From this video lecture how come the initial current values are different in two cases?

I think they're assuming that the switch resistance is small in the first case. However how about if they're the same switch and hence same resistance, just different capacitance to make one slow switching and the other fast swtiching?

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  • \$\begingroup\$ Losses are proportional to the magnitude of ripple. When switching period is small compared to RC, the capacitor gets charged more frequently and the ripple will be small. \$\endgroup\$ – across Mar 20 '20 at 8:21
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    \$\begingroup\$ @beccabecca the lecture made it looks like increasing switch resistance will have higher efficiency but I doubt it \$\endgroup\$ – anhnha Mar 20 '20 at 8:28
  • \$\begingroup\$ In the expression RC, the R refers to the load resistor, not the switch resistance. \$\endgroup\$ – across Mar 20 '20 at 8:30

While the first diagram is pretty and simple, it's too simple to be worth simulating or drawing a graph for the theoretical current. It never occurs. It is a picture of a switch with zero resistance, connecting components with zero resistance. The initial current will therefore be infinite, which is non-physical. Think of it as that hoary old puzzle, an irresistable force pushing an immovable object. Given the terms as presented, it's not analysable.

The graph they present shows a decay to the current over time, which implies some resistance, contrary to the diagram. They say 'high losses', but where is the dissipative component? There is no resistor shown. When we draw schematics, all components are assumed to be ideal, a capacitor with no ESR, a battery, switch, and wires with zero internal resistance.

The second diagram rolls up the series resistance of the switch, the power source and the capacitor into one R, which then controls the initial current. For small R the initial current will be large. For large R, the initial current will be small.

Interestingly, for the same capacitor and voltage, the losses are independent of the value of R, once the capacitor has become fully charged. Although in theory, the capacitor never becomes 'fully' charged, in practice, the error becomes exponentially smaller over time, and after 5 or 10 RC time constants (depending on your need for accuracy) it is generally considered to be fully charged.

  • \$\begingroup\$ However, the lecture doesn't really imply the first one has zero resistance. They're trying to say that the first one has time constant much smaller than the switching period while the second one time constant is compared to switching period. What you said only apply for slow switching limit. My question above doesn't assume that the first one has zero resistance but small resistance so its time constant is much smaller than switching time. Assume that 1 and 2 have the same switch resistance but different capacitance. Then how can 2 help in reducing loss? \$\endgroup\$ – anhnha Mar 20 '20 at 8:16
  • \$\begingroup\$ If the first one has a non-zero resistance, then why doesn't it show an non-zero resistance? The second one shows the graph terminating at less than full charge, so the losses will be lower, as there has been less energy transferred. When charging a capacitor through a resistor to completion, the resistor always dissipates the same amount of energy as the capacitor stores, so 50% charge efficiency. Do the integration if you don't believe me. It's a really sloppy lecture that's OK for a quick look, but too many errors to analyse deeply and draw conclusions from. \$\endgroup\$ – Neil_UK Mar 20 '20 at 8:21
  • \$\begingroup\$ They didn't show the resistance in 1 because it's really small. The energy loss 50% energy loss is easy to calculate and only correct if the initial capacitor voltage is zero. Assume that 1 and 2 have the same switch resistance but different capacitance. Then how can 2 help in reducing loss assuming that they transfer the same amount of charge? \$\endgroup\$ – anhnha Mar 20 '20 at 8:26
  • \$\begingroup\$ is 'really small' finite, or zero? If zero, don't show it. If finite, show it. Perhaps the losses they talk about are the loss of a switch as it evaporates conducting an infinite current! With so many other errors, don't assume they know what they're talking about comparing high and low losses. Given the end points of the charge are shown differently, there will be slight differences in the charge efficiency. However, it's not until you compare charging from different starting voltages (consider a voltage doubler) that you get differences worth calling 'High' and 'lower'. \$\endgroup\$ – Neil_UK Mar 20 '20 at 8:30
  • \$\begingroup\$ In case you wonder, a voltage doubler (or any diode capacitor voltage multiplier for that matter, Cockcroft Walton, Dickson etc) charges the capacitors from 90% or so of their final voltage to 100%, and the resistive losses are fairly small, compared to the 50% efficincy of charging a cap from zero volts. (2) doesn;t 'help in reducing losses', (2) is a totally different type of circuit to (1), they cannot be compared. (2) is the only circuit you'll ever meet in real life, and be worth analysing. (1) is an irresistable force pushing an immovable object, if you know that paradox. \$\endgroup\$ – Neil_UK Mar 20 '20 at 8:34

If you charge a capacitor up to a voltage via a series resistor then the energy you use is \$CV^2\$. The energy obtained by the capacitor is \$\dfrac{CV^2}{2}\$ i.e. 50% of that energy is lost to heat. It makes no difference how small or big the resistor is; you get 50% energy transfer period.

However how about if they're the same switch and hence same resistance, just different capacitance to make one slow switching and the other fast swtiching?

It makes no difference to the energy losses; if the larger capacitor is charged to the same voltage as a smaller capacitor, the energy loss proportion is the same. This is why we need a combination of inductor and capacitor in switch mode power supplies; to get high efficiency. Resistors don't cut the mustard.

In an SMPS such as a flyback converter, an inductor is charged by applying a voltage across it and letting the current ramp up. The value the current rises to defines how much energy is stored by the inductor. It may be a surprise to find that there are no losses incurred; all the energy taken from the voltage supply are delivered to the inductor. This is because there is no collision involved.

I mention "collision" as an analogy because, when charging a capacitor from a voltage source there is a collision just as if two objects at different speeds were physically joined. The resulting momentum is the same (i.e. momentum, like charge is conserved) but, energy is lost. It's the same as when trying to start a car moving by quickly releasing the clutch; there is a risk of the engine stalling because of the collision. If you avoid stalling, the clutch will slip until the engine and wheels are up to speed.

If you charge a capacitor by applying it to a variable voltage source and lift the voltage from 0 volts to (say) 12 volts, there is 100% energy transfer with no losses. That is why we charge an inductor then discharge it into a capacitor to obtain high efficiency power supplies.

  • \$\begingroup\$ I understand what you mean but what you said only works for the case where time constant is much smaller than the switching period which is the case 1 above. However, I think the lecture wants to say something different. When the time constant is compared to or larger than switching period then the charging current ripple is reduced which reduces the loss. \$\endgroup\$ – anhnha Mar 20 '20 at 10:05
  • \$\begingroup\$ What I've said is independent of the time constant. What's the point of having an RC time longer than the switching period? What does it bring to the party in terms of useful functionality i.e. why would anyone consider this? I'm asking so that you can give me that answer. At the moment I see a question that you've raised and that's all. If you want more detail then please provide context. \$\endgroup\$ – Andy aka Mar 20 '20 at 10:19
  • \$\begingroup\$ Did you watch the video in the post above? He explained about slow and fast switching limits and how fast switching can help reduce the loss. What you said is only for slow switching \$\endgroup\$ – anhnha Mar 20 '20 at 14:08
  • \$\begingroup\$ I watched a few minutes and sure, for capacitive switching you can get better than 50% if the recharge is small but, I wasn't going to watch through the whole video to understand the context within which your question was framed. YOU must put that context into your question and definitely not rely on people watching a 17 minute video. If that is the context behind the question you raised then please make that clear in your question. \$\endgroup\$ – Andy aka Mar 20 '20 at 14:17
  • \$\begingroup\$ I was thinking that everyone knew that so I didn't include it. \$\endgroup\$ – anhnha Mar 20 '20 at 16:43

The linked video makes a statement that I think has been missed-- the Q transferred in both cases is the same. For this to be the case both resistance and capacitance in the larger tau case are larger than in the first case since the plot shows that there is not a full transfer of voltage at t-on. Putting this constraint into the analysis results in less loss for the larter tau case.


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