NPN Transistor - Sink or Source?

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SUMMARY

The discussion centers on the use of NPN transistors in LED circuits, specifically comparing their configurations in "source" and "sink" modes. It is established that while both configurations can function, the sink configuration provides a higher LED current due to lower voltage drop across the transistor. The user seeks to source current for a 3D LED cube application and is advised to consider using a PNP transistor to achieve this without significant voltage drop penalties. The importance of base voltage and current limiting resistors is emphasized for optimal LED brightness.

PREREQUISITES
  • Understanding of NPN and PNP transistor operation
  • Basic knowledge of LED forward voltage and current limiting resistors
  • Familiarity with microcontroller output and base driving techniques
  • Knowledge of circuit analysis and voltage drop calculations
NEXT STEPS
  • Research PNP transistor configurations for LED applications
  • Learn about voltage drop characteristics in transistor circuits
  • Explore current limiting resistor calculations for LED circuits
  • Investigate the use of MOSFETs in LED driving applications
USEFUL FOR

Electronics enthusiasts, circuit designers, and developers working on LED projects, particularly those involving multiplexing and current sourcing challenges.

tangodirt
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All electron flow nuances aside, let's look at simple circuit analysis. Hopefully you guys can clarify something for me. Say we have to drive an LED. We have a +5V source, a 3.4V forward voltage, and a 150 Ohm current limiting resistor:

+5V --> Resistor --> LED --> GND

Okay, so let's add an NPN transistor in here to turn the LED on and off (ignore the base for now). We have two possible configurations:

+5V --> NPN Trans. --> Resistor --> LED --> GND
or
+5V --> Resistor --> LED --> NPN Trans. --> GND

Of course, the first one would have the transistor in a "source" mode while the second one would have it in a "sink" mode.

Which one is better for driving the LED, or does it not matter?
 
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I think both configurations can be made to work but the keys to the question will be how the base is to be driven (I assume you plan to use it to turn the LED on and off) and what you mean by better. For example, does better mean a tighter tolerance on LED current?
 
es1 said:
I think both configurations can be made to work but the keys to the question will be how the base is to be driven (I assume you plan to use it to turn the LED on and off) and what you mean by better. For example, does better mean a tighter tolerance on LED current?

The base pin would be driven by a 5V output from a micro controller with a 560 Ohm current limiting resistor - same in both situations. Better simply means a brighter LED.
 
Hopefully the attached sketch will make things clear.

The transistor is off when the voltage between the base and emitter (Vbe) is less than 0.7 volts.
That is no current passes through the load via either the emitter or collector.

If Vbe is increased beyond 0.7 volts, the transistor is on and current passes through the load.

Since Vbe) + VL = supply voltage you can see where the base has to reside in the off condition.
 

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So I think in your case, with the base driven by a 5V 560Ω source, the sink is better as it can produce a higher LED current when we ignore the practical limitations of the semiconductors.

If we set the LED series R to 0Ω to maximize LED current we get:

Source => Iled = Bf*(5-Vbe-Vled)/560
Sink => Iled = Bf*(5-Vbe)/560

As Vled > 0 the sink can produce greater currents.

These equations fall out naturally from Studiot's sketch, and he provides an explanation for them.

But I think in practice setting the series R to 0Ω would likely produce currents that exceed the Ic rating of the NPN and the If rating of the LED. And since those are the limiting terms for the circuit in both cases I would say for your base driver and definition of better they are equivalent.

To see how a slight change in parameters can make things different, consider the case where the microcontroller output was 3.3V with the same 560Ω current limiting resistor.
 
But I think in practice setting the series R to 0Ω

Perhaps I should have mentioned that my 'load' includes series limiting resistors as well as LEDs.

Some LED types have a built in resistor.
 
Last edited:
Okay, so I did some measurements today and got some interesting results.

With the transistor placed before the LED and resistor (sourcing), I found a voltage drop across the transistor of about 1V. This reduced the current flowing through the system (with 8 LED's) to about 83 mA.

With the transistor sinking the LED and resistor, a voltage drop of about 0.1V was had. The current flowing through the same circuit (8 LED's) was about 140 mA.

I am in a position now where I need the transistor to source the current, so what would you guys recommend? I believe a MOSFET will dissipate even more voltage, so I think that's out of the question. I had a EE recommend that I use a PNP transistor instead of an NPN, but I am not sure if that will solve my problem.

Basically, I want all of the advantages of the sinking NPN transistor, but I want to have it source the current (other limitations require this). Recommendations?
 
tangodirt said:
Basically, I want all of the advantages of the sinking NPN transistor, but I want to have it source the current (other limitations require this). Recommendations?
Use a pnp transistor. You may need to add a pull up resistor.BTW. Why the need for source, do your LED's have a common cathode?
 
uart said:
Use a pnp transistor. You may need to add a pull up resistor.


BTW. Why the need for source, do your LED's have a common cathode?

This is for a 3D LED cube and the transistor is used to select the current layer for muxing. The shift registers I am planning to use can both source and sink, but in the case that they don't work, I want to have the ability to use an LED driver (which as far as I can tell, only sink current). I really don't want to be flipping LED's around on the cube in the case that I have to flip the anode and cathode.

A PNP transistor will allow for sourcing current without the voltage drop penalty? I am much more familiar with NPN transistors than PNP.
 

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