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It only takes a minute to sign up. How to measure high and low trigger thresholds? One way to build a Schmitt trigger oscillator using a CD is to connect one input pin of one gate to a capacitor while connecting the same pin to a resistor that connects to the output of the same gate. You then connect the other side of the capacitor to Vss.
Let's say you are using the first gate on the CD The inputs are pins 1 and 2 and the output is pin 3. You could connect pin 1 to Vdd or to a "control signal" to make the Schmitt trigger oscillator output its square wave. I built that circuit. I connected pin 1 to Vdd to make the circuit oscillate.
Here is the schematic of the circuit:. If you connect the "enable pin" of the gate pin 1 in this case to Vss ground or zero volts in my case , the output of the gate is held high near Vdd. Another way to create a Schmitt trigger oscillator using the CD is to simply tie both input pins of a gate together. You then connect the two tied-together pins to the junction of the resistor and the capacitor the RC network.
For this version of the circuit, all you do is apply power to the IC pin 14 and the circuit starts oscillating as soon as power is applied. If power is disconnected to the circuit, there is no output at all. Since both inputs are the same, the output will simply be the opposite of the input - an inverter a Schmitt trigger "NOT" gate. Both circuits measured the same for all of these characteristics. SO, after much introduction, my question is this: In what situation would it be preferable to use the "enable pin" version of the circuit versus the "inputs tied together" version of the circuit?
I am an electronics hobbyist, not an electronics engineer, and my primary use for the CD Schmitt trigger oscillator is to generate an audio frequency square wave to feed to amplifiers and other projects. For me, the second version of the circuit, with the inputs tied together, seems to work fine.
In addition, since I don't want a steady "HIGH" output when the circuit is not oscillating, the second version of the circuit seems preferable. I guess I could add a second inverter to the "enable pin" version of the circuit to make it output a steady "LOW" output when the enable pin is not "enabled", but that seems like unnecessary extra complexity. I have not built circuits using microcontrollers, so I don't have experience sending a "control" signal to the "enable pin" version of the circuit.
Couldn't one just send a control signal from a microcontroller that turns on power to the circuit at pin 14? I guess you would need to use a transistor or something else as a switch.
Maybe that is the advantage of the "enable pin" version of the circuit: You can make it output a square wave, or not, without having to create another, intermediary circuit, to turn the power on and off? The previous version of the schematic for the "inputs tied together" version of the circuit had a typo.
It should have been labeled Vdd. Here is the schematic with the typographical error corrected:. The enable input provides a low-energy way to start and stop the oscillator. If you don't need the enable input it can be connected to the positive supply, or you could build the other circuit. However, the second circuit is not a "preferred " way to use any logic IC or one used in a semi analog mode as you are switching the power supply to the whole IC in order to control one section.
This means that all other gates in the IC are on or off at the same time as the oscillator, which is rather more "wasteful" than using another gate for gating. Also, the power supply current for the IC must be provided by the gating signal.
In most cases this is not a problem as many logic level outputs have enough power and voltage to drive the IC - but if you were using the oscillator in a way that gave a current draw near the upper allowed limit then some IC pins could not supply it.
When the diode conducts it swamps the feedback via VR1. If a low current enable line is required such that VR1 is too much of a load, drive another gate section and then use that gate to drive a diode disabler. Using a diode allows you to have a high or low disable line as desired - reverse diode polarity to change control polarity. I do like the "Forrest Mims"-style diagrams. Sign up to join this community. The best answers are voted up and rise to the top.
Home Questions Tags Users Unanswered. Ask Question. Asked 1 year, 2 months ago. Active 8 months ago. Viewed 2k times. The resistor and capacitor form an RC network that sets the frequency of the output. See page 4 of the datasheet.
The other pin of the gate connects to a "control signal" or Vdd. Here is the schematic of the circuit: If you connect the "enable pin" of the gate pin 1 in this case to Vss ground or zero volts in my case , the output of the gate is held high near Vdd. I built that version of the CD Schmitt trigger oscillator, too. Any advice on which circuit is preferable to use in different situations is appreciated! Brock R. Wood Brock R. Wood 9 9 bronze badges.
Well, OK, many things. Where precision is adequate it can be used for oscillators, delays, retriggerable and non retriggerable monostables, ADPCM modulator, analog amplifier now, there's a trick feed it AC coupled audio and DC bias it at mid point, adjust input level and see what happens - scope on output.
Add RC output filter and! And more. Note that most have the same hysteresis range but one CD? Properly drawn and labelled, as yours are, they do a good job. Active Oldest Votes. Jasen Jasen Thank you. I will try the circuit variations you mention. As for the Forrest Mims style schematics - thank you for the kind words! I have to admit that Mr. Mims is a hero of mine.
I find that hand drawing a schematic helps me learn a circuit's principles better than creating the schematic on a computer. Not sure why. Wood Mar 17 '19 at Sign up or log in Sign up using Google. Sign up using Facebook. Sign up using Email and Password. Post as a guest Name. Email Required, but never shown.
Help.... i'm newbie, ic 4093, CD4093, HCF4093 library
The IC may not have complicated specifications and attributes yet it proposes many useful utilities. It consists of some fundamental blocks which can be configured according to personal preferences and used for numerous different applications. It consists of 14 pins and has four CMOS blocks internally embedded inside its package. Just think them as an electronic component having a couple of inputs and a single output, quite like a transistor, but these gates are embedded inside a package and are not individual components like transistors. However the above explained gates are entirely different with their characteristics compared to linear devices like transistors.
How to Understand and Use IC 4093 NAND Gates, PinOuts