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Author Topic: Task of the circuit  (Read 628 times)
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biomed12
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« on: May 21, 2017, 10:42:27 22:42 »

Hello everyone,

I need to understand the task of circuit, please help me gurus of electronics.

4011 is a nand-gate IC
4017 is a decade counter
4046 is a PLL IC

When i draw the circuit in the proteus, buzzer is beeping brokenly and I think, simulation can not reach to the speed of circuit.

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h0nk
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« Reply #1 on: May 21, 2017, 11:09:31 23:09 »


Hello biomed12,

the circuit is playing a tune of nine notes.

IC1
G1&G2: RS-FF does start and stop the sequence
G3&G4: clock generator for the sequencer

IC2
Mod10 Counter with decoded outputs - the sequencer of the system

IC3
PLL(/VCO) - this one makes the noise.


Best Regards


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biomed12
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« Reply #2 on: May 21, 2017, 11:17:40 23:17 »

Hello biomed12,

the circuit is playing a tune of nine notes.

IC1
G1&G2: RS-FF does start and stop the sequence
G3&G4: clock generator for the sequencer

IC2
Mod10 Counter with decoded outputs - the sequencer of the system

IC3
PLL(/VCO) - this one makes the noise.


Best Regards




Dear h0nk,

Thanks for the reply. Can you give me a bit detail about IC3 and roles of R2,R3,C2,TR1.

Thanks.

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h0nk
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« Reply #3 on: May 22, 2017, 12:14:29 00:14 »


Hello biomed12,

The 4046 iss a PLL-IC. To do his duty he needs a converter between a voltage and a frequency.
This part of of the 4046 is used. With the voltage at Pin 9 You can control the generated frequency.
Its called "VCO", voltage controlled oscillator.

R2, R3 and TR1 are doing a nasty work. They "implement" a analog behaviour in a digital circuit.
With the inverter of G3 they form a inverting amplifier and at the same time they deliver the
current to charge and discharge the capacitor C2. C2 adds hysteresis to the amplifier.
Pin 2 of the 4011 forms a gate to start and stop this generator.
With TR1 You can adjust this behaviour to control the period of the resulting pulse.

Prior to start the output of G3 is high and the output of G4 is low.
Then C2 is charged over TR1+R3. When the voltage of C2 passes the switch point of G3,
G3's output switches to low and G4 to high.

...

You should try to draw the waveforms by Yourself and then compare it with Your simulation.


Best Regards
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M@X77
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« Reply #4 on: May 26, 2017, 11:56:44 11:56 »

To calculate the output frequency you need to use this formula:

F=1/(2,2x(TR1+R3)xC2)

in your case if TR1 is 0Ω F = 1/(2,2x100KΩx220nF) ≈ 20Hz otherwise if TR1 = 1MΩ F = 1/(2,2x1,1MΩx220nF) ≈ 2Hz

R2 ideally can be a short circuit, but to avoid high current into gate's input protection diodes and to have a more stable frequency is better that it must be much larger then minimum value of R3+TR1. (R2=~10xR3+TR1)
If you don't want R2 your frequency's calculation is little bit different:
F=1/(1,4x(TR1+R3)xC2)

It is possible have high value of capacitor and high value of resistors, or viceverse, this choose depends from the gate family caratteristic, to have more stable frequency is better use the buffered CMOS  that end with B, eg 4011B.

To have best result with buffered CMOS is better choose R3+TR1 between 10KΩ and 1MΩ and C2 never low then 100pF.
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HackAndCrack
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« Reply #5 on: May 26, 2017, 01:11:16 13:11 »

In Proteus use ACTIVE Speaker model and don't use transistor to drive the speaker like in buzzer circuit. Don't use Buzzer model.
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M@X77
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« Reply #6 on: May 26, 2017, 02:11:10 14:11 »

Just to know... in proteus, the 4011's model does not consider the input protection diodes and the output frequency is different from the real.
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