so would a 2kohm resistor work just as well if I don't have 1.91kohm

In your case, you should go with 1.8k because you're already on the edge of the minimum base current to guarantee saturation.

You could also use a larger current limiting resistor with a 2k base resistor because 20mA is sometimes the maximum continuous forward current for LEDs and you wouldn't be able to notice a difference in brightness between 20mA and 18mA or so. When I use LEDs as indicators, I often operate them at a few mA.

 

so would a 2kohm resistor work just as well if I don't have 1.91kohm

Actually, to nit pick just a tiny bit, i have to wonder how you came up with 1.9k in the first place.
If you assume the base emitter voltage is 0.7 volts and the supply is 5.0 volts, then the voltage across the resistor is:
vRb=5.0-0.7=4.3 volts.
Now using Ohm's Law i think you will find that 2k is even closer to the value you had in mind for the base current, but a little lower is fine.
See if you could come up with the exact theoretical value in order to get 0.002 amps into the base.
Perhaps you used a different base emitter voltage.

 

on the data sheet it says that the max saturation voltage is 1.2 and if the collector emmiter current is 20mA then the base current should be 2mA. using these values you can find the reistor vaule by using the formula (5V - 1.2V)/.0002A = 1900ohms = 1.9kohm ~ 1.8kohm

 

Vbe will never be anywhere near that saturation figure. I am not sure what is the purpose of that spec, but it may be achieved only at very high temperatures.

Normal practice is to use .6 or .7 as the Vbe when calculating the base resistor, and I am willing to be it will be closer to that than 1.2.

Bob

 

on the data sheet it says that the max saturation voltage is 1.2

That voltage is only for relatively high collector current (150mA) and it's 2.0V for a 500mA collector current. This graph shows the base emitter saturation voltage for a typical device:


At 20mA collector current, the drop will be closer to 0.7V.

 

The base-emitter saturation voltage in the datasheet is 1.2V max when the collector current is a massive 150mA and the base current is high at 15mA. Then the collector-emitter max saturation voltage is 0.3V.

With your 5V supply then the base resistor is (5V - 1.2V)/15mA= 253 ohms and the collector resistor is (5V - 0.3V)/150mA=31.3 ohms which is a completely different very high current circuit with no LED.

The datasheet "On Voltages" graph shows a typical base voltage of 0.75V when the saturated base current is 2mA and the typical collector voltage is 0.02V when the collector current is a saturated 20mA.

Then your base resistor will be (5V - 0.75V)/2mA= 2125 ohms (use 2.2k) and the collector resistor is 5V - 0.02V)/20mA= 249 ohms but with a 2V red LED in series the collector resistor is (5V - .02V - 2V)/20mA= 149 ohms (use 150 ohms).

 

Hi again,

Yes it depends on what the CE saturation voltage is and what the highest BE voltage is.
The "gain of 10 to keep in saturation" rule only applies when the CE voltage goes less than the BE voltage. However, that is assuming that you need low CE saturation voltage. 20ma is not much CE current so it should be easy to saturate,

If we want to comply with the "gain of 10 saturation" rule, then the following resistor values would be needed:
With a BE voltage of 0.7 volts the base resistor is 2150 Ohms.
With a BE voltage of 1.0 volts, the base resistor is 2000 Ohms.
With a BE voltage of 1.5 volts the base resistor is 1750 Ohms.

So with worst case BE voltage 1800 Ohms would be good, but it is unlikely it will be that high because of the low input current required (0.002 amps).

If you need to draw the least current from the power source then try a higher value, but if not try a lower value.
In any case, measure the voltages once you get it hooked up.

 

I looked in Digikey to see if the antique P2N2222A costs less than a modern 2N4401 that has similar spec but there is no price for the P2N2222A because it is obsolete.