RE: [IBIS-Users] How is Vdiff defined ? Does Vdiff=250mV mean +- 125mV or +-250mV?

From: Lenski, Eckhard (NSN - DE/Munich) <eckhard.lenski_at_.....>
Date: Fri Feb 27 2009 - 01:17:24 PST
Hello Johann,
 
Vdiff is the voltage difference between the pos and the neg input.
 
Vdiff =  Vpos - Vneg
 
suppose you have a signal voltage on the pos pin of 1.6V  and the neg pin of 1.2V
this would result in a Vdiff of 0.4V
 
suppose a signal voltage of  the pos pin of 0.125V and the neg pin of  -0.125V
this would result in a Vdiff of 0.25V
 
hope this makes it clear
 
regards
Eckhard
 
 

Mit freundlichen Grüßen / Best regards / Cordiali saluti / ystävällisin terveisin

Eckhard Lenski 

Nokia Siemens Networks GmbH & Co. KG


COO RA GERD BTS BR PDev CAE Lib&IBIS

CAE libraries and models
Balanstr. 59 

81541 München 
Germany 
phone : +49 89 636 79002
fax : +49 89 636 78895
email: eckhard.lenski@nsn.com <mailto:eckhard.lenski@nsn.com> 

http://www.nokiasiemensnetworks.com/ibis 


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________________________________

	From: owner-ibis-users@eda.org [mailto:owner-ibis-users@eda.org] On Behalf Of ext Nittmann, Johann
	Sent: Thursday, February 26, 2009 5:56 PM
	To: ibis-users@eda.org
	Subject: [IBIS-Users] How is Vdiff defined ? Does Vdiff=250mV mean +- 125mV or +-250mV?
	
	

	Dear all,

	 

	Could somebody explain how the value of Vdiff is defined? Does Vdiff=250mV mean +-125mV or +-250mV!

	 

	 According the IBIS cookbook-v4, at page 73 where it says

	 

	"For example, an active-high buffer with a vdiff

	of 100 mV will be considered high if the non-inverting pin is at least 100 mV above the inverting pin. "

	 

	I would expect it to be defined as +-125mV but one our SI tool provider is defining it at +-250mV?

	 

	Thanks,

	 

	Johann

	 

	 

	Johann Nittmann

	Principal Engineer

	Cavium Networks

	100 Nickerson Road

	Marlborough, MA 01752

	Tel: 508 683 8869 

	 


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Received on Fri Feb 27 01:20:42 2009

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