# Vce saturation voltage confusion

I finally finished my first transistor circuit and looks as follows:

For the LED to light up properly, I assumed a forward current of 70mA and a voltage drop of 1.3V. The datasheet of the PN2222A states that at 70mA collector current, the saturation voltage $V_{CE}$ will be about 0.06V. So $$R2 = \frac{5V - 1.3V - 0.06V}{0.07A} = 52\Omega$$ Lowest value of $\beta$ equals 10, resulting in a total minimum current gain of $$\beta_{Total} = \beta^2 + 2\beta = 120$$ $I_{B}$ is therefore $$I_{B} = \frac{I_{C}}{\beta} = 583\mu A$$ So $$R1 = \frac{3.3V - 1.62V}{583\mu A} = 2882\Omega$$ When I plugged in both voltage sources, the LED only glimmed lightly. I measured $I_{C} = 42.4mA$ which is definitely too low. For troubleshooting purposes I checked every voltage drop and one was far away from its theoretical value: $V_{CE} = 0.7V$.

Why is the real saturation voltage way higher than stated in the datasheet?

• The LED only glimmed lightly in the wavelength area that your eyes are sensitive to, seeing as it is an IR transmitter. – MrGerber Jan 9 '18 at 13:46
• Well I couldn't even see it as its peak wavelength is at 940nm. However, my phone camera can see it and it was way less bright than my previous circuits without transistors. – Overblade Jan 9 '18 at 13:51
• But your camera is just marginally sensitive at 940nm, so the intensity you see with your phone camera is not by any means representative of what you would expect from a visible LED at similar power levels. – MrGerber Jan 9 '18 at 14:06
• It is sensitive enough to see a big difference between 40mA and 70mA. I know I can't judge the LED's power based on a camera. – Overblade Jan 9 '18 at 16:00
• The IR sensitivity is not dependent on the power of the light, but the wavelength (see example. You can of course see relative differences in signal, but my original statement was that whatever light intensity you observed was not an absolute indication on what was going on in the circuit. – MrGerber Jan 9 '18 at 16:13