Help with non-overlapping clock generator

Roff · Oct 13, 2010

andrea22 said:
I tried everything you said,but I always get overlapped clocks.
I don't where is my mistake.I am desperate.
I can't get non-overlapping clocks.
Can you help me with that,I need that for tomorrow?

I'm out of ideas.

ericgibbs · Oct 14, 2010

Roff said:
I'm out of ideas.

hi Ron,
I have been following with interest this thread, I applaud your persistence.

I think what the OP is doing is he is adding the propagation delay at the start and at the end of the square wave pulses, so he always finishes up with an overlapping clock pulse train.

IMO he should consider the propagation delay as acting on the leading and trailing edges [ high to low and low to high transition] at the gate inputs.

example:
Say the propagation delay is 10uSec thru the gate for the leading edge, the leading edge will change the state of the gate output 10uS later, so the pulse width has 'shrunk' in width by 10uS.
The same applies to the trailing or falling edge, it takes 10uS to propagate thru the gate, so the pulse width has 'shrunk' by a further 10uS.
As this happens thru every gate the pulse width will get shorter and at the outputs the two pulse streams will not overlap..

Regards.

Roff · Oct 14, 2010

ericgibbs said:
hi Ron,
I have been following with interest this thread, I applaud your persistence.

I think what the OP is doing is he is adding the propagation delay at the start and at the end of the square wave pulses, so he always finishes up with an overlapping clock pulse train.

IMO he should consider the propagation delay as acting on the leading and trailing edges [ high to low and low to high transition] at the gate inputs.

example:
Say the propagation delay is 10uSec thru the gate for the leading edge, the leading edge will change the state of the gate output 10uS later, so the pulse width has 'shrunk' in width by 10uS.
The same applies to the trailing or falling edge, it takes 10uS to propagate thru the gate, so the pulse width has 'shrunk' by a further 10uS.
As this happens thru every gate the pulse width will get shorter and at the outputs the two pulse streams will not overlap..

Regards.

Eric, if leading and falling edges have the same delay (through an inverter, or whatever), the entire pulse will be delayed by that amount, but the pulse width will not change.
The key to that circuit is the S-R flip-flop with delays added in the feedback legs.

ericgibbs · Oct 14, 2010

hi Ron,
IMO the response time of the gate should be a better way to explain the shortening on the pulse width rather than the gate delay.

Its been my observation that when the pulse edge response time of a gate increases the pulse width is reduced.

I cannot understand why the OP is having problems, when following your description, how he gets an over lapping clock.???

Roff · Oct 14, 2010

Even if the gates and inverters had zero rise and fall times, but finite delay, the circuit would still create nonoverlapping clocks.

andrea22 · Oct 17, 2010

Ron,
Rise time and fall time are actually delay time.
You know that delay time is usually: tdelay=1/2*(t_rise+t_fall).
This equation is for CMOS inverter anad every logic circuit realised in CMOS technology can be equivalent
with CMOS inverter.
I draw timing diagrams on my way.
Anyway,thanks for your help.

Andrea

Roff · Oct 17, 2010

andrea22 said:
Ron,
Rise time and fall time are actually delay time.
You know that delay time is usually: tdelay=1/2*(t_rise+t_fall).
This equation is for CMOS inverter anad every logic circuit realised in CMOS technology can be equivalent
with CMOS inverter.
I draw timing diagrams on my way.
Anyway,thanks for your help.

Andrea

Who told you that? Can you provide a link?
Propagation delays are separate from output transition times, and may or may not match the equation you posted. It all depends on the internal circuitry and the complexity of the gate.

crutschow · Oct 17, 2010

I agree with Ron. Rise and fall times may contribute to delay, but a circuit can certainly have a much larger delay time then the rise and fall times, it depends largely upon the amount of circuitry been the input and output. Look at the difference been an inverter delay and a flip-flop delay, for example.

andrea22 · Oct 17, 2010

tpdr : rising propagation delay
– From input to rising output
crossing VDD/2
tpdf: falling propagation delay
– From input to falling output
crossing VDD/2
tpd: average propagation delay
– tpd = (tpdr + tpdf)/2
tr: rise time
– From output crossing 0.1 VDD to 0.9 VDD
tf: fall time
– From output crossing 0.9 VDD to 0.1 VDD
tpdr=1/2*tr
tpdf=1/2*tf

Roff · Oct 17, 2010

Simple CMOS logic gates and inverters can approach the performance of your diagrams. More complex gates, like 8-input gates, e.g. CD4068, exclusive ORs, etc. have internal circuitry that contributes to delay. Even NANDs and NORs have internal buffering that contribute to delay.
TTL gates have delay that is due to storage time, which is totally independent of transition times.
Your drawing is much too simplistic.

andrea22 · Oct 18, 2010

I tell you that I don't know type number of cmos logic gates,
and I don't need that.
I 've done simulation in ORCAD PSpice.
I draw CMOS inverter with 1 p-channel MOSFET and 1 n-channel MOSFET.
CMOS nand gate is consist of two serial n-channel MOSFETs,and two parallel p-chanell MOSFETs.

crutschow · Oct 18, 2010

Certainly a simple two transistor inverter has very little delay time as compared to the rise and fall times. But delay times can be much larger than rise and fall times in complex circuits. They are not the same parameter.

andrea22 · Oct 18, 2010

I don't need complex circuit.
I told you that my circuit is quite simple.
Thanks for help.

Roff · Oct 18, 2010

andrea22 said:
I don't need complex circuit.
I told you that my circuit is quite simple.
Thanks for help.

If you were implementing this circuit within an IC, your gate schematics are exactly what I would use.

andrea22 · Oct 19, 2010

Do yo know why in dynamic shift register non-overlapping clocks are used?
Or why this circuit doesn't work with overlapped clocks?

Roff · Oct 19, 2010

andrea22 said:
Do yo know why in dynamic shift register non-overlapping clocks are used?
Or why this circuit doesn't work with overlapped clocks?

There are parasitic capacitors at the gate inputs that act as hold capacitors. Each clock samples the output of the previous gate. If the clocks overlapped, data could shoot through all the inverters during the overlap time, or at least degrade the logic levels held on the capacitors to the point where errors would occur.

andrea22 · Oct 20, 2010

Do you have any literature about pass transistors?
These are my diagrams.The third diagrams is at the output of pass transiastor.
Why is happening when Phi1=0,Vin=Vdd and Phi1=0 and Vin=0?

Roff · Oct 20, 2010

andrea22 said:
Why is happening when Phi1=0,Vin=Vdd and Phi1=0 and Vin=0?

That is the result of the clock coupling through the parasitic capacitance of the pass transistor, transferring a little charge to the hold capacitor. It's a capacitive voltage divider.

andrea22 · Oct 20, 2010

These are diagrams for voltage in points:A0,A1,A2,A3.
This is ok?

Roff · Oct 20, 2010

If the inverters are powered with 3 volts, they look good.

Help with non-overlapping clock generator

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