Dim LED with Transistor Switch - Why?

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SUMMARY

The discussion centers on the use of a 2N3904 transistor to switch an LED powered by a 3V coin battery. It concludes that the LED appears dimmer when powered through the transistor due to the voltage drop across the transistor, specifically the 0.7V required to turn it on and an additional 0.3V drop across the collector-emitter junction. To mitigate this issue, the recommendation is to use a MOSFET with a low RDS(on) to minimize voltage loss and improve LED brightness.

PREREQUISITES
  • Understanding of transistor operation, specifically the 2N3904 model.
  • Basic knowledge of LED circuitry and voltage requirements.
  • Familiarity with MOSFET characteristics, particularly RDS(on).
  • Experience with basic electronic components and circuit design.
NEXT STEPS
  • Research the specifications and applications of MOSFETs, focusing on low RDS(on) options.
  • Learn about the differences between BJTs and MOSFETs in switching applications.
  • Explore methods to measure voltage drops in transistor circuits.
  • Investigate alternative transistor models for LED switching with lower saturation voltages.
USEFUL FOR

Electronics enthusiasts, hobbyists experimenting with LED circuits, and engineers seeking to optimize switching performance in low-voltage applications.

brodywx
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Hello,

I've just been playing around with a 2n3904 transistor and 3V coin battery. I noticed when powering an LED with a transistor switch the LED is significantly dimmer than when powering the LED directly.

Why is this? Is it because 0.7 volts is siphoned off just to turn the transistor on?
 
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The saturation voltage. Even if you drive the base hard, their is a 0.3V drop across the CE of the transistor so only get 2.7V instead of 3V. If you want to get rid of this, get a MOSFET with low rds.
 
Thank you for the reply.

I
am
learning
 

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