[SI-LIST] Re: Driver Output Impedance am I doing this correctly?
- From: steve weir <weirsi@xxxxxxxxxx>
- To: "Dennis Han" <Dennis.Han@xxxxxxxxxxx>, <ericsilist@xxxxxxxxx>, <si-list@xxxxxxxxxxxxx>
- Date: Sun, 27 Feb 2005 10:36:14 -0800
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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>
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