As per my comment, you can't assume that the capacitors are open circuits. On the rising edge of the step C1 initially gets all of the current, so it starts to charge up with the R*C1 time constant. But as it charges it's impedance increases and it no longer gets 100% of the current. It forms a voltage divider with the other circuit elements. This vastly complicates the transfer function. I had to use MATLAB to help me.

Take a look at the following picture:

V_C1 initially charges faster than V_C1_Test, but then as C1 becomes a larger impedance the voltage starts to slow down it's climb. You no longer have a nice pure 1-e^-t type of equation like you do for the V_C1_Test case. If it was the V_C1 voltage would either be strictly larger or smaller than V_C1_Test Voltage.

All I did in the MATLAB code above is find the total impedance and use voltage division to find the V_C1 voltage.

I made a function; 'll(x,y)' which returns x*y/(x+y). Kinda checky IMO.

Below I compare the step response of my analytical answer with LTSpice Simulation. The left waveform is my MATLAB output for V_C1, the right waveform is my LTSpice output for V_C1 (I normalized the voltage to 1V). As far as I can tell they are identical.

For fun, here's the transfer function. Oh how far it is from 1st order!

symbolic to TF code: https://www.mathworks.com/matlabcentral/answers/310042-how-to-convert-symbolic-expressions-to-transfer-functions

As per my comment, you can't assume that the capacitors are open circuits. Each cap acts as a short when presented with a high frequency. On the rising edge of the step C1 initially gets all of the current because R2+R5 is a large impedance in series with the rest of the circuit. C1 starts to charge up with the R1*C1 time constant. But as it charges it's impedance increases and it no longer gets 100% of the current. It forms a voltage divider with the other circuit elements. The other caps also act as low impedances initially until steady state is approached. This vastly complicates the transfer function. I had to use MATLAB to help me.

Take a look at the following picture:

V_C1 initially charges faster than V_C1_Test, but then as C1 becomes a larger impedance the voltage starts to slow down it's climb. You no longer have a nice pure 1-e^-t type of equation like you do for the V_C1_Test case. If it was the V_C1 voltage would either be strictly larger or smaller than V_C1_Test Voltage.

All I did in the MATLAB code above is find the total impedance and use voltage division to find the V_C1 voltage.

I made a function; 'll(x,y)' which returns x*y/(x+y). Kinda checky IMO.

Below I compare the step response of my analytical answer with LTSpice Simulation. The left waveform is my MATLAB output for V_C1, the right waveform is my LTSpice output for V_C1 (I normalized the voltage to 1V). As far as I can tell they are identical.

For fun, here's the transfer function. Oh how far it is from 1st order! I also have a case where C3 = 100nF which emphasizes the changing response.

symbolic to TF code: https://www.mathworks.com/matlabcentral/answers/310042-how-to-convert-symbolic-expressions-to-transfer-functions

As per my comment, you can't assume that the capacitors are open circuits. On the rising edge of the step C1 initially gets all of the current, so it starts to charge up with the R*C1 time constant. But as it charges it's impedance increases and it no longer gets 100% of the current. It forms a voltage divider with the other circuit elements. This vastly complicates the transfer function. I had to use MATLAB to help me.

Take a look at the following picture:

V_C1 initially charges faster than V_C1_Test, but then as C1 becomes a larger impedance the voltage starts to slow down it's climb. You no longer have a nice pure 1-e^-t type of equation like you do for the V_C1_Test case. If it was the V_C1 voltage would either be strictly larger or smaller than V_C1_Test Voltage.

All I did in the MATLAB code above is find the total impedance and use voltage division to find the V_C1 voltage.

I made a function; 'll(x,y)' which returns x*y/(x+y). Kinda checky IMO.

Below I compare the step response of my analytical answer with LTSpice Simulation. The left waveform is my MATLAB output for V_C1, the right waveform is my LTSpice output for V_C1 (I normalized the voltage to 1V). As far as I can tell they are identical.

For fun, here's the transfer function. Oh how far it is from 1st order!

symbolic to TF code: https://www.mathworks.com/matlabcentral/answers/310042-how-to-convert-symbolic-expressions-to-transfer-functions

As per my comment, you can't assume that the capacitors are open circuits. On the rising edge of the step C1 initially gets all of the current, so it starts to charge up with the R*C1 time constant. But as it charges it's impedance increases and it no longer gets 100% of the current. It forms a voltage divider with the other circuit elements. This vastly complicates the transfer function. I had to use MATLAB to help me.

Take a look at the following picture:

V_C1 initially charges faster than V_C1_Test, but then as C1 becomes a larger impedance the voltage starts to slow down it's climb. You no longer have a nice pure 1-e^-t type of equation like you do for the V_C1_Test case. If it was the V_C1 voltage would either be strictly larger or smaller than V_C1_Test Voltage.

All I did in the MATLAB code above is find the total impedance and use voltage division to find the V_C1 voltage.

I made a function; 'll(x,y)' which returns x*y/(x+y). Kinda checky IMO.

Below I compare the step response of my analytical answer with LTSpice Simulation. The left waveform is my MATLAB output for V_C1, the right waveform is my LTSpice output for V_C1 (I normalized the voltage to 1V). As far as I can tell they are identical.

For fun, here's the transfer function. Oh how far it is from 1st order!

symbolic to TF code: https://www.mathworks.com/matlabcentral/answers/310042-how-to-convert-symbolic-expressions-to-transfer-functions