Transistor as Switch

TURNING THE LED ON

1. The objective is to turn on the LED. So we have to start with the forward voltage drop across the LED. Forward voltage is a term that will come up a lot when working with LEDs. This number will help you decide how much voltage your circuit will need to supply to the LED. If you have more than one LED connected to a single power source, these numbers are really important because the forward voltage of all of the LEDs added together can’t exceed the supply voltage.

2. LED only can function at Vf 1.9V to 2.6V. Maximum power rated for LED is (Pmax=100mW). Some datasheet is even helpful enough to suggest a stable current rangeusually 16-18mA. The maximum current through LED, IC is should be not more than:

Pmax=IC X V

IC=Pmax/V; let say the voltage for the LED is 2V

ICmax= 0.125A

So we know that the LED will broken if ICmax=0.125A.

3. Check the transistor 2N3904 datasheet. PD max= 625mW. Ic max=200mA, hfe(beta)=100-300, VBE sat=0.65V-0.85V, VCE sat=0.2V. But for most transistor generally we can use VBE=0.7V and VCE=0V.

4. Make first assumption which is the value of IC to turn the LED on. IC surely must be lower than IC max of the transistor to ensure the transistor not blowing up. Let IC=15mA and Vcc=9V.

5. Then calculate the value of RC to limit the current flowing through collector to 15mA.

RC=VCC-VLED/IC

RC=9-2/15mA=466.67ohm

6. To make sure the resistor not broken power dissipation of resistor has to be lower than maximum power rating of the resistor.

P=IV=(VCC-VLED-VCE)xIC=(9-2)(15mA)=0.105W; whereby VCE=0V

7. To require IC of 15mA to flow through the collector, we must provide the sufficient base current, IB. According to datasheet min hfe=100, so required transistor base terminal to drive LED ON is:

IC=hfe X IB

IB=IC/hfe=15mA/100=0.15mA

8. Calculate the value of RB to limit IB. If you really want to drive the transistor into its fully saturate mode the value of RB must be lower than:

RB=(VCC-VBE)/IB=(9-0.7)/0.15mA=55.33kohm

TURNING THE LED OFF

Troubleshooting:

• Not sufficient amount of IC running through LED because of IB is not sufficient. IB might be equal to 0 or not achieve the min IB required to drive min IC required to turn the LED ON. Usually in cut off state it said that IB=0A.
• Transistor is blow up: Because the IC driving through the transistor exceeding the maximum power rating of transistor. Since IB is too large driving IC to be too large until exceeding Pmax.
• Resistor blow up: because current through resistor consuming power more than power rating of the resistor.