Thank you very much Wookie.

Operating a transistor deeply saturated causes turn-off time to increase. However, it also minimizes the power dissipation. If turn-off times are not significant, you're better off to simply use Ib=Ic/10.

Yes, I want it completely saturated.
How do you tell how much current goes from VIN to Rbe, and how much goes to the Base? (When it's saturated and when it's ON).

 

The current through Rbe will be Vbe/R(Rbe). If Ic is ~<= 1/3 of the maximum rated Ic, a Vbe of ~0.7v can be assumed to keep things simple.

So, if you need Ic=20mA for a 2N3904, 2N4401 or 2N2222 NPN transistor with Vin being 5v:

Rbase = (5v-0.7v)/(20mA/10) = 4.3/0.002 = 2150 Ohms, rounded down to 2.1k.
We'll use Rbase * 4 for Rbe, or 8.4k. This is not a standard value of resistance, but 8.2k is.

So, what's 0.7v/8.2k? 85.4 microamperes, or 0.0854mA.
This current is subtracted from I(Rbase).
So, 4.3v/2.1k = 2.048mA; 2.048mA - 0.0854mA = 1.9626mA base current. This is within 2% of the Ib=Ic/10 formula.

 

Using the circuit values from sgtwookie's solution and using Thevenin's Method as suggested by Bill Marsden and I, the calculation will look like this.


\(R_{th}\ =\ R_{in}||R_{be}\ =\ 2.1K||8.2K\ =\ 1.672K\)

\(V_{th}\ =\ \frac{V_{in}*R_{be}}{R_{in}\ +\ R_{be}}\ =\ 3.981\ Volts\)

\(I_b\ =\ \frac{V_{th}\ -\ V_{be}}{R_{th}}\ =\ \frac{3.281}{1.672K}\ =\ 1.9623\ mA\)

As you will see this result is in very good agreement with sgtwookie's approach.

hgmjr

 

Hey,

It was mentioned in this thread (by Marshall) that:

However, it doesn't say how from the first place, without the voltage divider, the BE diode could be turned on by 0.5V.

Don't look at me, it was someone else that had one turn on at 0.5V. Only way I can see that happening in silicon is if the junction is leaky.

 

Thank you very much for your answers.

About turn-off current,
When VIN goes LOW, Is the current that goes through R_be (and Rin if VIN is grounded) exponential as in an RC circuit?