555 as a PLL?

[mstechca]

Is there anyway I can use 1 or 2 555's to make a PLL circuit?

I want to take the LED output of my receiver, and compare it with a set of pulses coming from one 555 timer at a certain frequency. Probably 1Hz - 1Khz. I want to be able to obtain a high or low output based on whether or not the pulses are in sync or not.

I think 555's will be helpful because it has a reset pin which can reset the pulses.

can anyone help me out on this?

thanks.

 

[Russlk]

You can use the 555 as a VCO using pin 5 as the control input. Compare the signals in an XOR, filter and apply to pin 5. I will try to come up with a circuit. Hmmm - 1 Hz to 1kHz is too wide a range, why do you need that?

 

[audioguru]

At Digikey.ca, a CD4046 PLL with detectors IC costs $.71 and does everthing you want and more.
LM555 ICs cost $.63 and you need other stuff with them to make a frequency detector.
An LM565 frequency detector IC costs a small fortune but will also do what you want.

 

[Russlk]

The 565 VCO is non-linear, so it will not cover a wide range. The 4046 VCO is linear and will tune a range of 20:1. You will need a different circuit to tune a wider range.

 

[audioguru]

Hi Russ,
I think MStechca wants to decode frequency-shift keying that has different audio frequencies of no more than 3:1.

 

[Russlk]

Hi Russ,
I think MStechca wants to decode frequency-shift keying that has different audio frequencies of no more than 3:1.

That is a different story. The 565 was made for that.

 

[mstechca]

I just want to detect when two pulses are completely in sync. I would rather use 555 timers, or even transistors, just because I want the smallest PLL. A digital PLL is ok. I specified a range because the input frequencies can be between 1Hz and 1Khz.

For example, Let's say I have a squarewave signal of 1Hz into one input, and a squarewave signal of 1Khz into another input. I want to create an output that indicates that these signals are not equal.

Also, Let's say I have a squarewave signal of 1Hz into one input, and a squarewave signal of 1Hz into another input. I want to create an output that indicates that these signals are equal.

The output could be as simple as an LED.

 

[audioguru]

I just want to detect when two pulses are completely in sync.

Why? Two pulses will never be exactly the same unless they come from the same oscillator, or if the oscillator in a PLL locks onto the input. Two separate oscillators will be in sync only for a moment then there could be a fairly long wait for them to be in sync again as their frequencies and phasing vary.

I want the smallest PLL. A digital PLL is ok. I specified a range because the input frequencies can be between 1Hz and 1Khz.

1Hz is ridiculous. It would take a very long time to detect it and for a PLL to lock onto it.

For example, Let's say I have a squarewave signal of 1Hz into one input, and a squarewave signal of 1Khz into another input. I want to create an output that indicates that these signals are not equal.

A PLL can quickly detect that the 1kHz signal is within its lock range of frequencies but the 1Hz signal will take a very long time to detect and for another PLL to determine if it is within its lock range.

Also, Let's say I have a squarewave signal of 1Hz into one input, and a squarewave signal of 1Hz into another input. I want to create an output that indicates that these signals are equal.

They won't be equal. One will be high when the other is low and vice-versa. Their frequencies and phasing wil be slightly different. Only for a moment they will be in sync.

 

[akg]

I just want to detect when two pulses are completely in sync. I would rather use 555 timers, or even transistors, just because I want the smallest PLL. A digital PLL is ok. I specified a range because the input frequencies can be between 1Hz and 1Khz.

For example, Let's say I have a squarewave signal of 1Hz into one input, and a squarewave signal of 1Khz into another input. I want to create an output that indicates that these signals are not equal.

Also, Let's say I have a squarewave signal of 1Hz into one input, and a squarewave signal of 1Hz into another input. I want to create an output that indicates that these signals are equal.

The output could be as simple as an LED.

just thought of a method .
assuming a sq: signal

differentiate the i/p signal on one side say low-to-high transition, so u get a spike for each transition , charge a cap: using this spike (thru a buffer),
after a fixed time , compare the voltages of the two caps , same freq , same voltage , (high precision needed for all componets)
this is a very crude method . but if there is a very large variation in freq, this could work .

But definitely and FFT would work

 

[mstechca]

They won't be equal. One will be high when the other is low and vice-versa. Their frequencies and phasing wil be slightly different. Only for a moment they will be in sync.

Wouldn't making use of a reset function of the 555 somehow override this and make the signals more in sync?

Let's think about it.

The two inputs are the signal input and the local oscillator (LO) input.
the signal input cannot be changed and must be detected.
The LO input is the signal I am providing from the 555. I think that making use of the reset function will make the signals more in sync when the frequencies are the same.

I think I almost found the answer. I'll continue to research.

 

[audioguru]

Using the reset function of the 555 will reset it with any input frequency, even noise.
The 1st cycle from a reset 555 oscillator is longer (lower frequency) than the rest.

 

[Russlk]

Since you have the input signal, why do you need another to be the same? Why not use the input signal?

 

[mstechca]

Since you have the input signal, why do you need another to be the same? Why not use the input signal?

Because one signal is remote, and the other is local. I want to check to see if they are the same. You know how they make those lost airplane model detectors.

 

[audioguru]

Since you have the input signal, why do you need another to be the same? Why not use the input signal?

Because one signal is remote, and the other is local. I want to check to see if they are the same. You know how they make those lost airplane model detectors.

The lost airplane transmitters aren't crystal controlled so aren't the same.
Some transmit slightly different beeps and others transmit an unmodulated carrier which is heard as pulses of silence on an FM radio.

I don't know why you want to see if two unrelated signals are the same. They will be slightly different.

With frequency-shift-keying you don't need good frequency accuracy of the tones. The high-pitched tone is a logic high and the low-pitched tone is a logic low. The phase detector in a PLL will be high with the high-pitched tone and will be low with the low-pitched tone if both tones are within its lockin range. The phase detector normally drives the VCO so that it locks on the input frequency. You might need to amplify the output of the phase detector so that it is high and low enough for logic.

 

[Russlk]

If you are looking for a lost airplane, a directional antenna will be more useful.

 

[mstechca]

A regular PLL could probably detect two signals, but the problem is that when the frequency of the two signals are the same, the output does not stay the same. This can be because both square wave signals are not in sync (Like square wave A is rising about 1 second sooner than square wave b). The reason why I want to use a 555 is because I want to reset it so that I can shift the wave and eventually have wave A rise at the same time as wave B.

Here is an example. Say I started the local clock, and the remote clock (that is to be compared with the local one) starts 1/2 a second later. If both run at 1Hz, the PLL's output will not stay the same all the time. I want to somehow obtain the same output, just because the frequencies are the same.

I think either Resetting the 555 or triggering it will help.

I'll give you a diagram of what I am trying to achieve if you need it.

 

[Russlk]

The CD4046 using the type 1 phase detector (XOR) maintains a 90 degree phase angle but you can get an in phase signal with a divide by 4 Johnson counter. You will need several CD4046 to cover the frequency range, or perhaps a frequency detector and switch to change range.

 

[audioguru]

Aren't you using frequency-shift-keying so the higher frequency is a logic high and the lower frequency is a logic low? Like an old-fashioned modem.
Then frequency is important, not phase nor sync. Also, then you don't need a clock in the receiver.

What is the nonsence about using a clock frequency of only 1Hz?

 

[Nigel Goodwin]

Aren't you using frequency-shift-keying so the higher frequency is a logic high and the lower frequency is a logic low? Like an old-fashioned modem.
Then frequency is important, not phase nor sync. Also, then you don't need a clock in the receiver.

What is the nonsence about using a clock frequency of only 1Hz?

I don't think he's ever mentioned what he's trying to do?, however, as usual, it's probably complete nonsense?.

 

[mstechca]

I will show you a diagram to avoid all confusions. (see attachment)

The diagram is divided into 2 groups. A and B.

Group A shows completely equal waveforms in every aspect from two oscillators.

Group B shows waveforms of the same frequency, but they are not in sync.

Now here is my request in the simplest terms:

I want to make a circuit that has two inputs and one output. The inputs will be the waveforms coming from each oscillator. The output will be a logic high if the oscillators produce the waveforms like group A (everything synchronized). The output will be a logic low if nothing is in sync. For example, group B.

A PLL will not work for this task because the output oscillates when the inputs to it are equivalent to group B.

Do you get what I am saying yet?

It's more like a wave matching circuit, that works better than a PLL.