A basic Schmitt trigger RC oscillator the scope shows 0.4V positive.

That circuit with a standard TTL circuit will work poorly or not at all in practice since the feedback resistor is too large to source the 1.6mA needed to pull the TTL input low.

 

Since the antique "7414" works with the very high value for its input resistor, then maybe it is actually a more modern Cmos 74C14 or 74HC14?

 

Since the antique "7414" works with the very high value for its input resistor, then maybe it is actually a more modern Cmos 74C14 or 74HC14?

It would seem it would have to be, if it actually works.

 

Try a CMOS gate (CD40106, or CD4093), instead of a 74xx gate. A high value resistor between the gate input and a supply rail can be used to adjust the duty-cycle.

As I read the request the TS was asking for a way to obtain a zero-centered square wave, that is, equal positive and negative excursions. Duty cycle was not mentioned.

 

As I read the request the TS was asking for a way to obtain a zero-centered square wave, that is, equal positive and negative excursions. Duty cycle was not mentioned.

What about the 50%?
Max.

 

What about the 50%?
Max.

The waveform that I saw in the one display looked close to 50%.

 

Thats what I said!
Read the OP.
Max.

 

Hey...don't forget to consider LSTTL for more modern deisngs

 

Hi

Try a CMOS gate (CD40106, or CD4093), instead of a 74xx gate. A high value resistor between the gate input and a supply rail can be used to adjust the duty-cycle.

Alec_t has a winner, an almost perfect 50% by going to the CD4093BP !!! equal it is beautiful. The resistor value had to change to 26.3K also like having the other gates available. sorry the other idea did not work. The old part was standard 7414N

 

You have one input of the 4093 floating. That's a no-no with CMOS circuits. Connect the two inputs together. Since the 4093 is a quad gate IC, in the real world all unused inputs must be connected to something; usually ground or the +ve rail as appropriate.

 

You have one input of the 4093 floating. That's a no-no with CMOS circuits. Connect the two inputs together. Since the 4093 is a quad gate IC, in the real world all unused inputs must be connected to something; usually ground or the +ve rail as appropriate.

This is most certainly true in the real world, but it may not matter in a simulator. That is important for folks to understand.

 

A basic Schmitt trigger RC oscillator the scope shows 0.4V positive. How do I add offset control so positive and negative are equal ?
Is it better to move to an op amp circuit ?
View attachment 206195

Yes on OpAmp.

The problem with CMOS gates, even ones with Schmidt front ends, is severe
T & V dependency of thresholds, they vary all over the map. Same for many caps
generated timing.

The problem of exact 50% duty cycle easily solved (if you stay with RC gate solution) running osc at 2X and followed
by a flip flop /2.


Regards, Dana.

 

I thought about that. I could turn the voltage regulators to +/- 6 or 7 volts no problem.

I was just wondering about whether the 555 will also need the ground as a reference? If not, happy days!

As above, this will work if you are careful about the connections of control signals and external components. Note that the 555 decoupling cap still should go between its VCC (or Vdd) and GND pins.

The simplest solution is an output coupling capacitor. At 22 kHz and a 10 K load, a 0.1 uF coupling capacitor will create a high-pass network with a corner freq of 160 Hz. This is *way* below 22 kHz, and wont affect the squareness of the output waveform.

ak

 

I would make the Schmitt Trigger Nand gate an inverter by connecting the unused input to the positive supply, then the slightly different threshold voltages of both inputs won't fight over which one triggers first.

 

Thanks, outstanding specific information. I will leave a jumper and a blank op amp section with room for adjust.
The old 4069 hex inverter circuit is in a shoe box that my wife wanted to throw away. The circuit after that was a 7490 decade counter.

 

An antique 7490 needs an input low current of 1.6mA but a CD4069 with a 5V supply can provide only 0.51mA so it won't work unless you parallel a few of the inverters.

 

Thank you audio guru and others, that is so true. The projects from the 60's early 70's will stay in the shoe box. The Alec t cd4093 are on order. As mentioned the simulator and the real circuit could still have issues but I have a good feeling that the 4093 will work with the output of Arduino Uno. How exactly I am not sure? I ordered 6N137 opto coupler. The output of the 4093 is for audio range. The concern about symmetry to help check before and after inserting this output stage. The opto isolation is an attempt to improve the purity of the square wave at least attempting to get the best signal from the Arduino. I believe that a low pass filter will remove noise. Not sure but I think that if clean edges are transmitted by the 6N137 the 4093 might give a clean square wave.