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Single pulse from vco

Arvorsteve

New Member
HI
This is my first post so thanks for letting me join the group.
I am trying to work out the best way to obtain a single pulse from a VCO that is to be triggered from a footswitch. Basicaly a non retriggerable monostable. The idea is to drive a power mosfet and Interrupt the supply to a CDI unit .The duration of the pulse is determined by the speed of the engine. The output must only deliver one period cycle of the VCO regardless of when the footswitch is pressed or how long it is kept closed for.
I built a fixed timing monostable with a high sided mosfet output using 4013 cmos that seems to work, but now I want to make one that has a variable length of pulse determined by the engine revs, and VCO output, hence the proposal is to use the tachometer signal to control a VCO.

The part I am struggling with is the logic which will deliver the same duration of output pulse as the free running VCO is outputting, but only one pulse until the footswitch is operated again, regardless of where in the cycle the footswitch is pressed. I thought of latching both the footswitch input and the VCO output and then resetting the latches after the VCO waveform falls again but I keep getting a varying length output pulse when I work it through on paper, which is not what I want. I haven't got a circuit design analyser setup on my PC and don't know how to use one anyway.

I was going to use a 555 timer for the VCO and one idea I had was to trigger the 555 VCO when the footswitch was pressed but I don't know what settlement time I should expect from the 555 in VCO mode as I haven't used them much?
The VCO output pulses vary in length between 40 and 150mS depending on the vco

It's probably a simple issue to solve and might just require gating the VCO trigger with the footswitch input but I am now a pensioner and haven't done any electronics for over 40 years, so a bit rusty!
I am just trying to help out a friend and after initial success, struggling a bit with the best way forward.
Any help greatly appreciated.
Thanks
Steve
 

AnalogKid

Well-Known Member
Most Helpful Member
Three flipflops:

#1. Clocked by the footswitch. Output enables (gates) the input signal to the #2 clock.

#2. Enabled by #1, clocked by the external signal. #2 output goes high on the first positive edge after the footswitch is pressed, and gates the input signal to the #3 clock.

#3. Enabled by the #2 output, clocked by the next input positive edge. #3 output resets #1, ending the output pulse.

Another signal is required to reset the circuit, so it waits for the next footswitch input. OR, the circuit can be reset by releasing the footswitch.

I say the ff's are "clocked", but they do not have to be J-K or D types; they can be simple Set-Reset flipflops.

ak
 

AnalogKid

Well-Known Member
Most Helpful Member
What is the minimum frequency of the incoming signal? If it is so fast that you cannot get your foot off the footswitch before a cycle goes through, that eliminates one ff.

IOW - is there any chance that the footswitch can be pressed and released within the period of one cycle?

ak
 

Arvorsteve

New Member
I don't think a 555 timer VCO is the best way forward now ak.
The device on the motorbike needs to work between approximately 6000 and 12000RPM, when gear changing takes place, which I estimate equates to pulse lengths into the VCO from the tacho of between approximately 10mS and 5mS. I haven't measured it yet.
The output of the VCO would need to run slower as the input voltage drops and faster as it rises, with the 150mS being generated at the 10mS input and the 40mS being generated at the 5mS input. From what l have seen a 555 in VCO mode will actually work backwards to that and generate a lower frequency as the input voltage rises, which is not what I want. Then I have the problem of generating a linear response to arrive at two disproportioned 40mS and 150mS signals.
The more I think about this the more I am warming to the idea of just dividing up the incoming tacho signal to generate the required length of pulses I need something like a divide by ten to give me 50mS and 100mS signals for the mosfet section.
If I gate the footswitch with the tacho signal , divide it by ten then I will always be looking at the start of the pulse from a divider and have less of a problem than if I work with the slower VCO output . I think it would be simpler to get something working, even if I cannot achieve exactly 40mS and 150mS.
To answer your original question. The footswitch can of course be operated at any part of the timing cycle and potentially only generate a part pulse hence the need to gate it but gear changes only occur at intervals of 200mS plus, so if I use the incoming tacho signal a 10mS wait would be a worse case scenario and hopefully hardly noticeable especially as the delay is longer anyway at the slower RPM.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
At the risk of, as usual, suggesting a micro-controller solution - a simple little 8 pin PIC or AVR and a little bit of programming would be MUCH simpler, cheaper, and make the design much more versatile and effective. Personally I'd use a slightly larger device, at least for protyping, so you've got spare pins for easy ICSP.

Assuming a commercial system was available?, it WOULD use a micro-controller - unless it came from the 1970/80's?.

I presume the purpose of this is to save having to close the throttle when changing up?, what are you doing about changing down, where you need the exact opposite (blip the throttle to equalise engine and gearbox speeds for a smooth change). Presumably the switch only operates on upward changes?, and so doesn't affect down shifts?.

Does the bike have a 'kill switch'?, if so you could utilise that circuit for killing the engine briefly - although it probably does what you're suggesting anyway?.
 

Arvorsteve

New Member
Hi Nigel
I guess it's a case of working with what you are comfortable with for me. I did some dabbling with an Arduino but couldn't get to grips with it. I have seen the value of PICS though as we used them a lot at work. I am a child of the 1970's and didn't get to grips with them personally.
I haven't seen the bike in action yet so I can't answer your question but I take your point.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Hi Nigel
I guess it's a case of working with what you are comfortable with for me. I did some dabbling with an Arduino but couldn't get to grips with it. I have seen the value of PICS though as we used them a lot at work. I am a child of the 1970's and didn't get to grips with them personally.
I haven't seen the bike in action yet so I can't answer your question but I take your point.
What kind of bike is it?, as an ex-biker I like bikes :D
 

rjenkinsgb

Well-Known Member
Most Helpful Member
Try this:

circuit_600.png

The 555 represents the VCO, whatever that turns out to be.
The switch is the gear pedal(?) trigger.

Unused inputs should be grounded on the 4013s.

In the idle state, all three Ds are reset, so the /Q from the third is high.
Triggering the switch clocks that in to the first D, setting the Q output.

When the next edge from the VCO occurs, the second D is set high - and the output connection to the right.
The second positive VCO edge clocks the high state into the third D, which resets the first two. The third D is then itself reset by the next VCO edge.

It's self-debouncing, as long as any switch bounce duration is less than two VCO cycles at the highest frequency.

[Diagram drawn using this site: https://www.circuit-diagram.org/editor/ ]


Edit - looking back, it's almost what Analogkid suggested - but the second D must also be reset by the third, and they must be synchronous registers; SR could not work, in any way I can think of at least.

I think an appropriate shift register IC that has a synchronous reset could work in place of two, two section ICs.
 
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