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Re: Incorrect Rise and Fall time calculations with LTSpice .MEAS command


 

--- In LTspice@..., "Helmut" <helmutsennewald@...> wrote:

--- In LTspice@..., "kpotter999" <kpotter999@> wrote:

Dear Group Members,

I have been attempting to use the following commands
to calculate the output rise and fall times of an inverter
and the propagation delay with the following voltage source:

PULSE(0 3.3 1n 1.5n 1.5n 5n 10n)

.MEAS TRAN fall_delay
+ TRIG V(Vout) VAL=3.3*0.9 TD=0 FALL=1
+ TRIG V(Vout) VAL=3.3*0.9 TD=0 FALL=1

.MEAS TRAN rise_delay
+ TRIG V(Vout) VAL=3.3*0.1 TD=0NS RISE=1
+ TRIG V(Vout) VAL=3.3*0.9 TD=0NS RISE=1

.MEAS TRAN propagation_fall
+ TRIG V(Vin) VAL=3.3*0.5 TD=0 RISE=1
+ TRIG V(Vout) VAL=3.3*0.5 TD=0 FALL=1

.MEAS TRAN propagation_rise
+ TRIG V(Vin) VAL=3.3*0.5 TD=0 FALL=1
+ TRIG V(Vout) VAL=3.3*0.5 TD=0 RISE=1

Unfortunately the results from the .MEAS text do not
agree with manual plotting and reading of the waveforms.

i.e.

From Waveform, Rise time = 118ps, Fall time = 152ps

From text, Rise time = 6.7ns, Fall time = 1.6ns

Clearly, I am doing something fundamentally wrong with
the .MEAS command but I have been unable to resolve
this further. I would really appreciatte some feedback
and can supply the circuit schematic and inverter model
if required.

Many thank, Ken Potter.

Hello Ken,

I have uplaoded an example for you.

Files > Tut > MEASURE > TRAN >RC_measure_rise_fall_delay.asc



I have used the following MEASURE-commands.

.MEAS t1r WHEN V(out)=0.1*5 TD=0 RISE=2
.MEAS t2r WHEN V(out)=0.9*5 TD=0 RISE=2

.MEAS t1f WHEN V(out)=0.9*5 TD=0 FALL=2
.MEAS t2f WHEN V(out)=0.1*5 TD=0 FALL=2

.MEAS tdr1 WHEN V(in)=0.5*5 TD=0 RISE=2
.MEAS tdr2 WHEN V(out)=0.5*5 RISE=2

.MEAS tdf1 WHEN V(in)=0.5*5 TD=0 FALL=2
.MEAS tdf2 WHEN V(out)=0.5*5 TD=0 FALL=2

.MEAS trise PARAM t2r-t1r
.MEAS tfall PARAM t2f-t1f
.MEASURE tdelay_rise PARAM tdr2-tdr1
.MEASURE tdelay_fall PARAM tdf2-tdf1

Best regards,
Helmut
Dear Helmut,

Thank you very much for your reply, I will go and figure this out.

Regards, Ken.

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