likephysics
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How do you determine the output impedance of cmos gates?
likephysics said:How do you determine the output impedance of cmos gates?
likephysics said:Exactly! I am trying to calculate Zout to fix the series termination resistor value.
Are you sure about (Vcc-Voh)/Ioh.
Voh/Ioh gives the correct value.
likephysics said:Guys, I tried to calculate the Zo of some ICs.
For a clock buffer - CDCVF25081
http://focus.ti.com/general/docs/lit/getliterature.tsp?genericPartNumber=cdcvf25081&fileType=pdf
Zo = (Vdd-VOH) / IOH. This gave me (datasheet pg.5) Zo = 0.9/12mA = 75 ohms
From the IOH row, Z0 = Vo/IOH = 1.65/30mA = 55 Ohms
VOL/IOL = 0.8/12mA = 66.67Ohms
Simulation gave me close to 50 Ohms. (55 rising edge and 51 falling edge).
Now For another part CY25811
http://www.datasheetcatalog.org/datasheet2/e/0lajeyeqd5a8q0ytgz70kd4gfcky.pdf
Zo = 2.97-2.4/4mA = 142.5 Ohms
Zo = VOL/IOL = 0.4/4mA = 100 ohms
Now for one more - P2811B (onsemi)
http://www.datasheetcatalog.org/datasheet2/e/0lajeyeqd5a8q0ytgz70kd4gfcky.pdf
Zo = 3.3-2.5 /15mA = 53 Ohms
Zo = VOL/IOL = 0.4/15mA = 26 Ohms (This is given as output imp in datasheet)
Last one
Cyclone FPGA
http://www.altera.com/literature/hb/cyc/cyc_c51004.pdf
Pg 2 LVTTL
Zo = (3-2.4)/24mA = 25 Ohms
Zo = VOL/IOL = 0.45/24mA = 18.75 Ohms
The Zo from simulation is much less (7 ohms)!