# LM3915 to drive more LEDs / basic questions about PNP transistors in saturation

As a few others here, I'm trying to drive a few (10) LEDs from each output of a LM3915 vu-meter chip. I've read different posts, and it seems the universal recommendation is this great answer.

I have, however a few difficulties understanding it in details to adapt it to my configuration (V+ = 9v, 5 rows of 2 leds). How are there R1 and R2 resistors choosen ? I see you want to reach 2mA, but I'm not sure about how to translate this to the resistor values.

I also guess the LM3915 is a specific beast here, as its output current can be configured (to drive LEDs directly without resistors for example). So in my naive mind, are these two resistors necessary at all, if the output current of the chip is configured properly ?

Also, is R1 only there to ensure that the transistor based is positive (instead of floating) when nothing is happening ? In that case, shouldn't its value be larger thaOr has it an additional purpose ?

I think I also have to ensure that (base current * minimal Hfe) > needed output current, or is this not applicable when the transistor is used in saturation ?

• Thanks, but I don't understand your comment ! Can you clarify ? – niconoe Oct 2 '12 at 8:53

You'll have to dimension R2 such that $I_B \times h_{FE}$ > required $I_C$, like you say. Note that for a transistor in saturation $h_{FE}$ is much lower than the value specified in the datasheet, so a base current of a couple tens of mA is no luxe. Check the transistor's datasheet for the maximum allowed base current.
• @niconoe - Remember that T1 also drops 0.7 V and I would also leave some headroom, like 300 mV, then at 5 V supply R2 will be 390 $\Omega$ (which is an E12 value). Power is 4 V x 10 mA = 40 mW, so a 1/4 W resistor is OK, and the 3915 will dissipate 0.3 V x 10 mA = 3 mW per output, so 30 mW in total, also fine. – stevenvh Oct 2 '12 at 11:14