Marechiare 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".
> 

Using the word "obviously" doesn't make it so. Show me the math.

If you can't drive your chip's input with 5 volts, you either use a
divider or some form of logic conversion from the 3.3V supply.

A divider from 5V logic using 2k/3K resistors is very close to a 1.2K
resistor from 3.3V logic. Using a FET or other active device for
pulldown and a resistor for pullup will speed the fall time, but the
rise time will be the same.

I'm not trying to be cranky or fight about any of this, I just want to
make sure you are clear on the electrical models we're discussing. Olin
is on very solid ground.

Cheerful regards,

Bob

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