Electronics > Beginners

RC Circuit Natural Response Implementation

**Mechatrommer**:

ok ok i know, this is the 1st semester 1st year ee subject. but i'm the slow learner. so to fellow mentors, pls verify my method...

the objective is to measure the value of C from the decaying graph (picture) using the basic rc natural response formula thats on the page i'm opening infront of me right now (red text below).

assuming my DMM is good and accurate enough to measure the resistant of a resistor. so i'll work out the C value.

so from the graph...

Vo = 3.48V, V(t) = 0.56V, t = 0.25s,

DMM measured R = 130.8KOhm

so using the formula, i got C = 1.0462uF, this is pretty close to DMM reading,

but earlier i was testing the "said to be" 1nF smd capacitor, DMM confirm the value is close to that,

but using this method, using DMM measured 10.568MOhm, i calculated the C as 0.4-0.6nF, quite far from DMM reading.

so pls advice on any possible error source of this method of finding C using the formula below.

**jahonen**:

I think that the scope (Ri=10 M) is loading the circuit so that decay is faster than expected. Or, you can take it into account, in fact, it seems to be half of the expected value which seems to be right.

Regards,

Janne

**Mechatrommer**:

ok now with 1MOhm built in Rigol DSO, i dont have to use external R for small F. i measured and confirmed the input impedance of the Rigol is 1.001MOhm. so working it out

from the graph below and formula above, the capacitance is calculated to be:

if R 1.000M... C = 1.027nF, but

if R 1.001M... C = 1.026nF, close!

so i calculated how much small the Rigol DS1052E can measure a capacitance...

say its sampling at 1GSa/s = 1ns/Sa, we take 10 points to be comfortable = 10ns = t

working out R = 1MOhm, t = 10ns, and say Vt/Vo = 0.1... C is... 4.34fF

4.34 femto Farad, or 0.00434 pF! is this for real? sounds like a science fiction?! ???

edit: yes its a delussional science fiction :(

**Mechatrommer**:

i setup pic12f509 program to make latched switch as my finger is far from good for a stable switching OFF (natural respond). the diagram as below. ie when pin 2 is high, pin 5 will transit cleanly from low to high vice versa (as shown in ch1 graph, high to low).

procedure:

1) all measurement is done at node 1 and ground.

2) measure impedance using DMM while dso and pic are turned ON and connected to circuit (dso to node 1), pic is output low at pin 5 (i assume tied to ground), you'll get Xi (internal impedance)

3) calibrate (top graph) by removing C (DUT) and measure. using natural response formula, you'll get Ci (internal capacitance, incl cable etc)

4) measure the DUT capacitance... by placing it as in the schematic, using the formula and subtract the Ci, you'll get roughly the DUT C value

5) redo (4) several time to get more confidence in C reading, as it can jump up and down a little bit. more answers, you'll better picture which one is more reliable.

two graphs below is my measurement on a 2.2 pF smd cap (smallest i have). from calculation...

Xi = 127,390 Ohm

Ci = 22.67 pF

after subtraction, the DUT is measured as 2.6 pF. not very close but can be used as rough estimation. my Uni-T read 13pF after several minutes of wait, hopeless!

**saturation**:

This is beautify done work, and great photos. You have it now and answered the question raised here:

https://www.eevblog.com/forum/index.php?topic=3126.msg41723#msg41723

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