Need help to calculate time in an RC circuit with a AD converter and a multiplexer!

adcmultiplexer

I am not sure how to describe this in english, but i will give it a try.

I have a unipolar AD converter with Vref=3.3 V and 12 bits. The digital output is d=969 so that gives me the Vin=780.879 mV.

I also know that the LSB=805.664 µV.

So now a analog multiplexer (with inner resistance) is connected to the ADC. It looks like the image shows.

Circuit

Before the AC starts i want the voltage to reach 1/4 of the LSB value. C is uncharged and the multiplexer pluggs in a sender that gives the voltage 0-Vref and has a inner resistance of 7 kΩ.

I need to calculate the time to reach ¼ LSB in the worst case scenario if RMUX = 3 kΩ, C = 12 pF . Can someone help me with that?

I have a formula i found but i don't know if it is correct, and i am not completely sure on which U values to put in: Uc=Uc∞-(Uc∞-Uc0)e^-t/RC

Thanks!

Best Answer

You need to know (or estimate) your source impedance.

Assuming you turn on the switch at \$t=0\$, at which point the cap has a voltage \$V_C(0)\$, ans the source voltage and impedance is \$V_S\$ and \$R_S\$, the voltage \$V_C\$ across the capacitor as a function of time is: \$V_c(t) = V_c(0) + (V_S - V_C(0))\times \left[ 1 - e^{\cfrac{-t}{C*(R_\text{mux}+R_\text{S}) }} \right]\$

The worst case time is for Vc(0)=0 and Vs=3.3V and maximum source impedance. Calculate the time at which Vc reaches 3.3V - 1 LSB and thats the time you need to wait.

Related Topic