Phase angle of RC circuit

I could not really tell what you were trying to say. Nevertheless, I think I can explain how these probes work. This assumes that inductance (including strays) is not significant.

The resistor divider enables slow input signals to be divided accurately. To take care of high frequencies, a capacitive divider with the same ratio as the resistive divider is used. The biggest capacitance item usually is the input capacitance of the scope. To this is added the capacitance of the cable from the probe to the scope input. To get the capacitance ratio correct, a variable capacitance is usually placed in the probe. This is adjusted to the correct ratio while observing a square waveform on the scope. This division ration increases the input resistance and decreases the parallel capcitance at the probe compared to that at the input connector. This increased input impedance is achieved by reducing the signal level at the scope input connector.

The signal can be distorted by using a probe by its capacitance shunting out high frequencies. However, distorted as it may me, a good probe will present a smaller copy of the waveform to the scope as it exist in the circuit. This distortion or detuning if you wish, is why you try to keep probe capacitance small. Thus, a 100:1 probe will present less capacitance to the probed point than a 10:1 probe. The cost again, is a reduced signal at the scope.

You can go through the detailed circuit analysis, but it will not give you much greater understanding.

Bill

Thanks Bill. The probes were just an example being used while learning the circuit analysis though.

--------------------------- You have, in the formula, expressed things in polar form (magnitude and angle rather than real and imaginary). Go back to the rectangular form and work the Z2/Z1+Z2 relationship out then put both numerator and denominator in polar form to derive the formula. It is a good exercise in learning.

What is the phase angle of 1+jx? What is the real part and what is the imaginary part?. Now consider 1/1+jy). What happens to the phase angle? How do you treat the phase angle in polar form division (A @a)/B @b= ( Magnitude of A/B) @ phase? In other words, basic complex number manipulation.

By the way it would help to define T and Te because it would be nice to know which R-C is related to T and which to Te and do you mean R2/{(R1+R2)* .....} or {R2/(R1+R2)}*.....

Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer

If A@a = B@b and I divide (A @a)/B @b= ( > Magnitude of A/B) @ phase) =1

Is this correct?

Then if I remember correctly in polar notation you divide the magnitude by the magnitude and the angle by the angle to get the answer.

Is this correct?

Thanks Professor. peace dawg

It's been awhile but I seem to recall that the angles use the same rules as exponents, IE: subtract for division, add for multiplication, multiply for raising to a power etc .... bg

------------ bg has it right. the angle in this case would be a-b Remember that the complex number usage is based on De^jd =D (cos(d)

n message

I'm trying to understand an analysis of a oscilloscope probe

Thanks Don, and everyone else.

Considering 1/1+jy made things "click" for me. For some reason the idea of converting the numerator to polar, converting the denominator to polar, and then dividing the two of them didn't cross my mind. Now things are making sense.

Thanks again.