Evening All,

I've had a search, but can't find an answer to this.

I want to build a sound activated camera trigger - my previous effort leaves a lot to be desired. Straight forward mic > amp stage, but what I'm aiming for here is to drive a series of LED's as a sort of VU meter, and be able to select to level at which triggering occurs.

The circuit I have in mind is attached, the output of which would be then fed to an adjustable delay, and then via an optoisolator to the camera.

The bit I'm most unsure of is if the 3914's outputs going low would draw sufficent current from the transistor's base, or if the current through the LED would prevent this - and if so should I look at making the 3914 sink more current and add a resistor between the LEDs and selector switch?

An advise appreciated!

Regards,
Ian.

 

You might want to use an LM3915 because it has a logarithmic scale, like human ears.
These chips are designed to sink at least 7 ma of current. That means the output pin will be 1 LED worth of voltage away from the V+ supply at 7 ma to 13 ma. Use that voltage to figure your base resistor.

 

If you want to using the LM3914, you can adjust the brightness of LED through the resistor on pin 6,7.

LM3914 datasheet, at the second circuit diagram on the page 10, refer to pin 6,7.

You can also refer to the internal structure of LM3914 on the page 5.

 

Thanks,

Yes, I've used the 3914 & 3915 as straight voltmeters and VU type meters before - what concerns me about this configuration is if the current drawn in by the 3914 would be enough to light the LED *AND* draw sufficent current from the transistor base to saturate it - or if current would be drawn mainly through the LED as a path of least resistance.

Given what #12 has said about the base being at +ve less what is dropped over the LED I wondered if I needed to add a resistance between each LED and +ve so that more of the current that the 3914 drew was pulled from the transistor base?

Or is the resistance of the collector/base even less than that of an LED?

Or am I missing something fundamental? Forgive me if this is obvious, but the help given on this forum is very much appreciated!

 

Just expand the drawing with lots of labels and the answer will become apparent.