Output impedance of 74HC14 oscillator, how to compute it?

[wnorcott]

Hello friends.  I have  a simple Schmidt trigger square wave oscillator using a CD40106 hex inverter VDD = 3 Volts.  Powered by a pair of AA batteries and using it as a signal generator for audio projects. The oscillator is buffered by a second inverter on the 40106, see schematic.  The feedback R  is wired in with a suitable pot so it produces audio frequencies 80 Hz to 4,500 Hz and then a 100K voltage divider pot on the output of the 2nd inverter, to attenuate the signal.  [Useful signal  voltages  for this application are in the range of 100mV to 400mV RMS AC. ]   There is  no series resistor on the output.  I would like to know the output impedance .  Nothing on the datasheet specifies the output impedance of a gate.



Question:   What is the output impedance?  I  cannot find this info  on the data sheet. I assume from poking around the web it is around 50 ohms which is fine for this application it is not critical.  If it varies that's fine too just trying to get a ballpark  estimate.
For 5V operation,  it says I_OH = 0.88 mA  but I am using 3V so  if V_OH is 2.95, by Ohm's law that is 3346 , for DC resistance.    That does not seem right either and I had heard these gates are about 50 impedance.  Not only that I am attenuating the signal on the way out, to half a volt or less typically.  If it is indeed 3346 ohms that is still fine.  Please correct my arithmetic methodology of how to derive output impedance from the datasheet.   Thanks.

[Benta]

In the TI CD40106 data sheet, you use the graphs in Fig. 1, 2, 3 and 4. The slope of the curves will give you the approximate output impedance.

[wnorcott]

Thank you.  Related, since it is a square wave fluctuating from 0 -> 3 V at 50% duty cycle and VDD is 3V, then the average voltage or AC voltage should be 1.5V -- should I use 1.5V  AC as the average voltage where to read the slope on the graphs?    I know on a graph like that there is just an instantaneous voltage versus instantaneous current, no concept of AC versus DC, (I think).  Or should I just treat it as a DC 3 volts mark.    Just need to know to read it at the 3V mark or the 1.5V mark.  Anyway, thanks a lot that was very useful to me.

[duak]

If you have the circuit built, another way is to measure the voltage across a variable resistance load.  Reduce the resistance until the amplitude is half what it was open circuit.  Measuring the resistance will give actual output impedance.  I would bet that it's a few hundred ohms if not more.

[Benta]

Thank you.  Related, since it is a square wave fluctuating from 0 -> 3 V at 50% duty cycle and VDD is 3V, then the average voltage or AC voltage should be 1.5V -- should I use 1.5V  AC as the average voltage where to read the slope on the graphs?    I know on a graph like that there is just an instantaneous voltage versus instantaneous current, no concept of AC versus DC, (I think).  Or should I just treat it as a DC 3 volts mark.    Just need to know to read it at the 3V mark or the 1.5V mark.  Anyway, thanks a lot that was very useful to me.

Running the CD40106 at 3 V is at the limit, and the graphs do not show behaviour at this voltage, the lowest is 5 V.
But generally, use the output curves as they are for deriving output impedance. Full VEE to VDD swing divided by current to get the impedance. Remember, only one output transistor is on at any time.

[magic]

Thank you.  Related, since it is a square wave fluctuating from 0 -> 3 V at 50% duty cycle and VDD is 3V, then the average voltage or AC voltage should be 1.5V -- should I use 1.5V  AC as the average voltage where to read the slope on the graphs?    I know on a graph like that there is just an instantaneous voltage versus instantaneous current, no concept of AC versus DC, (I think).  Or should I just treat it as a DC 3 volts mark.    Just need to know to read it at the 3V mark or the 1.5V mark.  Anyway, thanks a lot that was very useful to me.

I'm not sure what you are doing and what exactly you need to know, but usually when people speak of output impedance they mean dynamic impedance, which is how much the output voltage changes with load current. 1Ω dynamic impedance is 1mV drop for each 1mA pulled or 1V rise for each 1A pushed into the device.
So obviously you take it at the point the chip will be operated. If it normally drives 3V high while sourcing 10mA of current, you look at the slope at 10mA output sourcing current and deduce how the output will change if you vary it from 9mA to 11mA - that's your output impedance. Similarly, you may want to sink 10mA while driving low so you check again the same, but on the sinking plot.
In your case, if the only load is a 100k pot to ground, sink current will be zero and source will be 30µA, so pretty much zero too. So you look at the slope right in the corner. You will get some number much lower than 100k and probably conclude that it doesn't really matter and the result isn't much different from a perfect 0Ω source driving that pot.

And, of course, the output impedance of the entire circuit, output pot included, will be dominated by the resistance of the pot, most of the time.

[wnorcott]

Hello again.  I used the method describe by duak (thanks) of actual measurement at the output, and the impedance is 575 ohms.

To answer magic's question:  I built a little signal generator as a cheap bench measurement tool, to generate an AC audio frequency that can be test  amplifiers and guitar effects, and other analog preamps I am interested in.  I used the Schmitt trigger because it is really cheap and easy way to get square waves, pretty stable, and the whole thing has a very low parts count and easy to build.  The 100K pot acts as the volume control.  I can turn it down to match the output of a guitar or bass pickup, which is in the range of 100 mV to 400 mV AC.

Thanks again to everybody.