When last we left our little tone generator, we noted that it was “on” continuously, but what we really wanted was the “beep-beep” of a true alarm (tonus interruptus). Not a problem.
As stated last month, the period of an op-amp oscillator is determined by the feedback resistor from output to (-) inverting input and the capacitor from the (-) input to ground. Something about t = 1/(1.4 x R x C) where t is the period of the waveform (in seconds), R is the feedback resistor, and C is the capacitor from the (-) input to ground. Most of the time our capacitor values are fixed to what we have on the shelf, and we know what the time period we want is going to be.
Redo the Equation
Let me rearrange the equation above so that we can calculate the resistor value knowing the capacitor value and the time. The equation becomes R = 1/(1.4 x t x C). Lets do a quick example. We have a 10-f (microfarad) capacitor and we want each “beep” to be a quarter of a second long, or a total on-off period of half a second. By inserting these values into the equation, I find that the resistor I need to use to make an oscillator with a half-second period is exactly 142 kilohms. The closest standard value to this exact number is 150k, so an op-amp oscillator with these values should give us an on-off cycle every half second.
So what? Now we have two oscillators, the one from last month that is on continuously at 714 Hz and the one from this month that is on continuously at two cycles per second, or 2 Hz. So far I haven’t given you the connection between
Heres the deal. See the connection on last months tone oscillator from R5 to the +9 supply? If that supply connection is removed, the oscillator shuts off. If I connect R5 to the output of this months 2-Hz oscillator, it will turn on for half a second, turn off for half a second, turn on for half a second, and so on. Beep-beep-beep, just like we wanted it to. All we do is connect R5 in Schematic B from last months article to the output of Schematic C on this months article, and beep-beep.
There is one more refinement you can make, then well set about trying to figure out how to make the trigger for the alarm. Suppose just for the moment that you wanted a tone that was on more than it was off, or that was mostly off with just a “tick” every now and again. Within some rather wide limits, you can use the circuit in Schematic D to give you a “duty cycle” from nearly all-on to nearly all-off and anywhere in between. Adjusting R13 to nearly all-off gives you something that sounds like tick-tick and R13 to nearly all on sounds like beeeeepbeeeeepbeeeep. (Note: If you are just testing this part of the circuit, connect what I call the trigger to the +9 volt supply.)
Set the Trigger
Now for the trigger. Remember I said I wanted to have the alarm go off with either a sensor that gave a “high” voltage near the positive battery supply or a “low” voltage near airframe ground? I can do that with one more section of LM324 op-amp.
Refer to Schematic E: R14 and R15 set the initial condition for op-amp section D. Because R14 is lower in value than R15, the (-) input of the op-amp will predominate and the output of U1D will be low. However, if a sensor ground input to the “ground to alarm” input is present, the voltage divider R14 – R16 will allow the (+) input to predominate and the output will go high, enabling pulse circuit U1B and thus the “beep-beep” output from U1A.
If a sensor “high” battery voltage is input to the “+ battery to alarm” input, then the combination of voltage coming from R15 and the voltage coming from R17 will predominate, the output of U1D will once again go high, and the beeper will be activated.
Caution: If both inputs are “activated” (i.e., a ground to “ground to alarm” and battery voltage to “+ battery to alarm” are done at the same time), the output is not well defined; it could be either high or low. Don’t do this. It wont hurt the op-amp, but you cant be sure of the results.
For next month, Ill answer a question Ive heard a lot lately: “Can you sell me a 12-volt Light Emitting Diode (LED)?” No, I cant. LEDs aren’t voltage rated, but current rated. Got you curious? Good, see you next month.