Hi Folks,

Re the attached Cct .. considering that the transistor is a BC548 (url) is it true that to fully turn the transistor ON and use it as a switch the Emitter to Base current needs to be 5.0mA in order to fully saturate and allow a Collector current of 100mA (Max allowed). IF this is correct I belive there will be 0.7v drop accross B to E and therefore R2 should be calculated given that the I flowing through it is 5mA and Vdrop is 8.3V so therefore a value of 1660Ω is required.

Does that make sense ..?

If the transistor is saturated and the LED is on what is the VDrop accross TP5 to TP4 ..?

Many thanks to all helping, really appreciate the advice/inputs.

 

You probably don't need that much current in the base circuit. Calculate R1 to allow the desired current through the LED (probably 10 - 20 ma). After that, R2 only needs to allow enough current in the base for the LED to be fully on. Once the collector current is up to the R1 linit, the transistor may be considered as being in saturation in this circuit. You may only need 1 - 2 ma base current, depending on the transistor gain.

 

When BJT is in saturation
Vce=Vce(sat) typical 20mV...0.2V= "Vdrop accross TP4 to TP5"

R1=( Vcc - Vf - Vce(sat) ) / Id

Id=10mA...20mA= LED forward current
Vf= LED forward voltage drop Vf=1.8V...4V dependent on LED color.

R2=[ (Vcc-Vbe)*hfe_min ] / ( Id*K)

K= 2...5

OR
R2 < R1/(10..50)

 

Transistors are all different even if they have the same part number.
All BC548 transistors are guaranteed to saturate with a saturation voltage loss of 0.6V or less when the collector current is 100mA and the base current is 5mA. Some transistors will have a lower saturation voltage loss and could saturate pretty well with less base current.

Notice that the base voltage is typically 0.7V when the collector current is only 10mA and the base current is only 0.5mA and is more than 1V at 100mA/5mA.

100mA will blow up most ordinary LEDs.