Let's say you're given a circuit diagram as follows:

See the

2N3904 datasheetI want to find R1 and R2 so that forward voltage drop across the LED is 2V with an overdrive factor of at least 2.

Current of 25mA from a source which outputs 0 or 5V through an output resistance of 610 is given. Beta ranges from 50 - 300

So here's my approach. Since we want to find beta forced with ODF of 2,

β

_{forced}=β

_{min}/2=25

Since source current = 25mA, we know this is the current going into the base, so Ib = 25mA.

Then I can find Ic at saturation which is given by

β

_{forced}=Ic

_{sat}/I

_{b} which leads to Ic

_{sat}=β

_{forced}I

_{b}βforced=IcsatIb which leads to Icsat=βforcedIb

Ic

_{sat}=25∗25mA=0.625A

Since we want a forward voltage drop of 2 across the LED, voltage at Vn can be found by

Vn=5−2=3V

We know at saturation V

_{ce} = 0.2v, which is V

_{c}We can then simply find R2 by ohms law since we know the current and voltage at each end of R2

Vn−Vc/R2=Ic

_{sat}R2=3−0.2/0.625=4.48Ω

That seems like a really small resistance to put at collector, but moving on, we can write a KVL in the lower loop to find R1 like so

5V−610I

_{b}−I

_{b}R

_{1}−0.7V=0

Solving the above equation gives me R1 = -438 ohms.

Something is definitely not right, I can't have negative resistance here. Where did I go wrong?