KSP44BU Overview
In this device, the DC current gain is 50 @ 10mA 10V, which is calculated by taking the ratio of the base current to collector current and dividing transistor by two.Single BJT transistor has a collector emSingle BJT transistorter saturation voltage of 750mV, which allows maximum flexibilSingle BJT transistory in design.A VCE saturation (Max) of 750mV @ 5mA, 50mA means Ic has reached its maximum value(saturated).A high level of efficiency can be achieved if the base voltage of the emitter remains at 6V.Normally, a fuse's current rating refers to how much current it can carry without deteriorating too much, which in this case is 300A.A maximum collector current of 300mA volts is possible.
KSP44BU Features
the DC current gain for this device is 50 @ 10mA 10V
a collector emitter saturation voltage of 750mV
the vce saturation(Max) is 750mV @ 5mA, 50mA
the emitter base voltage is kept at 6V
the current rating of this device is 300A
KSP44BU Applications
There are a lot of ON Semiconductor KSP44BU applications of single BJT transistors.
- Inverter
- Interface
- Driver
- Muting
