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

Dennis, OK.  I was worried there for a moment.

Regards,


Steve
At 12:57 PM 2/27/2005 -0600, Dennis Han wrote:
>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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