KSD5041RBU Overview
In this device, the DC current gain is 340 @ 500mA 2V, which is calculated by taking the ratio of the base current to collector current and dividing transistor by two.The collector emitter saturation voltage is 1V, which allows for maximum design flexibility.When VCE saturation is 1V @ 100mA, 3A, transistor means Ic has reached transistors maximum value (saturated).The emitter base voltage can be kept at 7V for high efficiency.The current rating of this fuse is 5A, which means that transistor can carry an indefintransistore amount of current wtransistorhout deteriorating too much.Single BJT transistor is possible for the collector current to fall as low as 5A volts at Single BJT transistors maximum.
KSD5041RBU Features
the DC current gain for this device is 340 @ 500mA 2V
a collector emitter saturation voltage of 1V
the vce saturation(Max) is 1V @ 100mA, 3A
the emitter base voltage is kept at 7V
the current rating of this device is 5A
KSD5041RBU Applications
There are a lot of ON Semiconductor KSD5041RBU applications of single BJT transistors.
- Inverter
- Driver
- Interface
- Muting
