escape path is through: +0V terminal -> (D1/Q1) -> left winding -> +12V terminal (thus, this conventional current will be charging the battery slightly). So this is exactly the purpose of diode D1 -- in case Q1 hesitates to turn on before Q6 turns off, D1 will provide the important path for magnetising current. \$\endgroup\$
    – akhmed
    Commented Jun 19, 2017 at 18:29

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