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How to determine the AGP voltage used



 
 
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  #1  
Old July 21st 04, 10:26 PM
Michael S.
external usenet poster
 
Posts: n/a
Default How to determine the AGP voltage used

As I understand it, an 8X AGP video card operates at 0.8 volts. How can I
determine the voltage being used by the AGP 8X video card by my P4C800E-D
motherboard??

I have looked everywhere in the BIOS, in AIDA32 and in Probe data, but
either have missed the AGP voltage or just do not know where to look.
Aida32 does show an 8X speed. Suggestions?

MikeSp


  #2  
Old July 22nd 04, 12:05 AM
Paul
external usenet poster
 
Posts: n/a
Default

In article , "Michael S."
wrote:

As I understand it, an 8X AGP video card operates at 0.8 volts. How can I
determine the voltage being used by the AGP 8X video card by my P4C800E-D
motherboard??

I have looked everywhere in the BIOS, in AIDA32 and in Probe data, but
either have missed the AGP voltage or just do not know where to look.
Aida32 does show an 8X speed. Suggestions?

MikeSp


AGP video chips use many voltages, but the voltage in question is
used to power the I/O pins on the AGP video card and also the
I/O pins on the Northbridge AGP interface. Matching the capabilities
of those two chips, so they can share the same voltage, is
what all this nonsense is about. That is why AGP slots have plastic
keys, and AGP video cards have slots cut in them, to prevent an
inappropriate mix of technology.

There are, in fact, only two I/O supply voltages. They are
3.3V and 1.5V. The 1.5V is necessary, as modern small geometry
chips no longer like the 3.3V voltage. They can only handle lower
voltages. Your P4C800E-D can only handle the lower of the two
voltages. There is a setting in the BIOS with options like this -
basically 1.5V, but with room to adjust if needed. Sometimes
a boost is needed for stability.

"AGP VDDQ Voltage" [1.5, 1.6, 1.7]

So, where does the 0.8V come from ? There are two ways to
terminate the AGP bus. Termination is any means used to
suppress signal reflection on an electrical interconnect.

There are two cases. Both electrical cases use VDD = 1.5V for the
silicon. One case doesn't use a parallel termination resistor
(it is AGP 4X). The VDDQ fed to the I/O pad results in a full
ampliude signal and we refer to that as 1.5V I/O.


VDD=1.5V Typical AGP 2.0 case VDD=1.5V
| Full 1.5V signal swing |
| |
Driver ------resistance-----the_bus-----+------Receiver
| |
| |
| |
GND GND


In the second case, a parallel termination resistor is used,
and as near as I can tell, this resistor is actually inside the
chip. The I/O pads at either end of the bus are still powered
by 1.5V, but there is a voltage divider action, due to the
use of the termination at the receiver.

AGP 3.0 case 0.8V swing
VDD=1.5V Parallel terminator | VDD=1.5V
| | |
| v |
Driver ------resistance-----the_bus-----+------Receiver
| | |
| resistance |
| | |
GND GND GND

Please note - I haven't build any AGP interfaces, and the
above figure is my interpretation of what I've been reading.
Start at PDF page 63 to do your own interpretation.

http://developer.intel.com/technolog...0_final_10.pdf

In conclusion, for an AGP 3.0 card, you cannot measure this
0.8V swing, because it only exists on a signal pin, and they change
too rapidly for any hardware to measure or make sense of them.
The driver and receiver pads are still powered by 1.5V.

The only thing you would be able to see, is the value of VDD.
The P4C800E-D doesn't connect the hardware monitor chip to the
VDD of the AGP. You can set the voltage in the BIOS, but the
hardware monitoring page in the BIOS doesn't show the result of
your change. Using a voltmeter on one of the AGP slot pins would
be one way to verify the voltage.

HTH,
Paul
  #3  
Old July 22nd 04, 12:32 AM
Michael S.
external usenet poster
 
Posts: n/a
Default

WOW Paul--what a nicely-written, detailed answer--THANKS!. I vaguely
remember running some program about a month ago that reported a number of
parameters of the computer and came up with 1.5 volts on the AGP--which is
why I was asking about the 0.8 volts. The general public is told that AGP
4x = 1.5 volts and 8x = 0.8 volts, so I was concerned that my Radeon x800
Pro was being fed too many volts because I overlooked some setting in BIOS
in this computer build, my first since 486's. From your detailed
explanation, I understand WHY the program I was running came up with 1.5
volts on the AGP.

MikeSp
-------------------------------
"Paul" wrote in message
...
In article , "Michael S."
wrote:

As I understand it, an 8X AGP video card operates at 0.8 volts. How can

I
determine the voltage being used by the AGP 8X video card by my

P4C800E-D
motherboard??

I have looked everywhere in the BIOS, in AIDA32 and in Probe data, but
either have missed the AGP voltage or just do not know where to look.
Aida32 does show an 8X speed. Suggestions?

MikeSp


AGP video chips use many voltages, but the voltage in question is
used to power the I/O pins on the AGP video card and also the
I/O pins on the Northbridge AGP interface. Matching the capabilities
of those two chips, so they can share the same voltage, is
what all this nonsense is about. That is why AGP slots have plastic
keys, and AGP video cards have slots cut in them, to prevent an
inappropriate mix of technology.

There are, in fact, only two I/O supply voltages. They are
3.3V and 1.5V. The 1.5V is necessary, as modern small geometry
chips no longer like the 3.3V voltage. They can only handle lower
voltages. Your P4C800E-D can only handle the lower of the two
voltages. There is a setting in the BIOS with options like this -
basically 1.5V, but with room to adjust if needed. Sometimes
a boost is needed for stability.

"AGP VDDQ Voltage" [1.5, 1.6, 1.7]

So, where does the 0.8V come from ? There are two ways to
terminate the AGP bus. Termination is any means used to
suppress signal reflection on an electrical interconnect.

There are two cases. Both electrical cases use VDD = 1.5V for the
silicon. One case doesn't use a parallel termination resistor
(it is AGP 4X). The VDDQ fed to the I/O pad results in a full
ampliude signal and we refer to that as 1.5V I/O.


VDD=1.5V Typical AGP 2.0 case VDD=1.5V
| Full 1.5V signal swing |
| |
Driver ------resistance-----the_bus-----+------Receiver
| |
| |
| |
GND GND


In the second case, a parallel termination resistor is used,
and as near as I can tell, this resistor is actually inside the
chip. The I/O pads at either end of the bus are still powered
by 1.5V, but there is a voltage divider action, due to the
use of the termination at the receiver.

AGP 3.0 case 0.8V swing
VDD=1.5V Parallel terminator | VDD=1.5V
| | |
| v |
Driver ------resistance-----the_bus-----+------Receiver
| | |
| resistance |
| | |
GND GND GND

Please note - I haven't build any AGP interfaces, and the
above figure is my interpretation of what I've been reading.
Start at PDF page 63 to do your own interpretation.

http://developer.intel.com/technolog...0_final_10.pdf

In conclusion, for an AGP 3.0 card, you cannot measure this
0.8V swing, because it only exists on a signal pin, and they change
too rapidly for any hardware to measure or make sense of them.
The driver and receiver pads are still powered by 1.5V.

The only thing you would be able to see, is the value of VDD.
The P4C800E-D doesn't connect the hardware monitor chip to the
VDD of the AGP. You can set the voltage in the BIOS, but the
hardware monitoring page in the BIOS doesn't show the result of
your change. Using a voltmeter on one of the AGP slot pins would
be one way to verify the voltage.

HTH,
Paul



  #4  
Old July 22nd 04, 03:03 AM
Paul
external usenet poster
 
Posts: n/a
Default

In article , "Michael S."
wrote:

WOW Paul--what a nicely-written, detailed answer--THANKS!. I vaguely
remember running some program about a month ago that reported a number of
parameters of the computer and came up with 1.5 volts on the AGP--which is
why I was asking about the 0.8 volts. The general public is told that AGP
4x = 1.5 volts and 8x = 0.8 volts, so I was concerned that my Radeon x800
Pro was being fed too many volts because I overlooked some setting in BIOS
in this computer build, my first since 486's. From your detailed
explanation, I understand WHY the program I was running came up with 1.5
volts on the AGP.

MikeSp


BTW - Apologies for the AGP doc link. It appears Intel has erased
history, by removing their AGP info. This is the first site I
could find that still has copies:

http://www.ac.uma.es/educacion/curso...I/PCI/agp/?N=A

Docs like that don't get backed up on web.archive.org, and I also
notice that a number of large companies have had archiving activities
stopped, making it easier for them to bury their past.

Paul
 




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