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 ?

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

R1 is there to make sure T1 won't conduct if the 3915's output is off. When off there may still be a small leakage current. This won't do any harm if the output drives LEDs directly, but if you drive an external transistor with it the leakage current will be amplified, and the LEDs may light up very slightly, though maybe only visible in the dark. In any case, it's easy to prevent that. Imagine the 3915 slowly increasing its current. At for instance 10 µA that current will go through R1, not T1's base. Why? 10 µA through 1 kΩ is 10 mV, and that's far below the 0.7 V T1 needs to begin conducting. So even 100 µA will only drop 100 mV, and still no base current. It's only when the current exceeds 700 µA that T1 begins to conduct. Because the emitter-base voltage will be 0.7 V the current through R1 will be limited to that 700 µA, what the 3915 will draw more comes from T1's base. In hindsight R1 may well be increased to 10 kΩ, then T1 will begin to conduct at 70 µA, which will well be above the 3915's leakage current.

Since the 3915's outputs are current sinks R2 is not really necessary to limit the current. But without R2 almost the full power supply voltage (minus 0.7 V) will be across the output, and then the device's dissipation may become too high. Hence R2, to take some of that voltage, and the power which goes with it.

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.

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  • \$\begingroup\$ Thanks a lot, that's VERY helpful. In my case, would be a 2N2907A a good choice for T1 ? It has an ICmax of 600mA, which is more than enough for me (5 LED branches, I should need something like 100mA). I can't find the maximum allowed base current in the datasheet... Does it sounds right if I ask my LM3915 to output 10mA (then I'll have to calculate R2 in consequence) ? Thx \$\endgroup\$ – niconoe Oct 2 '12 at 10:23
  • \$\begingroup\$ The 2N2907A's datasheet says maximum 400 mV saturation voltage at 15 mA base current and 150 mA collector current, so that looks OK: that's only 60 mW dissipation, and you still have a gain of minimum 10. \$\endgroup\$ – stevenvh Oct 2 '12 at 10:44
  • \$\begingroup\$ Great, I'm slowly getting it ! And I guess a 500Ω/1.4W resistor for R2 would allow to limit the voltage by dropping 5V before the 3915 output ? thanks \$\endgroup\$ – niconoe Oct 2 '12 at 11:06
  • \$\begingroup\$ @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. \$\endgroup\$ – stevenvh Oct 2 '12 at 11:14

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