# 555/556 timer for dual pulse output with 90º phase shift

Thread Starter

#### Catriona82

Joined Feb 15, 2017
47
Hi all,

I'm a newbie here and very rusty on electronic circuit design and theory so please bear with me!

I think i found what I was looking for in this thread (Wendy's second post): https://forum.allaboutcircuits.com/threads/variable-time-delay-using-555-timer.11593/.

I think what it is saying is the frequency is set by the first stage timer with the variable resistor and the polarised capacitor using the equation given. In this case I have selected 10Kohm and 0.01uF for a frequency of 7KHz.

What I am unclear about is how the second stage works. It appears that the capacitor/resistor arrangement creates an oscillator circuit but the capacitor is blocking DC so I don't understand how it creates an identical square waveform (with the shift). Can someone explain in simple (dummy) terms how this works?

Also, assuming it does work do i use the same t=RC equation to calculate the time delay required for the 90º phase shift?

thanks for your help!

Catriona.

#### AlbertHall

Joined Jun 4, 2014
12,385
I don't think that circuit will work.
You could use this circuit with the square wave input being from a '555 running at four times your desired output frequency.

Thread Starter

#### Catriona82

Joined Feb 15, 2017
47
Hi Albert, can you tell me how the series of JK flip-flops creat the 90º phase difference between the two output pulses?

Thread Starter

#### Catriona82

Joined Feb 15, 2017
47
i think I have figured it out by mapping out the counter outputs. Can you tell me why the diagram has the S and R inputs tied together on two of the flip flops? Is this required for it to operate, as it doesn't seemed to be tied to anything?

I don't think that circuit will work.
You could use this circuit with the square wave input being from a '555 running at four times your desired output frequency.

#### Bernard

Joined Aug 7, 2008
5,784
Here is an old outline of another 90 deg. shifter. Output 3 is a reset for counter.

output 3 is

#### AlbertHall

Joined Jun 4, 2014
12,385
i think I have figured it out by mapping out the counter outputs. Can you tell me why the diagram has the S and R inputs tied together on two of the flip flops? Is this required for it to operate, as it doesn't seemed to be tied to anything?
All the R & S inputs should be connected to ground.

Thread Starter

#### Catriona82

Joined Feb 15, 2017
47
I don't think that circuit will work.
You could use this circuit with the square wave input being from a '555 running at four times your desired output frequency.
All the R & S inputs should be connected to ground.
Thanks, is this for all R & S inputs on all the devices or just the ones shown connected on your sketch?

#### AlbertHall

Joined Jun 4, 2014
12,385
All R & S inputs on all flipflops. Incidentally, there will be one unused JK flip flop and all the inputs on that should be connected to ground.

Thread Starter

#### Catriona82

Joined Feb 15, 2017
47
All R & S inputs on all flipflops. Incidentally, there will be one unused JK flip flop and all the inputs on that should be connected to ground.
Great, thanks for your help!

#### EM Fields

Joined Jun 8, 2016
583
This will square up the 555's output - at any frequency the 555 can generate - to a precise 50% duty cycle, and output two square waves exactly in quadrature.

The outputs, Q and I, run at CLK/8, and U3B is spared out.

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#### ScottWang

Joined Aug 23, 2012
7,411
This will square up the 555's output to precisely 50% duty cycle, and give you two square waves out, exactly in quadrature.

View attachment 120658
The outputs, Q and I, run at CLK/8, and U2B is spared out.
The circuit can't be 50%, it just close to 50%, NE555 can't be reach that way, if you want then you can use cmos ic, and do some modify for the circuit.

#### AlbertHall

Joined Jun 4, 2014
12,385
The '555 output is fed through a divide by 2, U2A, which produces the 50% duty cycle.

#### ScottWang

Joined Aug 23, 2012
7,411
I just left the R2, C3, so you can try it to reach your goal.

#### EM Fields

Joined Jun 8, 2016
583
I just left the R2, C3, so you can try it to reach your goal.

View attachment 120660
Even with a perfect 50% duty cycle source at a single frequency, while it's possible to delay one signal edge and achieve a 90 degree displacement between, say, the leading edges, using that method makes it impossible to properly align the trailing edges.

The reason is due to the hysteresis in the HC14's front end which causes the rising edge's trip point to be different from the trailing edge's, as shown below.

Last edited:

#### ScottWang

Joined Aug 23, 2012
7,411
Even with a perfect 50% duty source at a single frequency, while it's possible to delay one signal edge and achieve a 90 degree displacement between, say, the leading edges, using that method makes it impossible to properly align the trailing edges.

The reason is due to the hysteresis in the HC14's front end which causes the rising edge's trip point to be different from the trailing edge's, as shown below.
View attachment 120674
I have had thought about that hysteresis could be caused that after I posted it, if using CD4069 this kind of inverter then it didn't had the hysteresis problem, or double frequency of the 555 and divided by CD4013 the end of the output.

I have another method to use pure logic to build it, but it can't use like 555 such ic.

#### EM Fields

Joined Jun 8, 2016
583
I have another method to use pure logic to build it, but it can't use like 555 such ic.
I'd very much like to see your "pure logic" method it if you'd care to post a schematic.

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#### ian field

Joined Oct 27, 2012
6,536
Hi Albert, can you tell me how the series of JK flip-flops creat the 90º phase difference between the two output pulses?
Search for quadrature clock (or similar) you should find example circuits.

#### ScottWang

Joined Aug 23, 2012
7,411
The circuit was designed with CMOS 555 and CD4069.

The circuit was designed with NE555, CD4069 and CD4013.

#### EM Fields

Joined Jun 8, 2016
583
Thread Starter

#### Catriona82

Joined Feb 15, 2017
47
OK, so this is what I've gone with.....

The idea is that I use a transistor to act as a switch contact.

First question is would this work as it is or would I need additional components?
Second question is, if i'm using a 9V battery to supply my pulse generator but the device i'm connecting it to uses a lower voltage, does this cause a problem for the transistor, having a base voltage higher than the collector voltage?

thanks!