Beta is irrelevant for switching (lowest collector-emitter voltage).
If you use a transistor as switch, then you should provide a base current of 5-10% of the collector current.
Not relevant for a 20mA LED, but it is when you switch several hundred mA.
Leo..
First time i read that Ib should be 5-10% of Ic. Interesting.
This is how I determine the how much Ib should be, in an common circuit where a transistor is acting as
a switch :
I choose the Ic. If i have a LED, then Ic should be 20mA. Than I do the following calculation
First i find the resistor value
Rc= (V supply - 1.83V (LED drop) - 0.3V) / 0.020A
Then I find the Ib
Ic = Ib x Hfe(Beta)
Ib = Ic / 160
Ib = 0,000125 A or 125 µA
Now the interesting part, i got this info in a bjt NPN transistor turorial
Since I want the transistor to be in the saturated zone (I wrote active, my mistake), I multiply the Ib with a number between 4 and 10.
Thats why Ib = 4 * Ib
And I supply the Transistor base with 500µA !
When you use a (bipolar junction) transistor as a switch the device will either be in cutoff (off) or saturation (on).
Standard transistor action only applies to the active operating region where current is flowing but the emitter-collector voltage is several volts or more. This is where the current gain figure applies, and the base-collector junction is reverse biased.
In full saturation the base current is about 5% to 15% of the collector current. There is no normal transistor action in saturation because the base-collector junction is forward biased. The amplification of current in saturation is an entropy effect due to the fact the emitter is doped about 10000 stronger than the collector in transistors, rather than due to charge carriers being injected into a reverse-biased BC junction and then being pulled across by the electric field.
To minimize power dissipation in the on-state you want full-saturation where the emitter-collector voltage is a fraction of a volt. If the current is small and the voltage drop between emitter and collector can be greater you can use the active region of operation where beta is large, but if the load is high current this will cook the transistor.
Dacha011:
2) Then I find the Ib
Ic = Ib x Hfe(Beta)
Ib = Ic / 160
Ib = 0,000125 A or 125 µA
No, Hfe only applies when there is some voltage left on the collector, say 4volt (transistor used as amplifier).
Hfe does not apply when the transistor is fully saturated (lowest voltage between CE).
If you want collector voltage to become lower than base voltage, then just pump 5% into the base.
Formula: 5% of 20mA is 1mA.
Collector voltage of a saturated transistor can become much lower (<0.2volt) than base voltage (0.65volt).
Important for a switch, because you don't want the transistor to get hot or 'steal' voltage from the load.
You can usually find switching specifications in a graph of the datasheet, with Ic:Ib ratios.
Leo..
Thanks Wawa, I will have this in mind next time I work with BJT transistors
For now I will just pump 5% of the Ic in the base when I'm using a BJT as a switch.
I understood the current flow of the transistor, but I have to read and learn more about the voltage drop that occurs in the BJT.
All i know is that the V drops approx. 0.7 when calculating the resistor value of Ib, and approx. 0.3 when calculating the resistor for Ic. But I am totally confused when we are talking about the relations between voltages on the C, B and E.
If you have a good tutorial at hand, I will be grateful if you share me the link.