Anyone knows if this experimental bypass could work?

[Crontox102098]

*COMMERCIAL* Jejeje

I'm trying with some ways to make a relay soft touch bypass and my final result is this... but i don't know if could work.

Can it work?  

[R.G.]

It could work.

I would like it better if the relay coil were between the transistor collector and the + supply, and there was a diode with anode to collector of the transistor to the + supply. As you have it now, the flyback voltage from the inductance of the relay coil will break over the base-emitter junction sometime, perhaps all the time, depending on the transistor. That will eventually ruin it.

On a bigger scale, although the CD4013 can be used for this, there is a simpler circuit in many ways. A CMOS hex inverter, like the CD4049 or CD4069, can be set up to use two sections as a flipflop, and these can be triggered by an external switch without the first transistor. You'll still need the second transistor to drive the relay coil.

[merlinb]

I would like it better if the relay coil were between the transistor collector and the + supply, and there was a diode with anode to collector of the transistor to the + supply. As you have it now, the flyback voltage from the inductance of the relay coil will break over the base-emitter junction sometime, perhaps all the time, depending on the transistor. That will eventually ruin it.

Actually, Crontox's configuration is quite safe because the flyback voltage pulls the emitter *down* not up, so it simply turns the transistor on for a bit longer. This configuration is sometimes preferred because it saves on a flyback diode.

Controntox, you don't actually need Q1; you can clock the 4013 directly from the switch network *provided* you use a 4013 that has a Schmitt triggered clock input (the Philips one does, but not all of them do). On the other hand, if the Schmitt trigger input is an issue for you, it might be better to do as RG says and simply use a hex inverter to do both the debouncing and toggling, ala:

[Crontox102098]

Here they don't have 4049 ¿I can use 40106?

[R.G.]

Yes.

But the CD40106 will not provide as much output current as the CD4049 if you try to use it to drive an LED directly.

If you use the transistor to drive both the relay coil and the LED, the 40106 will be fine.

[Crontox102098]

Yes.

But the CD40106 will not provide as much output current as the CD4049 if you try to use it to drive an LED directly.

If you use the transistor to drive both the relay coil and the LED, the 40106 will be fine.

Yeah, that's i'm going to make, so... some transistor "2N2222 or 2N3904" can support a relay coil? -Sorry, i'm a electronic ignorant-

[R.G.]

Yes. A 2N2222 or 2N3904 transistor can support a relay coil. 

The datasheet for the CD40106 will show the maximum output currents per section. I don't remember these exactly right now, but it's a few milliamperes. A relay coil will be in the 20-50ma range. That will be in the relay datasheet. And the collector current maximums for the transistors and their typical current gains will be in THEIR datasheets. So you can use the few milliamperes per gate of the CMOS device to drive the base current of the transistor and amplify that current to a much larger one.

The common thread here is datasheets. Data on parts is incredibly easy to find today. At one time you had to write or telephone the device manufacturers and wait while paper datasheets were mailed to you. Today your mouse gives you worlds of information. It's a Golden Age for electronic designers.

[Crontox102098]

Yes. A 2N2222 or 2N3904 transistor can support a relay coil. 

The datasheet for the CD40106 will show the maximum output currents per section. I don't remember these exactly right now, but it's a few milliamperes. A relay coil will be in the 20-50ma range. That will be in the relay datasheet. And the collector current maximums for the transistors and their typical current gains will be in THEIR datasheets. So you can use the few milliamperes per gate of the CMOS device to drive the base current of the transistor and amplify that current to a much larger one.

The common thread here is datasheets. Data on parts is incredibly easy to find today. At one time you had to write or telephone the device manufacturers and wait while paper datasheets were mailed to you. Today your mouse gives you worlds of information. It's a Golden Age for electronic designers.

And my last question is... this "bypass" could make pop/click or is it clickless?

[amptramp]

There is nothing inherently more "clickless" about this design than any using the standard 3PDT switch.  The way to make a design approach "clickless" is to keep the DC voltages the same for both active and bypassed conditions, usually done by a resistance to ground after the final coupling capacitor.  Note that the bypass output can be obtained straight from the input and not through another switch contact as you have drawn it.  Removing a contact not only improves reliability, it may reduce the effective difference in timing between switch sections so there is less of a click.  True "clickless" designs use electronic switching which is slowed down so the output appears to pan between the effect output and the input at a slow enough rate that the switching takes place below the minimum audio frequency.

[Crontox102098]

This is my last desing.

[R.G.]

This is my last desing.

I believe that will work. A couple of questions you need to answer to get two resistor values correct.

R3 needs to supply enough current to the base of the 2N3904 to be sure it's fully switched on. For most bipolar switching work, this means supplying a base current of 1/10th of the collector current, roughly. This condition makes the collector be as near the emitter voltage as is reasonable. To calculate this, you need to know the resistance of your relay coil. When you know that, you can calculate the current in the coil by dividing the 9V supply by the coil resistance. That will probably be something like 25ma. then to ensure that the base of the 2N3904 is getting enough current, you need to supply 2.5ma to it. The CD40106 will pull its output up to nearly its power supply with this magnitude current. So the resistor to get 2.5ma into the base of the 3904 is (9V-0.7V)/0.0025 = 3320 ohms. A 4.7K is a little small, but might work. There's no magic about the 1/10 number.

However, there is more collector current. You are powering the LED from the collector too. You need to decide how much current you want your LED to have, and adjust R4 to make that come true. That's done by subtracting the LED forward voltage (and the collector-emitter saturation voltage for the NPN, which I'm ignoring) from 9V. This changes depending on what color LED you use. If it's green, those are mostly about 2V. So say you want 5ma and are using a green 2V LED. The resistor R4 to do that is then 9V-2V = 7V divided by 0.005A, or 1400 ohms. 1K's not a bad starting place, but it lets through 7/1000 = 7ma of current.

This is added to the relay coil current. So if your relay needs 25ma when you check the datasheet, the 3904 needs to let through 30-32ma of current, and you need 3ma of base current (about). So the value of R3 needs to be smaller; about 8.3V/0.0032 = 2593 ohms. You could use a 2.4K or 2.7K and be fine.

[Crontox102098]

This is my last desing.

I believe that will work. A couple of questions you need to answer to get two resistor values correct.

R3 needs to supply enough current to the base of the 2N3904 to be sure it's fully switched on. For most bipolar switching work, this means supplying a base current of 1/10th of the collector current, roughly. This condition makes the collector be as near the emitter voltage as is reasonable. To calculate this, you need to know the resistance of your relay coil. When you know that, you can calculate the current in the coil by dividing the 9V supply by the coil resistance. That will probably be something like 25ma. then to ensure that the base of the 2N3904 is getting enough current, you need to supply 2.5ma to it. The CD40106 will pull its output up to nearly its power supply with this magnitude current. So the resistor to get 2.5ma into the base of the 3904 is (9V-0.7V)/0.0025 = 3320 ohms. A 4.7K is a little small, but might work. There's no magic about the 1/10 number.

However, there is more collector current. You are powering the LED from the collector too. You need to decide how much current you want your LED to have, and adjust R4 to make that come true. That's done by subtracting the LED forward voltage (and the collector-emitter saturation voltage for the NPN, which I'm ignoring) from 9V. This changes depending on what color LED you use. If it's green, those are mostly about 2V. So say you want 5ma and are using a green 2V LED. The resistor R4 to do that is then 9V-2V = 7V divided by 0.005A, or 1400 ohms. 1K's not a bad starting place, but it lets through 7/1000 = 7ma of current.

This is added to the relay coil current. So if your relay needs 25ma when you check the datasheet, the 3904 needs to let through 30-32ma of current, and you need 3ma of base current (about). So the value of R3 needs to be smaller; about 8.3V/0.0032 = 2593 ohms. You could use a 2.4K or 2.7K and be fine.

I had not seen that way, it is assumed that the LED is red this means that carries about 1.6 V and the resistance would 2Kohm, the relay that I will use takes 25ma.