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I am trying to reverse engineer and understand the following circuit found in an LED lighting controller.

The OUTPUT is basically a 12V supply where the polarity can be selected by the MCU (Q7/Q2/Q1 N-MOSFET gate inputs) - almost like an H bridge I guess. Strings of LEDs are connected to this output in alternating polarity.

When the Q7/Q2/Q1 N-MOSFET gates are low, the outputs are 12V/12V therefore 0V between them.

When the Q7 N-MOSFET gate is high, the outputs are 0V/12V therefore -12V between them.

When the Q2/Q1 N-MOSFET gates are high, the outputs are 12V/0V therefore +12V between them.

My question is, what is the point of the middle section (Q2 + zener diode) and why does it have separate MCU control (which always follows control of Q1). I imagine it is some form of protection?

Note: Q6/Q9 are P channel MOSFETs.

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  • \$\begingroup\$ Q6 and Q9 have their drain/source in the wrong way, like shown, the body diode will conduct. \$\endgroup\$
    – Bimpelrekkie
    Commented Jan 25, 2021 at 11:17
  • \$\begingroup\$ @Bimpelrekkie no, they're definitely connected like this in the circuit (they're the P channel MOSFETs) \$\endgroup\$
    – Warren Ashcroft
    Commented Jan 25, 2021 at 11:25
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    \$\begingroup\$ It all might make more sense if you draw the main components as an H bridge... \$\endgroup\$
    – user16324
    Commented Jan 25, 2021 at 11:48
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    \$\begingroup\$ they're definitely connected like this in the circuit OK, then explain how the body diode in Q9 will not be destroyed when Q1 is switched on as that would put 12 V across the body diode of Q9 in forward mode! Maybe the 12 V source is current limited? \$\endgroup\$
    – Bimpelrekkie
    Commented Jan 25, 2021 at 12:32
  • \$\begingroup\$ @Bimpelrekkie fixed ;) \$\endgroup\$
    – Warren Ashcroft
    Commented Jan 25, 2021 at 14:14

1 Answer 1

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This is probably used to discharge any capactitance that might be connected to the output. Imagine, that You have a capacitor in the led strip (or some other load), and it is not discharged to 0V. Then You switch the polarity and for the moment, capacitor charged with negative voltage is connected to 12V PSU. Q2 is turned on probably just before changing polarity (for some time, due to lack of current sensing through it).

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  • \$\begingroup\$ makes sense, I will have to hookup the oscilloscope and check whether there is a delay between Q2 and Q1 being turned on. \$\endgroup\$
    – Warren Ashcroft
    Commented Jan 25, 2021 at 11:29

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