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

#### AlbertHall

Joined Jun 4, 2014
12,396
You need two resistors between the JK's and the transistor bases to set the base current. The transistor base will be about 0.7V when it is turned on (when the JK Q is high). The resistors should allow a base current about one tenth of the expected collector current.

#### Catriona82

Joined Feb 15, 2017
47
You need two resistors between the JK's and the transistor bases to set the base current. The transistor base will be about 0.7V when it is turned on (when the JK Q is high). The resistors should allow a base current about one tenth of the expected collector current.
When you say a couple of resistor do you mean a series resistor between the Q pin of the JK and the base of the transistor?

I'm going to be connecting this device to either a galvanic barrier which will draw about 8mA when a switch contact is used. The PLC will draw 2.5mA when a switch is closed. I guess then I will need a selector switch on my generator so I can switch between two resistor values depending on which device i'm connecting to.

What resistor values would I need this this case (and what equation do you use).

#### AlbertHall

Joined Jun 4, 2014
12,396
Yes, series resistors as you describe. You don't need to switch them just use the value appropriate for the higher current. If the logic is being driven from 9V then there will be 8.3V across the resistor (allowing for the 0.7V base emitter voltage). The higher collector current is 8mA so the base current should be about 0.8mA (this is not a critical value). So the resistor should be about 8.3V/0.8mA ≈ 10kΩ

#### ian field

Joined Oct 27, 2012
6,536
OK, so this is what I've gone with.....

View attachment 120721

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!
A commonplace VCEsat is about 0.4V - enough current into the base to saturate the transistor will develop a Vbe of about 0.7V.

#### Bordodynov

Joined May 20, 2015
3,187
See

#### ScottWang

Joined Aug 23, 2012
7,413
@Bordodynov.
Could you post the .asc file?
Thank you.

#### Bordodynov

Joined May 20, 2015
3,187
See my file and example used HEF4000

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#### Kjeldgaard

Joined Apr 7, 2016
476
I have been following this thread, and had an idea that the task could be solved with a few D-Flip/Flop, but I was then in doubt if there were some steps that could lock in invalid sequences.

The following sketch on a two bits Johnson counter, can generate a quadrature signal from four clock pulses from eg. a 555 oscillator.

There are several posts with similar solutions, but here there is no additional active or passive components related to the counter circuit. And if we ignore differences in propagation delays between the two D-Flip/Flop, so Q1 will follow exactly 90 degrees after Q0.

#### Bordodynov

Joined May 20, 2015
3,187
Two timers:

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• 2 KB Views: 4

#### ScottWang

Joined Aug 23, 2012
7,413
Is this similar like ambulance sound effect?
As I remembered that I tried the circuit about 30 years ago during the poor ago, specially I was live in the country side, not that much ee stuffs..

#### Catriona82

Joined Feb 15, 2017
47
Ok. I've made my prototype circuit. I have a nice 90deg phase shift and a 50% duty cycle. However I am using the same signal generator circuit using the second channel of a 556 timer for a frequency of around 1Hz but the duty cycle is about 60%.

I used a 100nF capacitor between the trigger pin and GND, and 7M2 resistance between the output pin and the trigger pin to get the 1Hz frequency.
does anyone know what could be going on?

It is not of major importance as the PLC i'll be attaching it to will only be looking for the rising edge but i'd like to know why its not as i'd expect it to be.

thanks.

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#### AlbertHall

Joined Jun 4, 2014
12,396
If you are using NE555 (as opposed to the cmos version) then it is probably the leakage on the threshold and trigger pins that is causing the problem. Try using a lower resistor and bigger capacitor - 10uF, 72k, perhaps

#### Catriona82

Joined Feb 15, 2017
47
If you are using NE555 (as opposed to the cmos version) then it is probably the leakage on the threshold and trigger pins that is causing the problem. Try using a lower resistor and bigger capacitor - 10uF, 72k, perhaps
Albert, thanks, I am using the NE556N so hopefully this is what the issue is. thanks for your help.

#### ian field

Joined Oct 27, 2012
6,536
Albert, thanks, I am using the NE556N so hopefully this is what the issue is. thanks for your help.
In case you need it; the CMOS version is 7556, there may also be a LMC556 option.

#### Catriona82

Joined Feb 15, 2017
47
increasing the capacitor to 10uF didn't make a difference to the duty cycle. Its not important for my application so i will leave it as it is, but it would be nice to know whats going on with it. Do you think its still due to a leakage and if so why would the CMOS version not have this issue?

#### AlbertHall

Joined Jun 4, 2014
12,396
If you are using a much lower resistance that 7.2M then it's not leakage and I don't know what is causing it.

#### ian field

Joined Oct 27, 2012
6,536
increasing the capacitor to 10uF didn't make a difference to the duty cycle. Its not important for my application so i will leave it as it is, but it would be nice to know whats going on with it. Do you think its still due to a leakage and if so why would the CMOS version not have this issue?
Was that an aluminium electrolytic? They're known for leakage - its what replenishes the oxide layer dielectric against the caustic electrolyte.

A tantalum should be better as long as its never seen reverse polarity (even small signal).

Mylar or polycarbonate metalised film capacitors might do the job without too much leakage error.

The programmable unijunction transistor was developed to solve the problems where these kind of C/R values are involved.

A PUT is basically a thyristor with the gate at the wrong end, there are/were SCS devices which are the same basic 4 layer device with a gate at *BOTH* ends. The PUT is getting a bit rare these days, but the complementary asymmetry of the 4 layer device means you can turn the timing circuit upside down and use a common TO92 SCR like the 2N5061 etc.

#### Catriona82

Joined Feb 15, 2017
47
Hi all,

I got my prototype working on the breadboard then got a PCB fabricated. Here is the PCB bare and populated.

I've built it into a wee box with an on/off switch, power LED and a DPDT switch to change which signal leads the other. This is simulating a shaft encoder so the switch will simulate the drum changing direction.

P.S. The company I used to make the PCB was Ragworm/Stickleback in the UK. The single board cost me £21.50 including VAT with 1 week delivery. it would have been cheaper if I had ordered more but the next cheapest quote I got was £80 for 1 and two weeks delivery!
http://ragworm.eu/.

#### EM Fields

Joined Jun 8, 2016
583
I have been following this thread, and had an idea that the task could be solved with a few D-Flip/Flop, but I was then in doubt if there were some steps that could lock in invalid sequences.

The following sketch on a two bits Johnson counter, can generate a quadrature signal from four clock pulses from eg. a 555 oscillator.

There are several posts with similar solutions, but here there is no additional active or passive components related to the counter circuit. And if we ignore differences in propagation delays between the two D-Flip/Flop, so Q1 will follow exactly 90 degrees after Q0.
View attachment 120989
Very nice.