Electronics > Projects, Designs, and Technical Stuff
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 IOH = 0.88 mA but I am using 3V so if VOH is 2.95, by Ohm's law that is 3346 \$\Omega\$ , for DC resistance. That does not seem right either and I had heard these gates are about 50 \$\Omega\$ 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:
--- Quote from: wnorcott on March 19, 2019, 08:37:37 pm ---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.
--- End quote ---
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.
Navigation
[0] Message Index
[#] Next page
Go to full version