on-on-on pushbutton?

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ardarvin

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Hello,

I'm trying to make a guitar pedal with 3 inputs, one output, and a single pushbutton switch to switch between inputs. An LED will tell which line is active, and each input will have a volume control.

Since I don't think there is such thing as a on-on-on pushbutton, does anyone have any ideas on how to make one?

Here's my diagram, any suggestions on what sort of, uhmm...transistor(? see e-switch boxes...yes I'm very noob) I should use to do the switching between lines would be great. I'm trying to go for pop-free switching.

**broken link removed**

Thanks!
 
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The 3 step sequence can be done by a Johnson type counter such as the CD4017B or CD4022B. You connect the output of the fourth stage (output "3") to the reset line so the counter resets, after the third position, back to position one ('0"). It will then output "0", "1", "2", "0", "1","2"... etc. in sequence. You connect the pushbutton to the "Clock" input to trigger the counter. (Make sure you debounce the switch with a series resistor and capacitor to ground to avoid multiple triggers from one push, 100k ohm and 1µF should work. Also connect the "Clock Inhibit" line to ground).

The easiest way to switch between lines is to use an old-fashioned mechanical relay. It has low noise, high isolation, and high signal and transient handling capability. Use one that's designed to handle low level signals (gold contacts). You can use a double-pole relay to switch both the signal and the LED.

You will need to add a transistor driver (2N2222 or PN2222 work well) to drive the relay from the counter outputs. (Don't forget a diode across the relay coil, cathode to positive, to suppress the inductive transient, otherwise you'll fry the transistor.)
 
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Thanks! I'm hoping to be able to power it with a 9V battery. Do you know of any suitable low-power relays that would fit this bill?
 
instead of a relay

use a quad bi-lateral switch
I think it will do the deed
need to know how much voltage and current the switch is to control??
oh and yes the 4017 is perhaps the best choice to do the counting
 
ardarvin said:
Thanks! I'm hoping to be able to power it with a 9V battery. Do you know of any suitable low-power relays that would fit this bill?
Click to expand...
That's an additional requirement. Relays would rapidly kill the battery.

How long do you need the battery to last? A 9V alkaline battery has a capacity of 625mAh. Thus, for a 10mA load ), the battery would last 62.5 hours. (A typical LED takes 20mA although there are low current ones that operate at 2mA).

You could use a solid-state relay such as this Digi-Key - TLP592AF-ND (Toshiba - TLP592A(F)) which requires 5mA to turn on, much less than a mechanical relay.

A quad bilateral switch, such as a common 4066 has a maximum on-resistance of about 500 ohms. Is that low enough for your requirements? If not there are some out there with a lower resistance.

The 4066 switch takes essentially no power but it must be biased with a plus and minus voltage (or a split of the 9V) to carry the AC signal current. One way, using one battery, is to use two 10K ohm resistors (with capacitor decoupling) between the plus and minus leads of the 9V battery as a ground point for the AC signal voltage. The minus terminal of the battery is used for the counter, transistor, and LED grounds.

The LEDs can be driven by a separate transistor driver (2N2222).

Or power the unit from a wall-wort supply and you don't have to worry about changing the battery.
 
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here is the start of

a circuit using the 4017 and the 4066
you should maybe include 2n2222 transistors to drive the LEDs and the 4066 outputs
 

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Both of the clock debounce schemes recommended above ignore the clock maximum rise and fall time spec (see the datasheet). After the RC network, you will need a Schmitt trigger (CD40106) or a 555 configured as a Schmitt trigger. You could also use a 555 as a oneshot.
 
this should work

the 7555 is perhaps the best of the three?
 

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MrDEB said:
the 7555 is perhaps the best of the three?
Click to expand...
The 7555 is the only one that meets the rise and fall time specs of the 4017, but it needs a pullup resistor from pin 2 to +9V, and the input pulse has to be shorter than the output pulse. Either make the output pulse longer, or add a differentiator between the switch and pin 2, as shown below.

CD4017 will not reliably provide enough output current to light an LED and still have a good enough logic high level to switch a CD4066. Add an emitter follower for each LED, as shown.
 

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Hi ardarvin,

you might try this circuit. It offers all combinations (7 + OFF) to switch the pick-up coils of the guitar.

As the sound differs according to the pick-up location you might want to use any of those combinations for best sound.

There is no LED indication included yet.

The analog switches have terminating resistors on the output to prevent unwanted amplification of the unused pick-up.

Please check with audioguru if that part is OK for him.

Regards

Boncuk
 

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?? why the 10k resistors

the schmatic shows 10k resistors from the outputs to ground??
then I believe the 4066 needs a split power supply see above Crutschows post yoThe 4066 switch takes essentially no power but it must be biased with a plus and minus voltage (or a split of the 9V) to carry the AC signal current. One way, using one battery, is to use two 10K ohm resistors (with capacitor decoupling) between the plus and minus leads of the 9V battery as a ground point for the AC signal voltage. The minus terminal of the battery is used for the counter, transistor, and LED grounds.
the transistors to drive the leds but was wondering if a 4011 would be a better choice, fewer parts?? and only one would be required?? instead of 3 transistors. just a thought.?
 
revised outputs

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
 

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The 4011 outputs can only drive a couple mA. A better choice would be the 4049 which can typically sink about 15mA at 9V.
 
The 4066 control lines are inverted. Take them directly fom the 4017. The logic low voltage is marginal anyway if you take it from the junction of the LED and the resistor.
What was wrong with the emitter follower circuit I posted? It is much more efficient than using saturating switches to drive the LEDs. Even more efficient would be some small MOSFETs, like 2N7000.
 
The CD4017 from Texas Instruments and the MC14017 from Motorola/ON Semi have Schmitt-trigger clock inputs for unlimitied clock rise and fall times.
 
Your diode OR gates need pulldown resistors.
 
emitter follower circuit of 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
 

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