Right so 560 ohms for R1. It's just that based on the need for 3V for the channel indicator leds. I'm wanting to use that same 3V that the leds are getting and making that the signal voltage BEFORE going into the Pot. Then whatever voltage is left over after going through the Pot will be the signal voltage. And since there will be NO led's lit up at 0V or max resistance then there must be a small resistor before the pot.

9V -> Voltage Divider -> 3V -> Channel LED to ground. in parallel with -> small resistor -> Potentiometer -> signal in.

EDIT: See updated diagram to see what I mean. (i forgot the small resistor for infront of the POT)

 

The LM3914 has regulated current drivers for the LEDs. With a 5V supply, the LEDs can be 2V red ones or 3.5V blue ones and the LM3914 provides a regulated current that is set by R1. If the supply voltage is much too high then the LED driver transistors inside the LM3914 get hot. Then the LM3914 will overheat if it is set to the bar mode and many LEDs are lighted at the same time.

The LM3914 has an accurate Vref that is probably more accurate than your supply voltage so you might as well use it as the reference voltage for the LED trigger voltages and for your pot.

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Uncle $crooge

 

Ok I can see how I would rig up that Pot along pins 4 5 and 6...but as one of the features of this circuit there will be two leds of which one will lit up indicating which "channel" or pot is being used at the time. And using only a DPDT I can't see where I would put the indicator LED's so that they would have current. Would they be able to be driven by the 1.25V? I was under the impression that most led need like 2+ V to work.

 

You are using only half of the switch to switch the audio. Use the other half of the switch to light one of the two LEDs.

You are not switching two LM3914 circuits so you don't want to connect the two LEDs there.

Leds work with current, not voltage. An LED needs a resistor in series with it to limit the current or it might blow up.

Attached Images

balance contols.PNG (21.5 KB, 6 views)

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Uncle $crooge

 

There is only 1 LM3914. The switch changes which Potentiometer is going to be the input for it and the Indicators. (see my schematic)

 

Would I be ABLE to do it my method? Because it would save on not needing 2 LM3914's.

 

You don't need two LM3914 ICs. You just need a DPDT switch and a triple linear pot.
You won't be able to buy a triple pot.

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Uncle $crooge

 

wait why would I need a triple linear pot? One part for audio and one for the visual.

 

I don't know why you have two audio balance pots. If you use only one audio balance pot then it could be a dual pot so one part can do the audio balance and the other pot can apply the variable DC to the LM3914.

If you have two completely separate audio balance pots and if each is a dual pot for the variable DC voltage to the LM3914 then you need the switch to be a TPDT (3 poles) so it can switch the selected pot to the LM3914.

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Uncle $crooge

 

ohh ok the reason there are two audio pots is that there are two channels.

Like if the footswitch is up then channel a will have Audio Pot a which will also control the input for the IC. If the footswitch is down channel b is now Audio Pot b and control the input for the IC.

Each Pot will control an Audio and the input to the IC.

 

I don't know why you have left and right stereo channels and need a balance pot, and I don't know why you are switching channels and pots.

Your guitar is mono, isn't it?

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Uncle $crooge

 

I think using the word channel isn't helping my situation really.

Basically player plugs in the guitar (mono)....then the sound gets split into left and right by the Pot. And the position of the pot (left to right) is displayed on the led's. Then if the player pushes the footswitch the OTHER pot is now being used to split the sound to left and right, and is also being used to control the display.

 

Why do you want to switch two pots?
The player needs only a single pot to fade the sound from left through center to right.

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Uncle $crooge

 

yeah its so that lets say I'm playing all on the right...then I can switch over to either center or the other side...it just leaves more options for me.

 

Two hands play your guitar.
One foot controls the switch.
Do you have a tail to adjust the pot?

Get an old wah-wah pedal and replace its log pot for a linear pot and use your foot to fade left and right.

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Uncle $crooge