One more LM2907 question. (The last I hope)

[Glenn Ashmore]

OK, with everyone's help I have these things snapping on and off clean
within 3Hz of the targets. Now I have one last thing. The 2907 needs a
+ to - transition to trigger but the motion sensor has an NPN output.
Right now I am using a cookbook 555 circuit to produce a negative source
and switch it with the sensor through a 2N3906 but there must be a
simpler way.



--
Glenn Ashmore

I'm building a 45' cutter in strip/composite. Watch my progress (or lack
there of) at: http://www.rutuonline.com
Shameless Commercial Division: http://www.spade-anchor-us.com

 

[CFoley1064]

From: Glenn Ashmore [email protected]

Date: 2/16/2004 7:14 PM Central Standard Time
Message-id: <C0eYb.526$23.452@lakeread04>

OK, with everyone's help I have these things snapping on and off clean
within 3Hz of the targets. Now I have one last thing. The 2907 needs a
+ to - transition to trigger but the motion sensor has an NPN output.
Right now I am using a cookbook 555 circuit to produce a negative source
and switch it with the sensor through a 2N3906 but there must be a
simpler way.

Glenn Ashmore

Hi again, Glenn. Your voyage of discovery is almost done -- good. The world
needs more "contrivers of contrivances".

I'm not sure if picking on which transistion causes the charge pump to squirt
is that important -- the output response time is lots slower than the time
between the positive-going transition and the negative-going transition. As a
chief engineer I worked under once said, "There comes a time when you just have
to shoot the engineer and ship the product". Unless there's an important
reason for this, I'm not sure it's very important.

Congratulations, and good luck (which, as we all know, is the residue of hard
work)

Chris

 

[GPG]

OK, with everyone's help I have these things snapping on and off clean
within 3Hz of the targets. Now I have one last thing. The 2907 needs a
+ to - transition to trigger but the motion sensor has an NPN output.
Right now I am using a cookbook 555 circuit to produce a negative source
and switch it with the sensor through a 2N3906 but there must be a
simpler way.

I am beginning to despair of you ever reading my posts.
The LM2907 REQUIRES an AC coupled signal.
This is all that is needed.




VCC
o
|
.-.
pullup | |R1
| |
'-'
| ||C
o--||------o--------o LM2907, pin1
|/ || |
-| .-.
|> | |
| | | R2
| '|'
sensor NPN output o----------o
|
|
o
GND

Much like your original post with hex inverter removed as I have noted twice.
The input in the 8 pin version is ground referenced and only needs approx
100 mV p-p to operate (max 28V p-p). I would use non electro for C,
make R2 much higher than 1K shown and connect both pin1's together.
With the hex inv in circuit I can only assume that a little undershoot
enabled it to function at all.

 

[Glenn Ashmore]

I am beginning to despair of you ever reading my posts.
The LM2907 REQUIRES an AC coupled signal.
This is all that is needed.

I not only read it, I printed it out and put it in the notebook. You
missed the post to Fred where I gave you the credit for solving 75% of
the problems. John Fields solved the pressure switch and Winfield came
up with some other refinements.

Removing the hex inverter was the first thing I did. What is the
transistor in the circuit below?

VCC
o
|
.-.
pullup | |R1
| |
'-'
| ||C
o--||------o--------o LM2907, pin1
|/ || |
-| .-.
|> | |
| | | R2
| '|'
sensor NPN output o----------o
|
|
o
GND

Much like your original post with hex inverter removed as I have noted twice.
The input in the 8 pin version is ground referenced and only needs approx
100 mV p-p to operate (max 28V p-p). I would use non electro for C,
make R2 much higher than 1K shown and connect both pin1's together.
With the hex inv in circuit I can only assume that a little undershoot
enabled it to function at all.

Removing the hex inverter was the first thing I did. What is the
transistor in the circuit below?

--
Glenn Ashmore

I'm building a 45' cutter in strip/composite. Watch my progress (or lack
there of) at: http://www.rutuonline.com
Shameless Commercial Division: http://www.spade-anchor-us.com

 

[Fred Bloggs]

I am beginning to despair of you ever reading my posts.
The LM2907 REQUIRES an AC coupled signal.
This is all that is needed.




VCC
o
|
.-.
pullup | |R1
| |
'-'
| ||C
o--||------o--------o LM2907, pin1
|/ || |
-| .-.
|> | |
| | | R2
| '|'
sensor NPN output o----------o
|
|
o
GND

Much like your original post with hex inverter removed as I have noted twice.
The input in the 8 pin version is ground referenced and only needs approx
100 mV p-p to operate (max 28V p-p). I would use non electro for C,
make R2 much higher than 1K shown and connect both pin1's together.
With the hex inv in circuit I can only assume that a little undershoot
enabled it to function at all.

That circuit has absolutely no noise immunity whatsoever- the
differential thresholds are typically +/-25mV which when centered on GND
as for the LM2907-8 means single ended trip points of same magnitude.
The capacitor coupled circuit is essentially a high-pass to sensor cable
pickup and power supply noise, you would not want to use that around
motors. He should stick with his negative voltage drive- and make it as
much volts as practical.

 

[GPG]

That circuit has absolutely no noise immunity whatsoever- the
differential thresholds are typically +/-25mV which when centered on GND
as for the LM2907-8 means single ended trip points of same magnitude.
The capacitor coupled circuit is essentially a high-pass to sensor cable
pickup and power supply noise, you would not want to use that around
motors. He should stick with his negative voltage drive- and make it as
much volts as practical.

The transistor is your sensor's output. Did not consult original before posting.
Have altered numbering to suit. Did post earlier but seems to be lost in
space.
Good to see Mr bloggs has something positive to say.
The input is indeed a high pass filter, though if C4=1uF, R12=100K,
a very low corner.

 

[Fred Bloggs]

The transistor is your sensor's output. Did not consult original before posting.
Have altered numbering to suit. Did post earlier but seems to be lost in
space.
Good to see Mr bloggs has something positive to say.
The input is indeed a high pass filter, though if C4=1uF, R12=100K,
a very low corner.

It is not that simple- a low corner means greater susceptibility to
pickup noise. Also, the triggering amplitude about the LM2907 input
reference will be a function of the sensor output duty cycle- this
brings up a subtle application pitfall apparently not emphasized enough
in the data sheet, and that is the duration of any trigger pulse must be
at least Vcc*C1/(2* I2,min) to allow the charge pump time to produce a
full Vcc/2 transition across C1 and realize the full gain
I3,avg=Vcc*Fin*C1. This error could go either way, producing an
artificially high or low output, and is exactly the kind of thing that
catches up with a clueless user who writes the LM2907 off as "a simple
analog circuit"- usually with no comprehension whatsoever, no
understanding of component time constants and their relation to input
frequency, no sense of tolerance, and blindly struggles with the high
school algebra equations with no understanding of their range of
validity, fails badly, and resorts to blind random component selection
and coarse observation. I'm not exactly sure what this can be called or
if profanity can be avoided- but it has NOTHING to do with electronic
design- absolutely nothing.

 

[Glenn Ashmore]

It is not that simple- a low corner means greater susceptibility to
pickup noise. Also, the triggering amplitude about the LM2907 input
reference will be a function of the sensor output duty cycle- this
brings up a subtle application pitfall apparently not emphasized enough
in the data sheet, and that is the duration of any trigger pulse must be
at least Vcc*C1/(2* I2,min) to allow the charge pump time to produce a
full Vcc/2 transition across C1 and realize the full gain
I3,avg=Vcc*Fin*C1. This error could go either way, producing an
artificially high or low output, and is exactly the kind of thing that
catches up with a clueless user who writes the LM2907 off as "a simple
analog circuit"- usually with no comprehension whatsoever, no
understanding of component time constants and their relation to input
frequency, no sense of tolerance, and blindly struggles with the high
school algebra equations with no understanding of their range of
validity, fails badly, and resorts to blind random component selection
and coarse observation. I'm not exactly sure what this can be called or
if profanity can be avoided- but it has NOTHING to do with electronic
design- absolutely nothing.

OK Fred, you win. I am sticking with the negative power source. But,
if you will forgive my continued ignorance, your post brings up two
questions I am curious about. First, I figured out Ic but what is I2
and how do you calculate it? I have been using the 180uA from the data
sheet to calculate C2 but the value is so small it had little effect.

Pulse width was something I was wondering about. The only place I see
that formula for pulse width is in the App notes in fig 23 for the
output of a "two shot zero crossing detector". No reference to the
input pulse width but I do observe that the pulse width has a big
effects on the output voltage of pin3 when I vary it on the function
generator. The calculations in the data sheet only seem to hold true
for fairly even high/low pulse spacing. This is not going to effect my
installation because the pulses will be even but how do you handle
something like detecting detecting bolt heads in a spinning disk where
the pulse might be 10% of the cycle width?

--
Glenn Ashmore

I'm building a 45' cutter in strip/composite. Watch my progress (or lack
there of) at: http://www.rutuonline.com
Shameless Commercial Division: http://www.spade-anchor-us.com

 

[GPG]

Pulse width was something I was wondering about. The only place I see

that formula for pulse width is in the App notes in fig 23 for the
output of a "two shot zero crossing detector". No reference to the
input pulse width but I do observe that the pulse width has a big
effects on the output voltage of pin3 when I vary it on the function
generator. The calculations in the data sheet only seem to hold true
for fairly even high/low pulse spacing. This is not going to effect my
installation because the pulses will be even but how do you handle
something like detecting detecting bolt heads in a spinning disk where
the pulse might be 10% of the cycle width?

LM2907N-8
Ignoring the -vbe, (which cancels out) C1 (pin2) is charged to .75 Vcc
(pin6),
then discharged to .25 Vcc,ie .5 Vcc. During charge, a mirror of this
cuurent
is output at pin3. The current (I) is Vcc dependent,(see data) for
Vcc= 12V it will be abt 180uA. The time to change the charge on .1uF
by 6V it will take
T= C*V/I 3.3' mS = 300Hz. This is for a square wave and takes no
account
of tolerances. For 10% duty with the same components the max frequency
is 1/(3.3+9*3*3mS)=30.03 Hz. You will get one pulse of current per
zero crossing
so the ripple will be assymetric.
You still do not need a negative voltage

 

[Glenn Ashmore]

GPG wrote:

LM2907N-8
Ignoring the -vbe, (which cancels out) C1 (pin2) is charged to .75 Vcc
(pin6),
then discharged to .25 Vcc,ie .5 Vcc. During charge, a mirror of this
cuurent
is output at pin3. The current (I) is Vcc dependent,(see data) for
Vcc= 12V it will be abt 180uA. The time to change the charge on .1uF
by 6V it will take
T= C*V/I 3.3' mS = 300Hz. This is for a square wave and takes no
account
of tolerances. For 10% duty with the same components the max frequency
is 1/(3.3+9*3*3mS)=30.03 Hz. You will get one pulse of current per
zero crossing
so the ripple will be assymetric.
You still do not need a negative voltage

I see now. I missed that chart for I2. So you would have to drop C1 to
..001uF and raise R1 to get the same T? YOu got my curiosity up. I will
have to do some substitutions on the breadboard and see what happens.

--
Glenn Ashmore

I'm building a 45' cutter in strip/composite. Watch my progress (or lack
there of) at: http://www.rutuonline.com
Shameless Commercial Division: http://www.spade-anchor-us.com

 

[GPG]

GPG wrote:



I see now. I missed that chart for I2. So you would have to drop C1 to
.001uF and raise R1 to get the same T? Yes

YOu got my curiosity up. I will

have to do some substitutions on the breadboard and see what happens.

OOPS should be 1/2*3.3mS = 150 Hz so you are sailing close to the wind

 

[GPG]

I see now. I missed that chart for I2. So you would have to drop C1 to

YOu got my curiosity up. I will

OOPS should be 1/2*3.3mS = 150 Hz so you are sailing close to the wind

E-mailed you at site address and the one here, no response.
Add g to my address