Experiment 7 complete

With experiment 7 I was asked to construct a circuit with a 15V power supply, a 1K Ohms resistor at Rc and a 10K Ohms resistor at Rb and a BC547 NPN type transistor.

This is a Common Emitter, or open collector circuit because the Emittor has a common connection to both the input and output.

With the circuit connected to a power supply, I connected my voltmeter across the Base and Emitter and recorded .801V. This is because the Vbe only requires 0.7 volts to operate.

I then connected my multimeter across the collector and emitter and recorded 53.7mV. This is because the transistor is saturated and fully open, letting current flow through Vce.

In the saturated state, the transistor is fully on, meaning the switch is closed and the collector and emitter junctions are in forward bias.The Cut Off region is when the transistor is fully off, or the switch is open meaning the transistor is fully off. From this graph I can see that If the Vce was at 3V, The power dissipated by using Pd=Vce x Ic. If the Vce is 3v, then the current at Ic is 14mA. Then Pd will be 3 x 14 = 42mW

Using the graph I can also get the current gain (Beta) of this transistor. If the Vce was 2 volts, Ib = 0.8mA and Ic = 20mA. Beta = Ic/Ib, so 20/0.8 25Beta 3v=14/0.5=28Beta 4v = 5/0.2 = 25 Beta