[SI-LIST] Re: Driver Output Impedance am I doing this correctly?

Adjusting Z0 to match Zout is just to determine Zout in the simulation.  Z0 
= 50 is still the norm in the physical layout and the drivers are picked for 
their Zout to fit the situation.

Dennis

----- Original Message ----- 
From: "steve weir" <weirsi@xxxxxxxxxx>
To: "Dennis Han" <Dennis.Han@xxxxxxxxxxx>; <ericsilist@xxxxxxxxx>; 
<si-list@xxxxxxxxxxxxx>
Sent: Sunday, February 27, 2005 12:36 PM
Subject: [SI-LIST] Re: Driver Output Impedance am I doing this correctly?


> Dennis, well for a perfect Tx line and no load, each method better return
> identical results, or there is something that we haven't accounted for. 
> As
> far as adjusting the line to the driver, I can see such a technique if we
> are talking about tweaking out 10-20% to obviate the resistors. 
> Otherwise,
> I would be cautious about the side effects.  As much as the extra 
> resistors
> can be a nuisance, adjusting the line |Z| to match some low |Z| driver 
> like
> 20 ohms can make for big traces and a lot of power dissipation.
>
> Regards,
>
>
> Steve.
> At 12:24 PM 2/27/2005 -0600, Dennis Han wrote:
>>Yes, I do use a lossy line and the real length when I have an application,
>>but I use the lossless line to get an idea of what FPGA or ASIC driver to
>>assign to an output pin.  Then I have an idea of what the output impedance
>>is of the various drivers.  Your method would work well, too, and probably
>>better if the driver has an output impedance that is higher than Z0.
>>
>>I should have mentioned that another way to use my method would be to
>>change Z0 until the signal at the end of the line matches what was sent
>>from the Thevenin voltage source, i.e., a square wave with amplitude
>>VDD.  Then Zout = Z0.
>>
>>Dennis
>>
>>
>>
>>----- Original Message ----- From: "steve weir" <weirsi@xxxxxxxxxx>
>>To: <Dennis.Han@xxxxxxxxxxx>; <ericsilist@xxxxxxxxx>; 
>><si-list@xxxxxxxxxxxxx>
>>Sent: Sunday, February 27, 2005 11:02 AM
>>Subject: Re: [SI-LIST] Re: Driver Output Impedance am I doing this 
>>correctly?
>>
>>
>>>Dennis, I absolutely agree that the best method is to use an ACCURATE
>>>driver SPICE model.  I also agree that the simulation environment should
>>>be made to match the situation that we want to cover.  You are already
>>>half way there when you put in your arbitrary length, lossless TX line,
>>>so I recommend you just finish job with the real length, load and a lossy
>>>TX line.
>>>
>>>To adjust the value of  a discrete series termination resistor, I prefer
>>>to look at the two sides of the termination resistor.  This is less
>>>colored by what happens at the end of the line.  The belief system is
>>>that the cleanest results occur when we start out with the line as close
>>>to perfect as possible.  We can then adjust and compromise as called for
>>>to meet our goals.
>>>
>>>The series resistor decouples the driver somewhat, so I like to start
>>>with an approximate value, and just iterate once or twice.  Depending on
>>>what I think of the driver, I will start with 10 - 30 ohms for the
>>>initial value of the series terminator and go from there using the 
>>>formula:
>>>
>>>Rt_incremental = Zline * ( 2 - ( Vswing_driver_side_Rt /
>>>Vswing_junction_tx_line_side_Rt ) )
>>>
>>>The nice thing is that through the whole process, I get to use the same
>>>model, and just dial-in the coefficients.  Since I am a clerical disaster
>>>area, this works well for me.  You could use the far-end measurements
>>>easily enough, but then you will see coloration from the load.  I prefer
>>>to look at the line end only after I have the source termination set to
>>>match the line as well as possible.
>>>
>>>Regards,
>>>
>>>
>>>Steve.
>>>At 09:37 AM 2/27/2005 -0600, Dennis Han wrote:
>>>>The best method is to give the customer a SPICE model and let him figure 
>>>>out
>>>>the output impedance based on his frequency of operation and 
>>>>application.
>>>>The method I prefer ties an ideal open transmission line to the driver. 
>>>>I
>>>>like to make the length a few inches, but long enough so it is easy to 
>>>>see
>>>>the voltage transitions at the end of the line.  Then solve for Zout:
>>>>
>>>>Vout = 2 x VDD x Z0 / (Z0 + Zout)
>>>>
>>>>where Vout is the pedestal at the end of the line and assuming VDD for a
>>>>CMOS driver (for other driver types, use the voltage that would be used 
>>>>in a
>>>>Thevenin equivalent of the driver)
>>>>
>>>>Using a resistor to ground rather than an open transmission line keeps 
>>>>the
>>>>DC currents high, which can change the output impedance.  In most
>>>>applications, the DC currents are low, but use a topology that makes 
>>>>sense
>>>>for the application.  Hence, I prefer getting the SPICE model.
>>>>
>>>>Dennis
>>>>
>>>>
>>>>
>>>>----- Original Message -----
>>>>From: "eric steimle" <ericsilist@xxxxxxxxx>
>>>>To: <si-list@xxxxxxxxxxxxx>
>>>>Sent: Wednesday, February 23, 2005 6:44 PM
>>>>Subject: [SI-LIST] Driver Output Impedance am I doing this correctly?
>>>>
>>>>
>>>> > Hi,
>>>> > Thanks in advance for anyone taking the time to read
>>>> > this, I know how valuable everyone's time is.  I'm
>>>> > trying to calculate a number for our driver output
>>>> > impedance that I can give to someone working with our
>>>> > chip.  I'm relatively new at this so I was wondering
>>>> > if my approach was flawed.
>>>> >
>>>> > My configuration is:
>>>> >
>>>> > Driver --> BondWire --> Substrate Board Trace --> Ball
>>>> >
>>>> > My driver's rise time is 500ps and my goal is to
>>>> > calculate our total impedance value in order to allow
>>>> > others to choose the appropriate series terminating
>>>> > resistor.
>>>> >
>>>> > Driver
>>>> > 1.) I used spice with a 50 Ohm load on the output and
>>>> > Ohms law to get the real impedance of the driver
>>>> > Zdriver (14.5Ohms).
>>>> >
>>>> > BondWire
>>>> > 2.) I read through the SI archives and other docs and
>>>> > then I assumed the inductance of the bondwire was
>>>> > 1nH/mm on a 2mm bondwire.  That means:
>>>> >
>>>> > BondWire = 2nH
>>>> >
>>>> > Then I took the Frequency of the output to be:
>>>> > F = .35/Tr
>>>> > F =  .35/500ps
>>>> > F = 700Mhz
>>>> >
>>>> > Then I used the impedance formula for an inductor (can
>>>> > I do this with the F I calculated or does that not
>>>> > make sense?).
>>>> >
>>>> > Z = 2pi * f * L
>>>> > Z = 2pi * 700Mhz * 2nH
>>>> > Z = 8.8Ohms
>>>> >
>>>> > Finally
>>>> > So I figured that the Zdriver + Zbondwire +Zimag(I'm
>>>> > missing that) would equal the output impedance of the
>>>> > driver itself, and I know the impedance of the
>>>> > substrate trace.
>>>> >
>>>> > Does that make sense, or am I going about this all
>>>> > wrong?  I still don't know how to calculate the
>>>> > frequency dependant part of the driver output
>>>> > impedance but I'm researching like crazy.
>>>> >
>>>> > Any feedback or new reference material would be
>>>> > greatly appreciated.
>>>> >
>>>> > Thanks,
>>>> > Eric
>>>> >
>>>> >
>>>> >
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>>>>
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>>>
>>>The weirsp@xxxxxxxxxx e-mail address will terminate March 31, 2005.
>>>Please update your address book with weirsi@xxxxxxxxxx
>>>
>>
>
> The weirsp@xxxxxxxxxx e-mail address will terminate March 31, 2005.
> Please update your address book with weirsi@xxxxxxxxxx
>
>
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