Why are you still using 8 bit MCUs?

[AlfBaz]

Yes.

Z = v/i

Capacitor: i = C dv/dt

Z = v / (C dv/dt)

Z': Replace v by 2v; dv/dt becomes 2dv/dt:

Z' = (2v) / (2C dv/dt) = 2/2 v/(C dv/dt) = v/(C dv/dt) = Z

So doubling the input voltage does not change the impedance.

Edit: Sorry, mistake.

Thanks Mr P 
This may help me understand something I was looking at the other week

[dannyf]

am I close?

Probably not. a 10ns rise to (say 0-100%) on 1000v means that the signal gets to 1v much faster than 10ns -> the slew rate on a 10ns 1000v signal is much higher than that of a 10ns 1v signal.

The original assertion is simply wrong.

[AlfBaz]

am I close?

Probably not. a 10ns rise to (say 0-100%) on 1000v means that the signal gets to 1v much faster than 10ns -> the slew rate on a 10ns 1000v signal is much higher than that of a 10ns 1v signal.

The original assertion is simply wrong.

I'm all ears...
Please explain this then

4 sim runs

Here are 2 FFT's of a 1 volt 1kHz square wave. Top FFT has 1ns rise/fall time bottom is  10ns rise/fall time where we can see a substantial difference in the high frequency content



Here we have FFT's of a 1kHz square both with 10ns rise/fall times but the top one is 1000V and the bottom is 1V pk... remarkably similar in frequency content

[c4757p]

am I close?

Probably not. a 10ns rise to (say 0-100%) on 1000v means that the signal gets to 1v much faster than 10ns -> the slew rate on a 10ns 1000v signal is much higher than that of a 10ns 1v signal.

The original assertion is simply wrong.

Slew rate alone does not determine frequency... assuming a reference level 0dB being the amplitude of the fundamental, the frequency spectrum of the two signals will be exactly identical.

[nctnico]

IMHO this is one of the situations where something works counter intuitive.
The maximum amplitude for an amplifier (lets say a signal source) with a given bandwidth is: slewrate/(2*pi*bandwidth) (*)
If we take a look at the risetime we would need a 100kV/s slewrate for the 1V/10ns square wave and a 100MV/ns slewrate for the 1000V/10ns square wave.

If we use the formula to calculate the bandwidth for a given slewrate and amplitude then it becomes: bandwidth=slewrate/(2*pi*amplitude)
Fill in:
bw1=100k/(2*pi*1)=15.9kHz
bw2=100M/(2*pi*1000)=15.9kHz

So the amplitude of a signal has no influence on the bandwidth. I also did some simulations with a 10MHz signal (which is closer to the frequencies of interest) to check the math but there is no difference in the frequency spectrum.

Ofcourse the amplitude of the 1000V square is much higher so relative to the 1V square the amplitude of all the frequencies present in the signal will be 1000 times stronger.

(*) More math here:
http://www.ece.uprm.edu/~mtoledo/5207/2011/bw.pdf