TIP105TU Overview
This device has a DC current gain of 1000 @ 3A 4V, which is the ratio between the collector current and the base current.The collector emitter saturation voltage is 2V, which allows for maximum design flexibility.As a result of vce saturation, Ic reaches its maximum value (saturated), and vce saturation(Max) is 2.5V @ 80mA, 8A.If the emitter base voltage is kept at -5V, a high level of efficiency can be achieved.As defined by current rating, the maximum current a fuse can carry without deteriorating too much is -8A for this device.When collector current reaches its maximum, it can reach 8A volts.
TIP105TU Features
the DC current gain for this device is 1000 @ 3A 4V
a collector emitter saturation voltage of 2V
the vce saturation(Max) is 2.5V @ 80mA, 8A
the emitter base voltage is kept at -5V
the current rating of this device is -8A
TIP105TU Applications
There are a lot of ON Semiconductor TIP105TU applications of single BJT transistors.
- Driver
- Inverter
- Muting
- Interface
