Where to get high speed ADC's and DACs

[John Larkin]

On Mon, 02 Sep 2013 18:40:18 -0700, RobertMacy <robert.a.macy@gmail.com> wrote:

On Mon, 02 Sep 2013 13:02:10 -0700, Phil Hobbs
pcdhSpamMeSenseless@electrooptical.net> wrote:

...snip...
Sorry? Surely anybody designing high performance A/D subsystems knows
that 2**10 ~= 1000, that the RMS quantization noise of an ideal
digitizer is 1/sqrt(12) of the LSB, and that the noise is more or less
white? You can derive it yourself in about three lines.

All I'm saying is:

1. 1 LSB = FSR/2**N

2. 2**20 ~= 10**6.

Therefore, 1 LSB ~= 5 uV.

3. RMS quantization noise is 1/sqrt(12)* 1 LSB ~= 5 uV/3.46 ~= 1.4 uV,
spread out evenly over the Nyquist interval.

4. The Nyquist bandwidth is 50 MHz (not 100), so given that the noise is
white, the quantization noise PSD is 1.4 uV/sqrt(50 MHz) ~=200
pV/sqrt(Hz).

5. To get that many bits to stay reasonably still, your RMS input noise
has to be well below that. Even slow delta-sigmas are hard pressed to
reach a genuine 20 bits, and most of them actually crap out around 18 or
19, AFAICT.

6. IOW, good luck.

That's just engineering rules of thumb.

Cheers

Phil Hobbs




Sorry? LOL! I actually read that out loud for effect.

In the present 24 bit Data Acquisition system I get something like a
measured 22.5 bits, meaning not quite 23, but better than 22. And, I
couldn't believe I actually ran up to 91% of the Nyquist rate! but I
backed it off down to 89% so didn't see ANY limiting effects.

How are you measuring that 22.5 bits?


--

John Larkin Highland Technology Inc
www.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

 

[Jan Panteltje]

On a sunny day (Mon, 02 Sep 2013 19:34:12 -0700) it happened RobertMacy
<robert.a.macy@gmail.com> wrote in <op.w2tdfanm2cx0wh@ajm>:

On Mon, 02 Sep 2013 18:55:03 -0700, Phil Hobbs
pcdhSpamMeSenseless@electrooptical.net> wrote:

...snip...
So apart from posturing, what do you actually disagree with, and why?

Cheers

Phil Hobbs


Don't take umbrage. I don't think I disagree. I'm sure everything you said
was correct, and as you said, they are rules of thumb for engineering.
It's just that nothing there was useful for me. Well, at first glance
seemed like almost nothing. I couldn't see how to apply all you said to
what I'm doing.

In retrospect, not nothing, because you made me THINK. Your comment, 'good
luck with that' challenged my thinking. Perhaps, I was wrong in what I was
doing. So I went back to my basic system and had to recreate all my
equations from ground zero back up to try operating at 10-100MS/s rates.
Why ground zero? I don't even trust my own equations. I question
EVERYTHING. So four hours of intensive effort later, I verified a lot and
feel better, thanks. Plus, you triggered a topological approach using
'available' parts that had NEVER occurred to me until just now. And, all
those recreated equations allowed me to verify [not rigorously, but at
least a bit empirically] that the approach would be possible. Even better,
for my requirements, I might even be able to get more than 24 bits at
100MS/s.

Gone brain dead here, have I sent any images of results using the present
system to you? If not, send me an email address, and I'll send a couple.

Hey, I'd like to see some too, and I am sure more people here would.
Website?
If you cannot find a site I can put them on my server in /pub/

 

On Mon, 02 Sep 2013 16:02:10 -0400, Phil Hobbs
<pcdhSpamMeSenseless@electrooptical.net> wrote:

On 9/2/2013 11:36 AM, RobertMacy wrote:
On Sun, 01 Sep 2013 10:58:17 -0700, Phil Hobbs
pcdhSpamMeSenseless@electrooptical.net> wrote:

...snip...

Well, if we generously assume that an ADC that fast could have a 5V
input range, 1 LSB at 20 bits is 5 uV, and the quantization noise is
1/sqrt(12) times that, or 1.4 uV. In a 100 MHz bandwidth, that's
140 pV/sqrt(Hz). In real life, the input structure would have to be
several times faster than that in order to settle to that accuracy in
the time available, putting the maximum input noise down in the
50 pV/sqrt(Hz) range, not counting the effects of input capacitance.

Good luck with that.

Cheers

Phil Hobbs


Irritated yelling mode...You have just demonstrated once again WHY I
re-derive EVERYTHING! I NEVER trust 'cookbook' equations, especially
after getting severely burnt by an article in EDN showing 'cookbook'
values for a simple 'what's its name? filter [the simple 5-pole low pass
type using two 2N3904's in series]. Whereupon, I was forced to rederive
ALL the values for both Butterworth AND Tschebyshev(sp?) *and* using an
HP calculator with reverse polish input (spit, spit, curse begone!) I
'optimized' a response to obtain values and voila! worked. But that
little effort caught me on a late Friday [deadline Monday morning] to
make a filter that worked! All weekend!

I HATE PACKAGED FORMULAS!!! I have NO idea what these numbers you gave
me should mean to me.

Sorry? Surely anybody designing high performance A/D subsystems knows
that 2**10 ~= 1000, that the RMS quantization noise of an ideal
digitizer is 1/sqrt(12) of the LSB, and that the noise is more or less
white? You can derive it yourself in about three lines.

All I'm saying is:

1. 1 LSB = FSR/2**N

2. 2**20 ~= 10**6.

Therefore, 1 LSB ~= 5 uV.

3. RMS quantization noise is 1/sqrt(12)* 1 LSB ~= 5 uV/3.46 ~= 1.4 uV,
spread out evenly over the Nyquist interval.

If you have 1 LSB hum at 50/60 Hz it will definitely not spread nicely
around the Nyquist interval. Adding dithering noise artificially or
natural thermal noise about 1 LSB, then the quantization noise will
spread out.

4. The Nyquist bandwidth is 50 MHz (not 100), so given that the noise is
white, the quantization noise PSD is 1.4 uV/sqrt(50 MHz) ~=200 pV/sqrt(Hz).

5. To get that many bits to stay reasonably still, your RMS input noise
has to be well below that. Even slow delta-sigmas are hard pressed to
reach a genuine 20 bits, and most of them actually crap out around 18 or
19, AFAICT.

The thermal noise voltage from a 50 ohm resistor at room temperature
at 50 MHz BW would be 6 uV. If the ADC range is only +/-1V, that would
be about 19 clean bits.

6. IOW, good luck.

That's just engineering rules of thumb.

The OP claimed 22.5 bits with current system, I might even believe
him, if we are talking about measuring a steady state DC signal. The
thermal noise will dither the quantization noise and after heavy low
pass filtering (BW << 50 MHz) you should get quite decent figures.

In an ordinary CD system with 16 bit coding (nominally 96 dB SNR) and
44.1 k sampling rate feeding a low frequency about 1 kHz sine wave to
a variable attenuator and then recording it, while constantly reducing
the amplitude. In this "Fade to Noise" test, the tone is still
detectable at -110 dB to -120 dB levels thanks to the dither. Of
course one could view this as an oversampling example.

 

[John Devereux]

RobertMacy <robert.a.macy@gmail.com> writes:

On Mon, 02 Sep 2013 18:55:03 -0700, Phil Hobbs
pcdhSpamMeSenseless@electrooptical.net> wrote:

...snip...
So apart from posturing, what do you actually disagree with, and why?

Cheers

Phil Hobbs


Don't take umbrage. I don't think I disagree. I'm sure everything you
said was correct, and as you said, they are rules of thumb for
engineering. It's just that nothing there was useful for me. Well, at
first glance seemed like almost nothing. I couldn't see how to apply
all you said to what I'm doing.

In retrospect, not nothing, because you made me THINK. Your comment,
good luck with that' challenged my thinking. Perhaps, I was wrong in
what I was doing. So I went back to my basic system and had to
recreate all my equations from ground zero back up to try operating at
10-100MS/s rates. Why ground zero? I don't even trust my own
equations. I question EVERYTHING. So four hours of intensive effort
later, I verified a lot and feel better, thanks. Plus, you triggered a
topological approach using 'available' parts that had NEVER occurred
to me until just now. And, all those recreated equations allowed me to
verify [not rigorously, but at least a bit empirically] that the
approach would be possible. Even better, for my requirements, I might
even be able to get more than 24 bits at 100MS/s.

Gone brain dead here, have I sent any images of results using the
present system to you? If not, send me an email address, and I'll send
a couple.

I would like to see them too. Do they differ from a system where the
lower few bits are generated by random noise? A sigma-delta ADC might
have a 40MHz clock, so internally there is a 40MHz 1-bit ADC, that with
sufficient processing turns into a 24-bit result. But that does not make
a 40MHz 24-bit ADC either.


--

John Devereux

 

[Phil Hobbs]

On 09/03/2013 02:46 AM, upsidedown@downunder.com wrote:

On Mon, 02 Sep 2013 16:02:10 -0400, Phil Hobbs
pcdhSpamMeSenseless@electrooptical.net> wrote:

On 9/2/2013 11:36 AM, RobertMacy wrote:
On Sun, 01 Sep 2013 10:58:17 -0700, Phil Hobbs
pcdhSpamMeSenseless@electrooptical.net> wrote:

...snip...

Well, if we generously assume that an ADC that fast could have a 5V
input range, 1 LSB at 20 bits is 5 uV, and the quantization noise is
1/sqrt(12) times that, or 1.4 uV. In a 100 MHz bandwidth, that's
140 pV/sqrt(Hz). In real life, the input structure would have to be
several times faster than that in order to settle to that accuracy in
the time available, putting the maximum input noise down in the
50 pV/sqrt(Hz) range, not counting the effects of input capacitance.

Good luck with that.

Cheers

Phil Hobbs


Irritated yelling mode...You have just demonstrated once again WHY I
re-derive EVERYTHING! I NEVER trust 'cookbook' equations, especially
after getting severely burnt by an article in EDN showing 'cookbook'
values for a simple 'what's its name? filter [the simple 5-pole low pass
type using two 2N3904's in series]. Whereupon, I was forced to rederive
ALL the values for both Butterworth AND Tschebyshev(sp?) *and* using an
HP calculator with reverse polish input (spit, spit, curse begone!) I
'optimized' a response to obtain values and voila! worked. But that
little effort caught me on a late Friday [deadline Monday morning] to
make a filter that worked! All weekend!

I HATE PACKAGED FORMULAS!!! I have NO idea what these numbers you gave
me should mean to me.

Sorry? Surely anybody designing high performance A/D subsystems knows
that 2**10 ~= 1000, that the RMS quantization noise of an ideal
digitizer is 1/sqrt(12) of the LSB, and that the noise is more or less
white? You can derive it yourself in about three lines.

All I'm saying is:

1. 1 LSB = FSR/2**N

2. 2**20 ~= 10**6.

Therefore, 1 LSB ~= 5 uV.

3. RMS quantization noise is 1/sqrt(12)* 1 LSB ~= 5 uV/3.46 ~= 1.4 uV,
spread out evenly over the Nyquist interval.

If you have 1 LSB hum at 50/60 Hz it will definitely not spread nicely
around the Nyquist interval. Adding dithering noise artificially or
natural thermal noise about 1 LSB, then the quantization noise will
spread out.

4. The Nyquist bandwidth is 50 MHz (not 100), so given that the noise is
white, the quantization noise PSD is 1.4 uV/sqrt(50 MHz) ~=200 pV/sqrt(Hz).

5. To get that many bits to stay reasonably still, your RMS input noise
has to be well below that. Even slow delta-sigmas are hard pressed to
reach a genuine 20 bits, and most of them actually crap out around 18 or
19, AFAICT.

The thermal noise voltage from a 50 ohm resistor at room temperature
at 50 MHz BW would be 6 uV. If the ADC range is only +/-1V, that would
be about 19 clean bits.

The noise bandwidth will be a bit wider than that, because the input
structure has to settle to full accuracy to get a clean conversion. So
to be generous let's say 100 MHz bandwidth, i.e. 9 uV for a 50-ohm resistor.

If you want the bit to be vaguely clean, even assuming that the mean
input voltage is right in the centre of the interval, you need an LSB
of, say, 6 times the RMS.

That gets you a noise transition (false alarm) rate of

FAR = 2 B /sqrt(3)*exp(-0.5*[(0.5 LSB)/Vrms]**2) .

For Vrms = LSB/6, the exponent is -4.5, which for a 100 MHz bandwidth
comes out to a 1.3 MHz FAR, assuming that the mean value is exactly
halfway between transitions, which is the best case. (The formula is
from Rice, with an additional factor of 2 because there are two
equidistant thresholds. I had to go through this stuff in absolutely
gory detail some years back in my particle detection work.)

Even with a false alarm rate like that (more than 1% of the samples will
be wrong), you wind up with a 55 uV LSB, which is 15.2 bits at +-1V.

6. IOW, good luck.

That's just engineering rules of thumb.

The OP claimed 22.5 bits with current system, I might even believe
him, if we are talking about measuring a steady state DC signal. The
thermal noise will dither the quantization noise and after heavy low
pass filtering (BW << 50 MHz) you should get quite decent figures.

Lowpass filtering the daylights out of practically anything except 1/f
noise--including a PRBS--will make clean bits at the rate of one for
every two octaves of bandwidth reduction. That's not a 100 MHz ADC.

In an ordinary CD system with 16 bit coding (nominally 96 dB SNR) and
44.1 k sampling rate feeding a low frequency about 1 kHz sine wave to
a variable attenuator and then recording it, while constantly reducing
the amplitude. In this "Fade to Noise" test, the tone is still
detectable at -110 dB to -120 dB levels thanks to the dither. Of
course one could view this as an oversampling example.

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

hobbs at electrooptical dot net
http://electrooptical.net

 

[RobertMacy]

On Mon, 02 Sep 2013 23:44:21 -0700, Jan Panteltje
<pNaonStpealmtje@yahoo.com> wrote:

...snip...

Hey, I'd like to see some too, and I am sure more people here would.
Website?
If you cannot find a site I can put them on my server in /pub/

Thank you for your kind offer. For me, I've NEVER gotten Dropbox to work
nor most of those 'public post your picture' sites. Can't even pick up
pictures from Dropbox half the time.

 

[RobertMacy]

On Tue, 03 Sep 2013 00:24:20 -0700, John Devereux <john@devereux.me.uk>
wrote:

...snip...

I would like to see them too. Do they differ from a system where the
lower few bits are generated by random noise? A sigma-delta ADC might
have a 40MHz clock, so internally there is a 40MHz 1-bit ADC, that with
sufficient processing turns into a 24-bit result. But that does not make
a 40MHz 24-bit ADC either.

will do.

don't know.

 

[Jan Panteltje]

On a sunny day (Tue, 03 Sep 2013 10:09:51 -0700) it happened RobertMacy
<robert.a.macy@gmail.com> wrote in <op.w2uhyqcw2cx0wh@ajm>:

On Mon, 02 Sep 2013 23:44:21 -0700, Jan Panteltje
pNaonStpealmtje@yahoo.com> wrote:

...snip...

Hey, I'd like to see some too, and I am sure more people here would.
Website?
If you cannot find a site I can put them on my server in /pub/

Thank you for your kind offer. For me, I've NEVER gotten Dropbox to work
nor most of those 'public post your picture' sites. Can't even pick up
pictures from Dropbox half the time.

just email it to p a n t e l t j e @ y a h o o . c o m (n ospaces) and I will put them up
and give a link.

 

[Tim Wescott]

On Mon, 02 Sep 2013 12:50:52 -0700, Lasse Langwadt Christensen wrote:

On Monday, September 2, 2013 8:26:50 PM UTC+2, Tim Wescott wrote:
On Sun, 01 Sep 2013 13:57:16 -0700, John Larkin wrote:



On Sun, 01 Sep 2013 10:45:34 -0700, RobertMacy
robert.a.macy@gmail.com

wrote:



On Sun, 01 Sep 2013 09:44:40 -0700, John Larkin

jjlarkin@highnotlandthistechnologypart.com> wrote:



On Sun, 01 Sep 2013 09:33:15 -0700, RobertMacy

robert.a.macy@gmail.com

wrote:



Where do I get 24 bit high speed ADC and DAC systems out to 10MHz?



Or, turn around how much digitization can I get out to 100MHz
today?

20 bits?



This is about right:



http://www.linear.com/designtools/hsadcs.php





We use their 250 MHz, 12-bit LVDS ADC and it's pretty good.



https://dl.dropboxusercontent.com/u/53724080/PCBs/ESM_rev_B.jpg



A 20 or 24-bit ADC, at 100 MHz, probably isn't useful. Wideband
noise

would trash a lot of LSBs.









John,



You have no idea how much I respect Linear and their products. but I
was

talking about 20+bits not the insignificant 12 bit range.



They have 16 bits at 185 Ms/s.





[I recently did 18 bits at 500MHz - Jim Williams would have been
proud,

NEVER AGAIN!!!



From simulations, I need 20+ bits, else quantization noise eats me

alive!



At 500 MHz bandwidth (reasonable s/h bw for a 500 Ms/s ADC) a 50 ohm

resistor makes 20 uV RMS Johnson noise, and I doubt that any actual

front-end amp will be anywhere close to that, probably several times

worse. Seems to me that you'll have many LSBs of noise, which may be
OK

if downstream processing is essentially narrowband, like some RF
stuff.



John's nailed it.



The determining factor is the noise in the front end of the ADC, and

that's going to be much higher than Johnson noise for a monolithic
device.



There may be some ultra-boutique hybrid parts out there that can extend

this, but they'll still be limited by the noise in the comparator and
the

need for a high bit-count DAC.



If you really need a 144dB full scale to LSB ratio before the noise

starts interfering with your measurement, then at 10MHz you're probably

screwed. If there is some device that can do this, it's almost
certainly

not a chip.



But -- what do you really need? Are you really using the full
bandwidth

of this thing, or are you extracting some narrower-bandwidth signal out

of it? Averaging the output of an ADC can do wonders for the precision

of the measurement.


just like a deltasigma adc, but you just trade precision for bandwidth

Using a noisy ADC and filtering is worse than using a delta sigma,
because the noise isn't shaped away from the region you're interested in.

But the same precision vs. bandwidth tradeoff is there.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com