How is this use of a transistor?

Question

Im working with a LM3915 makeing a vu-meter. I dont think the LM3915 will directly drive more that 1 LED. I'm looking to find a way that I can hook up more that 1 led to the LM3915. I think that transistors will do the trick but I don't know enough about using them. I made this schematic in eagle, which I just learned, and was hoping someone could look at it and check out if this set up would work and then how I calculate the resistor's value if I'm using X LEDs.

LM3915 datasheet
Info on the LED

Schematic:

Answer 1

Why didn't you follow the suggestions from this answer?

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You have places your LEDs in parallel, which is a Bad Habit™. Each LED should have its own series resistor.

You also placed the LEDs in reverse, which may be caused by your placing of the GND on top of the schematic; if you would place it at the bottom, where it's usually placed, you might notice the error easier.

The datasheet doesn't say in so many words, but I presume that the outputs of the comparators are open collector NPNs. That means that to turn on LED the NPN will sink current, when the LED is off it's just open, i.e. there's no current. Your NPN transistors are in common collector. This could work if the LM3915's output would source current, but I wonder if you meant it that way. It would mean that the LEDs' function is reversed: a LED which would be on in the typical LM3915 application would be off, and vice versa. Did you mean to use PNPs? If so you have to add a base resistor (like I mentioned in the other answer).

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edit
I'll repeat the schematic from the other answer:

Since you only seem to have +5V you won't be able to place more than 1 LED in series: at 2.2V forward voltage 2 LEDs would leave too little voltage drop for the transistor + resistor. But you still have to use a resistor per "string"!
Maybe I wasn't clear enough about the type of transistor, but you're using an NPN, whereas my solution has a PNP. The difference in the symbol is in the direction of the arrow: for an NPN it points to the emitter, for a PNP it points from the emitter to the base. See also this question.
Also in the other answer I explained the function of R1 and R2. I don't know why you have omitted them.

edit
A PNP transistor will allow a higher current from emitter (the arrow's tail) to collector if a small current flows from emitter to base (following the arrow). R2 allows this base current if the LM3915 makes its output low. If you choose a 4k7 (that's shorthand for 4700\$\Omega\$, see also this question), you'll have about 1mA flowing from the base. The collector current is the base current multiplied by \$H_{FE}\$, a value you find in the transistor's datasheet. For the BC557 this is minimum 125, so we can have at least 125mA collector current. Can have, because we'll limit it to a lower value with the LEDs' series resistors. That's not only to protect the LEDs, but also the transistor doesn't allow more than 100mA collector current. That's why I advised for more than 4 LEDs (4 x 20mA = 80mA) to use a second transistor.
R1 is needed to make sure that the transistor won't conduct if the LM3915's output is off; there may be some leakage current.

On second thought the BC557 is not such a good choice. It's a small signal transistor, and we need more of a switching transistor, capable of a higher current. The current of the BC557 is also limited by its \$V_{CE(SAT)}\$, the voltage over the transistor when it's completely on. For the BC557 this can be nearly 1V, and a 1A current would cause a 1W dissipation, which is too much. We need a transistor with a lower \$V_{CE(SAT)}\$. Let's have a look at the NSS30070MR6T1G. This can source 700mA, more than enough to feed 20 LEDs in parallel at 20mA, and it still has an \$H_{FE}\$ of 150. Yet that means we need a bit more base current, so let's make R2 = 1k, that will give us at least 4mA base current.

afterthought:
In a comment you mentioned 20 LEDs per output. Does that mean that the bases of four of the transistors are connected to the same output of the LM3915? In that case provide a separate R1 and R2 for each transistor, and connect the lower end of the R2s together to the LM3915's output.

Note that you'll need a lot of current: 10 outputs x 20 LEDs per output x 20mA per LED = 4A (20W at 5V). If you could use a higher supply voltage like the 12V I used in my example you could place each time 4 LEDs in series and only require 10 x 5 x 20mA = 1A (12W at 12V). Note that you need much less power here. That's because of the voltage drop over your resistor. In the 5V version you lose 5V - 2.2V = 2.8V, representing 56% of your power in the resistor. In the 12V version you lose only 12V - (4 * 2.2V) = 3.2V or 27% of your power.

Answer 2

You have basically the right idea with two problems in your circuit and two in your schematic:

1. The LEDs are backwards.

2. Use a resistor for each LED instead of a single resistor per row. A small variation in forward drop between the LEDs of a row will cause large difference in current between LEDs. Think of the LED with the lowest forward drop hogging all the current. A resistor per LED fixes this.

3. Use dots on the schematic to show connections. This is especially necessary in your case where you have signal crossing each other in a grid.

4. Show the ground connections at the bottom of the columns, not at top. In general on a schematic, try to keep signal flowing left to right and high to low voltages top to bottom. Of course there are exceptions and everything can't be drawn that way, but basic stuff like putting the ground at bottom in your case is easy.

Answer 3

• Your LEDs are rated with a forward voltage of max. 2.2 V, so it is a good idea to connect them in parallel (like you do), not in series: Even two LEDs in series would leave only 5 V - (2 * 2.2 V) = 0.6 V for the transistor and resistor, which is not enough to take all the tolerances into account.

• Connect the emitters to GND. What makes a transistor work is the base-to-emitter voltage. With the emitter at GND, you know what you do to the base at all times. You can still put the load (LEDs + resistors) between 5 V and the collectors.

• The LEDs are applied the wrong way. You want the anode towards the 5 V rail and the cathode towards the GND rail.

• Use one resistor (each!) for each LED. Parallelling the LEDs without one resistor for each LED will not guarantee that the current is shared equally between the LEDs. Think of it this way: One LED may start to conduct first and take all the current, the others will stay dim or off.

• Use base resistors.

Answer 4

Your LEDs are backwards. The arrow points from positive to negative so that the side with the line on it should be on a lower potential.

Also, your inconsistent use of the dots on crossing wires to indicate a connection vs. a jump makes it hard to tell what is going on in your circuit.

Also, I would do it using a common emitter configuration of your NPN transistors. In this configuration the emitters are all connected to ground. The bases are controlled by your signals as you have it, but the collectors are very different. The negative side of your LED (side with the line) is connectetd to your collector and the positive side is connected through a current limiting resistor to your positive supply.