How does this modified H-bridge circuit work?
First of all, there is an error in your schematic. R4 needs to be connected between R3 and the base of Q1. Similarly for R8 on the other side. As shown, the circuit will not work — all of the transistors in the H bridge will be turned on simultaneously, shorting out your power supply. But even with this fix, you must never operate this circuit without the pulse generators &quot;A&quot; and &quot;B&quot; connected, for the same reason.
Getting back to your actual question, Q1 is switched on and off by the logic-level pulse coming from &quot;A&quot;. Obviously, if Q1 is conducting, then Q3 must not be. Q5 is being used as a common-base switch in order to make that happen.
D1 is supposed to be a zener diode (however, that's the wrong symbol in your schematic) with a rating that's roughly equal to your logic-high voltage on &quot;A&quot; — probably 4.7 or 5.1 V. Along with R5 and C2, it holds the base of Q5 at a constant voltage. When A is low, Q5 is turned on, which also turns on Q3. But when A goes high, Q5 is now cut off, and so is Q3.
Note that a separate circuit for the base of Q6 on the other side is not really necessary — the bases of Q5 and Q6 could be connected to the same zener diode. Or indeed, they could simply be connected directly to the same power supply that's driving your &quot;A&quot; and &quot;B&quot; pulse generators.