NPN LED Driver - Why an extra resistor to ground at the Base?

[Kian]

Hi Nigel,

I am using the IO/Pin from an Arduino (set to logic high), and I think it gives 40mA. The voltage of the IO pin at logic high is 3.3V. So in this case the base resistor is (3.3-0.7)/0.04 = 65 ohm?

Hmm, what about the very complicated calculator for the base resistor? What you proposed is a sufficient estimate?

 

[Nigel Goodwin]

Hmm, what about the very complicated calculator for the base resistor? What you proposed is a sufficient estimate?

It's an utter waste of time, and of no relevance to driving a transistor as a switch - you want the most base current possible, in order to switch the transistor ON as had as you can.

There's no 'estimate' involved, and no messing about with Hfe or anything else - which (as you've probably already noticed) is so wildly variable as to make no sense for calculations. Essentially you design to try and avoid such 'problems', and if you do consider Hfe you choose the lowest possible value.

 

[audioguru]

How can the green LED have a current of 300mA when the maximum voltage on its datasheet is almost 3.9V, the 2.7 ohm resistor will have a voltage drop of 0.81V and the transistor saturation voltage loss might be 1V or more? Then the total voltage is 5.7V from a battery that might produce only 4V?

1) While the transistor is lighting the green LED, measure the battery voltage. Is it 4.5V?
2) Measure the voltage across the 2.7 ohm current-limiting resistor and use ohm's law to calculate the LED current.
3) Measure the voltage across the series 200 ohms base resistor and use Ohm's Law to calculate the base current.
Is the base current 1/10th the collector current as shown on the datasheet for the BC817 transistor so that the transistor saturation voltage loss is less than 0.5V?

The graphs on a datasheet are for a "typical" transistor that you cannot buy. Some have minimum or maximum spec's. The maximum base-emitter voltage for a BC817 is 1.2V (not 0.7V) when the collector current is 300mA.