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Question on LT Spice equivalents...

P

Paul Burridge

Jan 1, 1970
0
Clearly, the AC gain of a buffered gate is a lot higher than that of an
unbuffered gate. But how exactly that leads to startup problems and
other trouble is not obvious to me. The application note Win has
mentioned does just note that without explanation. Another source is
Linear Technology's AN-12. Jim Williams writes there that the problem is
"the inability to reliably identify the analog characteristics of the
gates used as gain elements". He proceeds to describe the symptoms, but
without a deeper analysis of the causes.

I have since accepted it as common knowledge that unbuffered gates make
better crystal oscillators, but I admit that I still don't quite know
why it is so.

Hello, Stefan,

I've just read the app note Win pointed us to. It appears the argument
is that the buffered gates overdrive the crystal and can cause long
term damage to it. That's how I read it, anyway. Although it appears
from what is stated that such damage can be much mitigated if not
eliminated by the careful selection of the lower value resistor. I
wonder if ceramic resonators used in place of xtals in this type of
circuit are less susceptible to such damage?
I'm not altogether clear as to why the output of the 10Mhz oscillator
in the diagram is shown as being 0.5 - 40Mhz!
Incidentally, I just called Farnell to try to order some of these
devices but they've no record of the part number. :-(
 
W

Winfield Hill

Jan 1, 1970
0
Paul Burridge wrote...
Incidentally, I just called Farnell to try to order some of
these devices but they've no record of the part number. :-(

You can use any unbuffered inverter. They have part
numbers like 74HCU04, 74HCU04A, 74AHCU04 and 74VHCU04
for hex-inverter packages, and 74HC1GU04, 74VHC1GU04,
74V1GU04, 74VHC1GU04, TC7SU04, 74LVC1GU04A and NC7SZU04
for single-gate packages (these are part numbers from
data-sheet files stored in my computer). Farnell will
have some if not all of these parts. We have several
of these types in our parts inventory.

Thanks,
- Win

whill_at_picovolt-dot-com
 
S

Stefan Heinzmann

Jan 1, 1970
0
Paul said:
Hello, Stefan,

I've just read the app note Win pointed us to. It appears the argument
is that the buffered gates overdrive the crystal and can cause long
term damage to it. That's how I read it, anyway. Although it appears
from what is stated that such damage can be much mitigated if not
eliminated by the careful selection of the lower value resistor.

I would have thought that the argument for the series resistor applied
to both types of gate equally.
I wonder if ceramic resonators used in place of xtals in this type of
circuit are less susceptible to such damage?
I'm not altogether clear as to why the output of the 10Mhz oscillator
in the diagram is shown as being 0.5 - 40Mhz!

That appears to be a cockup. They probably meant to say that the circuit
can be used with frequencies in this range, but the crystal frequency
has to be selected accordingly, of course.
Incidentally, I just called Farnell to try to order some of these
devices but they've no record of the part number. :-(

The NL27WZU04? Make sure you read the datasheet before you order,
because they're tiny. You don't want to hand-solder them unless you have
to. Similar devices come from various manufacturers under different
names. Farnell may have some compatible ones from another manufacturer.
 
J

James Meyer

Jan 1, 1970
0
Now I'm completely baffled. There was me thinking "analogue" referred
to smoothly variable as opposed to quantised. What has being single or
multiple got to do with it? I'm obviously missing something here. :-|

Take a look at http://focus.ti.com/lit/an/scha004/scha004.pdf. Pay
especial attention to the I/O transfer curves for the two types of gates. The
smooth rounded curves for the unbuffered gates/inverters closely resemble the
curves for single transistor amplifiers where the output voltage is proportional
to the input voltage over a range. Buffered logic tends to switch rather
sharply at a threshold.

You can make digital logic circuits from single stage transistor
amplifiers. The same circuit could also be a linear amplifier. It just depends
on the signal levels involved. The circuit stays the same.

Oscillators need to be built around devices with gain. No configuration
of R, L, and C alone will ever oscillate without gain being introduced. Sine
wave oscillators are easiest to design and build using linear amplifiers as the
gain stages.

Relaxation oscillators are very easy to design with any sort of gain
stage, even buffered logic gates, but a relaxation oscillator is not a stable
sine wave oscillator.

Jim
 
R

Roy McCammon

Jan 1, 1970
0
Chris said:
Roy McCammon wrote:

Thus you can see, I have a pretty strong view about psychiatric drug
usage. <snip>


wow Chris

I have to admit that I haven't walked in your shoes. I might
feel the same way. If you are coping, then you are coping. A
person close to me also suffers from FMA; if she doesn't take
Prozac (vitamin P), its hard for us to cope with her.
 
P

Paul Burridge

Jan 1, 1970
0
wow Chris

I have to admit that I haven't walked in your shoes. I might
feel the same way. If you are coping, then you are coping. A
person close to me also suffers from FMA; if she doesn't take
Prozac (vitamin P), its hard for us to cope with her.

There's no "one-size-fits-all" remedy for depression. It's a complex
disorder, poorly-understood and its origin in any given individual can
depend on many factors both endogenous and exogenous. Having said
that, Chris is in the minority as something like 73% of patients
respond well to chemical prophyaxis whatever the aetiology. That's
fortunate, because New Scientist magazine estimates that by 2050,
we'll *all* be depressed!
 
K

Ken Smith

Jan 1, 1970
0
Jim Thompson said:
Wrong. 'HC04 *is* buffered, it has three inverters in series.
Besides
over-driving the crystal

I disagree with this part of the statement. Both the buffered and
unbuffered devices limit at the point where the output swings from ground
to VCC. Assuming that the crystal is AT cut it is easy to design an
oscillator where the drive is not excessive.
it is prone to spurious oscillations

This is the real killer with buffered gates. They have so much gain that
all sorts of funny modes of oscillation can get started in them.
plus
significant slowing of the oscillation frequency.

This is only true at higher frequencies. At 1 or 2 MHz the effect is
small.
 
J

Jim Thompson

Jan 1, 1970
0
I disagree with this part of the statement. Both the buffered and
unbuffered devices limit at the point where the output swings from ground
to VCC. Assuming that the crystal is AT cut it is easy to design an
oscillator where the drive is not excessive.

The unbuffered will look sinusoidal on the input side and
quasi-sinusoidal on the output side.
This is the real killer with buffered gates. They have so much gain that
all sorts of funny modes of oscillation can get started in them.


This is only true at higher frequencies. At 1 or 2 MHz the effect is
small.

--

74HC04: 8ns delay is 6.4% of the period of an 8MHz oscillator.

...Jim Thompson
 
K

Ken Smith

Jan 1, 1970
0
Stefan Heinzmann said:
I would have thought that the argument for the series resistor applied
to both types of gate equally.

I think it really comes down to "You have to design the oscillator
correctly."

Ascii art:
R1
----/\/\/\-----
! !
! !\ U1 !
!-------! O-----!-- Output
! !/ !
! [Z1]
! !! Y1 !
!----!!---------!
! !! !
--- ---
--- C1 --- C2
! !
GND GND


R1 serves mostly to bias the gate to its transition point. Its value can
vary widely so long as it is high enough to not degrade the Q and low
enough to provide a repeat in production.

U1 can be any type of inverting gate so long as it provides enough gain
and doesn't have too much phase lag. "Too much" depends on the purpose of
the oscillator and the nature of Z1. If U1 has no phase shift, the
tuned system sees an impedance of the same type as Z1 but the sign is
inverted. If the real part of all of the impedances seen in the tuned
system is negative, the circuit will oscillate.

Z1 controls the drive to the crystal. In most designs this is a resistor
but it can be a small (5pF) capacitor. Using a capacitor will allow
slower logic gates to be used. This is a good thing to remember if the
logic gate is a CD4000 series gate.

C1 and C2 are almost always made equal or nearly equal. C1 is actually in
parallel with the input capacitance of the logic gate. You can reduce C1
slightly because of this. You don't want the total capacitance on that
side of the crystal to ever be the smaller of the two however so don't
reduce it too much.
 
K

Ken Smith

Jan 1, 1970
0
The unbuffered will look sinusoidal on the input side and
quasi-sinusoidal on the output side.

The buffered will look sinusoisal on the input side (unless you look real
closely) and quasi-sinusoidal on the output side (if you call a near
square wave quasi-sinusoidal). Since you know the amplitude, you can
design for the right drive level.
74HC04: 8ns delay is 6.4% of the period of an 8MHz oscillator.

So at 1MHz it is less than 1%. I think you'd agree that this qualifies
for "the effect is small".
 
J

Jim Thompson

Jan 1, 1970
0
[...]
Wrong. 'HC04 *is* buffered, it has three inverters in series.

Besides
over-driving the crystal

I disagree with this part of the statement. Both the buffered and
unbuffered devices limit at the point where the output swings from ground
to VCC. Assuming that the crystal is AT cut it is easy to design an
oscillator where the drive is not excessive.

The unbuffered will look sinusoidal on the input side and
quasi-sinusoidal on the output side.

The buffered will look sinusoisal on the input side (unless you look real
closely) and quasi-sinusoidal on the output side (if you call a near
square wave quasi-sinusoidal). Since you know the amplitude, you can
design for the right drive level.
74HC04: 8ns delay is 6.4% of the period of an 8MHz oscillator.

So at 1MHz it is less than 1%. I think you'd agree that this qualifies
for "the effect is small".
--

It's even *smaller* for the 'HCU04 and the gain is down dramatically.

Don't trifle with me, young man... I design a new microchip with a
XTAL oscillator at least once every three months... and the clients
kiss my feet... in a recent RF chip *only* my cells came out working
to specification ;-)

...Jim Thompson
 
J

Jim Thompson

Jan 1, 1970
0
Stefan Heinzmann said:
I would have thought that the argument for the series resistor applied
to both types of gate equally.

I think it really comes down to "You have to design the oscillator
correctly."

Ascii art:
R1
----/\/\/\-----
! !
! !\ U1 !
!-------! O-----!-- Output
! !/ !
! [Z1]
! !! Y1 !
!----!!---------!
! !! !
--- ---
--- C1 --- C2
! !
GND GND


R1 serves mostly to bias the gate to its transition point. Its value can
vary widely so long as it is high enough to not degrade the Q and low
enough to provide a repeat in production.

U1 can be any type of inverting gate so long as it provides enough gain
and doesn't have too much phase lag. "Too much" depends on the purpose of
the oscillator and the nature of Z1. If U1 has no phase shift, the
tuned system sees an impedance of the same type as Z1 but the sign is
inverted. If the real part of all of the impedances seen in the tuned
system is negative, the circuit will oscillate.

Z1 controls the drive to the crystal. In most designs this is a resistor
but it can be a small (5pF) capacitor. Using a capacitor will allow
slower logic gates to be used. This is a good thing to remember if the
logic gate is a CD4000 series gate.

C1 and C2 are almost always made equal or nearly equal. C1 is actually in
parallel with the input capacitance of the logic gate. You can reduce C1
slightly because of this. You don't want the total capacitance on that
side of the crystal to ever be the smaller of the two however so don't
reduce it too much.

--

My rule of thumb is that Z1=R2 and that R2*C2 should produce 45° at
fo.

...Jim Thompson
 
P

Paul Burridge

Jan 1, 1970
0
It's even *smaller* for the 'HCU04 and the gain is down dramatically.

So everyone's agreed the 74HCU04 is a perfectly suitable active
element to use in a crystal oscillator?
I've come across another hex inverter in my stash of parts. It's a
SC74AC04PC. I seem to recall the "AC" has some significance. Does
anyone know if this would be equally suitable for oscillatory
purposes? Or how about a Schmitt trigger? Too much gain, I suppose?
Don't trifle with me, young man... I design a new microchip with a
XTAL oscillator at least once every three months...

Curious. You'd be better off sticking with PSpice. Should cut the
development time down significantly. ;-)
 
J

James Meyer

Jan 1, 1970
0
Ascii art:
R2
----/\/\/\-----
! !
! !\ U1 !
!-------! O-----!-- Output
! !/ !
! [Z1]
! !! Y1 !
!----!!---------!
! !! !
--- ---
--- C1 --- C2
! !
GND GND


--

My rule of thumb is that Z1=R2 and that R2*C2 should produce 45° at
fo.

...Jim Thompson

And the series value of C1 and C2 should come close to the value the
crystal manufacturer used when he specified the frequency of the crystal. If
you want the RF out to be close to the value marked on the crystal, that is.

Jim "The other one"
 
P

Paul Burridge

Jan 1, 1970
0
That appears to be a cockup. They probably meant to say that the circuit
can be used with frequencies in this range, but the crystal frequency
has to be selected accordingly, of course.

That does makes sense!
The NL27WZU04? Make sure you read the datasheet before you order,
because they're tiny. You don't want to hand-solder them unless you have
to.

Thanks!!! I *hate* those infernal SMDs. Think I'll stick with the good
old 74HCU04...
 
J

Jim Thompson

Jan 1, 1970
0
So everyone's agreed the 74HCU04 is a perfectly suitable active
element to use in a crystal oscillator?
Yes.

I've come across another hex inverter in my stash of parts. It's a
SC74AC04PC. I seem to recall the "AC" has some significance. Does
anyone know if this would be equally suitable for oscillatory
purposes?

No, it's buffered.
Or how about a Schmitt trigger? Too much gain, I suppose?

Schmitt triggers often don't start in XTAL oscillator applications.

That's just the chips with oscillators.
Curious. You'd be better off sticking with PSpice. Should cut the
development time down significantly. ;-)

Paul please place your head back up your ass, you look better that way
;-)

...Jim Thompson
 
W

Winfield Hill

Jan 1, 1970
0
Jim Thompson wrote...
Don't trifle with me, young man... I design a new microchip with a
XTAL oscillator at least once every three months... and the clients
kiss my feet... in a recent RF chip *only* my cells came out working
to specification ;-)

The question shouldn't be whether your oscillators work, but
whether they damage marginal crytals after extended use. :>)

Thanks,
- Win

whill_at_picovolt-dot-com
 
K

Ken Smith

Jan 1, 1970
0
Ascii art:
R1
----/\/\/\-----
! !
! !\ U1 !
!-------! O-----!-- Output
! !/ !
! [Z1]
! !! Y1 !
!----!!---------!
! !! !
--- ---
--- C1 --- C2
! !
GND GND
My rule of thumb is that Z1=R2 and that R2*C2 should produce 45° at
fo.

Yes, add the rule that (C1,C2) is the load capacitance and you usually
have a working design. Only if the gate is slow do you, often, have
trouble. In that case try Z1 = C3.
 
J

Jim Thompson

Jan 1, 1970
0
Jim Thompson wrote...

The question shouldn't be whether your oscillators work, but
whether they damage marginal crytals after extended use. :>)

Thanks,
- Win

whill_at_picovolt-dot-com

I have some GPS designs that have been working continuously for
probably 15 years without failure. (The *series* R sets the drive
level in inverter-style XTAL oscillators.)

I also have some NIC-based designs; and a number of serial-mode
designs; but they're *way* over the comprehension level of these
groups (require Algebra and thinking ;-)

...Jim Thompson
 
J

Jim Thompson

Jan 1, 1970
0
Ascii art:
R1
----/\/\/\-----
! !
! !\ U1 !
!-------! O-----!-- Output
! !/ !
! [Z1]
! !! Y1 !
!----!!---------!
! !! !
--- ---
--- C1 --- C2
! !
GND GND
My rule of thumb is that Z1=R2 and that R2*C2 should produce 45° at
fo.

Yes, add the rule that (C1,C2) is the load capacitance and you usually
have a working design. Only if the gate is slow do you, often, have
trouble. In that case try Z1 = C3.

That puzzles me... I'll have to run a Bode plot on that.

...Jim Thompson
 
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