on-on-on pushbutton?

[Roff]

your transistors need base resistors (the 4017 and/or the transistors will go poof!
the LEDs won't run to good with more than 25-30ma
see screen shot of TINA simulation

Go back and look at R3 in my schematic. You missed it. With R3, you don't need base resistors.

 

[MrDEB]

oups!! my mistake

I missed that resistor.
BUT isn't an npn more efficient with the load on the collector??
that's the way I see most of them connected.
I thought the base needed 1/10th of the load to achieve saturation?
other than that I think this circuit is a go??

 

[colin55]

The whole design can be simplified by taking each of the outputs of the 4017 and driving a transistor from one of the inputs of your guitar (no 4066 chip is required).

 

[crutschow]

BUT isn't an npn more efficient with the load on the collector??
that's the way I see most of them connected.
I thought the base needed 1/10th of the load to achieve saturation?
other than that I think this circuit is a go??

I take it you are not familiar with emitter-follower circuits? They may not be as common as a common-emitter circuit but they are still widely used.

Connecting the load in the collector is more efficient if you need maximum voltage to the load. But here we already have a resistor to drop some of the voltage, so dropping some additional voltage across the transistor does not affect the circuit operation, you don't need the transistor to be in saturation.

In a emitter follower the base only draws the current it needs, based upon the transistor beta and the emitter current.

 

[crutschow]

The whole design can be simplified by taking each of the outputs of the 4017 and driving a transistor from one of the inputs of your guitar (no 4066 chip is required).

Don't see that replacing one IC with three transistors and associated resistors particularly simplifies the design.

 

[colin55]

You are already using 3 transistors and one IC. My circuit removes the IC.

 

[audioguru]

The analog switch switches audio.
The CD4066 makes four excellent analog switches.
Transistors make lousy analog switches.

 

[Boncuk]

What are these e-switches on your circuit diagram?

Their inputs are labelled PU-I, PU-II and PU-III. Those are the inputs of the pick-ups.

Boncuk

 

[Boncuk]

this is a revised schematic of Boncuk with the 7555 circuit (added the pullup resistors. The pulse has a problem??)
the outputs from the 4017 can only supply a max of 10ma so I revised the output transistors
the base only draws 2 ma apx, the LED = 20ma, the data select linres only need .5+ - Vcc voltage so as I connected to the LED / resistor junction = 2v apx.
this is getting to look better all the time.Hope it dosn't end up like that bike alarm where the kid basically gave up because its tooo hard.
hopfully he will wake up and realize the pic method is more than he can chew

This circuit will hold the 4066s input high at all times. An active counter output will just light the LED with almost no change in switch behaviour. If a 3V voltage source is connected the output of the 4066 will drop to 2.8. With the counter output low the 4066 switches full 3V to its output.

I guess, that's not what the OP wanted to do.

Boncuk

 

[crutschow]

You are already using 3 transistors and one IC. My circuit removes the IC.

You can remove all the parts you like, but if the circuit doesn't do it's intended purpose, it's not of much use. The purpose is to route one of three analog signals to the output (the e-switch).

 

[Boncuk]

Here is a revised design of the circuit.

I played guitar for almost 40 years and I know that free selection of the pickups is very much desired by any guitar player.

E.g. playing country & western the player wants full acoustics of all strings, while playing jazz the high pitch sound is preferred. The sound depends very much on the pick-up selected.

The revision includes low current LED (2mA) indication of each selected pick-up.

Hi audioguru,

I used 10K terminating resistors at the outputs of the bilateral switches. Please have the kindness to suggest a good way to avoid cross talk.

Thank you

Hans

 

[audioguru]

Hi Hans,
The CD4066 analog gates must have their inputs biased at half the supply voltage.
The analog signal must be within its supply voltage.

The 10k output resistors add a little distortion but they kill any crosstalk.

 

[MrDEB]

isn't -50db enought??

in the data sheet it says -50db crosstalk typ.==not low enougth??
then .1% distortion??
I must be informed what is acceptable distortion etc.
thanks for the heads up on the transistors crutschow
learn something everyday.

 

[MrDEB]

they make a 2ma LED??

saw the 2700 ohm resistors and ???
ran it with TINA and yea 2ma
never knew there was such a beast.
must be pretty dim and/or small?

 

[Roff]

I missed that resistor.
BUT isn't an npn more efficient with the load on the collector??
that's the way I see most of them connected.
I thought the base needed 1/10th of the load to achieve saturation?
other than that I think this circuit is a go??

With the emitter follower circuit, all the current through the transistor, including base current, flows through the LED. In the saturating common emitter circuit, the base current is wasted in the sense that it does not flow through the LED.

 

[Boncuk]

Hi Hans,
The CD4066 analog gates must have their inputs biased at half the supply voltage.
The analog signal must be within its supply voltage.

The 10k output resistors add a little distortion but they kill any crosstalk.

The AF-output of the guitar pick-up delivers some 100mV only (Metal strings over a magnetic pick-up). Biasing the inputs will almost kill the AF-voltage because coupling caps will be required to keep the amplifier input free of DC.

I think it won't be much noticeable if a little distortion is added. (probably also wanted.)

 

[Boncuk]

saw the 2700 ohm resistors and ???
ran it with TINA and yea 2ma
never knew there was such a beast.
must be pretty dim and/or small?

Hi MrDEB,

low current LEDs have almost the same brightness as low cost 20mA standard LEDs. I used them a lot in switching cabinets for industrial applications.

Keeps the power supply at low current too, if many LEDs are involved.

In this particular case I assume battery operation.

Boncuk

 

[audioguru]

The AF-output of the guitar pick-up delivers some 100mV only (Metal strings over a magnetic pick-up). Biasing the inputs will almost kill the AF-voltage because coupling caps will be required to keep the amplifier input free of DC.

.
Most guitar pickup circuits on the web say that the output of the pickup is in Volts, not millivolts and a gain of less than only 3 is needed.

I think it won't be much noticeable if a little distortion is added. (probably also wanted.)

As much distortion and fuzz as is possible so it sounds horrible.
(maybe most electric guitar players are deaf to the distortion harmonics).

 

[audioguru]

saw the 2700 ohm resistors and ???

I looked at all 54 schematics in this thread (it took a while) and didn't see a 2700 ohm resistor.

Why don't you attach the schematic that you are talking about with your discussion about it?

 

[MrDEB]

the 2700 ohm resistors

t5he 2700 oh
m resistors are on the collector of the transistors.
54 schematics you mean 4??
https://www.electro-tech-online.com/attachments/pickup-sch-pdf.26815/