What is the point of the following capacitors? - Page 1
Forum section: electronics / beginners

author: Shaun_G
author_rank: Newbie
timestamp: November 03, 2021, 10:30:43 pm
content: Sorry if this question is dumb, I'm still a student...Anyways, I was looking at a design  of a high voltage differential probe on the web and was wondering what is the purpose of the capacitors seen in C1~C8, as well as C14~C15.I'd assume that the capacitors C9~C12 are meant to be used to achieve a low pass filter in conjunction with R1~R8, and that R18 is meant to be used to control the current feedback on the op-amps' gain. (Is there a reason the four resistors on R18 are in one package? and that two resistors are connected in parallel? why not just a 0.5k resistor?)However I don't understand what is being achieved by placing these capacitors (C1~C8,C14~C15) in parallel with a resistor in the signal path. Is there a formula that describes the signal after going through the circuit?…I looked up information on similar circuitry on google and only found one website saying that it is a "pre-emphasis" circuit . (is it a high-pass filter?)One last question that I have is how does the attenuation selection circuitry seen near SW18 works. Isn't U1A and U2A amplifying two different signals?Also, does anyone have resources for low voltage differential probe designs? If so, it's greatly appreciated.:https://circuitcellar.com/research-design-hub/high-voltage-differential-probe/:http://www.repeater-builder.com/tech-info/pdemph-post.html

author: bob91343
author_rank: Super Contributor
timestamp: November 03, 2021, 11:44:39 pm
content: Those are compensation capacitors. The ones at the input are intended to swamp the stray capacitance and offer flat frequency response at the expense of input impedance. They are calibrated by VC1 and VC2 to obtain flat response, usually by adjusting for flat top of a square wave input signal.The others are either compensation for feedback (avoiding oscillation) or integration; probably the former.

author: Manul
author_rank: Super Contributor
timestamp: November 03, 2021, 11:57:07 pm
content: I will try to explain simple. Imagine you use a resistor divider to achieve 10:1 voltage output. All is good in DC, but what about AC? Your divider output goes into some circuits which has input capacitance, there is also parasitic capacitances all over the place. At some high enough frequencies, capacitances begin to present a significantly low impedances in parallel with resistor divider arms. Divider starts to deviate from 10:1 at AC. So it equals to measurement error if it is an instrument meant to measure.On the other hand, you could make a voltage divider out of two capacitors at the ratio 10:1, just like resistors. It will be accurate for AC, but obviously will not pass DC. What you see in this circuit is a resistor divider combined with capacitor divider (in parallel). It is accurate at DC and it is accurate for AC (within some reasonable bandwidth). It is called compensated. In practice it is not always so simple, but thats the idea anyway.Edit: And for amplifier it looks like "intrumentation amplifier" configuration. Look it up and study.

author: David Hess
author_rank: Super Contributor
timestamp: November 04, 2021, 04:41:42 pm
content: C1 through C8 create a controlled AC attenuation which matches the DC attenuation of the resistor chain. Without them, the AC attenuation will depend on unmatched stray capacitances. VC1 and VC2 trim the AC attenuation to match the DC attenuation.C14 through C17 do the same thing for the differential amplifier stage. The result is a flatter AC response over frequency.