Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Simple question about LED driving by transistor

Status
Not open for further replies.

Mac7978

Member level 2
Joined
Oct 17, 2003
Messages
45
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,286
Activity points
377
Dear all

If uses a single NPN transistor to drive LEDs, can anyone tell me what is the advantage and disadvantage to put those LEDs at collector (graph A) or at emitter terminals (graph B)?

Thanks
Regards
 

Attachments

  • LED QUEST.JPG
    LED QUEST.JPG
    47 KB · Views: 110

The turn on voltage for the collector connected LEDS can be supplied by a logic level output from something like a micro.
 
Introducing a load in the emitter leg requires you give more attention to the bias. You must raise your bias V so as to turn on the transistor sufficiently.

Therefore your B circuit will require a higher bias than your A circuit. The amount will need to be 2.0 to 2.5 V higher. (I'm assuming 1.8v for red LEDs, plus a few tenths V drop across your current-equalizing resistors.)

This can be done by reducing bias resistor (R80) to 100 ohms. A 3V control pulse will deliver a milliamp or two of bias current.

In comparison the A circuit has a more sensitive bias input which will operate on a lower control voltage. This is an advantage of putting the load in the collector leg.

By the way your equalizing resistors can be as low as 100 ohms per LED (for both A and B circuits). Any oddball LED will still be within a milliamp or two of the others.
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top