How to get Cs and Rs?

Discussion in 'General Electronics Discussion' started by z0d, Sep 7, 2015.

1. Hello,

I'd like to know how to get Cs and Rs. . Here I've substituted arbitrary values for both Rs And Cs; only R1 is set at 100Ω[ I varied this from 10Ω - 150Ω ] . So from ltspice I got an output like this But if I hook this circuit up to an osci [with Cs and Rs unknown] I get this .

Any idea?

From that graph, I kinda got Cs through this - which is just a magnified pic of the blue output slope - by measuring the tau = 63.2% x Uoutput (max) and then Cs = tau/R1 = 0.380nF. How to get Rs? Not quite certain here. Think I'm doing it wrong somewhere.

Thanks!

2. LvW

604
146
Apr 12, 2014
At first, YOU have to answer some questions in order to let us know what you want.
1.) What are the purposes and the requirements of the circuit?
2.) What was the input signal for the LTSpice simulation?
3.) Input signal for oscilloscope measurements?

Old Steve likes this.

3. Hellow LvW,

ah, input is just a ±1V 4kHz signal. and Rs & Cs represent the unknown internal impedance of a sensor. SO what I did was add a known Resistor in the front so that the outputs shown on the Osci would represent the input voltage ( this being the YELLOW line ) in relation to the output voltage ( BLUE line ) and HENCE I can then calculate the unknown values from there.

I'm sorry for the lack of information. I just thought it might be a known formal procedure and if anyone who has done something like that could immediately see what I lack, be it in the circuit or calculation.

So, my bad. Is there anything else that I am missing out?

4. LvW

604
146
Apr 12, 2014
OK - what I see is the following (although I cannot find Rs and Cs in your circuit):
* You have a passive lowpass with a cut-off frequency of app. 4 MHz.
* Your input signal is squarewave with a frequency of app. 4 kHz.
* It is, therefore, no surprise that no real filtering takes place - and the output voltage is nearly identical to the input voltage.
* I think, the oscilloscope shows these signals (the "rounded" edges are due to the very weak filtering).

5. Hey,
thanks for your input! I went back and thought about it. Looks like I was taking a wrong approach.

6. Oops. Wasn't complete from above. Yeah so Rs and Cs represent the sensor and since I used a 100Ω outer resistance, which was more or less the same as that of the sensor, the I/O were more or less the same. I'll check further. Thanks again and have a good day!  