# Operation of an H-bridge with complementary transistors

I am trying to understand the operation of this H-bridge but one thing is not clear to me and I didn't find good explanations on the internet. When transistor Q5 is activated, how does it activate Q1 and Q4? The PNP transistor must activate with a logic 0 and the NPN transistor with a logic 1, how do these two transistors get these two values to activate? I don't understand where the logic values and current are coming from.

• Q1 does not need a logic 0 to operate. It only needs a base voltage of about 16.1 V (16.8 V - 0.7 V). Similarly, Q4 only needs a base voltage of 0.7 V to operate. And those two bases are separated by 2 resistors and Q1. So nothing inconsistent there. Commented Dec 14, 2023 at 20:23
• The conditions I mentioned above are met is you assume Q1 is saturated, and so ~11.3 mA flows through the emitter/base junction of Q1->R1->Q5->R4->BE jct of Q4. Commented Dec 14, 2023 at 20:40
• Further to SteveSh’s comments, a bipolar transistor base/emitter junction appears as a diode and this is where you get the current flow. If the devices were mosfets, this circuit wouldn’t work as a mosfet’s gate is isolated. Commented Dec 14, 2023 at 21:12

Assume Q5 is a switching transistor. When turned ON, its collector current and emitter current are nearly equal. So voltage drop across R1 (680) and voltage drop across R4 (680) are also nearly equal...about 7.7V.
To get Q5 ON, its base voltage should rise above 0.7 + 0.7 +7.7 volts...9.1V. Current to turn ON Q5 flows through opto-coupler and R5 (47k).
Since current through R1, Q5, R4 is similar, both Q1 and Q4 receive base current of about 11 mA, turning them both ON. This results in load current flowing.

You can simplify this circuit a little by eliminating a resistor:

simulate this circuit – Schematic created using CircuitLab

Now the turn-on voltage for Q5 is about 1.4V instead of 9.1V. R4 is not needed, and has been incorporated into R1.

Be aware that U1 optocoupler and U2 optocoupler should never be ON at the same time...that will smoke something.

• Why is the voltage drop across R1 and R4 almost equal and is approximately about 7.7V? Why for Q5 to turn on must its base voltage rise above 0.7 + 0.7 +7.7 = 9.1V? Commented Dec 14, 2023 at 21:37
• If you consider Q5 a saturated switch, its collector-to-emitter voltage ($V_{CE}$ is near zero). Q1's & Q4's $V_{BE}$ is roughly 0.7V. Current path of about 11mA takes the route: Q1 base-emitter, R1, thru Q5 (which is shorted collector-to-emitter), thru R4, Q4 base-emitter. Voltage drops on this path add to 16.8V. From GND, add Q4's base-emitter voltage (0.7V), $V_{R4}=7.7V$, Q5's base-emitter voltage (0.7V). Commented Dec 15, 2023 at 1:33

Keep in mind that transistors work on current. When there is a forward bias (Vbe) on the base-emitter path, the transistor will turn on. The Vbe junction is a diode, with a forward voltage of about 0.7V. The series resistors on the bases limit the current to what the transistor can handle.

Here's a sim to illustrate what's going on, and a possible simplification. Note the driver base currents (simulate it here):

(Note: I've omitted the diodes for clarity.)

Now, how does Q5 turn on Q1 and Q4? Q5 controls a current path to Q1 and Q4 bases. When Q5 is turned on it completes a current path starting from +16.8V, through Q1 emitter-base, through R1, then Q5 collector-base, then R4 and finally through Q4 base-emitter to ground.

Q1 and Q4, meanwhile, will have base-emitter voltages of either about 0.5V (when off) or 0.7V (when on.) The rest of the voltage drop is taken by R1 and R4.

The possible simplification shows the current path as well. Will it work? It depends on the optocoupler current transfer ratio, your required load drive, and the TIPxx gains.