#1 Capacitive charge sharing

This video describes charge sharing between two capacitors. Charge sharing between capacitors is a fundamental concept of circuit design. This concept is used in various types of circuits such as charge pumps and switched capacitor circuits. Video discusses various ways to solve the problems and common mistakes. This video is aimed at electronics graduates and entry level analog design engineers.

[@giuseppemacera5125]

Extremely clear and refreshing! Good to watch in all circumstances!

[@Digitisinglife]

A very nice concept. Quite useful. The best part is how not to solve a problem.

[@analogsnippets]

Thanks Hemant...

[@sajeshp3524]

Great, interesting. Will be watching other videos too. Thank u

[@arthurm7846]

Great video! Well explained and shows the pitfalls. Thanks!
p.s.: Nothing big, but maybe you can consider to write the units within [] or a bit more spaced to avoid one sec or two of misinterpretation of the equation in red [Minute 4:35].
Excellent video nonetheless! 😀

[@akshatsingh1435]

Amazing video,helped a lot

[@hispanicatthedisco2447]

This guy gets a big pog in my book

[@vpraj24]

Nice video

[@prashanthr3776]

Amazing 😉

[@Rajat0607]

Hello, Could you pls share, which software you are using to record these videos ? Thanks

[@analogsnippets]

Explain everything

[@98505177229850590818]

very useful video

[@sudhanshusingh1234]

Very nice explanation

[@analogsnippets]

Thanks Sudhanshu...

[@anandkumarpandey474]

Can you suggest some sources to practice more questions on these?

[@analogsnippets]

Not aware of any one good source. Questions at the end of text books should be good starting point.

[@yosefpatinio3669]

What is the sum of capacitors in series ,when the capacitors are charged?

[@Aadhyacedt]

Dear sir-amazing way of explaining. Could you please spend some time on familiar charge pumps like nakagome and other fly cap based circuits?

[@analogsnippets]

Dear Raja Sekhar, definitely something for future...

[@analogsnippets]

Dear Raj Sekhar. Thank you for providing your comment on the typo of current mirror mismatch video. I have corrected the typo.

[@venkatamanikantaprasadmand5793]

if switch is having on resistance of Ron,what will happen to steady state voltages

[@accidental.traveller]

Steady state voltage is independent of switch Ron. The effect of Ron is on settling time.

[@biprobhadra1102]

Nothing whenever the capacitor will settle down with a common voltage then current through R will be zero.

[@user-ep4hz6nq5y]

Sir at 2.14 ,
If initially the switch is open then How can we add charge 5+3

[@accidental.traveller]

5+3 is the total charge stored in these two capacitors. So we simply add the charges of individual capacitors.

[@sibidhayalan5919]

Can you explain how much time it takes to reach VF ? charge to discharge to capacitor? and can you explain different time interval how much will be the node voltage???

[@analogsnippets]

Charging and discharging time depends on switch or any other resistance in the path. Charging process is exponential and would take a few RC time constants. Here we are assuming that switches are on for several time constant so that Charging time is small.

[@sibidhayalan5919]

@@analogsnippets Hi, Thanks for quick reply!. let say RxC = 50us, at time 10us what will be node voltage on two capacitor positive terminals before reaching Vf

[@alterguy4327]

ThankYou

[@jazzjohn2]

Do you know where the missing energy is? Apply the stored energy formula to each cap using CV^2/2. The total energy stored after the caps take on the same voltage is LESS than the total energy before the switch is closed!

[@analogsnippets]

Rest of the energy is dissipated during charge transfer. Energy can be dissipated in switches, as heat or as radiation. Charge transfer between two caps with switches is always lossy.

[@sam-zm9yl]

👍👍

[@SAhellenLily]

So that’s it👍

[@yunusdification]

Capactive DACs also used this concept

[@vpraj24]

Good