KSP42TA Overview
In this device, the DC current gain is 40 @ 30mA 10V, which is calculated by taking the ratio of the base current to collector current and dividing transistor by two.A collector emitter saturation voltage of 500mV ensures maximum design flexibility.A VCE saturation (Max) of 500mV @ 2mA, 20mA means Ic has reached its maximum value(saturated).With the emitter base voltage set at 6V, an efficient operation can be achieved.Normally, a fuse's current rating refers to how much current it can carry without deteriorating too much, which in this case is 500mA.As a result, the part has a transition frequency of 50MHz.The breakdown input voltage is 300V volts.A maximum collector current of 500mA volts is possible.
KSP42TA Features
the DC current gain for this device is 40 @ 30mA 10V
a collector emitter saturation voltage of 500mV
the vce saturation(Max) is 500mV @ 2mA, 20mA
the emitter base voltage is kept at 6V
the current rating of this device is 500mA
a transition frequency of 50MHz
KSP42TA Applications
There are a lot of ON Semiconductor KSP42TA applications of single BJT transistors.
- Inverter
- Driver
- Muting
- Interface
