Problem
I have a circuit as depicted below, and after calculating the base resistor, I wanted to see when the base current would be too little to saturate the transistor. I kept increasing the resistance but even at \$560k\$ where the resistor had decreased the current to \$7.8\mu A\$, the transistor is still saturating. Shouldn't there be a base current threshold?
What I am trying
What I would like to be able to to, is to switch on and off a diode from an arduino pin. The supply voltage is 5 volts, the forward voltage drop across the diode is 3.4V and the transistor is a 2N3904.
According to this blog post, I should be able to find the base current by
$$ I_B = \frac{I_C}{h_{FE}} = \frac{20mA}{100} = 0.2 mA$$
with min \$h_{FE} = 100\$ from the datasheet
And then the base resistor value by
$$ R_B = \frac{V_{port}-V_{BE}}{I_B} = \frac{4.2V-0.7V}{0.2mA} = 17.5k \Omega $$
But the size of this resistor value seems to be irrelevant.
simulate this circuit – Schematic created using CircuitLab
Best Answer
The typical gain of a 2N3904 is 200-300 at a couple mA collector current (more as it warms up due to not being saturated)
Even with 560K, that's half a mA or so, which will give plenty of light from a modern LED, but you should be able to see that it's not as bright as when a 10K resistor is used.
Do not use the hfe for this calculation if you want the transistor saturated hard on, use a forced beta of something like 20 to 50, if the typical hfe is 200 or so and the minimum 100 or so. If you use, say 30, in your equation you get a resistor value of 8.8K, so you might use 10K or 8.2K.
The reason is that you won't likely have a guaranteed hfe for the current you're using, and the hfe decreases at temperature extremes. It's still only "wasting" a few percent of the LED current, so no big deal.
To prove this to yourself, take a voltmeter and measure Vce of the transistor when it is on. If it is something like 50-100mV it is saturated.