electrical interface problem

M

Melanie Nasic

Guest
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
 
Melanie Nasic 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
Click to expand...
You need about a 10dB pad. A suitable method, about 50 ohm in and out
(view with fixed font)

High drive output

R1 Low drive input
Out + --------- /\/\/\/\ --------- In+
/ /
R2 \ \ R3
/ /
Out - -------- /\/\/\/\ ------------- In-
R4

Where R2 = R3 = 62 ohm
R1 = R4 = 68 ohm

Vin -> 0.4Vp-p, Vout -> .125Vp-p, Rin = 47.2 ohm, Rout = 47.2 ohm

Cheers

PeteS
 
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
Click to expand...
May I presume that Vref is +2.2V and the CML is open-collector pulling
current through the 50 ohm terminations to make the ą400mV signals?
8mA alternating from each collector?

Can additional current be drawn from Vref?

...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 Wed, 07 Dec 2005 09:00:07 -0700, Jim Thompson
<To-Email-Use-The-Envelope-Icon@My-Web-Site.com> wrote:

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


May I presume that Vref is +2.2V and the CML is open-collector pulling
current through the 50 ohm terminations to make the ą400mV signals?
8mA alternating from each collector?

Can additional current be drawn from Vref?

...Jim Thompson
Click to expand...
I think CML is usually a unidirectional current sink, an open
collector, pulling about 16 mA for the full-swing version, basicly an
ECL sort of stage without the emitter follower. Some have an internal
termination to Vcc, some don't.

Somebody correct me if I'm wrong.


Melanie may be able to do a direct connection, depending on the
common-mode specs of the receiver. We need more detail.


John
 
On Wed, 07 Dec 2005 08:29:53 -0800, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 07 Dec 2005 09:00:07 -0700, Jim Thompson
To-Email-Use-The-Envelope-Icon@My-Web-Site.com> wrote:

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


May I presume that Vref is +2.2V and the CML is open-collector pulling
current through the 50 ohm terminations to make the ą400mV signals?
8mA alternating from each collector?

Can additional current be drawn from Vref?

...Jim Thompson

I think CML is usually a unidirectional current sink, an open
collector, pulling about 16 mA for the full-swing version, basicly an
ECL sort of stage without the emitter follower. Some have an internal
termination to Vcc, some don't.

Somebody correct me if I'm wrong.


Melanie may be able to do a direct connection, depending on the
common-mode specs of the receiver. We need more detail.


John
Click to expand...
ą400mV differential would imply an 8mA current source, steered by a
diff-pair. But sometimes people get balled up in confusion about
differential peak and peak-to-peak.

But 16mA _would_ give the conventional 800mV single-ended output.

I don't know what is the real spec, since I'm always on-chip and run
as little as 150mV P-P differential... gotta keep those swings down
when you're doing 3GHz ;-)

...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 PeteS,

thanks for your reply. I don't think that would work because the CML output
stage "must see" 50 ohm transmission lines and I guess it would not be
allowed to terminate them with 62 ohm in the middle?

Bye Mel


"PeteS" <ps@fleetwoodmobile.com> schrieb im Newsbeitrag
news:1133957459.889278.149180@f14g2000cwb.googlegroups.com...
Melanie Nasic 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

You need about a 10dB pad. A suitable method, about 50 ohm in and out
(view with fixed font)

High drive output

R1 Low drive input
Out + --------- /\/\/\/\ --------- In+
/ /
R2 \ \ R3
/ /
Out - -------- /\/\/\/\ ------------- In-
R4

Where R2 = R3 = 62 ohm
R1 = R4 = 68 ohm

Vin -> 0.4Vp-p, Vout -> .125Vp-p, Rin = 47.2 ohm, Rout = 47.2 ohm

Cheers

PeteS
Click to expand...
 
On Wed, 7 Dec 2005 17:52:33 +0100, "Melanie Nasic"
<quinn_the_esquimo@freenet.de> wrote:

Hi PeteS,

thanks for your reply. I don't think that would work because the CML output
stage "must see" 50 ohm transmission lines and I guess it would not be
allowed to terminate them with 62 ohm in the middle?

Bye Mel
Click to expand...
Is the signal at the end of these "transmission lines" your specified
ą400mV? Source and end termination _would_ account for the factor of
1/2 that John Larkin and I have been fretting over.

Does the far end have to terminate to Vref, or would 100 ohm
differential suffice?

...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 all,

thanks for your fast reply. Vref is in fact +2.5V and I'm not sure whether
CML is open-collector and how much current is alternating from each
collector. Maybe a look at http://www.xilinx.com/bvdocs/userguides/ug024.pdf
(page 103) will help but I am not so firm on those electrical specs. You
would help me a lot by explaining that to me, though. :)
As transmitter I am using Xilinx' RocketIOs but their swing is higher than
the maximum receiver swing (+/- 400mV instead of +/- 125mV) and the receiver
is NOT supposed to exept some other common mode range than the 1.15 Volt
(that's part of the specification: the goal is to achieve nearly exactly the
U(t) given in my first mail).
Many thanks in advance....

Regards, Mel


"Jim Thompson" <To-Email-Use-The-Envelope-Icon@My-Web-Site.com> schrieb im
Newsbeitrag news:r54ep151ai6kh9pus10uvejd8e7po4a060@4ax.com...
On Wed, 07 Dec 2005 08:29:53 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 07 Dec 2005 09:00:07 -0700, Jim Thompson
To-Email-Use-The-Envelope-Icon@My-Web-Site.com> wrote:

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


May I presume that Vref is +2.2V and the CML is open-collector pulling
current through the 50 ohm terminations to make the ą400mV signals?
8mA alternating from each collector?

Can additional current be drawn from Vref?

...Jim Thompson

I think CML is usually a unidirectional current sink, an open
collector, pulling about 16 mA for the full-swing version, basicly an
ECL sort of stage without the emitter follower. Some have an internal
termination to Vcc, some don't.

Somebody correct me if I'm wrong.


Melanie may be able to do a direct connection, depending on the
common-mode specs of the receiver. We need more detail.


John



ą400mV differential would imply an 8mA current source, steered by a
diff-pair. But sometimes people get balled up in confusion about
differential peak and peak-to-peak.

But 16mA _would_ give the conventional 800mV single-ended output.

I don't know what is the real spec, since I'm always on-chip and run
as little as 150mV P-P differential... gotta keep those swings down
when you're doing 3GHz ;-)

...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.
Click to expand...
 
On Wed, 07 Dec 2005 09:46:04 -0700, Jim Thompson
<To-Email-Use-The-Envelope-Icon@My-Web-Site.com> wrote:

On Wed, 07 Dec 2005 08:29:53 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 07 Dec 2005 09:00:07 -0700, Jim Thompson
To-Email-Use-The-Envelope-Icon@My-Web-Site.com> wrote:

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


May I presume that Vref is +2.2V and the CML is open-collector pulling
current through the 50 ohm terminations to make the ą400mV signals?
8mA alternating from each collector?

Can additional current be drawn from Vref?

...Jim Thompson

I think CML is usually a unidirectional current sink, an open
collector, pulling about 16 mA for the full-swing version, basicly an
ECL sort of stage without the emitter follower. Some have an internal
termination to Vcc, some don't.

Somebody correct me if I'm wrong.


Melanie may be able to do a direct connection, depending on the
common-mode specs of the receiver. We need more detail.


John



ą400mV differential would imply an 8mA current source, steered by a
diff-pair. But sometimes people get balled up in confusion about
differential peak and peak-to-peak.

But 16mA _would_ give the conventional 800mV single-ended output.

I don't know what is the real spec, since I'm always on-chip and run
as little as 150mV P-P differential... gotta keep those swings down
when you're doing 3GHz ;-)

...Jim Thompson
Click to expand...

Well, you don't have long tranny lines and remote terminations
on-chip! If you did, the chips would be toast.

Some of the CML parts, like a few of the GigaLogic things, have an
internal 50 ohms to Vcc, so that if you terminate again, externally,
you get logic levels of Vcc and Vcc-0.4.

I think "current mode logic" is a sometimes generic term, not precise
like, say "TTL". (Just a little joke here.)

A couple of the Giga parts have pure current sinks that are externally
programmable, all the way down to zero if you cheat a little, very
clean and linear, 40 ps edges. That can be real handy.

John
 
On Wed, 7 Dec 2005 18:04:51 +0100, "Melanie Nasic"
<quinn_the_esquimo@freenet.de> wrote:

Hi all,

thanks for your fast reply. Vref is in fact +2.5V and I'm not sure whether
CML is open-collector and how much current is alternating from each
collector. Maybe a look at http://www.xilinx.com/bvdocs/userguides/ug024.pdf
(page 103) will help but I am not so firm on those electrical specs. You
would help me a lot by explaining that to me, though. :)
As transmitter I am using Xilinx' RocketIOs but their swing is higher than
the maximum receiver swing (+/- 400mV instead of +/- 125mV) and the receiver
is NOT supposed to exept some other common mode range than the 1.15 Volt
(that's part of the specification: the goal is to achieve nearly exactly the
U(t) given in my first mail).
Many thanks in advance....

Regards, Mel


[snip]
Click to expand...

Well! That's certainly clear as mud ;-)

Are the 50 ohm resistors on-chip or off-chip?

...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.
 
Hello Jim,

thanks for your support so far. The source termination is both lines 50 Ohm
to Vref (CML) but the end termination should be 100 Ohm differentially.
Where does the factor of 1/2 accrue from? The far end must not terminate to
Vref but has to be 100 Ohm
differential. As transmitter I am using Xilinx' RocketIOs but their swing is
higher than the maximum receiver swing (+/- 400mV instead of +/- 125mV) and
the receiver is NOT supposed to exept some other common mode range than the
1.15 Volt (that's part of the specification: the goal is to achieve nearly
exactly the U(t) given in my first mail).
Many thanks in advance,

Mel



"Jim Thompson" <To-Email-Use-The-Envelope-Icon@My-Web-Site.com> schrieb im
Newsbeitrag news:715ep1hjntbvbuqmqmoc8mnccu04mema6r@4ax.com...
On Wed, 7 Dec 2005 17:52:33 +0100, "Melanie Nasic"
quinn_the_esquimo@freenet.de> wrote:

Hi PeteS,

thanks for your reply. I don't think that would work because the CML
output
stage "must see" 50 ohm transmission lines and I guess it would not be
allowed to terminate them with 62 ohm in the middle?

Bye Mel



Is the signal at the end of these "transmission lines" your specified
ą400mV? Source and end termination _would_ account for the factor of
1/2 that John Larkin and I have been fretting over.

Does the far end have to terminate to Vref, or would 100 ohm
differential suffice?

...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.
Click to expand...
 
On Wed, 7 Dec 2005 18:14:50 +0100, "Melanie Nasic"
<quinn_the_esquimo@freenet.de> wrote:

Hello Jim,

thanks for your support so far. The source termination is both lines 50 Ohm
to Vref (CML) but the end termination should be 100 Ohm differentially.
Where does the factor of 1/2 accrue from?
Click to expand...
Your observed signal swing. When you "source terminate" and then
terminate the far end of the line you lose 1/2 of the swing.

The far end must not terminate to
Vref but has to be 100 Ohm
differential. As transmitter I am using Xilinx' RocketIOs but their swing is
higher than the maximum receiver swing (+/- 400mV instead of +/- 125mV) and
the receiver is NOT supposed to exept some other common mode range than the
1.15 Volt (that's part of the specification: the goal is to achieve nearly
exactly the U(t) given in my first mail).
Many thanks in advance,

Mel



"Jim Thompson" <To-Email-Use-The-Envelope-Icon@My-Web-Site.com> schrieb im
Newsbeitrag news:715ep1hjntbvbuqmqmoc8mnccu04mema6r@4ax.com...
On Wed, 7 Dec 2005 17:52:33 +0100, "Melanie Nasic"
quinn_the_esquimo@freenet.de> wrote:

Hi PeteS,

thanks for your reply. I don't think that would work because the CML
output
stage "must see" 50 ohm transmission lines and I guess it would not be
allowed to terminate them with 62 ohm in the middle?

Bye Mel



Is the signal at the end of these "transmission lines" your specified
ą400mV? Source and end termination _would_ account for the factor of
1/2 that John Larkin and I have been fretting over.

Does the far end have to terminate to Vref, or would 100 ohm
differential suffice?

...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.
Click to expand...

...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.
 
What 50 ohm resistors (on-chip or off-chip) are you referring to, Jim? The
CML transmitters have an internal 50 ohms to Vcc. And a question to John: I
didn't understand why I get logic levels of Vcc and Vcc-0.4 if I terminate
again externally? Where does the 0.4V come from? Or is it my swing that Vcc
is high and Vcc-0.4 is low voltage level?

Thanks, Mel



"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> schrieb im
Newsbeitrag news:kf5ep1p3ir5euum0h02ogu4n7h8t72th42@4ax.com...
On Wed, 07 Dec 2005 09:46:04 -0700, Jim Thompson
To-Email-Use-The-Envelope-Icon@My-Web-Site.com> wrote:

On Wed, 07 Dec 2005 08:29:53 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 07 Dec 2005 09:00:07 -0700, Jim Thompson
To-Email-Use-The-Envelope-Icon@My-Web-Site.com> wrote:

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


May I presume that Vref is +2.2V and the CML is open-collector pulling
current through the 50 ohm terminations to make the ą400mV signals?
8mA alternating from each collector?

Can additional current be drawn from Vref?

...Jim Thompson

I think CML is usually a unidirectional current sink, an open
collector, pulling about 16 mA for the full-swing version, basicly an
ECL sort of stage without the emitter follower. Some have an internal
termination to Vcc, some don't.

Somebody correct me if I'm wrong.


Melanie may be able to do a direct connection, depending on the
common-mode specs of the receiver. We need more detail.


John



ą400mV differential would imply an 8mA current source, steered by a
diff-pair. But sometimes people get balled up in confusion about
differential peak and peak-to-peak.

But 16mA _would_ give the conventional 800mV single-ended output.

I don't know what is the real spec, since I'm always on-chip and run
as little as 150mV P-P differential... gotta keep those swings down
when you're doing 3GHz ;-)

...Jim Thompson


Well, you don't have long tranny lines and remote terminations
on-chip! If you did, the chips would be toast.

Some of the CML parts, like a few of the GigaLogic things, have an
internal 50 ohms to Vcc, so that if you terminate again, externally,
you get logic levels of Vcc and Vcc-0.4.

I think "current mode logic" is a sometimes generic term, not precise
like, say "TTL". (Just a little joke here.)

A couple of the Giga parts have pure current sinks that are externally
programmable, all the way down to zero if you cheat a little, very
clean and linear, 40 ps edges. That can be real handy.

John
Click to expand...
 
On Wed, 7 Dec 2005 18:21:17 +0100, "Melanie Nasic"
<quinn_the_esquimo@freenet.de> wrote:

What 50 ohm resistors (on-chip or off-chip) are you referring to, Jim? The
CML transmitters have an internal 50 ohms to Vcc. And a question to John: I
didn't understand why I get logic levels of Vcc and Vcc-0.4 if I terminate
again externally? Where does the 0.4V come from? Or is it my swing that Vcc
is high and Vcc-0.4 is low voltage level?

Thanks, Mel



[snip]
Click to expand...

We think it's likely "standard" CML with a switched 16mA current sink,
each 50 ohm would get pulled down 800mV. Add another 50 ohm in
parallel ( the input impedance of the transmission line) and you get
400mV.

...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.
 
Okay then, all I got is a "standard CML driver" as described in
http://www.xilinx.com/bvdocs/userguides/ug024.pdf and I want to interface to
the receiver that is NOT supposed to exept some other common mode range than
1.15 Volt and voltage swing of +/- 125mV. How can this be achieved? I'm not
a crack on this field so maybe you have to start be zero to explain to me.
But first of all a quick solution would be better... ;-) Can or SHOULD I use
a resistor network?

Regards, Mel





"Jim Thompson" <To-Email-Use-The-Envelope-Icon@My-Web-Site.com> schrieb im
Newsbeitrag news:cj6ep1100m5m4fij4i2si9lfh0vre5ueuj@4ax.com...
On Wed, 7 Dec 2005 18:21:17 +0100, "Melanie Nasic"
quinn_the_esquimo@freenet.de> wrote:

What 50 ohm resistors (on-chip or off-chip) are you referring to, Jim?
The
CML transmitters have an internal 50 ohms to Vcc. And a question to John:
I
didn't understand why I get logic levels of Vcc and Vcc-0.4 if I terminate
again externally? Where does the 0.4V come from? Or is it my swing that
Vcc
is high and Vcc-0.4 is low voltage level?

Thanks, Mel



[snip]

We think it's likely "standard" CML with a switched 16mA current sink,
each 50 ohm would get pulled down 800mV. Add another 50 ohm in
parallel ( the input impedance of the transmission line) and you get
400mV.

...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.
Click to expand...
 
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
Click to expand...
See...

Newsgroups: alt.binaries.schematics.electronic
Subject: electrical interface problem - MelanieCML.pdf
Message-ID: <6acep1d0gm5qbckb17t6506aof82lfbd60@4ax.com>

for a possible solution.

(If you can't access the binary group let me know and I will post to a
URL.)

This uses AC coupling.

If you happen to have the luxury of being able to adjust the current
source in the CML output, I think this can be done direct-coupled.

...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 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
Click to expand...
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
 
On Wed, 07 Dec 2005 09:46:04 -0700, Jim Thompson
<To-Email-Use-The-Envelope-Icon@My-Web-Site.com> wrote:


I don't know what is the real spec, since I'm always on-chip and run
as little as 150mV P-P differential... gotta keep those swings down
when you're doing 3GHz ;-)
Click to expand...
---
So, no help offered, just more of your "Look at me, Ma" bullshit.


--
John Fields
Professional Circuit Designer
 
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.


Melanie Nasic wrote:
What 50 ohm resistors (on-chip or off-chip) are you referring to, Jim? The
CML transmitters have an internal 50 ohms to Vcc. And a question to John: I
didn't understand why I get logic levels of Vcc and Vcc-0.4 if I terminate
again externally? Where does the 0.4V come from? Or is it my swing that Vcc
is high and Vcc-0.4 is low voltage level?

Thanks, Mel



"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> schrieb im
Newsbeitrag news:kf5ep1p3ir5euum0h02ogu4n7h8t72th42@4ax.com...

On Wed, 07 Dec 2005 09:46:04 -0700, Jim Thompson
To-Email-Use-The-Envelope-Icon@My-Web-Site.com> wrote:


On Wed, 07 Dec 2005 08:29:53 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:


On Wed, 07 Dec 2005 09:00:07 -0700, Jim Thompson
To-Email-Use-The-Envelope-Icon@My-Web-Site.com> wrote:


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


May I presume that Vref is +2.2V and the CML is open-collector pulling
current through the 50 ohm terminations to make the ą400mV signals?
8mA alternating from each collector?

Can additional current be drawn from Vref?

...Jim Thompson

I think CML is usually a unidirectional current sink, an open
collector, pulling about 16 mA for the full-swing version, basicly an
ECL sort of stage without the emitter follower. Some have an internal
termination to Vcc, some don't.

Somebody correct me if I'm wrong.


Melanie may be able to do a direct connection, depending on the
common-mode specs of the receiver. We need more detail.


John



ą400mV differential would imply an 8mA current source, steered by a
diff-pair. But sometimes people get balled up in confusion about
differential peak and peak-to-peak.

But 16mA _would_ give the conventional 800mV single-ended output.

I don't know what is the real spec, since I'm always on-chip and run
as little as 150mV P-P differential... gotta keep those swings down
when you're doing 3GHz ;-)

...Jim Thompson


Well, you don't have long tranny lines and remote terminations
on-chip! If you did, the chips would be toast.

Some of the CML parts, like a few of the GigaLogic things, have an
internal 50 ohms to Vcc, so that if you terminate again, externally,
you get logic levels of Vcc and Vcc-0.4.

I think "current mode logic" is a sometimes generic term, not precise
like, say "TTL". (Just a little joke here.)

A couple of the Giga parts have pure current sinks that are externally
programmable, all the way down to zero if you cheat a little, very
clean and linear, 40 ps edges. That can be real handy.

John
Click to expand...
 
John Fields wrote:
On Wed, 07 Dec 2005 09:46:04 -0700, Jim Thompson
To-Email-Use-The-Envelope-Icon@My-Web-Site.com> wrote:

I don't know what is the real spec, since I'm always on-chip and run
as little as 150mV P-P differential... gotta keep those swings down
when you're doing 3GHz ;-)

---
So, no help offered, just more of your "Look at me, Ma" bullshit.

--
John Fields
Professional Circuit Designer
Click to expand...

What else do you expect out of him, John? he has to beat his chest
at every chance to try to convince himself he's still worth something.

--
?

Michael A. Terrell
Central Florida
 
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