Quadrature signals help

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What is your supply voltage? How are you measuring the voltages on pins 6 and 7? A voltmeter will not give results that are very meaningful. Have you probed the circuit with an oscilloscope? If so, what results are you getting on pins 1, 3, 12, and 13 of U2?
You may have damaged the ICs if you plugged them in backwards. Try new ones. Do you have datasheets for CD4013 and **broken link removed**?
 
Hi Ron
We were just comparing results from working 555 circuits and yes we have more ICs than you could poke a stick at. Today we are going to try Johnson counter on known "working" 555 cct. I will let you know how we go.
Thanks
 
Hi Ron
It appears that the Johnson circuit may have been working after all but our system could not recognise the signals. We tested the frequencies with a fluke 87 with Hz test and found that it picked up the signals from basic 555 circuits and showed the same values on the display of the meter as were seen on the GUI screen of our simulator.
Now when basic 555 circuit was connected to one of the quadrature inputs it showed up as a ground speed on GUI display and a frequency in the IO screen however when the Johnson counter was connected no ground speed or frequency was displayed on the GUI. We did see a frequency deviation on the Fluke 87 (U2A pin 1 & U2B pin 13) that correlated to the frequencies shown in the schematic ????? Oh and U2A got hot fairly quickly, as in after about 30sec.
We do not have access to an oscilloscope unfortunately so cannot check the amplitude of the wave forms.
The system was tried with both 5V and 12V power supply.
Do we need to amplify Fout signals ???
Thanks
 
The famous Robert Pease from National once said " I always get nervous when the customer I am trying to help doesn't have an oscilloscope"
 
TheOne said:
The famous Robert Pease from National once said " I always get nervous when the customer I am trying to help doesn't have an oscilloscope"
Click to expand...

He obviously not 'that' famous, I've never heard of him, or 'National' :lol:

But I agree with the quote!, and why have an overly expensive multimeter and no scope?.
 
TheOne said:
The famous Robert Pease from National once said " I always get nervous when the customer I am trying to help doesn't have an oscilloscope"
Click to expand...
Point taken
We are currently in the going through the process of purchasing one as I guess trying to fault find without one would be akin to urinating into a stiff breeze
 
Hello guys
just purchased an oscilloscope to have a look at the circuit and what we found with Johnson counter was ;

Same frequency in to CLK (U2A & B) as out on Q (U2A & B).

No phase shift

Duty cycle is virtually non existant when it is adjusted below maximum frequencyof 5000Hz (I would say that waveform is high between 95 -99%). Frequency generated by 555 part of circuit app. 50% duty cycle.

I would also like to point out that this is great fun.
thanks
 
Seems like it would be even more fun if the circuit were working. You have something wired wrong, or your CD4013 is dead (most likely).
And yes, the duty cycle will be very high on the 555, but the Johnson counter will still work (when you get it wired correctly), because the CD4013 is edge triggered. Clock duty cycle is irrelevant.
 
If the 555 is putting out 50% duty cycle (not possible) it will be drawing BIG pulses from the power rail and probably screwing up the counter.
 
you can always use clock generator of your choice (555 or whatever) and 4017. add two OR or NOR gates and smile.
first gate would be connected to 4017 outputs "0" and "1" and the other to "1" and "2". make it reset on state "4" so you only count through 0,1,2 and 3. put LEDs on all outputs (one on 555, 4 on 4017 and two on the gates outputs) and add "test mode" switch that will allow adding of large capacitor in parallel with the one used in oscillator circuit. this way you can see circuit running at low frequency (1-5Hz) without oscilloscope.
 
Here is yet another way of getting two quadrature outputs using a tone decoder. Frequency set by R1,C1
 

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Hello All
Regarding the drawing posted up by Ron at the foot of page 1 of this thread.
We have now had three attempts at making up this circuit using up multiple ic's and have had the same results in that Fout (13 U2B) is spot on while Fout(9U2B) is only a fraction of the amplitude.
We are supplying this circuit with a 12V benchtop power supply.
Good signal has app. 8V amplitude while second signal is app 2V.
Signals appear to be quadrature.
Also U2B becomes very hot after only a couple of seconds
Any thoughts
 
And, of course, you can get 50% duty cycle from a CMOS 555 by connecting the timing resistor between pin 3 and pins 2 & 6 (tied together), leaving pin 7 unconnected.
I don't understand Russ' comment
it will be drawing BIG pulses from the power rail
Click to expand...
.
Edit:
Never mind - I think I get it. Russ, I believe you were inferring that the resistor from pin 7 to VCC would be a very small value. I agree with your conclusion.
 
It sounds like you are overloading the outputs. What are you driving into? What is the supply voltage of the part you are driving? If it is less than 12V, do you have series current limiting resistors? If so, what value(s) are you using?
 
I would just like to say thanks everyone for theyre help.
The problems I was having with Rons circuit were down to the fact that I am a goose. i have been an electrician now for 25yrs and when I look at a schematic always assume (the mother of all) when there are two components in a drawing, if they are part of the same assembly they are surrounded by dotted lines. So instead of using one CD4013 I used two and connected them up as U2A and U2B. Hence the second IC was missing supply voltage and much else besides.

Once this was rectified circuit worked like a dream so thanks again.

There is no such thing as a stupid question
 
I would just like to say thanks everyone for theyre help.
The problems I was having with Rons circuit were down to the fact that I am a goose. i have been an electrician now for 25yrs and when I look at a schematic always assume (the mother of all) when there are two components in a drawing, if they are part of the same assembly they are surrounded by dotted lines. So instead of using one CD4013 I used two and connected them up as U2A and U2B. Hence the second IC was missing supply voltage and much else besides.

Once this was rectified circuit worked like a dream so thanks again.

There is no such thing as a stupid question
 
That's kinda funny (but probably not to you). I thought of that, but then I thought - naaahhh. They are both labeled U2. I guess I should go with my gut next time. I'm happy you finally got it working.
 
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