1. I'm studying how the transistor works as inverter but in some books they have a small section where they talk about how to drive lamps and LEDs, but they don't use the same configuration. I would like to know what's the difference between these configurations and what it could be a better option when you need to drive a LED with a power above 1 Watt.

  2. If it's possible, someone can provide a guide or equations about how to design and choose the required parts of the circuit? I saw some videos on Youtube and some application notes but I would like to know how to choose the correct transistor, specially the hFE that it's needed in the transistor in order to ensure its proper functioning.

Configuration 1

Configuration 2

  • \$\begingroup\$ The first schematic is correct. What is forward voltage of led? What is Vcc? What is the transistor type? \$\endgroup\$ Commented Nov 12, 2023 at 3:35
  • \$\begingroup\$ Vf (led) = 2.6 V, Vcc = 10V \$\endgroup\$
    – A. V.
    Commented Nov 12, 2023 at 3:36
  • \$\begingroup\$ I_led=1W/2.6V=0.38A … Then Rc=(10V-2.6V)/0.38A~=22ohm \$\endgroup\$ Commented Nov 12, 2023 at 3:44
  • \$\begingroup\$ Ib=0.38/Beta … (use Beta=20 since you need drive fast because of pwm)= 20mA …Then Rb=(3.3V-0.7V)/20mA~=150ohm \$\endgroup\$ Commented Nov 12, 2023 at 3:55
  • \$\begingroup\$ if Beta is bigger means that the response of the transistor is slower? \$\endgroup\$
    – A. V.
    Commented Nov 12, 2023 at 4:14

2 Answers 2


Both of these schematics will work somehow.

In the second one, you can omit the base resistor. Current will be set by 2.8 V divided by RE.

The first schematic is the conventional way of doing it. You saturate the BJT and use it as a switch. RC sets the current as explained in the comments.

However, if you don't need very close control of the LED brightness, you could also omit RC. In that case, current will be given approximately by (2.8 V/RB) * beta. beta typically has a large spread from part to part of about a factor of 3. So current could vary by a factor of 3 with this method.

  • \$\begingroup\$ This is a good write-up on how to drive power LED with BJT. efficiencywins.nexperia.com/efficient-products/… \$\endgroup\$
    – liaifat85
    Commented Nov 12, 2023 at 13:47
  • 2
    \$\begingroup\$ Unless you are positive that the LED has built-in current limiting, do not omit RC. Relying on a transistor's beta to control the collector current is called dangle-biasing, and is a very bad idea outside of a classroom. Beta varies with temperature, collector current, collector-emitter voltage, aging, production lots, and just about everything else except the phase of the moon. Control the LED current by design, not by luck. \$\endgroup\$
    – AnalogKid
    Commented Nov 23, 2023 at 13:52

In the first schematic, the transistor is used as a switch. When it's on, it will be saturated, ensuring lowest possible wasted voltage on Vce. The resistor in series with the LED sets the current, simply by ohm's law: voltage on the resistor is (supply voltage) - (LED forward voltage) - (transistor Vcesat).

In the second schematic, the transistor is used as a constant current source. It is not intended to be saturated, so it acts as a follower, and voltage on the emitter resistor is (base voltage) - Vbe. So if the base is driven to 3.3V then you'll get about 2.8V on the resistor which sets the current. In order for the transistor to work as a constant current source it should not be saturated, which means its collector voltage should be higher than the base: this circuit wastes about 3.3V for the transistor and resistor.

If your supply voltage can vary widely and you want constant LED current, the second one is better, since LED current does not depend on power supply voltage. But the transistor may dissipate significant power, which has to be taken into account. For a high power LED it's not ideal. If your supply voltage is known, then the first circuit burns power in the resistor (which is cheaper) and wastes less voltage.


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