Edge detection

Remi · Apr 6, 2004

Dear all
I am looking for a circuit that gives one short impulsion
(10ns<T<50ns) 5V/TTL at the rising and the falling edge of a sinewave
at ~1.5MHz. The pulse should occur at the average voltage of the
sinewave, which have an amplitude of 5V peak to peak. The propgation
time between the detected edge and the pulse should be also as short
as possible.

Thank you for your help

Neil · Apr 6, 2004

Impulsion.........HMMMMM............

"Remi" <RemiB@netcourrier.com> wrote in message
news:75f4a790.0404061154.453068c1@posting.google.com...

Dear all
I am looking for a circuit that gives one short impulsion
(10ns<T<50ns) 5V/TTL at the rising and the falling edge of a sinewave
at ~1.5MHz. The pulse should occur at the average voltage of the
sinewave, which have an amplitude of 5V peak to peak. The propgation
time between the detected edge and the pulse should be also as short
as possible.

Thank you for your help

Bill Sloman · Apr 6, 2004

"Remi" <RemiB@netcourrier.com> wrote in message
news:75f4a790.0404061154.453068c1@posting.google.com...

Dear all
I am looking for a circuit that gives one short impulsion
(10ns<T<50ns) 5V/TTL at the rising and the falling edge of a sinewave
at ~1.5MHz. The pulse should occur at the average voltage of the
sinewave, which have an amplitude of 5V peak to peak. The propagation
time between the detected edge and the pulse should be also as short
as possible.

You want to use a comparator to compare the level of the sine-wave with a
low-pass filtered version of the sine-wave, which will produce a squared-up
version of the sine-wave.

The Linear Technology LT1016 is a reasonably fast comparator, with a
propagation delay of about 10nsec, which produces a TTL compatible output.

The Fairchild SPT9689 comparator is rather faster, at 650 psec, but produces
ECL-compatible output.

http://www.fairchildsemi.com/ds/SP/SPT9689.pdf

You could use the squared up version of the square wave to drive a pair of
mono-stables, one triggered by the rising edge, and the other by the falling
edge, to produce your pulses - you'd then have to OR or AND them together to
produce you pulse train.

Unfortunately, even 50nsec is pretty narrow pulse to get out of a TTL
compatible monostable (74121 or 74221).

There used to be an ECL monostable - the Motorola MC10198 -

http://www.onsemi.com/pub/Collateral/MC10198-D.PDF

which was a lot faster and could go down to 10nsec, but it is no longer in
production.

A neater alternative is to produce a delayed version of your square wave
with some kind of delay line - 10nsec is about two metres of coaxial cable
or two sections of a Newport 11ACB50012E lumped constand delay line - and
compare it with your un-delayed square wave with an XOR gate such as the
Motorola MC100EL107

http://www.onsemi.com/pub/Collateral/MC10EL07-D.PDF

Something of an over-kill for a 1.5MHz sine wave, but you did ask for the
shortest possible propagation delay, and this will get pretty close.

-------
Bill Sloman, Nijmegen

John Fields · Apr 6, 2004

On 6 Apr 2004 12:54:53 -0700, RemiB@netcourrier.com (Remi) wrote:

Dear all
I am looking for a circuit that gives one short impulsion
(10ns<T<50ns) 5V/TTL at the rising and the falling edge of a sinewave
at ~1.5MHz. The pulse should occur at the average voltage of the
sinewave, which have an amplitude of 5V peak to peak. The propgation
time between the detected edge and the pulse should be also as short
as possible.

---
The average voltage of a 5VPP sinewave centered about zero volts is
zero volts, so you must mean something else. Perhaps you'd like to
detect when the signal goes through 5V/pi = 1.59V?

OK, that's fine, but that transition occurs four times per cycle; the
first time on the way up to +2.5V from 0V, the second time on the way
down to 0V from +2.5V, the third time on the way down to -2.5V from 0V
and the fourth time on the way up to to 0V from -2.5V.

The solution to your problem isn't difficult; it sounds basically like
a window comparator followed up with a one-shot, but without a little
more detail it's impossible to tell.

--
John Fields

John Fields · Apr 6, 2004

On Wed, 7 Apr 2004 01:51:10 +0200, "Bill Sloman"
<bill.sloman@ieee.org> wrote:

Something of an over-kill for a 1.5MHz sine wave, but you did ask for the
shortest possible propagation delay, and this will get pretty close.

---
knowing that he's looking for a response when the 1.5MHz sine wave
goes through a particular voltage and knowing the detector's delay
should allow him to detect the voltage which occurs that much earlier
and have his pulses arrive with, essentially, zero delay.

After all, if it's a sine wave, there shouldn't be much trouble
adusting the windage...

--
John Fields

mike · Apr 7, 2004

Remi wrote:

Dear all
I am looking for a circuit that gives one short impulsion
(10ns<T<50ns) 5V/TTL at the rising and the falling edge of a sinewave
at ~1.5MHz. The pulse should occur at the average voltage of the
sinewave, which have an amplitude of 5V peak to peak. The propgation
time between the detected edge and the pulse should be also as short
as possible.

Thank you for your help

You don't mention what you're doin' or exactly what you mean by
"as short as possible". If you really mean zero and you really have
a sinewave, you might be better off with a PLL.

If the back of my envelope is correct, the slew rate is only 15mV/nS.
Doesn't take much noise or hysteresis to mess this all up. Your edges
are gonna be wandering all around. Your layout is gonna look more like
a microwave cavity than a circuit board.

Last time I tried to do this, I was more concerned with jitter than
delay. I ended up with zero hysteresis in my comparator but needed
latches after it to catch only the first transition in the inevitable
burst of noise that came thru at the threshold. But mine had to work at
much lower input frequencies too.

mike

--
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John Larkin · Apr 7, 2004

On 6 Apr 2004 12:54:53 -0700, RemiB@netcourrier.com (Remi) wrote:

Dear all
I am looking for a circuit that gives one short impulsion
(10ns<T<50ns) 5V/TTL at the rising and the falling edge of a sinewave
at ~1.5MHz. The pulse should occur at the average voltage of the
sinewave, which have an amplitude of 5V peak to peak. The propgation
time between the detected edge and the pulse should be also as short
as possible.

Thank you for your help

AC couple into a zero-crossing comparator. Take the comparator output,
run it directly to one input of an XOR gate, and run it into the other
input through a short delay: an RC or a few gate delays will work. The
XOR output will glitch up at each sine crossing.

You can tweak the time constant of the input AC-coupling network to
add a little phase lead to correct for comparator+gate delay, if you
want.

John

KR Williams · Apr 7, 2004

In article <q7j67052vi85egmc9kpba5dv3gru2avj3e@4ax.com>,
jfields@austininstruments.com says...

On Wed, 7 Apr 2004 01:51:10 +0200, "Bill Sloman"
bill.sloman@ieee.org> wrote:

Something of an over-kill for a 1.5MHz sine wave, but you did ask for the
shortest possible propagation delay, and this will get pretty close.

---
knowing that he's looking for a response when the 1.5MHz sine wave
goes through a particular voltage and knowing the detector's delay
should allow him to detect the voltage which occurs that much earlier
and have his pulses arrive with, essentially, zero delay.

After all, if it's a sine wave, there shouldn't be much trouble
adusting the windage...

If one has to be grossly complicated, a PLL could track/cancel
the "windage".

--

Keith

KR Williams · Apr 7, 2004

In article <shn670l6erec8qr8nvjik2mcrfd3o486mc@4ax.com>,
jjlarkin@highlandSNIPtechTHISnologyPLEASE.com says...

On 6 Apr 2004 12:54:53 -0700, RemiB@netcourrier.com (Remi) wrote:

Dear all
I am looking for a circuit that gives one short impulsion
(10ns<T<50ns) 5V/TTL at the rising and the falling edge of a sinewave
at ~1.5MHz. The pulse should occur at the average voltage of the
sinewave, which have an amplitude of 5V peak to peak. The propgation
time between the detected edge and the pulse should be also as short
as possible.

Thank you for your help

AC couple into a zero-crossing comparator.

I'd go the other way, assuming the jitter requirements weren't
harsh (then only a PLL would do). I'd use an integrator (RC
filter, perhaps) to average the signal and then compare the raw
signal to that. Of course there may be unstated start-up
issues...

Take the comparator output,
run it directly to one input of an XOR gate, and run it into the other
input through a short delay: an RC or a few gate delays will work. The
XOR output will glitch up at each sine crossing.

That works, after the crossing detector. A COMS XOR gate is
suggested.

You can tweak the time constant of the input AC-coupling network to
add a little phase lead to correct for comparator+gate delay, if you
want.

That's an idea too.

--
Keith

Glenn Gundlach · Apr 7, 2004

RemiB@netcourrier.com (Remi) wrote in message news:<75f4a790.0404061154.453068c1@posting.google.com>...

Dear all
I am looking for a circuit that gives one short impulsion
(10ns<T<50ns) 5V/TTL at the rising and the falling edge of a sinewave
at ~1.5MHz. The pulse should occur at the average voltage of the
sinewave, which have an amplitude of 5V peak to peak. The propgation
time between the detected edge and the pulse should be also as short
as possible.

Thank you for your help

Delay with an XOR gate looking and delayed and undelayed. Delay can be
a silicon delay (a bit jittery), RC or LC. I've done multiple units of
this in 22V10 using RC delays. For the sinewave I would first use a
comparator to square things up and then detect edges.
GG

Ban · Apr 7, 2004

RemiB@netcourrier.com (Remi) wrote in message
news:<75f4a790.0404061154.453068c1@posting.google.com>...
Dear all
I am looking for a circuit that gives one short impulsion
(10ns<T<50ns) 5V/TTL at the rising and the falling edge of a sinewave
at ~1.5MHz. The pulse should occur at the average voltage of the
sinewave, which have an amplitude of 5V peak to peak. The propgation
time between the detected edge and the pulse should be also as short
as possible.

Well, to get the correct value of the average value you have to rectify and
average the sine wave and then use an analog comparator followed by an
impulse shaping circuit. Here is a short diagram:

o-+-----------------|___|-+-|___|-+---------+
| 2k | 2k | Vrect| |\ 100p
| ___ ___ ___ | |\ | +-|+\ || to Schmitt
+-|___|-+-|___|-+-|___|-+-|-\ | ___ | >-||-+--+--o
1k | 1k | 1k | >--+-|___|-+---|-/ || | |
| OP1 V +-|+/ 1k |Vav|/ | .-.
| |\ - D1 | |/ --- Comp1 - | |
+--+-|-\ | === OP2 1n5--- ^ | |150R
| | >--+ GND | | '-'
| +-|+/ | === | |
| | |/ V D2 GND +--+
| === - |
| GND | ===
+----------+ GND
precision rectifier Comparator
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
view\font\fixed
OP1: fast opamp with high slew rate is needed. AD8052 dual
Diodes: schottky. BAT54
Comp: fast comparator AD8561
If the following digital input doesn`t have a schmitt-trigger input, you
need an additional schmitt-gate.
The output pulse has the length 150R*150p = 22.5ns.
The delay is dependent on the comparator speed. You can compensate for it by
lowering the average threshold. this can be accomplished by paralleling the
1n5 with a resistor 10k...100k.
--
ciao Ban
Bordighera, Italy

budgie · Apr 7, 2004

On Wed, 07 Apr 2004 06:44:22 GMT, "Ban" <bansuri@web.de> wrote:

Well, to get the correct value of the average value you have to rectify and
average the sine wave and then use an analog comparator followed by an
impulse shaping circuit. Here is a short diagram:

To get the correct average value of a sine wave, you only need provide DC
blocking, and the value is zero.

Jim Thompson · Apr 7, 2004

On Wed, 07 Apr 2004 08:05:21 -0700, John Larkin
<jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote:

On Wed, 07 Apr 2004 06:44:22 GMT, "Ban" <bansuri@web.de> wrote:

RemiB@netcourrier.com (Remi) wrote in message
news:<75f4a790.0404061154.453068c1@posting.google.com>...
Dear all
I am looking for a circuit that gives one short impulsion
(10ns<T<50ns) 5V/TTL at the rising and the falling edge of a sinewave
at ~1.5MHz. The pulse should occur at the average voltage of the
sinewave, which have an amplitude of 5V peak to peak. The propgation
time between the detected edge and the pulse should be also as short
as possible.

Well, to get the correct value of the average value you have to rectify and
average the sine wave and then use an analog comparator followed by an
impulse shaping circuit. Here is a short diagram:

o-+-----------------|___|-+-|___|-+---------+
| 2k | 2k | Vrect| |\ 100p
| ___ ___ ___ | |\ | +-|+\ || to Schmitt
+-|___|-+-|___|-+-|___|-+-|-\ | ___ | >-||-+--+--o
1k | 1k | 1k | >--+-|___|-+---|-/ || | |
| OP1 V +-|+/ 1k |Vav|/ | .-.
| |\ - D1 | |/ --- Comp1 - | |
+--+-|-\ | === OP2 1n5--- ^ | |150R
| | >--+ GND | | '-'
| +-|+/ | === | |
| | |/ V D2 GND +--+
| === - |
| GND | ===
+----------+ GND
precision rectifier Comparator
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
view\font\fixed
OP1: fast opamp with high slew rate is needed. AD8052 dual
Diodes: schottky. BAT54
Comp: fast comparator AD8561
If the following digital input doesn`t have a schmitt-trigger input, you
need an additional schmitt-gate.
The output pulse has the length 150R*150p = 22.5ns.
The delay is dependent on the comparator speed. You can compensate for it by
lowering the average threshold. this can be accomplished by paralleling the
1n5 with a resistor 10k...100k.

Good grief, just AC couple and look at the zero crossings.

John

What's the OP's definition of "average" ?

480)460-2350 | |
| E-mail Address at Website Fax

Ban · Apr 7, 2004

John Larkin wrote:

On Wed, 07 Apr 2004 06:44:22 GMT, "Ban" <bansuri@web.de> wrote:

RemiB@netcourrier.com (Remi) wrote in message
news:<75f4a790.0404061154.453068c1@posting.google.com>...
Dear all
I am looking for a circuit that gives one short impulsion
(10ns<T<50ns) 5V/TTL at the rising and the falling edge of a
sinewave at ~1.5MHz. The pulse should occur at the average voltage
of the sinewave, which have an amplitude of 5V peak to peak. The
propgation time between the detected edge and the pulse should be
also as short as possible.

Well, to get the correct value of the average value you have to
rectify and average the sine wave and then use an analog comparator
followed by an impulse shaping circuit. Here is a short diagram:

o-+-----------------|___|-+-|___|-+---------+
| 2k | 2k | Vrect| |\ 100p
| ___ ___ ___ | |\ | +-|+\ || to Schmitt
+-|___|-+-|___|-+-|___|-+-|-\ | ___ | >-||-+--+--o
1k | 1k | 1k | >--+-|___|-+---|-/ || | |
| OP1 V +-|+/ 1k |Vav|/ | .-.
| |\ - D1 | |/ --- Comp1 - | |
+--+-|-\ | === OP2 1n5--- ^ | |150R
| | >--+ GND | | '-'
| +-|+/ | === | |
| | |/ V D2 GND +--+
| === - |
| GND | ===
+----------+ GND
precision rectifier Comparator
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
view\font\fixed
OP1: fast opamp with high slew rate is needed. AD8052 dual
Diodes: schottky. BAT54
Comp: fast comparator AD8561
If the following digital input doesn`t have a schmitt-trigger input,
you need an additional schmitt-gate.
The output pulse has the length 150R*150p = 22.5ns.
The delay is dependent on the comparator speed. You can compensate
for it by lowering the average threshold. this can be accomplished
by paralleling the 1n5 with a resistor 10k...100k.

Good grief, just AC couple and look at the zero crossings.

John

It is a bit overkill, if only a fixed frequency is used, but actually works
for all frequencies up to 10MHz. And also the waveform can be arbitrary.
Gives precise results.
--
ciao Ban
Bordighera, Italy

Apr 7, 2004

In article <0368705ksvc7qamcvh37sjbkm6k8ro15sk@4ax.com>, John Larkin says...

RemiB@netcourrier.com (Remi) wrote in message
news:<75f4a790.0404061154.453068c1@posting.google.com>...
Dear all
I am looking for a circuit that gives one short impulsion
(10ns<T<50ns) 5V/TTL at the rising and the falling edge of a sinewave
at ~1.5MHz. The pulse should occur at the average voltage of the
sinewave, which have an amplitude of 5V peak to peak. The propgation
time between the detected edge and the pulse should be also as short
as possible.

Good grief, just AC couple and look at the zero crossings.

I'd think that would meet 99.9% of all requirements for something
like this, but the original post contained those wonderful
phrases "should occur at the average" and "should be as short
as possible". This lets us run wild with all sorts of solutions
emphasizing "average" (in at least three different definitions so
far) and "as short as possible" (witness the ECL suggestions).

I'm still waiting for the question which I can answer with neon
bulbs and Cadmium Sulfide light-sensitive resistors

Tim.

Tim.

Dave VanHorn · Apr 7, 2004

<shoppa@trailing-edge.com> wrote in message
news:c51bra04mr@drn.newsguy.com...

In article <0368705ksvc7qamcvh37sjbkm6k8ro15sk@4ax.com>, John Larkin
says...
RemiB@netcourrier.com (Remi) wrote in message
news:<75f4a790.0404061154.453068c1@posting.google.com>...
Dear all
I am looking for a circuit that gives one short impulsion
(10ns<T<50ns) 5V/TTL at the rising and the falling edge of a sinewave
at ~1.5MHz. The pulse should occur at the average voltage of the
sinewave, which have an amplitude of 5V peak to peak. The propgation
time between the detected edge and the pulse should be also as short
as possible.

Isn't this self-conflicting?

The only "edges" in a sine wave, if you can stretch the point to call them
that, are the peaks, and the zero crossings, which don't occur at the
average voltage, unless you take the average voltage to be zero volts.

As far as prop delay, does it really matter, as long as it's constant?
Are we working on single cycles here?

If it's that critical, then take the sine wave through some coax, and delay
it so that the pulse comes out wherever you want it.

The Real Andy · Apr 8, 2004

<snip>

Use one of them dps thingies, that should sort out your impulsions.

Jim Thompson · Apr 8, 2004

On Thu, 8 Apr 2004 19:53:40 +1000, "The Real Andy"
<i_luv_beer_especially_pilsners@yahoo.com.au> wrote:

snip

Use one of them dps thingies, that should sort out your impulsions.

"dps"? "Department of Public Service", otherwise known as the Arizona
Highway Patrol ?

480)460-2350 | |
| E-mail Address at Website Fax

Edge detection

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