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I am using an LM3915 LED driver in dot-mode. I would like to drive 5 blue LEDs (3.3V forward voltage @ 20mA per LED) per output. What type of transistor should I use to make this happen? And how do i put the transistors in the circuit i was thinking to use 18 volts as power supply I don't think the IC can handle 5 LED's by itself thanks

Original question: what i want to do is 5 leds all blue per output with the lm3915 in dot mode the leds have a voltage of 3,30 V the current of the leds each are 20 mA i want to put them in a analog vu shape <<< something like you get like simulated needle movement but with leds thanks

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    \$\begingroup\$ Please rewrite this question to be more coherent...otherwise it's going to be (rightfully) closed. \$\endgroup\$ – dext0rb Jun 25 '12 at 23:06
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    \$\begingroup\$ I Did rewrite the question i hope it be more clear thanks \$\endgroup\$ – Danny Jun 25 '12 at 23:42
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The LM3915 can drive LEDs at up to 30mA on each output, so it is capable of doing what you want without the transistors. The outputs are open collector, which means they sink the LED current through them to ground.

Having 5 LEDs on each output isn't a problem as long as your supply voltage is high enough to cover the total voltage drops from the LEDs, and low enough not to go over the maximum recommended supply voltage.
Just to check, 5 x 3.3V = 16.5V total drop from the LEDs, and the maximum supply is 25V so you should be fine. To reduce the power the LM3915 has to handle, use a supply just a bit over 16.5V, say 18-20V. If you use say 20V and 20mA per pin, the power consumed per pin will be (20V - 16.5V) * 0.020 = 70mW.

If you use all 10 outputs (with 5 LEDs @20mA on each) then the total power dissipation will be 70mW * 10 = 700mW (plus ~10mA for the rest of the IC operation), so it will get a bit warm. The LM3915 is rated for a maximum of 1365mW so it's within limits if you are running it at normal temperatures (e.g. ~0 - 40 deg C).

To limit power dissipation further you can drop the supply voltage to e.g. 18V, or put a series resistor in place on the supply line. A single 0.5W 5 ohm resistor would reduce the LM3915 dissipation by around a third.

Read the datasheet carefully, all the stuff you need to know is in there, plus plenty of example circuits.

EDIT - To clarify things, in a series circuit the current flowing through each component is the same, and the sum of the (can be different) voltages across each component adds up to the supply voltage.
In a parallel circuit, the voltage across each component is the same, and the (can be different) currents add up to the total supply current.

So with your 5 3.3V LEDs, if we put them in series and apply 20mA (it doesn't matter about how we limit the current for these examples), then 20mA will pass through each one. The total voltage drop will be 3.3 * 5 = 16.5V. The total power is 16.5 * 0.020 = 0.33W.

If we place them in parallel and apply 20mA then we know the total current will be 100mA, and the voltage drop will be 3.3V. The total power will be 0.1 * 3.3 = 0.33W. So you can see the same power is dissipated in each case.
For further reading Wiki has a good page on Series and Parallel circuits.

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  • \$\begingroup\$ well it is 20mA per LED so it is 100 mA per output so that is over the 30 mA of the IC thanks \$\endgroup\$ – Danny Jun 27 '12 at 0:24
  • \$\begingroup\$ @Danny - If the LEDs are in series (as inferred above) then the same current flows through all of them, so it's 20mA. They would only draw 100mA if they were in parallel. The same amount of power is dissipated in either case. \$\endgroup\$ – Oli Glaser Jun 27 '12 at 1:18
  • \$\begingroup\$ ok thanks what about the brightness ? \$\endgroup\$ – Danny Jun 27 '12 at 1:30
  • \$\begingroup\$ The brightness for each LED will be whatever it gives in the datasheet for 20mA. Obviously you can choose to run them at a different current if you want a different brightness - the datasheet has an example of an adjustable brightness setup. \$\endgroup\$ – Oli Glaser Jun 27 '12 at 1:34

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