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I understand that for an RC circuit acting as a load of a driver will afftect the delay of the output after the RC load.

rc circuit So as long as RC for different values of R,C are equal, delays are equal, is this right? For example R = 8 Ohm, C = 0.25p, RC = 2 and R = 5 Ohm, C =0.4p, RC=2

I simulate with these above 2 configurations of RC values but i get different delay values after the RC load, and also different rise/fall time.

I think that delays in these 2 cases should be equal and so do rise/fall times.

Are they any constraints on the R, C for to compare delays based on different RC time constant.

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  • \$\begingroup\$ \$ \tau = RC = 8 \times 0.25p = 2~ps \$ (pico-seconds, not seconds). \$\endgroup\$ – Transistor Aug 3 '16 at 10:45
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    \$\begingroup\$ An very very ugly circuit diagram - My eyes are hurting. \$\endgroup\$ – JIm Dearden Aug 3 '16 at 10:50
  • \$\begingroup\$ You can click on the image to enlarge it \$\endgroup\$ – vqtuyen Aug 3 '16 at 10:52
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So as long as RC for different values of R,C are equal, delays are equal, is this right? For example R = 8 Ohm, C = 0.25p, RC = 2 and R = 5 Ohm, C =0.4p, RC=2

In theory yes but in practice no. The output driver has a finite output impedance that will add to the external series resistor and as that external resistor value gets lower in value it will be swamped by the residual output resistance of the driver.

This output resistance might be 10 ohms to 100 ohms hence it will swamp the values of 8 ohm and 5 ohm you used.

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  • \$\begingroup\$ Can you be more specific please. I think that in 2 cases, driver resistance will be the same, and so does the RC time constant \$\endgroup\$ – vqtuyen Aug 3 '16 at 10:48
  • \$\begingroup\$ I don't know how to state this more clearly. \$\endgroup\$ – Andy aka Aug 3 '16 at 11:11
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    \$\begingroup\$ With 10R inside the chip it adds to the 8 ohm to make 18 ohm and it adds to the 5 ohm to make 15 ohm - this MUST mean the RC time constants are no longer the same. \$\endgroup\$ – Andy aka Aug 3 '16 at 11:19

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