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mjakubec
modified 5 years ago

LED Driver with Two NPN Transistors for Stronger Loads

5
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03:02:27
This stronger driver is able to drive LEDs by applying a high logic level sourced from output pin of a logic IC belonging to common 74xx logic families (HC, HCT, LS, etc.) or CMOS microcontrollers. The output pin of IC is simulated by a voltage supply and a switch. General rule of thumb is that the output pin of the IC has to be able to source at least 1 mA of current when it is set to high logic level. As a real NPN transistor for the first stage, you can use almost any "jellybean" small-signal one, such as BC54x or 2N3904, which are cheap and widely available. For the second stage, you have to choose a type with greater current handling capability, such as BC337 or 2N2222.
published 5 years ago
wyoelk
5 years ago
Seems generally a resistor is needed at the base to protect the transistor. Is there a reason this circuit wouldn't need one?
hurz
5 years ago
you mean the first transistor. Its a voltage follower, as long the base current is limited by the emitter resistor, the negative feedback does follow the voltage but 0.7V the base voltage, everything is ok.
wyoelk
5 years ago
Yes the first transistor

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