[SI-LIST] Re: Driver Output Impedance am I doing this correctly?
- From: "Dennis Han" <Dennis.Han@xxxxxxxxxxx>
- To: <ericsilist@xxxxxxxxx>, <si-list@xxxxxxxxxxxxx>, "steve weir" <weirsi@xxxxxxxxxx>
- Date: Sun, 27 Feb 2005 12:24:51 -0600
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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