The H-bridge

BrugThis design requires the supply voltage to the steering logic to be the same as the voltage on the H-Bridge (5v to 18v). The circuit provides BRAKE feature when the output of both gates are the same (either HIGH or LOW). There is a “shoot through” current during the time when the inverters change state and this occurs as follows:
When the output of the microcontroller port is low, the bottom transistor is not turned on but the top transistor is fully turned ON. When the output of the inverter rises, the top transistor is ON and the lower transistor is also turned on. When the inverter is HIGH, the top transistor is turned OFF. During the time when the inverter is changing from LOW to HIGH, both transistors are turned ON.
The HIGH on the motor will be rail voltage minus the collector-emitter voltage (about 0.3v). The total voltage-drop to the motor will be about 0.6v.
There is also a leaking current through the PN junctions of the transistors The current will flow via the PN junction of T3 throug R1, through R2, to the base-emitter junction to earth. The same will happen at the right side.
This circuit does not have the “shoot through” current during the time when the inverters change state but it does not have the same performance as the circuit above. The voltage on the steering logic and H-Bridge must be the same. The transistors are EMITTER FOLLOWERS and the voltage on the motor will be less than the voltages on the circuit above because the HIGH on the motor will be determined by the output voltage of the IC, minus the slight drop across the 1k and the voltage drop across the base-emitter junction of the transistor (a total of about 1v). The total voltage drop to the motor (due to both sides of the bridge) will be about 2v.
The circuit is only suitable for a low-current motor as the the 1k base resistors will limit the current available through the transistors. The resistors can be reduced to 470R.