Because the the outputs of the 4050 are directly connected to the base of the darlington transistors. To the 4050, the load is like two diodes in series and therefore cannot go higher than 1.3V.(1)the output of 4050 hex buffer should be same as its inputs, that is, about 5V.but when i measure the output from the 4050, the outputs are only 1.3V only.i double check the connection but its correct.whats wrong with it?
Yes, the pins will remain high until reset. However your code only sets the the outputs to HIGH but does not reset them to LOW. Eventually all of the outputs will be in the HIGH state. One excitation sequence for the stepper drive is as follows:(2)is it when i make a pin HIGH,the pin will continue HIGH until i make it LOW?it is becasue when i measure the output pin(when i try to make the motor forward and reverse), all the pins are at HIGH level.
Because of the reason above. Also, step motors are rated in Amps/phase. There are motor excitation methods that energize 2 of the 4 windings at the same time and therefore consume double the rated current. Finally, the current will depend on the voltage you apply to the windings.(3)the circuit seems to draw more than 3A from the DC power supply.but the motor rating is only 2A.why?