Operation of an H-bridge with complementary transistors

Question

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.

Answer 1

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.

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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.

Answer 2

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.