What relation exists between rise/fall time and frequency?

Hi members,

Please answer these queries:
1- What relation exists between rise/fall time and frequency ?
2- What frequency is used for 65 nm technology gates ?

Thanks in advance.

Re: Signals

master_picengineer,
Rise time (tr) is a function of bandwidth (BW). An approximate relationship is as follows:
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tr = 0.35/BW
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Where tr and BW are in constistent units, e.g, microseconds/mHz.
Regards,
Kral

Signals

Hi everybody,

Thanks Kral for the reply. Could you please upload a Document that detail this issue. The relation that youprovided seems to be not reasonable.
In fact for a 2 Ghz Clock the tr would be 0.175 ns while the period is T=0.5 ns
If we suppose taht t_fall=t_rise we have T=3 t_rise+ 2 pulse_width.
Alone 3 t_rise= 0.525 ns. So pulse_width<0 which is impossible.

Re: Signals

master_picengineer,
I'll see if I can find a reference that derives this relation. In the meantime, don't confuse bandwidth with clock frequency. The bandwidth is the frequency at which the frequency response of the system is down by 3db. It is independent of the clock frequency. In the example you give, it is apparent that the bandwidth of your system is not sufficently wide to support a 2 GHz clock. The relatively slow rise time (I'm assuming that these are measured or simulated results) indicate an insufficient bandwidth.
regards,
Kral

Signals

Thanks Kral,
I understand the difference between BW and frequency. In fact I ment that In the best case the BW in terms of bit/s is equal to the frequency in Hz this if each bit takes one clock cycle to be transmitted in a serial interconnect. Sure the relation that you provide le is applicable for a parallel interconnect with a certain Width (perhaps you forgot to mension it). The relation can't be used for serial interconnect.

I the exemple that I gave

BW is proportional to the frequency. In ideal I considered the best case BW= Frequency.

Re: Signals

master_picengineer,
I couldn’t find an article that develops the equation, so here’s my version:
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Assume the following:
The system is defined as a single pole lag filter with a time constant of Tau.
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Rise time is defined as the time required for the output puls to go from 10% to 90% of its final values.
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The equation for the step response of a single pole lag is:
Vo = Vin[1-e^(-t/Tau)]
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Solve for the time reqired for the output to reach a specified voltage Vo
Vo = Vin-Vin e^(-t/Tau)
Vin e^(-t/Tau) = Vin – Vo
Take the natural log of both sides:
ln(Vin) –t/Tau = ln(Vin-Vo)
-t/Tau = ln(Vin-Vo) – ln(Vin) = ln[(Vin-Vo)/Vin]
-t = -Tau ln[(Vin-Vo)/Vin]
Solve for the time required to reach 10%:
For a single pole lag, Tau = 1/w, where omega is the cutoff frequency in radians/second.
.
SO
T10 = (-1/w) ln [(Vin-.1Vin)/Vin] = .105360515/w
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Solve for the time required to reach 90%
T90 = (-1/w) ln [(Vin-.9Vin)/Vin] = 2.3202585093/w
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Time required to go from 10% to 90% = t90-t10 = 2.2/w
Converting w to Hz,
.
T90-t10 = 2.2/2 pi f = .35/f
Regards,
Kral