Experiment #7

Experiment #7

Transistor as a switch

Constructing a circuit using a 15v power supply, an NPN type transistor, a 1k/ohm resistor and 10k/ohm resistor.

A voltage drop reading across Vbe (Base to Emitter) of only 0.7v is produced. This transistor only needs this slight amount of voltage to operate. A reading across Vce (Controller to Emitter) is 0.05v, thus confirming that our transistor is in the 'saturation zone' on a transistor behaviour graph.

Ic = (Vs - Vce) / R
    = (15v - 0.05v) / 1000 ohm
    = 14.95 / 1000
    = 0.01495
    = 14.95mA

Ib = (Vs - Vbe) / R
    = (15v - 0.7v) / 10000 ohm
    = 14.3 /10000
    = 0.00143
    = 1.43mA

 Saturation zone refers to the transistor being fully open or 'maxed out'. The Active zone refers to the transistor working to the desired specifications or in its 'prime'. The Cut-off zone refers to the transistor not working or not flowing at all.

Just breifly, if the voltage at Vce is at 3v and we are trying to find the power dissipated (Pd) by the transistor, then;

Pd = Ic x Vce

If Vce = 3v, then we find using the transistor behaviour graph that Ic will = 13mA (0.0013A), so;

13 x 3 = 39W

Finding the Beta (B) gain at different Vce voltages;

B @ 2V = Ic/Ib

= 0.02/0.0008

= 25

B @ 3V = Ic/Ib

= 0.0125/0.0005

= 25

B @ 4V = Ic/Ib

= 0.005/0.0002

= 25