hi

i'm wondering how important the grounding is on the output (VR3 and VR4) of the two oscillators here:




i would like to build this circuit, but only have one pot to control the mixing, like so:

but in that case, i don't ground the output...

from breadboarding it, it does seem to work just fine. with the pot centered i get some good frequency beating going on, but maybe it is a bit noisy...

should i be using a dual-pot in order to ground it?

hmmmmmhmm...

- bob

 

Assuming the circuit is driving a high impedance input,

The difference in your circuit
is the impedance seen by the oscillators varies as you change the mix.

At one end of the pot travel, an output is shorted to ground via the diode, not good,

 

ah.. i see...

i suppose a stereo-pot would do the trick then?

 

As long as it's has a logarithmic taper just kidding please apply Two linear taper pots, your breadboarding the circuit so you will be amazed what you see in a difference if you hook it up to a scope,

 

You can mix the two oscillator outputs with a single linear pot.
If the pot wiper feeds a high impedance load, such as an amplifier input stage, the wiper position should have no significant effect on the oscillators.
Here's a sim. (I used a voltage-controlled pot model to sweep the wiper from one end to the other).


R1 prevents the oscillator outputs being shorted to ground accidentally.
Note that the mixer output amplitude varies somewhat, depending on the relative values of the resistances of the pot and load.

 

thank you so much! i love this forum.

i was gifted an oscilloscope ages ago, i just never got around to use it. i think now is the time.

side-question: what software do you use for simulation?

 

what software do you use for simulation?

LTspice (a free download from Analog Devices). Many members here use it. A steep learning curve, but lots of tutorials online.

 

i would like to build this circuit, but only have one pot to control the mixing, like so:
but in that case, i don't ground the output...

Yes, you do. Or, you should. Actually, you must.

For any circuit, "ground" is the reference potential. It is the point against which all voltages are measured (we'll leave out "floating" signals for now). Voltage is the potential difference between *two* points. No two points, no volts.

In your second schematic, the output signal is between the pot wiper and the circuit ground - pin 7 of the hex inverter, the - return wire to the power supply, etc. You might not have a direct connection between your measurement device and the circuit ground, but there is one anyway.

With very high input impedance test equipment such as a DMM with a 10 M input impedance or a scope with a 10x probe, things can be deceiving - you touch one test lead to something and get a reading. In this case, the return path for the signal current (note, there always is signal current through the test equipment) might be capacitance through the air, or capacitance through the scope cable from the center conductor to the shield, which is tied to the scope chassis, which is tied to the AC plug third pin, which is tied to the power supply, which is tied to the output negative terminal, which is connected to the IC ground pin. You mention that the output signal is noisy. Maybe that's because there is no clean, tight connection for the signal return path.

ak

 

Yes, you do. Or, you should. Actually, you must.

For any circuit, "ground" is the reference potential. It is the point against which all voltages are measured (we'll leave out "floating" signals for now). Voltage is the potential difference between *two* points. No two points, no volts.

thanks a lot for taking the time to give such a detailed answer!


i knew there was something not right about not grounding the signal........ i think i get it (a little bit more) now.

 

To be clear, you are not "grounding the signal". To us, that means tieing the signal directly to GND, as in a dead short. 0 V means no noise, but, alas, no signal.

If you mean adding a ground connection between your circuit and whatever is downstream of the pot (other circuits, test equipment, whatever), then, yes - providing a return signal path is the correct way.

ak

 

Ah, yes. Exactly.

I'm still (also) just learning the lingo. As everbody clearly can see...

 

Tyr the single pot without the diodes.