Maker Pro
Maker Pro

Simple pulse stretcher

F

Frank Miles

Jan 1, 1970
0
My ‘scope is only 200MHz, so I’m not really sure I even believe the
5ns... what’s that rule of thumb for ‘scope bandwidth/ rise time? tau=
1/(3*BW) ?

Very close. Tek always gave the formula: 0.35/BW ; it's based on the
semi-mythical gaussian bandpass characteristic. If aberrations are
significant, neither can be trusted.

You probably know the formula for estimating risetime when the
measurement approaches the instrument limits:
tr= sqrt(tmeas^2 - tinstr^2)
which of course only works when you already know that the pulses are
clean - which you probably can't tell with that scope.
 
G

George Herold

Jan 1, 1970
0
Very close.  Tek always gave the formula: 0.35/BW ; it's based on the
semi-mythical gaussian bandpass characteristic.  If aberrations are
significant, neither can be trusted.

You probably know the formula for estimating risetime when the
measurement approaches the instrument limits:
   tr= sqrt(tmeas^2 - tinstr^2)
No I didn't know that. Thanks, I guess mostly I like to have minimal
instrumental effects :^)

George H.
 
G

George Herold

Jan 1, 1970
0
The tinylogic one-shot that John mentioned is guaranteed to trigger
off 2.5ns pulses (with a 5V 'Vcc'). Available in a nice friendly
0.65mm pitch 8-pin package (also a 0.5mm pitch or 1 x 2mm BGA if you
are in need of pain).

Sweet, I think a nice one shot is the right medicine.
It looks like I'll stretch it out to ~200ns, not a big deal
with 1us bins, the first time bin is ~20% 'short'.
We'll document it in the manual and specs,
70% of our users will never notice.

Hey, we should use the 200ns pulse as the monitor output!
maybe they'll notice if it's on the 'scope.

I'm not sure why I didn't have a one-shot in to begin with.
(there's been some 'vibe' that one shots are bad.(?)
It all seems so obvious in retrospect.

George H.
 
B

Bill Sloman

Jan 1, 1970
0
It's a one-shot. It has no internal trigger. It generates no pulses.

And neither you nor Jim have a clue as to how this might be done.

We have a customer who wants us to take this down to 10 ps pulses. At that
point, I'm not sure that I know how that might be done. We're thinking about it.

http://books.google.co.nz/books?id=...nepage&q="emitter-coupled" monostable&f=false

describes the emitter-coupled monostable. Put one together out of a
pair of wide-band tansistors - BFR92 or better - with 33R up against
each base, and you can cetainly get below 10nsec. Since the mechanism
depends on the change of base-emitter impedance with emitter current,
it isn't as easy as it might be to get a wide range of output pulse
widths.

Jim Thompson could probably remember a better solution for you. The
long-obsolete MC10198 ECL monostable

http://www.digchip.com/datasheets/parts/datasheet/343/MC10198-pdf.php

could just get down to 10nsec, but we used two of them when we wanted
to offer long pulses as well - being able to switch in bigger
capacitors put too much stray capacitance on the relevant input pin
for 10nsec operation.

Something boringly obvious with a constant current ramp and a fast
comparator would do the job, but - as with the MC10198, being able to
switch in bigger capacitors to generate much longer pulses is probably
incompatible with a 10nsec pulse width.
 
B

Bill Sloman

Jan 1, 1970
0
//www.highlandtechnology.com/DSS/T240DS.shtml[/URL]
---
That's not a pulse stretcher, cheater, that's a pulse _generator_.
It's a one-shot. It has no internal trigger. It generates no pulses.
And neither you nor Jim have a clue as to how this might be done.
We have a customer who wants us to take this down to 10 ps pulses. At that
point, I'm not sure that I know how that might be done. We're thinkingabout it.

describes the emitter-coupled monostable. Put one together out of a
pair of wide-band tansistors - BFR92 or better - with 33R up against
each base, and you can cetainly get below 10nsec. Since the mechanism
depends on the change of base-emitter impedance with emitter current,
it isn't as easy as it might be to get a wide range of output pulse
widths.
Jim Thompson could probably remember a better solution for you. The
long-obsolete MC10198 ECL monostable

could just get down to 10nsec, but we used two of them when we wanted
to offer long pulses as well - being able to switch in bigger
capacitors put too much stray capacitance on the relevant input pin
for 10nsec operation.
Something boringly obvious with a constant current ramp and a fast
comparator would do the job, but - as with the MC10198, being able to
switch in bigger capacitors to generate much longer pulses is probably
incompatible with a 10nsec pulse width.

I want 10 picoseconds, not 10 ns. I already sell boxes that go down to 100 ps
pulse width.

http://www.highlandtechnology.com/DSS/T240DS.shtml

I just need another factor of 10.

Interesting. If electromagnetic radiation propagates 20cm in 1nsec in
ordinary dielectrics, 10psec is 2mm, and it's going to be a bit tricky
to provide connections that don't introduce significant impedance
discontinuities. Are you planning on designing stuff that surface
mounts onto his printed circuit?

Since microstrip is intrinsicly dispersive, strip line might be
better, and he'd have to sandwich your contribution.

Clearly a fresh field to conquer, not exactly nano-engineering, but
definitely one for the tiny-minded.
 
F

Fred Abse

Jan 1, 1970
0
That used to be true. The high-end digital scopes that I've tested lately
don't seem to have gaussian response; they seem to be tweaked for
bandwidth bragging rights, and subsequently ring.

My "200 MHz" Tek DPO2024 rings a bit and has a rise time of 1.85 ns.

3dB bandwidth versus rise time used to be the test for 'scope spec
"cheats". It's dead easy to make things look better than they really are.

Tek service manuals used to be quite explicit about setting up Y
amplifiers for minimum overshoot and ringing. Properly set up, the
correlation between 3dB bandwith and rise time, using the 0.35 rule of
thumb was quite close. Nearly every vertical amplifier I've come across,
apart from the 50 ohm ones, has been overcompensated at the HF end.
 
F

Fred Abse

Jan 1, 1970
0
Very close. Tek always gave the formula: 0.35/BW ; it's based on the
semi-mythical gaussian bandpass characteristic.

Lowpass. DC to whatever.

ln(9)/(2*pi), 0.3496
 
B

Bill Sloman

Jan 1, 1970
0
On Mar 9, 3:54 am, John Larkin
//www.highlandtechnology.com/DSS/T240DS.shtml[/URL]
---
That's not a pulse stretcher, cheater, that's a pulse _generator_.
It's a one-shot. It has no internal trigger. It generates no pulses..
And neither you nor Jim have a clue as to how this might be done.
We have a customer who wants us to take this down to 10 ps pulses. At that
point, I'm not sure that I know how that might be done. We're thinking about it.
http://books.google.co.nz/books?id=-pi4vP6xMOQC&pg=PA571&lpg=PA571&dq...
describes the emitter-coupled monostable. Put one together out of a
pair of wide-band tansistors - BFR92 or better - with 33R up against
each base, and you can cetainly get below 10nsec. Since the mechanism
depends on the change of base-emitter impedance with emitter current,
it isn't as easy as it might be to get a wide range of output pulse
widths.
Jim Thompson could probably remember a better solution for you. The
long-obsolete MC10198 ECL monostable
http://www.digchip.com/datasheets/parts/datasheet/343/MC10198-pdf.php
could just get down to 10nsec, but we used two of them when we wanted
to offer long pulses as well - being able to switch in bigger
capacitors put too much stray capacitance on the relevant input pin
for 10nsec operation.
Something boringly obvious with a constant current ramp and a fast
comparator would do the job, but - as with the MC10198, being able to
switch in bigger capacitors to generate much longer pulses is probably
incompatible with a 10nsec pulse width.
I want 10 picoseconds, not 10 ns. I already sell boxes that go down to100 ps
pulse width.
http://www.highlandtechnology.com/DSS/T240DS.shtml
I just need another factor of 10.
Interesting. If electromagnetic radiation propagates 20cm in 1nsec in
ordinary dielectrics, 10psec is 2mm, and it's going to be a bit tricky
to provide connections that don't introduce significant impedance
discontinuities.

Naturally.

Are you planning on designing stuff that surface
mounts onto his printed circuit?

No, onto mine. My output would be a coax connector.

Do you have a specific connector in mind? SMA crapped out at 18GHz.
Somebody has obviously introduced something better since then, but
Farnell wasn't stocking whatever it is when I last looked,
You say that like it's a bad thing.

It's true that there is a deliberate ambiguity there.
 
B

Bill Sloman

Jan 1, 1970
0
Sweet, I think a nice one shot is the right medicine.
It looks like I'll stretch it out to ~200ns, not a big deal
with 1us bins, the first time bin is ~20% 'short'.
We'll document it in the manual and specs,
70% of our users will never notice.

Hey, we should use the 200ns pulse as the monitor output!
maybe they'll notice if it's on the 'scope.

I'm not sure why I didn't have a one-shot in to begin with.
(there's been some 'vibe' that one shots are bad.(?)
It all seems so obvious in retrospect.

One-shots are really chunks of analog electronics, and don't fit
perfectly into all-digital systems. Decoupling the 5V supply at the
monostable to the wrong 0V track can be a real problem.

If the track running from the monostable to the timing capacitance is
too long, or runs close up against an active trace, cross-talk can
make the output pulse lengths erratic. Analog designers tend to be
conscious of stuff like that. Digital designers tend to over-simplify
and find out the hard way.
 
G

George Herold

Jan 1, 1970
0
One-shots are really chunks of analog electronics, and don't fit
perfectly into all-digital systems. Decoupling the 5V supply at the
monostable to the wrong 0V track can be a real problem.

Yup, I'll do the schematic and look over the layout.
I was drawing on the white board and explaining the circuit to 'the
boss' and I drew a line between the comparator and first HC14, this
sides analog, that's digital, The diode RC thing is on the analog side
of the circuit, I'll put the one shot in the same place.

George H.
 
G

George Herold

Jan 1, 1970
0
One-shots are politically incorrect, the legacy of days past when far toomany
people designed asynchronous, hazardous, glitchy hairballs. Moto sold a DTL part
that was absurdly noise sensitive, and that contributed to the bad reputation.

They can be handy, used carefully.

--

John Larkin                  Highland Technology Incwww..highlandtechnology.com  jlarkin at highlandtechnology dot com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME  analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators- Hide quoted text -

- Show quoted text -

The last time I used a one shot was back in '93. (real TTL)
Two of them set the pulse lengths in a NMR spectrometer.

George H.
 
G

George Herold

Jan 1, 1970
0
There is? It seems that is a result only for normal distributions.

Yup.  From the derivative theorem, the variance is proportional to the
second derivative of the transform, evaluated at zero frequency.  To
within a constant factor that depends on your Fourier transform definition,

          int(-inf to inf) [t**2 * h(t)]
var(h) = ------------------------------- = H''(0)/H(0)
              int(-inf to inf) [h(t)]

Because the function is real, its transform is Hermitian, i.e. the real
part is even and the imaginary part is odd.  Even-order derivatives of
an odd function vanish at the origin, as do odd-order derivatives of an
even function.

When you convolve g(t) and h(t), the transform is G(f)H(f).  The
variance of this is proportional to the normalized second derivative at
the origin, as before.  Ignoring a possible constant of proportionality
that we don't care about,

d(GH)/df = GH' + HG' so

            d**2/df**2[GH]|         H(0)G''(0) + G(0)H''(0) + 2G'H'
var(g*h) = --------------|      =  -------------------------------
                 GH       |f=0                 H(0)G(0)

       G''(0)    H''(0)    2G'(0)H'(0)
   =  ------- + -------- + ------------
        G(0)      H(0)       H(0)G(0)

The first term is the variance of G, the second is the variance of H,
and the third is zero if either G or H is an even function.

There is a slightly more subtle condition that will make the third term
zero for any choice of g and h: that the first moment of either h(t) or
g(t) is zero, i.e. that at least one of them has its centroid at t=0,
which makes its first derivative zero at f=0.

It is always possible to satisfy this condition by an appropriate choice
of the time origin, so if you'll allow me to slide over that rather
trivial issue(*),  variances add under convolution.

Cheers

Phil Hobbs

(*) The reason for this is the effect of a shift of origin on the
variance.  If you convolve a function centred at t=5 with one centredat
t=3, the convolution's centroid is at t=8.  If you're computing the
variance as the second moment about t=0, this will make a big
difference, but it doesn't change the shape of the resulting convolution
function.  Forcing one of them to have its centroid at 0 gets rid of
this shift of time origin.

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510 USA
+1 845 480 2058

hobbs at electrooptical dot nethttp://electrooptical.net

Awesome... (Be careful what you ask for.)

Phil, If I mix two lasers together on a photodiode and look at the
bandwidth of the beat note to measure the laser bandwidth.. Is the
measured bandwidth the quadrature sum of the individual lasers (as
above for rise times, noise..) or does the power law nature of the PD
put a twist on it?

George H.
 
F

Fred Abse

Jan 1, 1970
0

Using the "0.35" rule, 1.865ns equates to 187.7MHz BW.

Not far below 200MHz, which is 1.75ns.

I can think of a few reasons:

Were you using a hi-Z probe, or a 50 ohm (internal or external?)
terminated measurement?

Interestingly, a 200MHz scope with a 500MHz probe, theoretically, would
show 1.88ns risetime (186MHz BW).

I've seen some commercial BNC through terminations start to get
sticky around 200MHz, and be quite useless at 500MHz. An SMA terminator on
a tee with adaptors outperforms quite dramatically.

If your scope provides 50 ohm inputs, I'd be inclined to TDR them.

Another point; how does that scope measure risetime? Does it calculate the
10% and 90% from the peak of the overshoot (wrong), or from the flat
portion (correct)?
 
G

George Herold

Jan 1, 1970
0
On 3/9/2013 6:51 PM, [email protected] wrote:
On Friday, March 8, 2013 3:59:36 PM UTC-5, Phil Hobbs wrote:
It's actually more general than that--there's a Fourier transform
theorem that variances add under convolution, ...
There is? It seems that is a result only for normal distributions.
Yup.  From the derivative theorem, the variance is proportional to the
second derivative of the transform, evaluated at zero frequency.  To
within a constant factor that depends on your Fourier transform definition,
           int(-inf to inf) [t**2 * h(t)]
var(h) = ------------------------------- = H''(0)/H(0)
               int(-inf to inf) [h(t)]
Because the function is real, its transform is Hermitian, i.e. the real
part is even and the imaginary part is odd.  Even-order derivatives of
an odd function vanish at the origin, as do odd-order derivatives of an
even function.
When you convolve g(t) and h(t), the transform is G(f)H(f).  The
variance of this is proportional to the normalized second derivative at
the origin, as before.  Ignoring a possible constant of proportionality
that we don't care about,
d(GH)/df = GH' + HG' so
             d**2/df**2[GH]|         H(0)G''(0) + G(0)H''(0) + 2G'H'
var(g*h) = --------------|      =  -------------------------------
                  GH       |f=0                 H(0)G(0)
        G''(0)    H''(0)    2G'(0)H'(0)
    =  ------- + -------- + ------------
         G(0)      H(0)       H(0)G(0)
The first term is the variance of G, the second is the variance of H,
and the third is zero if either G or H is an even function.
There is a slightly more subtle condition that will make the third term
zero for any choice of g and h: that the first moment of either h(t) or
g(t) is zero, i.e. that at least one of them has its centroid at t=0,
which makes its first derivative zero at f=0.
It is always possible to satisfy this condition by an appropriate choice
of the time origin, so if you'll allow me to slide over that rather
trivial issue(*),  variances add under convolution.
Cheers
Phil Hobbs
(*) The reason for this is the effect of a shift of origin on the
variance.  If you convolve a function centred at t=5 with one centred at
t=3, the convolution's centroid is at t=8.  If you're computing the
variance as the second moment about t=0, this will make a big
difference, but it doesn't change the shape of the resulting convolution
function.  Forcing one of them to have its centroid at 0 gets rid of
this shift of time origin.
Awesome... (Be careful what you ask for.)
Phil, If I mix two lasers together on a photodiode and look at the
bandwidth of the beat note to measure the laser bandwidth.. Is the
measured bandwidth the quadrature sum of the individual lasers (as
above for rise times, noise..) or does the power law nature of the PD
put a twist on it?

Hmm, no good deed goes unpunished. ;)

The PD is square law, but it isn't E**2, it's |E**2|, so to be
completely safe you have to stay in real-valued functions (sin and cos)
and not complex exponentials. (At least at first.)

Assuming that the relative phase of the two beams depends only on time,
i.e. that they're both really really single transverse mode, then
shining E1(x, y, t) and E2(x, y, t) on a sufficiently large photodiode
will get you

i_photo(t) = int(-inf < x < inf) int(-inf < y < inf)[ R|(E1(x,y,t) +
E2(x,y,t)|**2]dxdy

= R int int[ |E1|**2 + |E2|**2 + 2 Re{E1 E2*}]dxdy

where R is some constant (R is the responsivity if E**2 has units of
watts per square metre, but it doesn't matter here).

With a large enough beat frequency, we can regard E1**2 and E2**2 as DC
and filter them out.  (Since the negative frequencies are twice as far
away as that, we can use the complex exponential notation with no
worries.)  So at AC, all we get is

i_AC = 2R Re {int int[ E1(x,y,t)E2*(x,y,t)]dxdy }

If both lasers are single transverse mode, this integral is proportional
to Re{E1(0,0,t) E2*(0,0,t)}.  Since the beat frequency is large, we
sample all relative phases of E1 and E2 much more rapidly than their
fluctuations, so we pretty much get an honest multiplication.
Multiplying them in the time domain is convolving them in the frequency
domain, so yes, the frequency variances add, *provided they're computed
around the nominal beat frequency*.  Of course nobody in his right mind
would do anything else, but a computer might. ;)

The only thing that modifies this significantly is the frequency
response of the photodiode and the electrical measurement system.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510 USA
+1 845 480 2058

hobbs at electrooptical dot nethttp://electrooptical.net- Hide quoted text -

- Show quoted text -

Cool, thanks Phil, (In practice I just quote the laser bandwidth as
the beat note bandwidth and ignore the above convolution.)
Ahh, what's the line about the PD size "on a sufficiently large
photodiode" refer too?
(I used this tiny EOT photodiode ET-2030.)

George H.
 
There is? It seems that is a result only for normal distributions.



Yup. From the derivative theorem, the variance is proportional to the

second derivative of the transform, evaluated at zero frequency. To

within a constant factor that depends on your Fourier transform definition,



int(-inf to inf) [t**2 * h(t)]

var(h) = ------------------------------- = H''(0)/H(0)

int(-inf to inf) [h(t)]



Because the function is real, its transform is Hermitian, i.e. the real

part is even and the imaginary part is odd. Even-order derivatives of

an odd function vanish at the origin, as do odd-order derivatives of an

even function.



When you convolve g(t) and h(t), the transform is G(f)H(f). The

variance of this is proportional to the normalized second derivative at

the origin, as before. Ignoring a possible constant of proportionality

that we don't care about,



d(GH)/df = GH' + HG' so



d**2/df**2[GH]| H(0)G''(0) + G(0)H''(0) + 2G'H'

var(g*h) = --------------| = -------------------------------

GH |f=0 H(0)G(0)



G''(0) H''(0) 2G'(0)H'(0)

= ------- + -------- + ------------

G(0) H(0) H(0)G(0)



The first term is the variance of G, the second is the variance of H,

and the third is zero if either G or H is an even function.



There is a slightly more subtle condition that will make the third term

zero for any choice of g and h: that the first moment of either h(t) or

g(t) is zero, i.e. that at least one of them has its centroid at t=0,

which makes its first derivative zero at f=0.



It is always possible to satisfy this condition by an appropriate choice

of the time origin, so if you'll allow me to slide over that rather

trivial issue(*), variances add under convolution.



Cheers



Phil Hobbs



(*) The reason for this is the effect of a shift of origin on the

variance. If you convolve a function centred at t=5 with one centred at

t=3, the convolution's centroid is at t=8. If you're computing the

variance as the second moment about t=0, this will make a big

difference, but it doesn't change the shape of the resulting convolution

function. Forcing one of them to have its centroid at 0 gets rid of

this shift of time origin.



--

Dr Philip C D Hobbs

Principal Consultant

ElectroOptical Innovations LLC

Optics, Electro-optics, Photonics, Analog Electronics



160 North State Road #203

Briarcliff Manor NY 10510 USA

+1 845 480 2058



hobbs at electrooptical dot net

http://electrooptical.net

Okay, thanks, don't think I've run into that kind of general analysis before. There is a more thorough rehash here:
Circuits, Signals, and Systems, Volume 2
By William McC. Siebert

Chap 16, but they omit the pages getting into the blasted second moments analysis:

http://books.google.com/books?id=zB...ms, Volume 2 By William McC. Siebert&f=false
 
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