discharging a capacitor to a capacitor - Page 1
Forum section: electronics / beginners

author: Capernicus
author_rank: Frequent Contributor
timestamp: February 27, 2021, 07:00:24 pm
content: I just did an energy loss experiment.I charged an electrolytic capacitor, to 28.9 volts.I then discharged it to an empty, equal rated, same volts, same farads, capacitor, then read the volts on both at the end,I made sure I got them the right way around.it was about 11 point something volts, on both.that means I lost about 25% of the energy, does this sound right?

author: PartialDischarge
author_rank: Super Contributor
timestamp: February 27, 2021, 07:20:46 pm
content: Energy goes with the square of voltage, seems more energy loss, what are the values of the capacitors and what type are them?

author: Capernicus
author_rank: Frequent Contributor
timestamp: February 27, 2021, 07:23:03 pm
content: Thanks for reply, I dont understand what the square of the volts means... to do with energy...Wouldnt it be if they were both ~14.5 volts each that would be 0% loss?they are 50 volt.100 microfarad.Electrolytic.I actually want to do it through a spark gap with higher volt caps, then measure the loss through the spark gap after this.

author: PartialDischarge
author_rank: Super Contributor
timestamp: February 27, 2021, 08:13:33 pm
content: You should play a bit with this calculatorhttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html

author: Capernicus
author_rank: Frequent Contributor
timestamp: February 27, 2021, 08:21:31 pm
content: So why was it 11 volts on both capacitors, instead of 7 volts?Isn't the voltage stored on the capacitor how much it is filled?I didnt use a resistor, maybe I should try it, and check the loss. BRB.

author: TimFox
author_rank: Super Contributor
timestamp: February 27, 2021, 08:23:30 pm
content: The energy stored in a capacitor isE= (1/2)CV2.If your capacitor value for each cap isC, then the initial energy is (1/2)Cx (28.9 V)2.The total energy in both units after connection is (1/2) x (2C) x (11 V)2.What is the tolerance on the capacitance values?

author: Capernicus
author_rank: Frequent Contributor
timestamp: February 27, 2021, 08:25:17 pm
content: Hmm... Im getting very confused.When I used the 1k resistor they were 13 volts and 12 volts. but maybe i didnt hold it long enough on.Its as if it had no effect on energy loss, it just slowed the transfer process down.I dont know the tolerance, but if you hit the amount of volts, doesnt mean the charge got into the capacitor?I tested them with an led, and yes it was quite a strong blip on both of them.Since both capacitors are the same farads, doesnt that take it out of the variables for the experiment?

author: jpanhalt
author_rank: Super Contributor
timestamp: February 27, 2021, 08:35:22 pm
content: Explained here, if you really did that experiment:https://en.wikipedia.org/wiki/Two_capacitor_paradox

author: Capernicus
author_rank: Frequent Contributor
timestamp: February 27, 2021, 08:37:27 pm
content: Im really confused now.What I'm thinking now, is why did I even lose the 25%???!?!?Its even one of great paradi, on the internet, amazing...

author: jpanhalt
author_rank: Super Contributor
timestamp: February 27, 2021, 08:39:24 pm
content: Because the paradox is framed with "perfect" components and zero resistance. If you did it with real components, there are other losses.

author: Capernicus
author_rank: Frequent Contributor
timestamp: February 27, 2021, 09:15:09 pm
content: How the hell can an LC oscillator have 99% efficiency (they do ring for a fair amount of time.) when a capacitor loses 25% or 50% when it goes to another capacitor.How could it be any different than discharging to an inductor?here:https://www.giangrandi.ch/electronics/ringdownq/ringdownq.shtmlthis is clearly more than 75% efficient per rebounce.Doesnt this mean if you bounce off an inductor to the next capacitor, you get 98% efficiency or something decent?

author: rfclown
author_rank: Frequent Contributor
timestamp: February 27, 2021, 09:42:34 pm
content: The answer, even in a perfect world with ideal lossless components, is very surprising.Like the other poster said, the energy in a capacitor is E = (1/2)CV2Suppose I have a capacitor charged with voltage V, and I connect it through a resistor (R) to another capacitor of the same value with initially no voltage. The total energy in the system before connecting the second capacitor is (1/2)CV2.At some time, the voltage between the capacitors is the same, and current stops flowing. The energy in the system is then 2 x (1/2)CVT2, where VTis the terminal voltage. There are several ways to figure VT,but here's one. The voltage of the first capacitor will decrease with a time constant of RC/2 (since the two capacitors in series will have a combined capacitance of C/2). So the voltage equation for the first capacitor is V x e-t/(RC/2). The second capacitor will similarly charge with the equation V x (1-e-t/(RC/2)). If you figure out where these two equations are equal, you will find that it is at t=-(RC/2) x ln(1/2), and if you plug that back into the voltage equations, will find that the final voltage is V/2, and isindependentof R and C.So the energy in the system in the final state is 2 x (1/2)(V/2)2= (1/4)CV2. So we have lost half of our energy. Where did it go? If you do the math, you will find that the energy is disipated in the resistor. But the really crazy thing is (and I encourage you to do the math), as is the final voltage state of the system, it isindependentof R. When you are through manipulating the equations, you find that R falls out, and the energy lost is purely a function of V and C. So even if the resistance is zero, the energy is still lost in the resistor.Try it (calculating the energy lost in the resistor). I=V/R. P=VI. E=integral of P over time. If you do the integration, you will get a result of E=(1/4)CV2(R drops out of the equation). It's bizzare.

author: rfclown
author_rank: Frequent Contributor
timestamp: February 27, 2021, 09:52:10 pm
content: Quote from: jpanhalt on February 27, 2021, 08:35:22 pmI hadn't looked at the link before I replied. Interesting. I hadn't run across an explanation of other possible energy loss mechanisms before. I went through the math many years ago when I was curious to know where the energy went. I was surprised at what I found.

author: TimFox
author_rank: Super Contributor
timestamp: February 27, 2021, 09:58:07 pm
content: The usual explanation for the zero-resistance case being the same as finite resistance is that there is always inductance in the loop formed by the two capacitors, therefore the circuit will "ring down" as shown in the CRO trace. During that ring-down, the circuit radiates out to the distant galaxies, even if the jumper wire is a superconductor.Also, since the capacitors are electrolytic, the ESR would be significant, although the system is independent of resistance.

author: penfold
author_rank: Frequent Contributor
timestamp: February 27, 2021, 10:01:12 pm
content: (apologies rfclown didn't spot your post at first, looks like you had the same explanation)One thing that may make it clearer, if you take a few different resistances and repeat the experiment, leaving it long enough to settle (or do it in Spice) you will find that as you use progressively smaller resistors, the energy loss stay the same.From the moment you connect the caps together, with 1k you will have a very small power loss (initial voltage x "low" current) but it lasts a long time as the current flows and the capacitor voltages equalises, at 1 ohm you have a very high initial power loss (initial voltage x "high" current) but it doesn't last very long.. resulting in an equal total energy loss by the time the voltages have fully stabilised. I wouldn't dare speculate where the energy goes.. but I'd always back Kirchhoff in a circuits based problem.If you mentally extend that to zero ohms between the caps, you have an infinite current flowing for zero time in zero ohms... interestingly if someone wanted (really, really wanted to) to solve the differential equations (properly) I'm about 99% sure (infinity)^2 x 0 x 0 will give the same answer of half the total energy.Also curiously, the mean energy (total energy of the two capacitors divided by 2) is equal before and after they are connected together... similar to the conservation of momentum in gasses (or many particles bouncing around) where the mean-squared velocity is a useful figure

author: Capernicus
author_rank: Frequent Contributor
timestamp: February 27, 2021, 10:36:12 pm
content: whatever you do, dont put a diode, cause then its 0%.and 2 of the volts didnt even travel, not that I know cause it was 0...

author: rfclown
author_rank: Frequent Contributor
timestamp: February 28, 2021, 12:16:40 am
content: Quote from: rfclown on February 27, 2021, 09:42:34 pmJust realized the error in my statement. If R=0, the equations I have for voltage blow up (aren't valid). I recant. If there is any resistance (as in any realistic case), the energy is lost in the resistor. If R=0I'm not going to try to figure it out now.

author: mawyatt
author_rank: Super Contributor
timestamp: February 28, 2021, 12:36:33 am
content: This was an assignment over 50 years ago in undergrad school. You can solve by integrating the current thru the resistor R connecting the equal valued capacitors from t = 0 to infinity, you find that half the initial energy is lost in R, but independent of R!!Or the simpler approach is based on conservation of charge where initial charge Q=CV, with two capacitors in parallel the charge divided between both capacitors, so Q=2CVfinal, and Vfinal=Q/2C, or Vfinal=V/2 since V=Q/C. Now the initial energy is simply (CV^2)/2 and the final energy is (C(V/2)^2)/2 + (C(V/2)^2)/2 for both capacitors, or (CV^2)/8 + (CV^2)/8, or [(CV^2)/2]/2. Thus half the initial energy (CV^2)/2 is lost regardless of method of connecting the two capacitors; Cu wires, carbon resistors, superconducting nanotubes, niobium, YBCO, or BSCCO wires, or whatever you wish to connect the capacitors together with!!Best,

author: mawyatt
author_rank: Super Contributor
timestamp: February 28, 2021, 01:20:06 am
content: The differential equation approach is a little more complicated, but straight forward, so here goes without Google help (like the charge model above, no Google help).The two caps C are connected by resistor R, thus the current flows thru R and the caps C are now inseries, not parallel wrt the current (very important point!!). A little elementary differential equations and calculus will show the current behaves as I(t) = (V/R)e^(-t/tau), where tau is the time constant. Since the capacitors are inseries, then tau = R(C/2), thus I(t) = (V/R)e^(-2t/RC).So the lose in R is the integral of R(I^2) from 0 to infinity:the integral of R(V/R)^2 (e^2(-2t/RC)) orintegral (V^2/R)(e^(-4t/RC)) evaluated from t=0 to infinity is,(V^2/R)(-RC/4)[e^(-infinity)- e^(-0)]or (CV^2)/4Since the initial energy is (CV^2)/2, then half the initial energy is lost, independent of R. Same result you get with charge conservation as it should be.Best,

author: pwlps
author_rank: Frequent Contributor
timestamp: February 28, 2021, 11:09:33 am
content: Quote from: Capernicus on February 27, 2021, 08:37:27 pmI understand, I had a look at some articles on the internet and explanations are often not very clear for me.First of all there is no paradox from the energy consideration standpoint alone. Since W=1/2Q^2/C, transferring a charge Q from a small to a big capacitance (here C -> two C in parallel hence C->2C) will yield some energy, in other words, since energy is conserved the charge flow will necessarily do some work in the process. It is the same as if you were transferring water from a small to a big tank: if the water level in the final state is lower than in the initial state then the energy of the system decreases and the excess energy must be released in one way or another (mechanical work, dissipation as heat etc.).A paradox arises ONLY when trying to model the processsolely using lossless discrete components so that there is no way the work done by the charge can be accounted for within the model. For me talking about losses by radiation etc. is pointless and confusing even more without this being stated, in the first place.

author: Siwastaja
author_rank: Super Contributor
timestamp: February 28, 2021, 11:45:10 am
content: Efficiency is always 50%. Loss happens in all system resistances. Energy loss depends only on capacitance and voltage difference. Amount of resistance only defines the time constant. When R approaches 0, t approaches 0 and P approaches infinity.From practical engineering viewpoint, the problem is what could be calledlinear regulation; due to voltage mismatch, excess voltage is burned away in resistances because there is nopower conversionstep like a DC/DC converter. This is pretty obvious if you allow yourself to draw a discrete resistor, then run time-step analysis on pen&paper, for example dividing the charging event to ten timesteps, doing normal Kirchoff stuff to calculate current and voltages, and energy stored in capacitors.Initially, when one capacitor is at empty and another at full voltage, loss is 100%.Finally, when the capacitors are at nearly equal in voltage, loss is nearly 0%.Average loss is 50%.DC/DC converters solve this problem for example by storing energy in the magnetic field of an inductor so that the same energy can be released at different voltage.Some people completely "shut down" at this problem, some of them start an infinite forum shitstorm as a psychological reaction to their brain not accepting this.

author: pwlps
author_rank: Frequent Contributor
timestamp: February 28, 2021, 11:48:21 am
content: Quote from: Siwastaja on February 28, 2021, 11:45:10 am50%

author: Terry Bites
author_rank: Super Contributor
timestamp: February 28, 2021, 01:18:14 pm
content: Q is conserved. Q=CV . After charging to Q coulombs the first capacitor has V volts on it. The charge it holds is CV. Connecting the second capacitor distributes the charge between them- the voltage drops. Assuming that the second capcitor has no reisdual charge, and the capacitors are identical, the voltage will be: V= Q/(C1+C2). But its unlikely that the two electolytic capacitors have exactly the same C so the expected and acutal voltages differ. Power losses are I.IR come from the capacitors internal resistance. Arcs and sparks also consume energy from the circuit. Q is lost so energy is lost. The total enrgy in the system is determined by the total charge. Energy is accordingy conserved. E=QV/2

author: TimFox
author_rank: Super Contributor
timestamp: February 28, 2021, 02:36:01 pm
content: This assumption of charge conservation is key to the calculations. It should hold unless there is parallel resistance (leakage) across one or both capacitors. Of course, electrolytics have more leakage than film capacitors. Energy is dissipated in series resistance, but charge is not lost thereby.

author: Jeff1946
author_rank: Regular Contributor
timestamp: February 28, 2021, 09:54:00 pm
content: Analogy. Two cars collide and stick together. Car 1 goes 10 mph and car 2 0 mph. After collision both going 5 mph by conservation of momentum. But only half the original kinetic energy (1/2 mv^2). This is an inelastic collision. In an elastic collision car 1 stops and car 2 goes 10 mph. This would happen if a giant spring were between them. You can see this if you strike one coin with another or even a row of coins where only last coin moves after the collision.Cars: p=mv E= 1/2 mv^2Capacitors: Q=CV E= 1/2 CV^2

author: amyk
author_rank: Super Contributor
timestamp: February 28, 2021, 09:57:16 pm
content: People were doing this with Leyden Jars almost 200 years ago... and discovered radio waves along the way.

author: Zero999
author_rank: Super Contributor
timestamp: March 01, 2021, 11:37:39 am
content: If an ideal diode were used and the whole system had virtually no resistance and just a little inductance, the efficiency would be close to 100%.I tried it with LTSpice and it generated some odd results, which are overunity.

author: Capernicus
author_rank: Frequent Contributor
timestamp: March 01, 2021, 11:42:37 am
content: The question is, why does it need a diode? what the hell is happening when there isnt a diode?

author: Zero999
author_rank: Super Contributor
timestamp: March 01, 2021, 11:55:14 am
content: Quote from: Capernicus on March 01, 2021, 11:42:37 amThe diode stops the capacitor from discharging back into the source.

author: Capernicus
author_rank: Frequent Contributor
timestamp: March 01, 2021, 12:01:16 pm
content: Yes of course, the energy loss isphase out, Its not off to the 4th dimension.Thanks, I needed a tap on the head for that one, I was close to working it out...What happens if you discharge a 100 microfarad into a 50 microfarad, there might be less loss because small capacitors short to in series from large capacitors.I need to test that yet.But still if the volts of the discharging capacitor are always above the charging capacitor, it actually wouldnt discharge at all, I dont understand fully yet...

author: Zero999
author_rank: Super Contributor
timestamp: March 01, 2021, 12:18:16 pm
content: Quote from: Capernicus on March 01, 2021, 12:01:16 pmDownload LTSpice and try the circuit I attached above, but with a 50µF capacitor.QuoteThe inductance of the wiring will keep the current flowing, like a flywheel.

author: Capernicus
author_rank: Frequent Contributor
timestamp: March 01, 2021, 12:44:13 pm
content: I just did it at my desk. strange, so you can end up with more volts...im just getting ltspice so i can see ur creation.

author: Capernicus
author_rank: Frequent Contributor
timestamp: March 01, 2021, 01:20:21 pm
content: With LtSPICE, im not sure what im looking at.I clicked the little running man, and then this black window came up, nothing happened,so I clicked the switch and then I got a result, but im not sure what im looking at...

author: Zero999
author_rank: Super Contributor
timestamp: March 01, 2021, 02:04:01 pm
content: Quote from: Capernicus on March 01, 2021, 01:20:21 pmLTSpice doesn't work like that. It's not a realtime, interactive simulator. The component values and states need to be set, before the simulation in run.he switch on my circuit is really a behavioural resistor. It just uses the switch symbol to improve readability. The expression R = if (time>20m,0.1p,1000T) is similar to a spreadsheet if function. R = if(expression, true_value, _false_falue), R is the value of the resistor and time is a global SPICE variable, which ticks upwards, through time. When the time is >20ms the resistor value will be 0.1p Ohm, a dead short, and when the time is <20ms it will be 1000T Ohm, an open circuit.

author: mawyatt
author_rank: Super Contributor
timestamp: March 01, 2021, 03:09:05 pm
content: Here's a quick note on this topic. It's based on Charge Conservation as I mentioned much earlier in this thread, which makes the solution rather simple, no DE or Calculus required.You have two capacitors C1 and C2, each charged to some voltage V1 and V2 respectively. They are connected by an open switch S and resistor R at time zero. The switch is closed and the final voltage is Vf at time infinity, so now each capacitor has the same Vf voltage regardless of R. Since the initial energy in the system is the sum of energy in C1 and C2, or (C1V1^2)/2 + (C2V2^2)/2, and the final energy is simply (C1Vf^2)/2 + (C2Vf^2)/2 = ((C1+C2)Vf^2)/2 since both capacitors have the same voltage Vf. So it's easy to see the final energy in the system is always less than the initial since the final voltage Vf will always be less than initial capacitor voltages V1 or V2. A few simple examples are shown to illustrate the results.Please check my algebra, I'm noted for making dumb arse mistakesBest

author: Siwastaja
author_rank: Super Contributor
timestamp: March 01, 2021, 07:45:05 pm
content: Quote from: Zero999 on March 01, 2021, 12:18:16 pmAs I stated earlier, it's all about voltage mismatch, which is burned away in all circuit resistances.But indeed, if the circuit uses inductance instead of resistance as you have drawn in Spice, excess voltage (which accounts for 50% of the original energy) is not burned into heat, but stored into magnetic field. Given the diode, energy stored in that magnetic field can be returned to the capacitor-to-be-charged.With such simple circuit though, it's hard to arrange conditions where L is large enough, and R is small enough, and you'd need that ideal diode as well. The problem with said combination of large L and small R is, to make large L, you need a lot of wire, which then will have large R. Or you can add a magnetic core to bring L up with way fewer turns of wire, but then your saturation current comes down.You might be able to demonstrate the circuit with superconducting wire.The sameideais utilized in switch-mode converters and they work very well; it's easy to demonstrate capacitor-to-capacitor energy transfer exceeding 90% efficiency with a synchronous buck or boost converter. The key is to utilize the energy storage properties of the chosen inductor by chopping the time in suitable periods, alternatively charging the inductor then letting it discharge. In other words, nothing prevents you from transferring energy from a 10V-charged capacitor at high efficiency into magnetic field of an inductor, then transfer that into a 1V-charged capacitor, again at high efficiency.

author: Capernicus
author_rank: Frequent Contributor
timestamp: March 02, 2021, 04:01:39 am
content: I extra volts I got, was because it was 150 farads total not 200, from downsizing one of the caps, just to make that clear, I worked it out later on. (im getting a bit slow these days) I had to go to bed and I didnt even have the strength to post again... head is pretty bad.How realistic is LTSpice? Its a good one?, I didnt want to trust circuit sims, when getting real results never lies. (Unless god is playing with your mind.)

author: CatalinaWOW
author_rank: Super Contributor
timestamp: March 02, 2021, 04:32:09 am
content: Quote from: pwlps on February 28, 2021, 11:09:33 amI really hope radiation losses aren't pointless. I actually like radio. And there are always radiation losses. They may be small relative to other losses, but never zero.For another way to create a conundrum by thinking only of an ideal case, go to the water tank analogy mention earlier. Simple ideal case the water levels equalize and energy is lost without explanation. So add a little realism and add friction to the pipe walls and you have an explanation of where the energy went. Now add a water powered noise maker to the plumbing. Sound is broadcast out of the system. But the water levels at the end are the same. Where did that extra sound energy come from?The problem is stated to make it mysterious. A hint to the answer is how does pipe friction change with flow rate.

author: Capernicus
author_rank: Frequent Contributor
timestamp: March 02, 2021, 05:39:00 am
content: Quote from: mawyatt on March 01, 2021, 03:09:05 pmThat is some pretty impressive mathematics, if its involving the loss from the wire inductance...

author: mawyatt
author_rank: Super Contributor
timestamp: March 02, 2021, 02:22:37 pm
content: Quote from: Capernicus on March 02, 2021, 05:39:00 amThanks, but just some simple algebra based upon conservation of charge.The more involved math that includes all cases is to use an RLC circuit, thus C1 and C2 with initial charge voltage V1 and V2, Switch that closes at time zero, and series R and L. Calculate the circulating current wrt time using Laplace or differential equation techniques, then integrate the resistive power loss R*I^2 result from t=0 to infinity to get the energy lost. Using the initial energy of the system in C1 and C2 due to V1 and V2, then subtract the energy lost from the integration, yields the resultant final energy in the system. Since the system is steady state at time infinity, then the final voltage on C1 and C2 must be equal and straight forward to evaluate based upon the final energy. If done correctly, the result will be the same as the conservation of charge solution.You can solve this without the use of Google. The solution will have a decaying exponential function times a sinusoidal function. Note the sinusoidal function will only exist if the roots of the quadratic are partially imaginary, or under-damped, if critically or over-damped then no sinusoidal just exponential. If L/R is such that the system is under-damped then a decaying sinusoidal current will be the result with a natural frequency related to L, C1 and C2.If I was still an adjunct Prof, I'd give this to a grad student to solve, but retired now...so no grad students available!!Best,

author: mawyatt
author_rank: Super Contributor
timestamp: March 02, 2021, 07:42:08 pm
content: As a note about the practical nature of the capacitive switching and energy loss, about 40 years we inherited a project that had been attempted by two very well known and respected technology companies. The project required the use of many very long custom CCDs (math sine and cosine type convolvers) that had multiple clocks with very high capacitance (>1000pF) and needed to be driven to 15 volts at 10MHz. If you follow the above discussions you have a >1000pF chip clock capacitor per clock phase that must be charged and discharged to 15 Volts every 100ns, so the CFV^2 power required is >2.25 watts per clock phase. There are 4 phases per CCD chip, so total power per chip was >9 watts, and the system had multiple CCD chips and other electronics but had a power goal of less than 10 watts (battery powered) for the entire system!! One of the previous technology companies had patented an idea to use a stepped transformer like we see in our modern lab power supplies to create the stair-step clock signals for each clock phase, and we then knew why!!So not to infringe on the previous technology company (Don't tick off the Wokie!!) we set out to find another solution. We came up with a Reactive Clock (we should have patented but didn't) which would charge the CCD clock capacitance to 15 volts thru an inductor, then hold the voltage and 15 volts, then discharge thru the same inductor to ~ zero volts and hold at ground. The "holding" at 15 volts and ground was accomplished by a small shunt FETs, and the charging up and down was with a series larger FET which interrupted the inductor current crossing zero. The clock voltage waveform was an offset sinusoidal and interrupted at the upper peak, held then allowed to ring down toward ground and interrupted at the lower peak. This half sinusoid rising and falling edge didn't bother the CCDs, in fact this waveform was preferred for enhancing CCD performance rather than the usual fast clock edge.A funny thing happened during testing, we were running the Reactive Clock from a +7.5V supply, which would allow the clock waveform to ring up 2 times to the required +15 volts, then back to ground. I accidentally knocked the +7.5V supply lead off, but the system kept working and the clock waveform still was on the scopeAfter some investigation we noted the system power level dropped without the +7.5V clock power, +15V supply current droppedWhat had happened is the +7.5V supply voltage was being created by the Reactive Clocking scheme and stored on the decoupling caps. After some detailed analysis it was found this ~7.5V supply voltage would be naturally created to minimize the clocking overall power consumption, and self correcting for timing and component errors, including temperature and agingAs much as I would like to say we knew that, and this was the way it was designed to work, it was discovered purely by accident...then the analysis was done to fit what was observedAnyway, the Reactive Clocking scheme saved over 90% of the conventional clocking power, the total system consumed under 4 watts, and it worked beautifully!!Best,

author: Capernicus
author_rank: Frequent Contributor
timestamp: March 02, 2021, 07:56:41 pm
content: A capacitor charge isn't slower than a square wave generated by a transistor.That's one thing I'm going to proove before I die. (which will probably be soon via accidental lightning bolt.)If you short a capacitor (I mean charging it), and switch on a transistor, the capacitor just flat lines, and the transistor is just a big slow slurpee.I want to see that someday, cause it should beat the nyquist frequency of the most high end oscilloscope on the planet, in some quantum scientist's lab.

author: mawyatt
author_rank: Super Contributor
timestamp: March 03, 2021, 01:21:47 am
content: Hang in there because there's a new type ADC coming that could make some very interesting Digital Oscilloscopes. This ADC quantizes the input signal in both amplitude and time and lends itself to advanced Digital CMOS processes. The latest Digital CMOS allows very precise high resolution timing, which this ADC directly benefits from. One of the unique properties of this ADC is that the waveform creates it's own Nyquist limit, not the fixed sampling rate like in conventional ADCs. Because of this property no conventional analog pre-ADC anti-aliasing filter is required, the anti-aliasing filter is introduced post ADC conversion!! This was part of our DARPA program a decade ago, and Dr Mike Chen at USC was the emphasis behind the Non-Uniform Sampling ADC.Fun stuff ahead indeed!!Best,