electrical interface problem

Oh, thank you so much.... everything became quite clear as I had a look at
your drawings. I'm not sure if AC coupling like proposed by Jim Thompson in
MelanieCML.pdf is relly necessary. In my oppinion a "plain" resistor network
will do the job, so no need for AC coupling, right?

Regards, Mel




"qrk" <SpamTrap@spam.net> schrieb im Newsbeitrag
news:p5jep1t0jm99oupiciml5shum6ecqbq647@4ax.com...
On Wed, 7 Dec 2005 11:56:31 +0100, "Melanie Nasic"
quinn_the_esquimo@freenet.de> wrote:

Hi there,

I'm currently working on a board design and I have to interface two VLSI
chips to each other that don't have exactly compatible level
characteristics. To be more precise I have a differential CML output of
U(t)
= 1.8V +/- 0.400V * rect(t) being terminated 50 Ohm to Vref and a
non-standard differential Input which assumes U(t) = 1.15V +/- 0.125V *
rect(t) being 100 Ohm differentially terminated.
I first thought about AC coupling but that only kills my DC part but
doesn't
come to solve the problem of incompatible swings. Maybe a resistor network
would be the right choice but I am not so firm on that topic. Any help,
suggestions and calculation examples would be appreciated.

Many thanks in advance and best regards,

Melanie

See Message-ID: <0bjep1hbrm067tksqjgtoqanvme5c310tl@4ax.com> in
alt.binaries.schematic.electronic, subject "electrical interface
problem" for a PDF on this discussion.

See if the schematic in abse is what you are thinking of. If so, then
you might get by with an attenuator and a DC voltage divider. R4-R8 is
the attenuator which provides 10.1 dB of attenuation. R7, R8, and R10
form the DC voltage divider. R9 is the terminator for your mystery
device.

You can see in the simulation that your input common mode level is
1.15V and the signal level to the input (v(inp,inn))is 0.125Vpp. The
equivalent power supply voltage running your 50 Ohm source resistors,
VTTX, is 2.14 VDC which meets the Xilinx VTTX minimum supply of 1.8V.

Is this close to what you are thinking of?

---
Mark
 
Could someone please do me a favor and post something for me at Newsgroups:
alt.binaries.schematics.electronic.

I think I found a solution myself but I am not sure whether it will work. I
made a drawing and would like to discuss that but I'm not allowed to post at
Newsgroups: alt.binaries.schematics.electronic.

Thanks, Mel



"Melanie Nasic" <quinn_the_esquimo@freenet.de> schrieb im Newsbeitrag
news:dn92g0$r8v$1@mamenchi.zrz.TU-Berlin.DE...
Oh, thank you so much.... everything became quite clear as I had a look at
your drawings. I'm not sure if AC coupling like proposed by Jim Thompson
in MelanieCML.pdf is relly necessary. In my oppinion a "plain" resistor
network will do the job, so no need for AC coupling, right?

Regards, Mel




"qrk" <SpamTrap@spam.net> schrieb im Newsbeitrag
news:p5jep1t0jm99oupiciml5shum6ecqbq647@4ax.com...
On Wed, 7 Dec 2005 11:56:31 +0100, "Melanie Nasic"
quinn_the_esquimo@freenet.de> wrote:

Hi there,

I'm currently working on a board design and I have to interface two VLSI
chips to each other that don't have exactly compatible level
characteristics. To be more precise I have a differential CML output of
U(t)
= 1.8V +/- 0.400V * rect(t) being terminated 50 Ohm to Vref and a
non-standard differential Input which assumes U(t) = 1.15V +/- 0.125V *
rect(t) being 100 Ohm differentially terminated.
I first thought about AC coupling but that only kills my DC part but
doesn't
come to solve the problem of incompatible swings. Maybe a resistor
network
would be the right choice but I am not so firm on that topic. Any help,
suggestions and calculation examples would be appreciated.

Many thanks in advance and best regards,

Melanie

See Message-ID: <0bjep1hbrm067tksqjgtoqanvme5c310tl@4ax.com> in
alt.binaries.schematic.electronic, subject "electrical interface
problem" for a PDF on this discussion.

See if the schematic in abse is what you are thinking of. If so, then
you might get by with an attenuator and a DC voltage divider. R4-R8 is
the attenuator which provides 10.1 dB of attenuation. R7, R8, and R10
form the DC voltage divider. R9 is the terminator for your mystery
device.

You can see in the simulation that your input common mode level is
1.15V and the signal level to the input (v(inp,inn))is 0.125Vpp. The
equivalent power supply voltage running your 50 Ohm source resistors,
VTTX, is 2.14 VDC which meets the Xilinx VTTX minimum supply of 1.8V.

Is this close to what you are thinking of?

---
Mark
 
I think I found a solution to the post "electrical interface problem" but I
am not sure whether it will work. I made a drawing and would like to discuss
that but I'm not allowed to post at Newsgroups:
alt.binaries.schematics.electronic. Could someone help me, please?
I'm sorry for bothering you with this question but I'm out of ideas...

Best Regards,

Melanie Nasic
 
Melanie Nasic wrote:
Could someone please do me a favor and post something for me at Newsgroups:
alt.binaries.schematics.electronic.

I think I found a solution myself but I am not sure whether it will work. I
made a drawing and would like to discuss that but I'm not allowed to post at
Newsgroups: alt.binaries.schematics.electronic.
Email them to me, along with the thread title you want them labeled with.
 
On Wed, 07 Dec 2005 21:49:19 -0800, none <""doug\"@(none)"> wrote:

Since this is a xilinx related issue you might try to post this
question to comp.arch.fpga where the xilinx folks hang out.
Others have run into this and generally are very happy to give
answers. There are people from xilinx there as well to answer
questions.


[snip]

Also make sure of the rigidity of the 1.15V receiver common-mode spec.

This sounds like LVDS. This number is what the LVDS _transmitter_
puts out. Most of the _receivers_ (Fairchild, for instance, some
parts of which I've designed) can tolerate almost rail-to-rail input
common-mode.

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.
 
John, would you please be so kind to post my message with the binaries?!

Thanks a lot.

Mel


"John Popelish" <jpopelish@rica.net> schrieb im Newsbeitrag
news:XMadnSSYtdKx0QXeRVn-tg@adelphia.com...
Melanie Nasic wrote:
Could someone please do me a favor and post something for me at
Newsgroups: alt.binaries.schematics.electronic.

I think I found a solution myself but I am not sure whether it will work.
I made a drawing and would like to discuss that but I'm not allowed to
post at Newsgroups: alt.binaries.schematics.electronic.

Email them to me, along with the thread title you want them labeled with.
 
Melanie Nasic wrote:
John, would you please be so kind to post my message with the binaries?!

Thanks a lot.
Your message? Which one?
And I haven't received the binary files from you, yet.

Please email whatever binary files you wish to be posted, to me, along
with whatever text you want included, and I will post the whole thing
on A.B.S.E for you.

Mel


"John Popelish" <jpopelish@rica.net> schrieb im Newsbeitrag
news:XMadnSSYtdKx0QXeRVn-tg@adelphia.com...

Melanie Nasic wrote:

Could someone please do me a favor and post something for me at
Newsgroups: alt.binaries.schematics.electronic.

I think I found a solution myself but I am not sure whether it will work.
I made a drawing and would like to discuss that but I'm not allowed to
post at Newsgroups: alt.binaries.schematics.electronic.

Email them to me, along with the thread title you want them labeled with.
 
Hi John,

I've send them to you yesterday along with my email. I did it again for 3
minutes, could you please check? Maybe your mail program thought of me like
spam?! ;-)

Bye, Mel



"John Popelish" <jpopelish@rica.net> schrieb im Newsbeitrag
news:HvKdnbD-bLOwHQTeRVn-jA@adelphia.com...
Melanie Nasic wrote:
John, would you please be so kind to post my message with the binaries?!

Thanks a lot.

Your message? Which one?
And I haven't received the binary files from you, yet.

Please email whatever binary files you wish to be posted, to me, along
with whatever text you want included, and I will post the whole thing on
A.B.S.E for you.

Mel


"John Popelish" <jpopelish@rica.net> schrieb im Newsbeitrag
news:XMadnSSYtdKx0QXeRVn-tg@adelphia.com...

Melanie Nasic wrote:

Could someone please do me a favor and post something for me at
Newsgroups: alt.binaries.schematics.electronic.

I think I found a solution myself but I am not sure whether it will
work. I made a drawing and would like to discuss that but I'm not
allowed to post at Newsgroups: alt.binaries.schematics.electronic.

Email them to me, along with the thread title you want them labeled with.
 
Melanie Nasic wrote:
Hi John,

I've send them to you yesterday along with my email. I did it again for 3
minutes, could you please check? Maybe your mail program thought of me like
spam?! ;-)

Bye, Mel
Yes, they were in the trash. One of my filters caught them. Sorry.

Posted.
 
Hi,

I wonder what is the characteristic impedance of the circuit proposal
CMLmystery.pdf? I think this is the easiest way to do it but I want to be
sure that the CML transmitter still "sees" 50 Ohm line impedance. Any
comments on that?

Bye, Mel.




"John Popelish" <jpopelish@rica.net> schrieb im Newsbeitrag
news:78adnQe3mYkiGgTenZ2dnUVZ_tidnZ2d@adelphia.com...
Melanie Nasic wrote:
Hi John,

I've send them to you yesterday along with my email. I did it again for 3
minutes, could you please check? Maybe your mail program thought of me
like spam?! ;-)

Bye, Mel

Yes, they were in the trash. One of my filters caught them. Sorry.

Posted.
 
On Fri, 9 Dec 2005 16:27:44 +0100, "Melanie Nasic"
<quinn_the_esquimo@freenet.de> wrote:

Hi,

I wonder what is the characteristic impedance of the circuit proposal
CMLmystery.pdf? I think this is the easiest way to do it but I want to be
sure that the CML transmitter still "sees" 50 Ohm line impedance. Any
comments on that?

Bye, Mel.

[snip]

qrk/Mark implies that the transmitter can be operated with less output
common mode (as I suspected), so just use my version without caps, use
no divider to set output CM, just twiddle the pull-down R, as in
Mark's version until you get the +1.15V at the receiver. (My values
were line-match based.)

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.
 
Hi Jim,

don't the changes you suggest effect the line match (twiddling the pull-down
Rs)? Nonetheless I would like to calculate the impedance of Mark's circuit.
Any suggestions how I can find out about the impedance?

Bye Mel


"Jim Thompson" <To-Email-Use-The-Envelope-Icon@My-Web-Site.com> schrieb im
Newsbeitrag news:dq8jp1hkpvvr9l2og58amk4m08jdrht3nf@4ax.com...
On Fri, 9 Dec 2005 16:27:44 +0100, "Melanie Nasic"
quinn_the_esquimo@freenet.de> wrote:

Hi,

I wonder what is the characteristic impedance of the circuit proposal
CMLmystery.pdf? I think this is the easiest way to do it but I want to be
sure that the CML transmitter still "sees" 50 Ohm line impedance. Any
comments on that?

Bye, Mel.

[snip]

qrk/Mark implies that the transmitter can be operated with less output
common mode (as I suspected), so just use my version without caps, use
no divider to set output CM, just twiddle the pull-down R, as in
Mark's version until you get the +1.15V at the receiver. (My values
were line-match based.)

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.
 
On Fri, 9 Dec 2005 16:42:15 +0100, "Melanie Nasic"
<quinn_the_esquimo@freenet.de> wrote:

Hi Jim,

don't the changes you suggest effect the line match (twiddling the pull-down
Rs)? Nonetheless I would like to calculate the impedance of Mark's circuit.
Any suggestions how I can find out about the impedance?

Bye Mel

[snip]

Oooops! In Mark's version they do affect impedance. Use a single
midpoint pull-down... my divider without the upper resistor.

Insert transmission line into your drawing, then you can visualize
source and termination easier.

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.
 
On Fri, 9 Dec 2005 16:42:15 +0100, "Melanie Nasic"
<quinn_the_esquimo@freenet.de> wrote:

"Jim Thompson" <To-Email-Use-The-Envelope-Icon@My-Web-Site.com> schrieb im
Newsbeitrag news:dq8jp1hkpvvr9l2og58amk4m08jdrht3nf@4ax.com...
On Fri, 9 Dec 2005 16:27:44 +0100, "Melanie Nasic"
quinn_the_esquimo@freenet.de> wrote:

Hi,

I wonder what is the characteristic impedance of the circuit proposal
CMLmystery.pdf? I think this is the easiest way to do it but I want to be
sure that the CML transmitter still "sees" 50 Ohm line impedance. Any
comments on that?

Bye, Mel.

[snip]

qrk/Mark implies that the transmitter can be operated with less output
common mode (as I suspected), so just use my version without caps, use
no divider to set output CM, just twiddle the pull-down R, as in
Mark's version until you get the +1.15V at the receiver. (My values
were line-match based.)

...Jim Thompson

Hi Jim,

don't the changes you suggest effect the line match (twiddling the pull-down
Rs)? Nonetheless I would like to calculate the impedance of Mark's circuit.
Any suggestions how I can find out about the impedance?

Bye Mel
See posting in abse, "electrical interface problem - Melanie_attn.gif"

I'm not quite sure of your electronics knowledge, so I'll show the
steps to finding the equivalent input impedance of the network. This
will work for both mine and Jim's circuits since they are essentially
the same.

Item 1, in Melanie_attn.gif, is the full network. R7 has no effect on
the differential impedance. R7 is only to set the DC bias point for
your receiver. In Jim's schematic (MelanieCML.pdf), the bias parts R8,
R9, C3 have no effect on the differential impedance.

Item 2 shows the network without R7 since the node R7,R4,R5 is at zero
AC potential.

Items 3, 4, and 5 show the step by step simplification of the network
to get the equivalent input impedance of the network. The double bars
(||) mean parallel, i.e. R1||Rb means R1 in parallel with Rb.

Your Rocket IO wants to drive an impedance of 100 Ohms differential,
not 50 Ohms.

The difference between Jim Thompson's circuit and mine:

Jim thinks you need to add termination resistor at the input to the
receiver (the pair of 49.9 Ohm resistors). Jim's circuit gives 17.6 dB
attenuation which might be too much.

I think that the termination resistor is included on the receiver
silicon die. That would be R6 in the schematic. My circuit gives 10.1
dB attenuation. I also use non-standard resistor values only because
I'm too lazy to look up standard values. Boy, it's nice being
unemployed and lazy!

Before this discussion goes on, we really need to know what your
receiver device is. John Larkin brings up some valid points about
simplification of the interface which could null this discussion of
level shifting and attenuator pads.

---
Mark
 
Hi Mark,

thank you so much for your elaborations. Everything became pretty clear for
me now. So I got approx 100 Ohm for the resistor network now. But as I want
to transmit data in the Gigahertz range aren't I supposed to take a complex
HF impedance value into account? Are there any further calculations I have
to make to gain a conclusion whether this circuit will work for signals with
data rates of 3 Gbps.

Regards, Melanie



"qrk" <SpamTrap@spam.net> schrieb im Newsbeitrag
news:gopjp1hc8rc6s34urakcth28fah5qhqlb7@4ax.com...
On Fri, 9 Dec 2005 16:42:15 +0100, "Melanie Nasic"
quinn_the_esquimo@freenet.de> wrote:

"Jim Thompson" <To-Email-Use-The-Envelope-Icon@My-Web-Site.com> schrieb im
Newsbeitrag news:dq8jp1hkpvvr9l2og58amk4m08jdrht3nf@4ax.com...
On Fri, 9 Dec 2005 16:27:44 +0100, "Melanie Nasic"
quinn_the_esquimo@freenet.de> wrote:

Hi,

I wonder what is the characteristic impedance of the circuit proposal
CMLmystery.pdf? I think this is the easiest way to do it but I want to
be
sure that the CML transmitter still "sees" 50 Ohm line impedance. Any
comments on that?

Bye, Mel.

[snip]

qrk/Mark implies that the transmitter can be operated with less output
common mode (as I suspected), so just use my version without caps, use
no divider to set output CM, just twiddle the pull-down R, as in
Mark's version until you get the +1.15V at the receiver. (My values
were line-match based.)

...Jim Thompson

Hi Jim,

don't the changes you suggest effect the line match (twiddling the
pull-down
Rs)? Nonetheless I would like to calculate the impedance of Mark's
circuit.
Any suggestions how I can find out about the impedance?

Bye Mel

See posting in abse, "electrical interface problem - Melanie_attn.gif"

I'm not quite sure of your electronics knowledge, so I'll show the
steps to finding the equivalent input impedance of the network. This
will work for both mine and Jim's circuits since they are essentially
the same.

Item 1, in Melanie_attn.gif, is the full network. R7 has no effect on
the differential impedance. R7 is only to set the DC bias point for
your receiver. In Jim's schematic (MelanieCML.pdf), the bias parts R8,
R9, C3 have no effect on the differential impedance.

Item 2 shows the network without R7 since the node R7,R4,R5 is at zero
AC potential.

Items 3, 4, and 5 show the step by step simplification of the network
to get the equivalent input impedance of the network. The double bars
(||) mean parallel, i.e. R1||Rb means R1 in parallel with Rb.

Your Rocket IO wants to drive an impedance of 100 Ohms differential,
not 50 Ohms.

The difference between Jim Thompson's circuit and mine:

Jim thinks you need to add termination resistor at the input to the
receiver (the pair of 49.9 Ohm resistors). Jim's circuit gives 17.6 dB
attenuation which might be too much.

I think that the termination resistor is included on the receiver
silicon die. That would be R6 in the schematic. My circuit gives 10.1
dB attenuation. I also use non-standard resistor values only because
I'm too lazy to look up standard values. Boy, it's nice being
unemployed and lazy!

Before this discussion goes on, we really need to know what your
receiver device is. John Larkin brings up some valid points about
simplification of the interface which could null this discussion of
level shifting and attenuator pads.

---
Mark
 
On Fri, 16 Dec 2005 20:41:02 +0100, "Melanie Nasic"
<quinn_the_esquimo@freenet.de> wrote:

"qrk" <SpamTrap@spam.net> schrieb im Newsbeitrag
news:gopjp1hc8rc6s34urakcth28fah5qhqlb7@4ax.com...
On Fri, 9 Dec 2005 16:42:15 +0100, "Melanie Nasic"
quinn_the_esquimo@freenet.de> wrote:

"Jim Thompson" <To-Email-Use-The-Envelope-Icon@My-Web-Site.com> schrieb im
Newsbeitrag news:dq8jp1hkpvvr9l2og58amk4m08jdrht3nf@4ax.com...
On Fri, 9 Dec 2005 16:27:44 +0100, "Melanie Nasic"
quinn_the_esquimo@freenet.de> wrote:

Hi,

I wonder what is the characteristic impedance of the circuit proposal
CMLmystery.pdf? I think this is the easiest way to do it but I want to
be
sure that the CML transmitter still "sees" 50 Ohm line impedance. Any
comments on that?

Bye, Mel.

[snip]

qrk/Mark implies that the transmitter can be operated with less output
common mode (as I suspected), so just use my version without caps, use
no divider to set output CM, just twiddle the pull-down R, as in
Mark's version until you get the +1.15V at the receiver. (My values
were line-match based.)

...Jim Thompson

Hi Jim,

don't the changes you suggest effect the line match (twiddling the
pull-down
Rs)? Nonetheless I would like to calculate the impedance of Mark's
circuit.
Any suggestions how I can find out about the impedance?

Bye Mel

See posting in abse, "electrical interface problem - Melanie_attn.gif"

I'm not quite sure of your electronics knowledge, so I'll show the
steps to finding the equivalent input impedance of the network. This
will work for both mine and Jim's circuits since they are essentially
the same.

Item 1, in Melanie_attn.gif, is the full network. R7 has no effect on
the differential impedance. R7 is only to set the DC bias point for
your receiver. In Jim's schematic (MelanieCML.pdf), the bias parts R8,
R9, C3 have no effect on the differential impedance.

Item 2 shows the network without R7 since the node R7,R4,R5 is at zero
AC potential.

Items 3, 4, and 5 show the step by step simplification of the network
to get the equivalent input impedance of the network. The double bars
(||) mean parallel, i.e. R1||Rb means R1 in parallel with Rb.

Your Rocket IO wants to drive an impedance of 100 Ohms differential,
not 50 Ohms.

The difference between Jim Thompson's circuit and mine:

Jim thinks you need to add termination resistor at the input to the
receiver (the pair of 49.9 Ohm resistors). Jim's circuit gives 17.6 dB
attenuation which might be too much.

I think that the termination resistor is included on the receiver
silicon die. That would be R6 in the schematic. My circuit gives 10.1
dB attenuation. I also use non-standard resistor values only because
I'm too lazy to look up standard values. Boy, it's nice being
unemployed and lazy!

Before this discussion goes on, we really need to know what your
receiver device is. John Larkin brings up some valid points about
simplification of the interface which could null this discussion of
level shifting and attenuator pads.

Hi Mark,

thank you so much for your elaborations. Everything became pretty clear for
me now. So I got approx 100 Ohm for the resistor network now. But as I want
to transmit data in the Gigahertz range aren't I supposed to take a complex
HF impedance value into account? Are there any further calculations I have
to make to gain a conclusion whether this circuit will work for signals with
data rates of 3 Gbps.

Regards, Melanie
Yes, you need to consider complex impedances, or more likely how to
minimize the reactive part. You don't say what sort of path your
signal must travel between driver and receiver. If only a couple cm on
the circuit board, this should be manageable with careful layout
practices. If your signal goes through connectors, difficulties can
arise. I suggest getting a copy of Howard Johnson's book, "High-Speed
Digital Design: A Handbook of Black Magic" (ISBN: 0133957241). It has
lots of practical information. Good WC reading. Your uni library may
have a copy. Look for application notes on routing high-speed lines on
printed circuit boards. You usually find these articles for 10 Gb/s
I/O.

If you have access to simulation tools like Mentor's HyperLynx I would
suggest learning how to use these tools. You can simulate your signal
path and see what sort of problems you may run into. Xilinx has Mentor
binary models of the output structure and packaging.

To model the differential impedance of the traces on your circuit
board, you can use ATLC <http://atlc.sourceforge.net/>. It's a
graphically instructive program, well worth looking at the results.
Version 4.4.4 is compiled for Windows. It is a user unfriendly
program, but free and multi-platform.
There are other programs that calculate differential impedance like
Polar's very expensive programs <http://www.polarinstruments.com/>.
Many printed circuit board manufacturing houses use the Polar
programs. You can call your PCB fabricator and ask them for the trace
dimensions. Once you know your PCB stackup, you can talk to your PCB
fabricator. This is normal communication between the designer and
fabricator when dealing with high-speed signals.

---
Mark

---
Mark
 

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