Electronic – How to calculate base-emitter current

I had to read this question several times to try and get my head around what is being asked.

(1) Forget about the saturation voltage of a 2n2222 transistor - this has nothing to do with the operation of the H bridge circuit.

(2) The voltage is not ACROSS the output (bridge) transistors. It is measured between OUT1 and OUT2 and will be just less than the supply voltage.

They operate in opposite PAIRS. One transistor connects the load to the POSITIVE SUPPLY and the other connects it to the GROUND (0) SUPPLY through the current sense resistor. In other words they act as SWITCHES (fully saturated transistors).

What appears ACROSS the load is almost the full supply voltage. The circuit can switch the DIRECTION of the voltage by using the opposite PAIR of transistors in the bridge output.

(3) The maximum load current supplied through the L298 to 2A per bridge. You can parallel connect the outputs to give a maximum of 4A but be sure to connect the inputs the correct way around.

The CURRENT output depends on the LOAD. In the case of a SHORT CIRCUIT the output current should be limited to 2A per bridge using the SENSE pin outputs and external circuit.

The current for any particular load can be calculated using OHM's LAW I = V/R where V is the output voltage (OUT1 -> OUT2) volts from the bridge and R is the load resistance in ohms.

The diode is in the best position, and is of an appropriate type.

It conducts when the input is negative, the same as the transistor base conducting when the input is positive. The 47K resistor is about 1/10 of a normal RS-232 load. One could also block the voltage, but then a -100V spike (ESD say) could break down the 1N4148 and break down the E-B junction, causing irreversible damage.

Also, a 1N4148 is an appropriate diode for this application. It's a "switching diode", low capacitance and fast reverse recovery. A 1N4001 would also likely work okay, at least at slow baud rates. The 200mA rating means that even if a very high voltage were to appear at the input the transistor is fully protected, at least up until the resistor arcs over,.