| 摘要 |
PROBLEM TO BE SOLVED: To make slow the turn-on speed of a drive transistor(TR) which is to be turned on, while the other drive TR is turned on reversely. SOLUTION: When the outputs of a switching circuit 107 and a 2nd switching circuit 111 are turned off, while a drive TR T1 is off, a driving TR T2 is turned off and an inductive load L1 generates a counterelectromotive force. With this counterelectromotive force, the potential at a point P in the figure rises, and when the potential at the point P exceeds the value obtained by adding the forward drop voltage of a diode 11 to the potential at the base terminal of the drive TR T1, the drive TR T1 is turned on reversely. When the outputs of a switching circuit 107 and a 2nd switching circuit 111 are turned on (t4 in Fig. 3) in this state, the drive TR T2 begins to be turned on quickly. When the output of the 2nd switching circuit 111 is turned off after a delay time tD2 (t45 in Fig. 3), the current supplied to the base terminal of the drive TR T2 decreases to make the turn-on speed slow.
|
