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

Transistor Circuit

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.

  • \$\begingroup\$ Use Darlington pair configuration (2 x PN2222A) and treat like a single BJT with massive gain and Vbe = 1.2V. \$\endgroup\$ Jan 6, 2018 at 19:23
  • \$\begingroup\$ So I have a new beta and base-emitter voltage but currents and collector-emitter voltage stay the same? \$\endgroup\$
    – Overblade
    Jan 6, 2018 at 19:27
  • \$\begingroup\$ The base current would change. Yes. The ice would not. \$\endgroup\$
    – Passerby
    Jan 6, 2018 at 20:47
  • \$\begingroup\$ The other option is that you measure the actual base current and gain with a test circuit, instead of using the lowest hfe listed. \$\endgroup\$
    – Passerby
    Jan 6, 2018 at 20:48

2 Answers 2


In order to get the base current IB, I looked for DC current gain in the datasheet. The lowest value is β=10 as shown in Figure 11.

Figure 11 uses Ic/Ib = 10 to guarantee the lowest possible saturation voltage, but the transistor will still have a low saturation voltage at higher ratios. In figure 4 you can see that a low saturation voltage is typically maintained for Ic/Ib ratios as high as 50 at Ic=150mA.

enter image description here

The PN2222A is guaranteed to have an Hfe of at least 50 at Ic=150mA and Vce=1V, and typical Hfe is ~200. To switch 70mA with 2mA you need at least 35. So even in the worst case it should be able to do the job, just with a higher saturation voltage than you used in your calculations.


"Please don't tell me to buy other transistors" ...that's what I wanted to suggest: Use a MOSFET instead of a NPN, rated at 150 mA (or more). The gate/base of a MOSFET draws less (usualy much less) than 1 mA. Otherwise the darlington pair is another solution if for some valid reason you have to keep these specific transistors. Your calculation about R2 is correct.

  • \$\begingroup\$ Ok thank you very much! I've heard about MOSFETs but first I want to give NPNs a try. But now I know what I have to look for :) \$\endgroup\$
    – Overblade
    Jan 6, 2018 at 20:47

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