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I've seen tutorials aimed at beginners suggest the way to drive an LED from something without enough current drive is this:

schematic A
(option A)

but why not this:

schematic B
(option B)

Option B seems to have some advantages over option A:

  • fewer components
  • the transistor does not saturate, leading to a faster turn-off
  • the base current is put to good use in the LED, instead of making the base resistor warm

and the advantages of option A seem to be few:

  • brings the load closer to the supply rail

but when Vcc is significantly greater than the forward voltage of the LED, this hardly matters. So, given these advantages, why would option A be preferred? Something I'm overlooking?

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    \$\begingroup\$ This is a invalid question because it is based on a faulty assumption, or at least no evidence for the premise the question is based on. I often do put the LED in the emitter leg. When enough voltage is available, I put the resistor on the emitter and the LED on the collector. That makes a current sink such that the supply voltage doesn't matter as long as it is high enough for the total voltage and not so high to cause excessive dissipation. It's a good way to deal with a supply that can vary. Fix and I'll undo the -1. \$\endgroup\$
    – Olin Lathrop
    Feb 12, 2013 at 22:18
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    \$\begingroup\$ @OlinLathrop I think I'll back up Phil here and say that I can't remember the last time I saw an online schematic for a LED drive circuit that was an emitter follower. For ancedotal evidence, doing a google image search for "LED driver schematic" yields a combination of common-emitter and switch-mode solutions. \$\endgroup\$
    – W5VO
    Feb 12, 2013 at 23:08
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    \$\begingroup\$ @W5VO: Like I said, I often don't do it that way. What random people suggest on the internet isn't much evidence of anything useful. Asking why a bunch of uknown people are posting a particular type of answer isn't really a useful question, but I guess I'll go answer it anyway. \$\endgroup\$
    – Olin Lathrop
    Feb 12, 2013 at 23:59
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    \$\begingroup\$ @OlinLathrop You have to write an answer because of this \$\endgroup\$
    – Kortuk
    Feb 14, 2013 at 16:05
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    \$\begingroup\$ There's one unmentioned advantage of circuit A: It makes do with a much lower logic voltage to drive the transistor. You could successfully drive it from 1.2 V logic, whereas depending on the LED forward voltage, circuit B wouldn't even work with 3.3 V logic. That doesn't diminish the merits of circuit B, which have been described well. \$\endgroup\$
    – sh-
    Apr 11, 2019 at 14:32

5 Answers 5

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I would argue that there are fewer "gotcha's" with option A. I would recommend option A to people of unknown electronics skill because there's not a lot that can keep it from working. For option B to be viable, the following conditions must be true:

  • \$V_{CC_{LED}}\$ must be equal to \$V_{CC_{CONTROL}}\$
  • \$V_{CC}\$ must be greater than \$V_{f_{LED}} + V_{BE}\$
  • It is a topology unique to BJT devices

These conditions are not as universal as they might first seem. For example, with the first assumption, this rules out any auxiliary power supply for the load that is separate from the logic power supply. It also starts constricting values of \$V_{CC}\$ for a single LED when you start talking about blue or white LEDs with \$V_f\$ > 3.0 V and a controller running off a supply less than 5.0 V. And I think the other thing is that you can't really replace the BJT in option B with a MOSFET if you wanted to eliminate that base current.

Additionally, it is more complicated (marginally, but still) to calculate your load resistance. With option A, you can use an analogy such as "consider the transistor to operate like a switch". This is easy to understand, and then you can use familiar equations to calculate \$R_{load}\$.

\$R_{load}=\dfrac{V_{CC}-V_{f_{LED}}}{I_{LED}}\$

Compare that to what is required for option B and there is the marginal increase in difficulty:

\$R_{load}=\dfrac{V_{CC}-V_{f_{LED}}-V_{BE}}{I_{LED}}\$

Couple that with the fact that the advantages of option B often are not needed. Aside from the reduced part count, the base current from option A shouldn't increase the power consumption by more than 10%, and LEDs are rarely (unsubstantiated qualitative guess) driven fast enough for BJT saturation to matter.

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    \$\begingroup\$ If you're going to include V_be in your second equation, then in all fairness, you need to include V_ce(sat) in your first equation. \$\endgroup\$
    – Dave Tweed
    Feb 12, 2013 at 23:33
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    \$\begingroup\$ @DaveTweed Sure, you still have Vce, but in saturation it can be less than 0.1 V. The forward drop of your LED, or your power supply can vary that much. I would argue that it is in the noise of the calculation and can be safely ignored. However, Vbe is significant when facing low Vf LEDs (red, IR) or low power supply voltages because it is much larger. I can think of situations where it would matter, but none where an emitter-follower would work as well. \$\endgroup\$
    – W5VO
    Feb 13, 2013 at 1:36
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    \$\begingroup\$ I don't know if you can say it's unique to BJTs -- a MOSFET works as a source follower also, but I suppose a BJT does it better, in most respects. \$\endgroup\$
    – Phil Frost
    Feb 13, 2013 at 2:40
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    \$\begingroup\$ @PhilFrost Perhaps it would be better to say it is uniquely suited to a BJT. A MOSFET would not give you the same behavior with the same basic input configuration and circuit topology. That's not to say you couldn't make it work, but it wouldn't be equivalent. \$\endgroup\$
    – W5VO
    Feb 13, 2013 at 9:35
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An even better variation on your option "B" is to put the LED in series with the collector, while leaving the resistor in series with the emitter.

schematic

simulate this circuit – Schematic created using CircuitLab

This turns the transistor into a controlled current sink, where the current is determined by the base voltage, minus VBE, across the resistor. The base voltage normally comes from a digital output of a microcontroller, which is fed from a regulator, so its value is tightly controlled. For example, if you're using 3.3V logic, and have a 270Ω resistor, you'll get a nice 10 mA through the LED.

The anode of the LED (or even a long string of LEDs) is fed from a higher voltage (which doesn't even need to be regulated), and whatever voltage drop that doesn't appear across the LED(s) appears across the transistor.

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  • \$\begingroup\$ I suppose I was considering cases where there is only a +5V supply available, but this is a good point, when a higher voltage than the logic is available. I suppose one could always add resistance to the base to make a voltage divider, too, and have the same part count as option A. \$\endgroup\$
    – Phil Frost
    Feb 12, 2013 at 22:40
  • \$\begingroup\$ @Dave Could you add a schematic showing your variation of option "B"? Would be helpful for the visual. \$\endgroup\$
    – JYelton
    Jun 20, 2013 at 22:16
  • \$\begingroup\$ @JYelton I just did. Hopefully I got it right. \$\endgroup\$
    – Phil Frost
    Jun 22, 2013 at 0:56
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Option B requires the control signal to be raised to a higher voltage than the LED drop voltage plus the base/emitter drop voltage. If your control driver is able to operate at a higher voltage than the LED drop voltage plus the transistor base/emitter drop voltage, then Option B would be valid.

Option A on the other hand can easily drive any LED drop voltage assuming your supply rail is high enough and you don't reach the base/collector breakdown voltage.

Also keep in mind if you intend to drive multiple LED's in series you have to add up all the drop voltages of the LEDs.

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    \$\begingroup\$ Given the limited ability for TTL outputs to pull high, option A was the safest at that time. Which was probably when today's educators were learning... \$\endgroup\$
    – user16324
    Feb 12, 2013 at 23:41
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Option A is a neat ON/ OFF switch. When BJT is saturated, LED current depends basically on Vcc and R3, so LED will have a constant brightness.

Option B is an "emitter follower" and makes LED current to depend on input voltage, as VE would be Vin -0.7.

Option B is good if you want to control LED current and brightness. But most of the times, it's better done with option A and a PWM scheme (more accurate)

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    \$\begingroup\$ Why is option B less suited to PWM control? I'd argue it's better suited. Among other things, option B does not exhibit storage delay. \$\endgroup\$
    – Phil Frost
    Jul 16, 2014 at 12:47
  • \$\begingroup\$ Phil, storage delay is usually negligible at common PWM frequencies, especially if what we want is control an LED brightness, a few kHz are fine. Second, a PWM driver is normally a microcontroller that may run at 3V3 or less (already few at 5V). You might not have enough voltage to drive the EF configuration. \$\endgroup\$
    – Joan
    Oct 28, 2017 at 7:29
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I am not convinced about your implied assumption that the usual way is to use a the common emitter configuration. However, let's assume that's true. It's not worth going into the merits of the various approaches since that's not your question anyway.

I think the reason is that the common emitter configuration is the conceptually obvious one, and there is little more to it than that. Keep in mind who writes this kind of advice you "see on the internet somewhere". The guy that uses whatever method is appropriate for the particular design without it occurring to him this is even a issue isn't going to think of writing up a web page about how to drive a LED. It's the person who just spent 2 days figuring out which legs of the transitor is the collectorator, emisser, and base-a-ma-thing, then a week getting the microcontroller code to blink the LED that's going to proudly post Looky me world, I done blinked me a LED!!! For those people, the common emitter configuration is the conceptually obvious one.

Common emitter is sortof the poster case of how to use a bipolar transistor. It is more obvious how the transistor provides amplification. For the newbie, emitter follower and even worse, using a bipolar as a controlled current sink, sound like advanced concepts.

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  • \$\begingroup\$ Could you please replace the 2nd paragraph with a (then third) paragraph that explains a little bit what makes a BJT a controlled current sink? This would make your answer way more valuable. Thank you. \$\endgroup\$
    – try-catch-finally
    Feb 4, 2017 at 8:06
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    \$\begingroup\$ @try: That would be getting off the topic of the question, which is why is one method more commonly "seen on the internet". \$\endgroup\$
    – Olin Lathrop
    Feb 4, 2017 at 14:22

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