Experiment 8

Testing a BC547 in it's Saturated switch and active amplifying region.

For this experiment I set up a circuit with a 470R at Rc, an LED and BC547 NPN transistor, and alternated 5 different value resistors 47K, 220K, 270K, 330K and 1M, to be used at Rb.

For this experiment I measured voltage drop and current through the transistor pins as I put different resistors in place and record them on a chart.

I can see from this chart that Vce changed dramatically as more resistance was added. This is because as more resistance was added, the transistor switched from being fully open (Saturated) to Amplifying (Active).

As I increased resistance from 47k to 1M the voltage across Vbe stayed the same. This is because Vbe only requires 0.7 volts to operate. Increasing the resistance will only decrease the current to Ib, it will not decrease the voltage.

Ib decreased from the increased resistance because as Ohms Law illustrates, V=IR. So if more resistance is added to a set voltage, the current will decrease.

Ic decreased from the added resistance, but not as much as Ib because only a 470R was used for Ic, whilst Ib had a much larger resistance rating varying from 47K-1M

This graph is used to calculate the DC current gain of the transistor. Ib is the current at the Base of the transistor and Ic is the current at the Collector. If the Vce was 3 volts, I could use the load line to find the gain or (Beta). We can see that Ib will be 0.5 mA and Ic will be 14mA. To calculate Beta I can use the formula Beta=Ic/Ib. 14/0.5=28.

The Active region is when the transistor is neither fully open or fully closed. The transistor is amplifying small signals

On this graph I can see two new regions. The Saturated region is best described when connected as a switching circuit. It is when the switch is fully open, and allowing current to flow through.

The Cut Off region is when there is not enough power at Vbe to allow current to go though, meaning the switch is closed and no current will flow through.