# Can't get 2N3904 transistor to allow the collector current I want, when used as a switch

I have a 3.7 V 18650 battery and I want to control a 3 V, 130 mA LED (high power CREE LED).

I have calculated the series resistor for the LED (0.7 V/0.130 A) = 5 Ohms (ish)

But when I apply any voltage or current to the base, I don't get the 130 mA at the collector like I want. Instead, the LED is dim and I'm getting considerable voltage drop over the transistor.

I'm just confused about how to get the transistor to become fully saturated and act like a switch.

From what I understand, the base needs at least 0.7 V and the current Ib needs to be calculated from Ic/gain.

Could someone help me out please?

• Part number for LED? – Brendan Simpson Mar 16 at 18:31
• Did you look at the datasheet? Even with 10mA of base current, the transistor will have trouble passing over 100mA of Ic current and Vce voltage is not lower than about 0.5V. It is not a good part for this, given your current requirement and Vce requirement. – Justme Mar 16 at 18:32
• Better use a MOSFET (500 mA is enough). Same principle, except that it's voltage depedent and requires only very minimal current. With a mosfet you can connect the base (here called the gate) with a 10 Kohm resistor and you are good. – Fredled Mar 16 at 20:50

The gain ($$\I_c/I_b\$$) of a BJT is lower in saturation than in forward active operation. A typical value used to define the onset of strong saturation is $$\I_c/I_b=10\$$.
Also notice that the $$\V_{ce(sat)}\$$ and $$\V_{be}\$$ for this transistor increase above our usual crude estimates for them when operated at currents above 50 mA: