# Powering an LED with transistor in saturation mode

I currently try to design my first transistor circuit to switch an LED on and off. So I came up with this circuit diagram: I have to use the following components:

What I need to figure out are the values of R1 and R2. So first I picked a desired current of 70mA from the 5V source to GND. According to the LED datasheet, 70mA of current induces a voltage drop of 1.3V. Figure 11 of the transistor datasheet shows that the saturation voltage at 70mA is roughly 0.06V. Now we can calculate R2 using Ohm's Law:

$$R2 = \frac{5V - V_{D1} - V_{CE(sat)}}{I_{C}} = \frac{5V - 1.3V - 0.06V}{0.07A} = 52\Omega$$

In order to get the base current $I_{B}$, I looked for DC current gain in the datasheet. The lowest value is $\beta = 10$ as shown in Figure 11. Therefore

$$I_{B} = \frac{I_{C}}{\beta} = \frac{0.07A}{10} = 7mA$$

Which is definitely a problem. I don't want to draw more than 2mA from the logic source as it can possibly be damaged. To stay on the safe side, I need to find a way to increase the lowest possible value of $\beta$ to reduce the base current.

Is this the way to go? And if yes, how would I accomplish an increase of $\beta$ ?

Please don't tell me to buy other transistors with higher DC current gain, although I will definitely do that later.

• Use Darlington pair configuration (2 x PN2222A) and treat like a single BJT with massive gain and Vbe = 1.2V. – JIm Dearden Jan 6 '18 at 19:23
• So I have a new beta and base-emitter voltage but currents and collector-emitter voltage stay the same? – Overblade Jan 6 '18 at 19:27
• The base current would change. Yes. The ice would not. – Passerby Jan 6 '18 at 20:47
• The other option is that you measure the actual base current and gain with a test circuit, instead of using the lowest hfe listed. – Passerby Jan 6 '18 at 20:48 