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When using a 2222 or 3904 NPN-type transistor to switch a simple LED, is it better to place it inline with the negative or positive rail of the load (LED)?

Also, which rail is best when working with darlington transistors or MOSFETs for high powered loads?

Soviero
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4 Answers4

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The two configurations are:

schematic

simulate this circuit – Schematic created using CircuitLab

The left side configuration is the common emitter

  • The input needs to be just 0.7v higher than the emitter in order to turn the transistor on.
  • A base resistor is required to limit the base-emitter current to safe levels (but also enough for the intended output current).
  • When the transistor turns on the load (led+resistor) gets the full power supply across it reduced just by the Vce saturation voltage.
  • A change in the power supply level also changes the voltage across the load.
  • The collector sinks current (provides ground) when the input (base) is positive, so the output level is inverted (the led on/off operation is not inverted).

The right side configuration is the common collector or emitter follower

  • The input needs to be 0.7v higher that the intended emitter voltage, so the proper drive level depends on the intended output level.
  • When the transistor turns on the voltage applied to the load (led+resistor) is 0.7v lower that the voltage of the base no matter if the collector supply level changes (assuming it doesn't become too low).
  • There is no need for a base resistor (it's optional), the base-emitter current is limited as a function of the required load current and the transistor.
  • The emitter sources current (provides positive supply) when the input (base) is positive, so the output is not inverted.
  • The power dissipated on the transistor can be high since it's the output current multiplied with the Vce voltage which changes as the collector supply level changes.

For a simple led you can use any of the two configuration assuming that a control voltage of led Vf (forward voltage) +0.7v is available.

In the common emitter circuit, because of the way that the base resistor is calculated (using a gain estimation) there will be some base-emitter current wasted (an emitter resistor is an option of course).
On the other hand with the common collector circuit the base current will be exactly as high as it needs to be based on the load requirement and the transistor gain.

If you want to keep the led voltage constant when the power supply level changes then the common collector is better suited.

If instead of a simple led you have a seven segment display then the common emitter suits better to drive a common cathode type and the common collector to drive a common anode display.

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alexan_e
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  • +1, but completeness, the BJT, being a three terminal device, has a *third* configuration: [common base](http://en.wikipedia.org/wiki/Common_base). Can't say I've ever thought to drive an LED with a common base BJT, though. – Phil Frost Jan 27 '14 at 12:58
  • For that matter, it's possible to put the transistor in the middle of the LED and the resistor, too. With the resistor on the emitter, and the LED on the collector, it makes the transistor into a current sink, which can be pretty handy if the supply voltage for the LED isn't regulated, but the supply voltage for the thing driving the transistor (microcontroller, etc) is. – Phil Frost Jan 27 '14 at 12:59
  • @PhilFrost Yes, the transistor can be placed in the middle but for simplicity in order to explain the two options I preferred to treat the resistor+led as a single load. I didn't think the common base was relevant for what was asked so I didn't mention it, maybe I'll add a couple of lines with possible uses for that. – alexan_e Jan 27 '14 at 14:20
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There is no universal "best" or "better" in either case. It depends on your requirements. These are all valid configurations:

schematic

simulate this circuit – Schematic created using CircuitLab

These all work with N-channel MOSFETs also, and if Vcc is made negative (or if you consider Vcc to be "ground"), then they work with PNP BJTs or P-channel MOSFETs. Each has unique advantages:

The LED current in Q1's circuit is dependent only on the base voltage, and not on Vcc, and it uses only one resistor.

In Q2, \$V_{ce}\$ is only about 0.2V, giving you more available voltage for the LED and resistor than the other configurations. It also has high voltage gain, making LED current largely insensitive to input voltage (besides being on or off).

In Q3, the base current helps light the LED, the transistor doesn't saturate and so can switch faster, and it uses only one resistor.

I can't think of a lot of reasons you'd use Q4 for blinking an LED, but it does work. This arrangement is called common base, and it finds application in high frequency amplifiers.

Of course, each arrangement also suffers from unique disadvantages which I won't elaborate. Point being, the answer to your question is "it depends."

For some elaboration on some of these possibilities, see Why would one drive LEDs with a common emitter?

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Phil Frost
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N-type devices (NPN BJTs, NMOSFETs) are for low-side (negative supply) switching. P-type devices (PNP BJTs, PMOSFETs) are for high-side (positive supply) switching. That's not to say that they couldn't be used for the other, but there are additional voltage level concerns to take into consideration when doing so.

Ignacio Vazquez-Abrams
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    That's only if you are using a common-emitter arrangement. Plenty of reasons you might want to do otherwise, even for something as simple as driving an LED: [Why would one drive LEDs with a common emitter?](http://electronics.stackexchange.com/q/57845/17608) – Phil Frost Jan 27 '14 at 12:56
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For an NPN transistor, a more effective circuit is achieved with the LED in the collector (in series with a resistor). The reason is that the transistor is easily switched on into saturation with a moderate base voltage of about 0.6V.

With an emitter follower configuration (LED in the emitter and collector tied to positive), the LED is turned on when the base is at about 2.6volts - it has to overcome the LED forward voltage (about 2 volts for standard LEDs) AND the base emitter "turn-on" voltage of 0.6V.

This means that the emitter follower configuration is out of reach for logic supplies less than about 2.4V.

Also, which rail is best when working with darlington transistors or MOSFETs for high powered loads?

This is too broad to answer as it totally depends on the type of load.

Andy aka
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