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A simple dc-ac converter in full bridge structure and the structure I created to drive the MOSFETs are given in the diagram below.The output of the converter is working properly.

The circuit consists of optocoupler, bootstrap and transistors to drive the high side mosfet.

The signal coming from And gate which controls the optocoupler (green trace), Q8's Ib (blue trace) and Q8's Ic (red trace) are given below as well.

Besides interesting fluctuations in the base of Q8, the base seems to always be open. Whether the optocoupler is on or off, the Q8 base always appears to be on. What is the reason for this?

Since the base of Q8 is always open, Ic is always open, as seen in the graph, it is flat.

Since Q8 is like this, I cannot take proper measurements on other transistors.

enter image description here

enter image description here

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  • \$\begingroup\$ What does the base voltage of Q8 look like? How much current is running into the optocoupler? Try first to add a series resistance to the base of Q8. The base current looks very high. \$\endgroup\$
    – Tyassin
    Commented Jun 24 at 9:14
  • \$\begingroup\$ when optocoupler is OFF, that means the Q8 is ON so R6 is already a series resistance to the base of Q8 . Dont you see Ib is ON during ON/OFF pulses (green trace) ? @Tyassin \$\endgroup\$
    – Mhan
    Commented Jun 24 at 9:30
  • \$\begingroup\$ Yes I see it, but it don't make sense, the reason I asked for the voltage of Q8 base. There must be some setup in the simulation or the optocoupler is not working as intended. Try to get that to work independently of the rest of the circuit. \$\endgroup\$
    – Tyassin
    Commented Jun 24 at 12:26
  • \$\begingroup\$ "Besides interesting fluctuations in the base of Q8, the base seems to always be open. Whether the optocoupler is on or off, the Q8 base always appears to be on." The words "open" and "on" are different concepts. In electronics and "open" circuit has 0A of current. "On" means the system or devices is powered up and working. So I can't figure out what you are asking. \$\endgroup\$
    – user319836
    Commented Jun 24 at 23:28
  • \$\begingroup\$ VE of the 6N137 should be connected to VCC. It especially should not be left open (not connected to anything) \$\endgroup\$
    – user319836
    Commented Jun 24 at 23:30

1 Answer 1

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Some general comments, and some tips to encourage a self-solution. (IMHO: the best questions are those that didn't need to be asked -- because one is equipped to figure them out themselves!)

  1. Why are Q8 and Q3 operated at hFE(sat) = 1? This greatly stretches out storage time, and greatly increases the current consumption (an important parameter for a bootstrap gate drive).
  2. Note that bootstrap only works if M4 is fired first, and regularly thereafter (or there's enough load current to forward-bias its body diode). The choice of isolated gate signal is a bit peculiar, without also isolating the DC supply; consider changing to a floating source instead. (Or, if bootstrap is adequate given inverter strays, don't bother with cooking up your own driver: just use a IR2110 style IC, perhaps boosted with BJTs if needed.)
  3. Is M1 a typo? You're driving a huge power transistor with itself? (M2 likewise, almost certainly doesn't have to be a 10A device. Certainly not if it's driving through 100Ω.)
  4. I don't know what voltage A1 is set for, but beware 6N137 specifies 5mA min. IF for operation, so it won't turn on (or reliably so) if this is a 5V gate. Simulation logic may not model supply current (correctly, or at all!), and use an assumed voltage (usually a global VCC or VDD net).
  5. U8 VE is floating. This should be acceptable per the datasheet, but leaving a logic input floating does create a noise hazard in practice, and it may not be modeled correctly in SPICE. Try tying it with VCC.
  6. Try things, in general -- when confronted with a potentially erroneous SPICE model, it is worthwhile to take a step back from the immediate problem (the inverter), and build a test fixture featuring just the model in question, and some sources and sinks to make it go. Measure the output voltage and current range, response time, everything on the datasheet you feel like checking. (Bonus: this reinforces researching what datasheet parameters are and how they're measured, and reproducing those tests virtually, or with the real thing for that matter -- sometimes you get parts that aren't quite right, and setting up a real test can be informative.)

From this, you might now imagine some things to look into; like the VO pin, if it has the specified current and voltage range; that VCC is adequate (or the bootstrap supply in general); that R13 is suitable; or generally that the model is behaving at all.

Key insight: SPICE models don't always work right (in any given simulator). For that matter, they aren't always right in general! Always have a feel for how your circuit should be working, given everything as specified, and spot-check that things are behaving accordingly. If you find something suspicious, consider taking a deeper look. Never assume that things are as they should be, always leave open the possibility that something is just wrong in some respect. Particularly for simulation models, only certain aspects may be modeled -- they are only models after all, not the real thing!

This advice, by the way, generalizes to all of life -- never take something for granted, at face value; don't take things too seriously, too literally; always keep a critical eye out, and cross-check that things make sense, and when they don't, look deeper into it, research what's going on.

Best of luck!

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  • \$\begingroup\$ Hello sir you are right, I go ever everything that you mensioned. I have a small question. The GND pin of 6N137 passes through the emitters of the transistors and connects to the midpoint of M3 and M4, that is, a floating node. GND can not be put onto that line, but for example if we want to measure the Vce of Q8 where emitter is connected to the GND pin of 6N137, this takes the measurement with respect to 240 V. because the midpoint of M3 is M4 is high voltage always . What is your opinion about this? @Tim Williams \$\endgroup\$
    – Mhan
    Commented Jun 26 at 13:13
  • \$\begingroup\$ I don't catch what this is in regards to. You are making a local ground reference for the gate driver; it doesn't matter that it's bouncing with inverter output voltage, all local voltages shall be measured with respect to that. You wouldn't usually measure it IRL without a proper differential probe, but the simulator at least can do it trivially. For basic testing, you can disconnect inverter supply and measure with respect to ground, then hope that it doesn't misbehave with voltage applied. \$\endgroup\$ Commented Jun 26 at 21:24

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