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I've been experimenting on Half Bridge converter and there is something that I can not understand. What is the purpose of the diodes and capacitor (red one in the picture) in base drive circuit. I check and realize most of base drive has it. why the emitters if Q3 and Q4 are not connected to the ground directly?enter image description here

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  • \$\begingroup\$ you have selected a very good circuit to learn on .The details of Q3 and Q4 and how it starts are very subtle .I do not know why R51 and R52 are not the same value .Is it a misprint? \$\endgroup\$
    – Autistic
    Commented Jul 15, 2017 at 8:28
  • \$\begingroup\$ This is an old AT PC PSU. TL494 is used here as open collector output. So, to make sure that TL494 open collector output will be able to cut-off Q3 and Q4 properly, the designer add D18, D19 diodes in the Q3 and Q4 emitters to rise turn-on voltage. \$\endgroup\$
    – G36
    Commented Jul 15, 2017 at 14:43

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This technique biases both emitters of Q3,Q4 at 2 diode drops or 1.4V while the base voltage toggles from 0 to 2V at some fast rate like 40~50kHz.

These Push-Pull drivers are never on at the same time but the secondary high voltage drivers Q1,Q2 could be if the the primary Q3,Q4 are slow to turn off.

It's all about Miller Capacitance and minimizing the turn off time when turning each of Q3,Q4 into the OFF state by reverse biasing the base emitter by 2 diode drops.

The base collector capacitance is much higher if the emitter is grounded and the base voltage goes to 0V while the collector voltage was saturated near 0V and you ought to know that diode capacitance or the base collector junction is maximum at OV and reduces sharply as it becomes reverse biased thus speeding up the rise in collector voltage. In fact, the collector voltage may be quite large, from V=LdI/dt depending on core inductance and slew rate but very short in duration. THe net effect is that the secondary turn off times are also reduced so they do not overlap with both Q1,Q2 being on at the same time across the high voltage DC bridge rectifier used to power the high power transformer.

This technique is to provide rapid switch with minimum deadtime needed to commutate (toggle) the 2nd stage bridge transistors, Q1,Q2 and avoid shoot-thru fatal effects.

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  • \$\begingroup\$ But in this circuit, Q3 and Q2 are ON at the same time (especially at low loads). But the "driver trafo"is wound on a core in such a way so that the flux cancels out. \$\endgroup\$
    – G36
    Commented Jul 15, 2017 at 14:41
  • \$\begingroup\$ THat's weird, then how does it work? like a Flyback transformer rather than a Push-Pull? Yet base drive non-inverted with center tap to 12V and primary tap windings inverted wrt schematic ( no dots shown) would add in flux , not cancel, leading to saturation. In any case the 2 emitter diodes +cap speed up turnoff times for Vbe=+0.7 to -1.4V \$\endgroup\$ Commented Jul 15, 2017 at 19:33
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    \$\begingroup\$ It is rather sloppy , but yet it should be connected to emitters and anode \$\endgroup\$ Commented May 27, 2019 at 23:05
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    \$\begingroup\$ I agree . D20,D22 must be TVS diodes (Zener) I chose 24V here with 20mH tinyurl.com/y6mtkslb \$\endgroup\$ Commented May 28, 2019 at 1:06
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    \$\begingroup\$ I don't yet know why the pull down R's are there or why they are different tinyurl.com/yykkajzh \$\endgroup\$ Commented May 28, 2019 at 1:12

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