What's wrong with this circuit?

[chebyshevF]

We have to make a two tone siren using a Schmitt Trigger (74HC14N) and my group and I came up with this circuit. Pretty simple really, but the tutor for our class said that it's not a very good design because: it's not a good idea to drive the two Schmitt Triggers at the same time, and because of the position of the diode, we'd have a high output impedance right after it, along with the low output impedance coming out of the top Schmitt Trigger. He mentioned that we'd have to justify in our report that these values meet the specs of the IC, since we could damage it, or other components. Just wondering if someone could provide a more clear explanation since I don't exactly understand just why/how our circuit composition could lead to a damaged component, or why it isn't a good idea to drive the two triggers at the same time?

This is the circuit (power is given to the IC through a 3V battery):

It produces the two tones (similar to an ambulance siren) btw. But that's also another thing that we're unsure of, just how it produces them. We understand that the configuration we have gives rise to a square wave, and we decided to allow the square wave of the top Schmitt Trigger to pass through (during which the capacitor is discharging, right?) and then the diode kicks in by allowing current to pass through from the bottom trigger (thus allowing the the top capacitor to charge, with the bottom one discharging) and thus allowing that tone to go through. Is this a correct interpretation?

(Btw, he said that to reduce these effects, we should just introduce a resistor before the diode)
Thanks.

 

[uart]

You tutor is 100% correct. When the output of the top 7414 is low, and that of the lower 7414 is high, then the two outputs are effectively shorted together through the diode. It is definitely bad design to have two outputs shorted together.

 

[yungman]

The 7414 is tolldem pole? output, which mean it drive active high and low. What you have will cause the two output to fight each other and may burn the IC.

Design is not good.

Look into 556 IC, you might have much better luck using it. Here is an application note:

http://www.doctronics.co.uk/pdf_files/555an.pdf

 

[chebyshevF]

Thanks fellas. So introducing a resistor before the diode should solve/mitigate the issue of a short?

Also, is my understanding of the operation of the circuit right?

 

[yungman]

Thanks fellas. So introducing a resistor before the diode should solve/mitigate the issue of a short?

Also, is my understanding of the operation of the circuit right?

Not really, in your circuit, one might over ride the other and I am not sure you really get a two tone... If I want a two tone, I'll use the 556 which has two 555. It is meant to do this kind of tricks. You can modulate the threshold and change the frequency, so all you have to do is set one of them at audio frequency to drive the speaker and the other set to toggle every half a second to modulate the threshold of the first one to change the frequency. I don't think I want to design it for you either. Download the data sheet and read the app notes I gave you and see whether you can do something about it.

 

[chebyshevF]

Well I ended up introducing a 1k resistor, and I still get the two tones.

 

[uart]

Well I ended up introducing a 1k resistor, and I still get the two tones.

Yes, but do you know why you get two tones?
Can you predict the frequency and $\Delta \, f$ of your tones?
Do you think it's a good design if you cannot do either of the above?

Anyway I'll let you know what's happening. With 1k resistors there is a relatively large current demand on the 7414 output, so in effect it's output resistance is not negligible in determining the capacitor charging (and discharging) duration.

Therefore when the second (the modulating) oscillator output is high then it provides a small amount of current boost to the charging of the first (the tone) oscillator, this reduces the output-high time for this output. It also however subtracts a small amount to the available discharge current for the tone oscillator, this increases the output-low time of this oscillator.

So at the very least you will be changing the duty cycle of the output waveform. In practice it is unlikely that the increase in low-time will exactly cancel the decrease in high-time so there probably will be some small overall change in frequency as well.

BTW. Even if there is no change in frequency, and only a change in duty cycle, then the ear will mostly likely still detect it as "two tone" due to the slightly greater harmonic content of the narrower duty cycle pulse train.

 

[yungman]

I think UART point out a lot of the issue already. Unless you are required to use the 7414 to do the project, look into the 555 and 556. This is one of the IC that is well explored, I have seen books on just this IC that have all the different applications. I designed a car alarm using this to generate the alarm sound, delay alarm after the door is open. You might impress the instructor using this and do it the right way.

I am not saying the 555 is the only right way, there must be 100 other right ways to do this circuit...your's not one of them...Sorry.

 

[chebyshevF]

^^Unfortunately we have to use the 7414. I was looking into the 555 IC, since I have a few at home.

Yes, but do you know why you get two tones?
Can you predict the frequency and $\Delta \, f$ of your tones?
Do you think it's a good design if you cannot do either of the above?

The data sheet has a formula to find the frequency of the individual relaxation oscillators:
f= 1/(0.8*R*C), for the top one (with the 1k R) I get 1250Hz, the bottom one is 1.25Hz.

Anyway I'll let you know what's happening. With 1k resistors there is a relatively large current demand on the 7414 output, so in effect it's output resistance is not negligible in determining the capacitor charging (and discharging) duration.

Therefore when the second (the modulating) oscillator output is high then it provides a small amount of current boost to the charging of the first (the tone) oscillator, this reduces the output-high time for this output. It also however subtracts a small amount to the available discharge current for the tone oscillator, this increases the output-low time of this oscillator.

So at the very least you will be changing the duty cycle of the output waveform. In practice it is unlikely that the increase in low-time will exactly cancel the decrease in high-time so there probably will be some small overall change in frequency as well.

BTW. Even if there is no change in frequency, and only a change in duty cycle, then the ear will mostly likely still detect it as "two tone" due to the slightly greater harmonic content of the narrower duty cycle pulse train.

So I was a bit off in my understanding. Thanks for clearing it up. I'm not happy with the circuit though, so I'm investigating other methods. But just to add, I removed the diode and just kept the resistor, and I still get the two tones.

 

[yungman]

I am really lazy to study today, so I can spent a little time here, but no guaranty it will work!. The higher frequency tone is generate with C1, the lower frequency tone generated with C1 parallel to C2. C3 is to control the speed of switching between the high and low tone.

This is a stupid design, but at least there will be no contention. The first transistor Q1 just to add C1 in and out according to the first oscillator and the second oscillator generate the tone. R1 is to keep C1 reference to ground when Q1 is off in half the cycle. You can change the tone and the switching rate by playing with the value of C1, C2 and C3. I am sure someone else can come up with a better design.
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[chebyshevF]

^^Thanks for your input. I'm trying to design another circuit now, that involves using two Schmitt triggers, yet outputting them to two transistors (so one each) and then outputting those transistors to a single transistor, which then goes to the speaker.

 

[hameroonz]

nice thank ^^

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[AlephZero]

Here's another idea. Start with a single tone oscillator. Then make the low frequency oscillator change its triggering voltage, or the "ground" potential of its capacitor.