LM3915--Dot mode Question

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vinniekins

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Hi there,

I am trying to hook up a LM3915 to 8ohm speaker outputs to get some dancing lights as attached (the transistors on the outputs are for attaching more than 1 LED per output). The bar mode worked, but when I switch to dot mode, LED1 will be on when there is a signal... ie: when LED7 is on in dot mode, LED1 is also on.

I have read that it maybe ground loops, but I am not certain. Is there anything I've missed on my schematic for starters?
 

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According to the **broken link removed**, the leakage from pin 1 can be as much as 450uA (see bottom of p.3). This will keep your PNP driver turned on. You need a 1k resistor from base to emitter of each transistor. This will keep the transistor turned off when it is supposed to be off. The leakage from the other outputs is 10uA max, but this could still result is several mA of collector current. For simplicity, make all the base-emitter resistors the same value.
 
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For voltage reference stability and predictability, you should run at least 1mA through the reference voltage divider. This is because the adjust pin current can be as high as 120uA. If you are trying to get ≈10V on pin 6, change the 2.7k resistor to 1k, and change the 18k to 6.8k.
 
The pull-up resistors worked! Thanks!

So with the pull-up resistors in place, would I be able to remove the 2.2k resistors on the base of the 3906 as attached, just like the attached schematic?
 

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You should leave the base resistors in.

From website Transistor Circuits

 
You don't need series base resistors for the transistors because the outputs of the LM3915 have current regulation.
 
You don't need series base resistors for the transistors because the outputs of the LM3915 have current regulation.
True, but I recommended that he run at least a milliamp through the reference divider. Sans resistors, this will give him ≈10mA of base current in his PNPs. Not excessive, of course. It just offends my sensibilities. Leaving the resistors out would not be harmful to devices or performance.
Higher than necessary base currents are deleterious in high speed switching circuits (which this is not), because they increase the amount of base charge that must be removed before the transistor can turn off.
 
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