Touch activated Switches

[Roff]

Here's the sim I ran.

 

[Space Varmint]

On the input side of the inverters after the 10k resistor I have 5.7v and 2.5v on the output side then it goes into the 47ohm resistor and drops to 0.7v

I have 9v on both of the 4069's pins 14 ?

Yeah, sounds like "fan out". What you want to do is not use the oscillator output directly. Run the output (pin 4) back into another inverter and use the output of that inverter into the parallel inverters and see if the voltage goes up at both points. If it does but not enough, then you may have to spit up the parallel inverters. Try this first though.

 

[killivolt]

probably have a "fan out" problem.

My measurements

On the input side of the inverters after the 10k resistor I have 5.7v and 2.5v on the output side then it goes into the 47ohm resistor and drops to 0.7v

I wonder if that is what's happening if your showing the single gate with that swing voltage at the Exciter ouput "Roff " I'm wondering if I should just drop it down to ( 1 ) gate and see what happens.

If so what would I want to do to the other Inputs and outputs tie them to ground? or float both sides

 

[killivolt]

Run the output (pin 4) back into another inverter and use the output of that inverter into the parallel inverters and see if the voltage goes up at both points.

If it's fanning out couldn't I just drop these out of circuit. ( A2's 5-6, 9-8,11-10,13-12 ? )

I'll try to do it though it'll take some time for me today. But I'll try.

 

[Space Varmint]

It would be a good way to test and see if that is the problem which I believe it is. I would also most definitely run the output of the oscillator (pin 4) back into another inverter gate and use it's output to drive the circuit. It's not good to put a load directly on the oscillator without some buffering which adding another gate will do.

 

[killivolt]

Should perform this right off pin (4) or after the resistor.

I'm going to take pin (4) either after or before the resistor and feed A2 pin (13) and take pin (12) out back to the

parallel input side of the Gates after the resistor. Correct ?

Edit: I have pins 13 and 12 isolated and have them ready to wire.

 

[killivolt]

Done, the voltage increased. Before the 47ohm resistor from 2.7v to 3.3v

Recap, I isolated A2 pin - 13 connected it to pin 4 of A1 after the 10k resistor, then I connected A2 pin -12 back into the parallel intputs directly.

However, the wave form is no longer shark tooth it's square on top.

 

[Roff]

On the input side of the inverters after the 10k resistor I have 5.7v and 2.5v on the output side then it goes into the 47ohm resistor and drops to 0.7v

I have 9v on both of the 4069's pins 14 ?

You need to use better punctuation. Does this mean you are seeing 5.7V DC on the input side, 2.5V DC on the output side, and 0.7V DC on the right side of the 47 ohm resistor, or are you seeing AC waveforms?

Did you see my simulation? it should be pretty close to what you should see.

 

[killivolt]

You need to use better punctuation. Does this mean you are seeing 5.7V DC on the input side, 2.5V DC on the output side, and 0.7V DC on the right side of the 47 ohm resistor, or are you seeing AC waveforms?

Did you see my simulation? it should be pretty close to what you should see.

Yes. The waveform is exactly that and Voltages are at those points.

 

[killivolt]

I just got this back from Dave.

Use your scope to look at the output of the oscillator first. Make sure it is a nice 20KHz square wave signal. Next, disconnect everything from the outputs. See if you get a triangle signal. Yes, you should see a triangle signal near 9v pp. The diodes take care of voltage transients on the long cable. If you have those diodes backwards then that is your problem.

 

[Roff]

Yes. The waveform is exactly that and Voltages are at those points.

I still don't understand whether you are seeing oscillating signals or DC. Am I thick, or whut????

 

[killivolt]

I'm seeing a saw tooth now at 6.2v with the diode in the right direction. Now I have 6.2v to 0v oscillation on the output side of the gates after the 47ohm resistor.

Still fuzzy or did I say it correctly.

 

[Roff]

I'm seeing a saw tooth now at 6.2v with the diode in the right direction. Now I have 6.2v to 0v oscillation on the output side of the gates after the 47ohm resistor.

Still fuzzy or did I say it correctly.

I got it.

 

[killivolt]

Ok, I've been tracing the voltage back and found my other mistake. I have the resistor on Gate of the Mosfet tied to the ground side of the .049 capacitor.

Very Close I can see the 2907 Jumping. A few more adjustments, I made sure I got 2 of everything so if I've damaged it I have the replacement.

Boy, it's hard to do this with so many interruptions. Thats just how it goes around here.

 

[Roff]

How are you testing this? I would do it first with the exciter (first tested separately, as Dave Johnson recommended) and one sensor and the receiver, all close-coupled (no cables). Make sure you have all the grounds of the 3 circuits connected together.

 

[killivolt]

New measurements I have a fluctuation of 4.5v to 6.5v at the input of the receiver. if I bridge my finger across the Touch Button surface and ground. I have a constant voltage of 2.2v at the Gate of the Mosfet it will reduce only to about 2.1v and then back to 2.2v

No response from the relay yet.

Edit: At this point I did find the Mosfet Leaking. I replaced it.

 

[killivolt]

Check it out

Got it.


Once it warmed up I was able to trim the resistance and with about 30 feet of cable un-sheilded.



I likes it........... yes we do.**broken link removed**




Just like a touch lamp.

Dave Johnson did a good job.

 

[killivolt]

Thank you, Roff and SV for hanging in their with me on this.

And the circuit Designer Dave Johnson. A Good Man.

 

[Space Varmint]

OK, that brought the voltage up some right? With having a saw tooth wave form you will get a different average voltage than if it was exactly square wave with 50% duty cycle.

WHere does the relay go? To the output of the MOSFET? One other thing you can do to increase the drive (exciter sig.) is to lower the 10K resistor to about 2.2K since the output of the oscillator is now buffered. Try to get that drive up to the full 9 volts peak output and then we work on that receiver circuit which I believe you have connected to the relay.

 

[Space Varmint]

Got it.


Once it warmed up I was able to trim the resistance and with about 30 feet of cable un-sheilded.




I likes it........yes I do.**broken link removed**




Just like a touch lamp.

Dave Johnson did a good job.

Gee. I guess I missed that. Well, CONGRATULATIONS my man! Alright