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I have to activate four LEDs at the same time when the signal varies from 0 V to 5 V. I still have not defined the transistor I will use, but before that I would like to know the best way to do this, with schematic 1 or with schematic 2: Activate all the LEDs with a single transistor or each LED to be activated by a dedicated transistor.

I know I should consider the current that will pass between collector and sender to determine which of the two methods to use, but I would like to know which one is the most correct.

schematic

simulate this circuit – Schematic created using CircuitLab

schematic

simulate this circuit

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    \$\begingroup\$ How do you feel about MOSFET in place of the BJT? \$\endgroup\$ – awjlogan Apr 13 '18 at 13:35
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    \$\begingroup\$ #1 is better unless you want to control the LEDs separately. \$\endgroup\$ – τεκ Apr 13 '18 at 13:36
  • \$\begingroup\$ When you say the signal 'varies' between 0v and 5v do you mean that it is either 0v or 5v or do you expect an LED response in the range 0.5v, 1v, 2v, etc? \$\endgroup\$ – nvuono Apr 13 '18 at 13:44
  • \$\begingroup\$ @awjlogan In this case I think it is better to use BJT because the cost is lower. \$\endgroup\$ – Eduardo Cardoso Apr 13 '18 at 13:53
  • \$\begingroup\$ @nvuono Digital signal, 0V or 5V. \$\endgroup\$ – Eduardo Cardoso Apr 13 '18 at 13:54
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Both will work well. However, the second circuit is obviously more expensive to build (and takes more time to solder manually), so you will probably choose the first option (single transistor).

The second option (multiple transistors) can be a better choice when there is a lot of current going through the LEDs. You can then share the current through multiple transistors, which can therefore be smaller (sometimes, 4 small transistors are cheaper than one big, and the heat can be dissipated more easily). But given the currents involved in your case, it doesn't matter.

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Either method works, but I would generally use the first because it is simpler and requires fewer parts.

If the signal is actively driven both directions, like from a CMOS digital output, then you don't need the resistor between base and ground. In other words, you can lose R2 in the first circuit, and R10, R11, R12, and R13 in the second.

As you say, the transistor in the first circuit needs to be able to handle the combined LED current. For four normal indicator LEDs, that is not much of a limitation.

Depending on the voltage of the LEDs, you might be able to drive two at a time in series with a single resistor. For example, if these are red with 1.8 V forward drop, and the transistor goes to 200 mV in saturation, then there is still 1.2 V left for a resistor to set the current. Doing that gets you the same LEDs lit with the same brightness, but with half the current used.

Green LEDs with 2.1 V drop are on the edge, but can work doubled up. Two LEDs would drop 4.2 V. Again, figuring 200 mV for the saturated transistor, that still leaves 600 mV for the current limiting resistor. Especially if you're not trying to run the LEDs at their limit, this can be a legitimate current savings.

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  • \$\begingroup\$ Thanks for the reply. I have chosen to use the pull-down resistors because I will use NPN transistors and I would like to make sure I always have 0V at the base of the transistor when the signal is not activated \$\endgroup\$ – Eduardo Cardoso Apr 13 '18 at 13:56
  • \$\begingroup\$ Not nitpicking, but I, for one, tend to avoid using in series because of the risk of a fault -- if one burns, the whole branch goes dark. It looks very "nice" in case of custom logos+walking text. :-) \$\endgroup\$ – a concerned citizen Apr 13 '18 at 14:42
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    \$\begingroup\$ @aconcernedcitizen Depending on the application, failure of a single LED may look bad enough anyway. (And just to note that "failure" is generally a problem with soldering; LEDs themselves will usually have a longer lifespan than any of the mechanical elements they're attached to.) \$\endgroup\$ – Graham Apr 13 '18 at 14:56
  • \$\begingroup\$ @Graham True, but, somehow, it's one thing to have one dark spot, compared to a few in a row/column (there may be more than two in series if you have 12V, for example). That's why I said it can look "nice".Still, it was a side comment. "Fault" is probably the result of "lost in translation". To me it means a fault, in general: faulty PCB, faulty LED, faulty everything (one's self included). \$\endgroup\$ – a concerned citizen Apr 13 '18 at 15:23
  • \$\begingroup\$ Whoever downvoted this, what exactly do you think is wrong? \$\endgroup\$ – Olin Lathrop Apr 13 '18 at 20:46
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Both circuits will work.

The first circuit is entirely adequate unless you value independent control of each LED.

The second circuit has the advantage that if desired you can turn individual LEDs on or off independently and the (obvious) disadvantage of requiring more components.

Almost any NPN transistor will work. If the LEDs are white they will have ABOUT 3v on voltage. So I per LED = V/R = (5-3)/560R = 3.6 mA per LED or about 15 mA for all 4 LEDs. Almost any small NPN transistor will work here.

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Assuming that your LED require 20 mA each, circuit A will require a transistor that can drive roughly 80 mA. Most transistors will be able to do the job without any problem.

On the second circuit, each leg of the circuit will require 20 mA. Again, this is not a problem for transistors.

From an engineering standpoint, circuit A is more cost effective since you need fewer components (thus reducing future board ''real estate'' and unit cost). The second one's main advantage is the ability to control each LED individually.

One way that you could perhaps improve on circuit one would be to pair up two LED per leg and reduce the resistor accordingly. That way, each leg will still be running at 20 mA, but you will reduce your total current by half and reduce the amount of voltage that is lost through each current limiting resistors.

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This circuit is more efficient on parts, saving two resistors:

schematic

simulate this circuit – Schematic created using CircuitLab

It's called an emitter follower. The voltage the LEDs see is the input voltage minus 0.6V. For this reason, you need a CMOS output which can drive close to 5V. A TTL output may need a pullup resistor to get it to go to 5V.

For a 5V supply, you want 220 ohm resistors for a decent brightness with red or green LEDs. 560 will work but it will be dim. Blue LEDs will need even lower.

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  • \$\begingroup\$ If it's driven by TTL, couldn't it be factored in by lowering the resistor values? What would be the power dissipation in the two cases? \$\endgroup\$ – Peter Mortensen Apr 14 '18 at 6:11
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From your questions, I get the feeling you misunderstand the circuits. Both are "electrically correct." If the question was: 1-which one is easier to build?, 2-which one has fewer parts (more reliable), 3-which one has more LED control?, etc. then you would be able to decide which circuit you should use. However, the amount of "current from the collector to the emitter," would NOT be a parameter to determine which circuit to use. This parameter is used to determine the current capability of the single transistor or of the 4 "equivalent" transistors.
If you decide to use a single transistor, it obvious that this transistor should be able to handle 4 times the current of each of the "four" transistors.

For example, each LED needs 15ma, then you need a single transistor that handles 60ma (min.), OR four transistors that each handles 15 ma (min.).

I hope that this makes it clear, that the current capability of the transistor(s) is NOT the deciding factor, to select circuit 1 or 2.

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