At 04:03 PM 10/4/2009, you wrote:
> >>>>> Exactly. But to get the same speed, meaning same signal
> >>>>> impedence, you would need 1.2Kohms since
> >>>>> 2Kohm // 3Kohm = 1.2Kohm.
> >>>> On rising edge I don't see how does mentioned in your calcs
> >>>> 3Kohm divider's bottom resistor help to speed the rising.
> >>> This is very basic electronics.  Look up something called
> >>> "Thevenin Equivalent".
> >>
> >> Interesting, obviously, removing the bottom resistor from the divider
> >> would increase the speed of rising (assuming inputs would tolerate
> >> it). But, according to your logic/formula removing the bottom resistor
> >> would increase impedance and, thus, decrease the speed of rising.
> >
> > Not exactly. With two resistors the impedance is 1.2k and with only one
> > it's 2k. But the equivalent voltages are 3.3 and 5 respectively. If you
> > are driving logic with TTL(1.2 volt) thresholds the rise times will be
> > very close.
>
>TI CC1101 is not TTL.
> >From its datasheet:
>4.8 DC Characteristics
>Logic "1" input voltage (VDD-0.7V) to VDD
>
>For your 3.3V it will be >=2.6V.
>Obviously the presence of the bottom resistor of the divider would
>slow the signal rising to that level, the times with/without the
>resistor should not be "very close".

Well, if you connect the same upper resistor to 300 DC it will have an even
faster rise to 0.7* 3.3 VDC, but it's precisely as pointless a comparison
as your example since it violates the abs. max. ratings of the chip and
thus is not a valid solution.

Do you know the simple equation that the v(t) follows?

>Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com



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