[SI-LIST] Re: Drivers on a 50Ohm line

Usually the specification for the output current
capability is a minimum for what can be delivered
when the driver is within some small voltage of 
the power rail.  This voltage is invariably closer
to the rail than the threshold voltage, especially
for CMOS devices and devices that reference the 
input to some VTT value.

David Lieby
Siemens Med, Ultrasound Division

--- "Bhagwath, Nitin" <nitin.bhagwath@xxxxxx> wrote:

> Steve, I'm afraid I don't see what part you're
> referring to when you say
> "No that is not it" since I'm not sure I found much
> difference from your
> reply and what I wrote.
> 
> Again assuming Vcc=3D2.5, and Zo=3D50Ohms,
> 
> 1) Parallel ground (What I called End termination to
> ground):- I didn't
> take the driving low case, because I'm assuming
> worst case, i.e. driving
> high, in which case as we both stated, Vcc/Zo=3D50mA
> is required.  - I
> agree this is a bad choice.
> 
> 2) Parallel Thevenin (What I called Thevenin end
> termination):-
> Regardless of termination resistor power
> requirements, the current drive
> requirements of the driver stay the
> same...Vcc/(2*Zo)=3D25mA in my =
> example
> (assuming equal pullup/pulldown output impedances,
> and Vth=3DVcc/2).=20
> 
> 3) I left out Parallel AC, because my understanding
> is that the driver
> current requirement is similar to regular parallel
> end termination
> requirements at high enough frequencies.
> 
> 4) Source Series:- I follow you, and agree with you
> for the round trip
> waveform you describe.  Now, for the average power,
> the equation you
> stated holds only for 2*Tflight_one_way >
> Tperiod_average. So:
>   a) If 2*Tflight_one_way < Tperiod_average, then
> it'd be difficult to
> make assumptions
>      regarding average current requirements, and the
> safest route would
> be to revert to
>      the worst case of Vcc/(2*Zo)=3D25mA.
> 
>   b) With 2*Tflight_one_way > Tperiod_average being
> met, there is still
> a portion of
>      each clock cycle equal to 2*Tflight_one_way
> where the driver would
> need to drive
>      Vcc/(2*Zo)=3D25mA (and ideally drive 0mA on the
> remaining =
> portions).
> 
> 
> So in any case, my original question still stands. 
> If we take the 50MHz
> 5V (twice the Vcc voltage of the above assumption)
> CMOS oscillator part
> I've mentioned below as an example, how can it drive
> a regular 50Ohm
> trace with a max drive spec of 25mA? =20
> 
> With Vcc=3D5V, Zo=3D50Ohms,
> 
> 1) In the end termination to ground schemes, Vcc/Zo
> =3D 5/50 =3D 100mA =
> is
> required
> 2) In the end Thevenin scheme, Vcc/(2*Zo) =3D 5/100
> =3D 50mA is required
> 3) In the source series scheme
>    a) 2*Tflight_one_way < Tperiod_average, worst
> case assumption 50mA is
> required
>    b) 2*Tflight_one_way > Tperiod_average, 50mA is
> required for portions
> of a cycle,
>       which the driver is unable to meet.  So the
> waveform must become
> distorted
>       for the time period 2*Tflight_one_way after
> each transition.
> 
> -Nitin
> 
> -----Original Message-----
> From: steve weir [mailto:weirsp@xxxxxxxxxx]=20
> Sent: Wednesday, July 28, 2004 10:27 PM
> To: Bhagwath, Nitin; tom@xxxxxxxxxxxxx;
> si-list@xxxxxxxxxxxxx
> Cc: Cherniski, Mike
> Subject: Re: [SI-LIST] Re: Drivers on a 50Ohm line
> 
> 
> No that is not it.  The classifications commonly
> found in texts are:
> 
> Source series
> Parallel ground, usually a poor choice.
> Parallel Thevenin, consumes more power than parallel
> Vtt Parallel Vtt,
> parallel termination of choice for performance.
> Parallel AC, has
> multiple caveats but can be used to reduce power
> consumption.
> 
> Skin resistance is only a factor in long or very
> high frequency traces.
> 
> The characteristic impedance does not appear until a
> low frequency
> cut-off=20
> that is a function of the trace length.
> 
> Your case of resistor to ground, the driver sees 50
> ohms and 2.5V / 50mA
> 
> when high and 50 ohms and 0V / 0mA ( more or less )
> when low.
> 
> Source series, you are closer, the driver sees
> Vdelta through Rterm +=20
> Zline.  For the simple case where everything is
> linear and Rinternal +=20
> Rterm =3D Zline ( the way it is supposed to be ) the
> driver sees +/-25mA
> for=20
> 2*Tflight_one_way.  Average current =3D 50mA *
> Tflight_one_way /
> Tperiod_average.
> 
> Most drivers are quite non-linear and drive high
> currents at a high
> Vdelta,=20
> and have weak current capability at a low Vdelta. 
> The current rating is
> 
> the continuous capability.
> 
> Steve,
> 
> At 08:32 PM 7/28/2004 -0700, Bhagwath, Nitin wrote:
> >Hi All,
> >
> >So the three significant point-to-point termination
> cases are Receiver=20
> >termination to ground, Receiver termination to a
> non-Vcc source=20
> >(usually
> >Vcc/2) and finally Source Series termination.  I'm
> again assuming
> >Vcc=3D3D2.5 and Zo=3D3D50Ohms.
> >
> >End termination to ground
> >------------------------------
> >Here, the driver will see the load more or less as
> a characteristic=20
> >impedance from DC to somewhere where skin effect
> kicks in (I'd rather=20
> >not go there for this discussion).  To drive 2.5V
> into a 50Ohm line=20
> >would require 50mA of driver strength.
> >
> >
> >Thevenin end termination (say to Vcc/2)
> >---------------------------------------
> >Whether implemented by placing directly to a second
> Vcc/2 source, or by
> 
> >placing a voltage divider from Vcc to ground, this
> setup still shows a=20
> >characteristic impedance of 50Ohms to the driver. 
> But since the driver
> 
> >only needs to drive to half the voltage, it now
> needs to be able to=20
> >drive 2.5/(2*50) =3D3D 25mA
> >
> >
> >Source Series termination
> >-------------------------
> >Assuming a series termination, which when added to
> the internal output=20
> >impedance of the driver equals the characteristic
> impedance of the=20
> >line, the driver will need to drive an effective
> load of 2*Char=20
> >Impedance =3D3D 100Ohms.  Now, I agree that "over
> time"/at DC, the=20
> >effective impedance of the transmission line
> changes, but I'm mainly=20
> >concerned about the situation over a frequency
> range where the=20
> >transmission line behaves as a resistive impedance
> (where Zo =3D3D=20
> >(L/C)^(1/2)).
> >
> >Here, if the combined output impedance =3D3D
> characteristic impedance =
> of=20
> >=3D the line, then the input to the line =3D3D
> 1/2Vcc, which reflects =
> at=20
> >the unterminated end of the line to yield a full
> Vcc swing.
> >
> >This situation requires the driver to be able to
> drive 2.5/100 =3D3D=20
> >25mA, similar to the end termination to Vcc/2.
> >
> >
> >
> >Where my confusion comes in is that this is that
> 25mA is a rather large
> 
> >requirement for a driver.  I know there are drivers
> which can drive=20
> >more than this, but there are devices which find
> this requirement to be
> 
> >at or above their drive limits.  An example is a
> KSS oscillator
> >(http://202.226.94.5/pdf/e/fxo31fe.pdf) which has
> 24mA as its drive=20
> >capability (it's shown for a 5V CMOS type part,
> with the 5V just making
> 
> >the current requirement more accute!).  I would
> assume that these=20
> >devices are not driven to/past their limits under
> normal situations (or
> 
> >are they?).  If we take a series terminated 5V
> driver oscillator (I'm=20
> >assuming say a 50MHz KSS part from the above
> datasheet) which can drive
> 
> >24mA, then how can it drive a normal 50Ohm
> transmission line, if the=20
> >required current is itself 5/100 =3D3D 50mA? 
> Wouldn't this seriously =
> =3D=20
> >affect the voltages at the load?
> >
> >Thanks
> >-Nitin
> >
> >
> >
> >-----Original Message-----
> >From: Tom Dagostino [mailto:tom@xxxxxxxxxxxxx]=3D20
> >Sent: Wednesday, July 28, 2004 12:52 PM
> >To: Bhagwath, Nitin; si-list@xxxxxxxxxxxxx
> >Cc: Cherniski, Mike
> >Subject: RE: [SI-LIST] Drivers on a 50Ohm line
> >
> >
> >Nitin
> >
> >The answer depends on the full net that the example
> you site consists=20
> >of. If it is a 1 nsec 50 Ohm trace unterminated
> then an open=20
> >transmission line effect comes into play.  The
> signal that is launched=20
> >into the line will have an amplitude of
> >
> >Vlaunch =3D3D (Vopen * 50 Ohms)/(Zout + 50 Ohms) or
> the classic voltage =
> 
> >divider where 50 Ohms is the line impedance and
> Zout is the output=20
> >impedance of the driver.
> >
> >If the line is unterminated when the signal reaches
> the line after 1=20
> >nsec it reflects and doubles in amplitude.  The
> reflection gets back to
> 
> >the driver after another 1 nsec and the full open
> Voltage Vopen will be
> 
> >seen at both ends of the line.
> >
> >If the line is terminated in the characteristic
> impedance then the=20
> >above equation applies at all times.
> >
> >See the good explanation of this in Howard
> Johnson's book "High Speed=20
> >Digital Design, a Handbook of Black Magic" chapter
> 4 or any book that=20
> >covers transmission line theory.
> >
> >
> >Tom Dagostino
> >Teraspeed Consulting Group LLC
> >503-430-1065
> >tom@xxxxxxxxxxxxx
> >www.teraspeed.com
> >
> >-----Original Message-----
> >From: si-list-bounce@xxxxxxxxxxxxx=20
> >[mailto:si-list-bounce@xxxxxxxxxxxxx]On Behalf Of
> Bhagwath, Nitin
> >Sent: Wednesday, July 28, 2004 11:23 AM
> >To: si-list@xxxxxxxxxxxxx
> >Cc: Cherniski, Mike
> >Subject: [SI-LIST] Drivers on a 50Ohm line
> >
> >
> >Hi All,
> >
> >I have a simple question, for which I'm yet to find
> a complete answer.
> >
> >A standard PCB trace has an impedance of 50Ohms. 
> Does this imply that,
> 
> >say, for a 2.5V signal, a driver must be able to
> drive 2.5/50 =3D3D3D=20
> >50mA =3D =3D3D of current on this signal to allow
> for a full swing of =
> the=20
> >signal?
> >
> >If this is the case, then how can a driver, such as
> a clock oscillator,
> 
> >which has a drive strength in the order of 20-30mA
> drive a buffer=20
> >through a 50ohm line without the voltage being
> attenuated?  This would=20
> >get accentuated at 3.3V, or 5V where the current
> requirement would=20
> >increase.
> >
> >Any thoughts appreciated
> >
> >Thanks
> >
> >-Nitin
>
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