{"id":60,"date":"2006-11-22T19:17:02","date_gmt":"2006-11-22T19:17:02","guid":{"rendered":"http:\/\/www.olino.org\/us\/?p=60"},"modified":"2008-05-21T20:14:32","modified_gmt":"2008-05-21T20:14:32","slug":"charge-efficiency-capacitor","status":"publish","type":"post","link":"https:\/\/olino.org\/blog\/us\/articles\/2006\/11\/22\/charge-efficiency-capacitor\/","title":{"rendered":"Charge Efficiency Capacitor"},"content":{"rendered":"

This article explains how charging a capacitor can use a lot of energy, and how your charging method has a big influence on its efficiency. It becomes more important when considering the possible super capacitors that I spoke of in the article Has the supercapacitor been invented?<\/a><\/p>\n

\"plaatje
\n<\/p>\n

Charge Efficiency of the Capacitor<\/h2>\n

When charging and discharging a capacitor, we lose energy due to the already existing ESR (Equivalent Series Resistance). This resistance is always present since all of the charge has to be transported to the capacitor plates and, since the materials are not superconducting, it means that there is loss in transport that is converted into heat. We have a loss of energy, which we originally wanted to store or retrieve from the capacitor. We put the ESR into a scheme, and add all relevant series resistances in one R.<\/p>\n

\"plaatje<\/p>\n

By choosing the charge current wisely, we can have an influence on the losses in the ESR R. There exists a simple expression that computes the losses in R:<\/p>\n

\"formule<\/p>\n

This means that the energy losses in R (ER<\/sub>) are the integral (= sum) of the square of the current through it, multiplied by the resitance value R itself. The sum has to be taken for the complete charging or discharging period, whichever is considered.<\/p>\n

This loss factor we want to put in relation to the energy that is stored in the capacitor, which is:<\/p>\n

\"formule<\/p>\n

In the next chapters I will relate the results, depending on the method of charging the capacitor.<\/p>\n

Charging with a Fixed Voltage<\/h2>\n

In the next scheme will be shown how a capacitor is charged with a fixed voltage V.
\n\"plaatje<\/p>\n

The energy stored in the capacitor<\/strong><\/h3>\n

When we charge the capacitor with a fixed voltage source V, then after a time which is normally set at 5 RC, we have the capacitor charged up to V volt, and this corresponds to a stored energy amount of 1\/2CV2<\/sup> in the capacitor.<\/p>\n

The Energy Losses in the ESR<\/strong><\/h3>\n

To compute the losses in the ESR, we need to know the current during the charging cycle. This can be computed for the mentioned scheme:<\/p>\n

\"formule<\/p>\n

The amount of energy that we lose in the ESR R can be computed as follows:<\/p>\n

\"Formule<\/p>\n

You can see that the loss in the ESR is just as big as the energy stored in the capacitor itself! This is due to the very high current that flows at the beginning of the charge cycle (just after connecting the voltage source V on the empty capacitor C).<\/p>\n

Charging with a Constant Current Source<\/h2>\n

The next scheme shows charging with a fixed current source. This current source is connected to the capacitor (with its ESR R).<\/p>\n

\"plaatje<\/p>\n

The Energy Stored in the Capacitor<\/strong><\/h3>\n

When we charge the capacitor with a current source J, then, after a certain time, e.g. t1<\/sub> seconds, the capacitor will be charged to V volt which gives an energy stored of 1\/2CV2<\/sup>.<\/p>\n

We can compute t1<\/sub>:<\/p>\n

\"formule<\/p>\n

The charge in the capacitor QC<\/sub> is equal to the time period t1<\/sub> times the constant current J during this time period. This is also equal to the capacity value C times the voltage V over the capacitor.<\/p>\n

The Energy Losses in the ESR<\/strong><\/h3>\n

To compute the amount of lost energy in R during this charge cycle, we compute with the constant value J:<\/p>\n

\"Formule
\nThis equation is very different from the equation we found when charging with a constant voltage source.<\/p>\n

If we now want the loss to be no more than 10 % of the energy stored in the capacitor itself. Then we use the equations:<\/p>\n

\"formule<\/p>\n

So with a charge time of 20 RC, and that is 4 times longer than with the constant voltage source, we reduced the amount of lost energy over the ESR to 10 %!
\nOf course, when we allow for a longer charge time, we can reduce the energy loss over the ESR evern more!<\/p>\n

Other schemes<\/h2>\n

I showed two very different schemes to charge a capacitor and computed the results in energy loss over ESR. The most efficient one in energy loss is to use a current source and allow for a longer charge time. It is not always easy to make a constant current source, and perhaps a nice scheme that is between the two mentioned schemes is presented here:<\/p>\n

\"Plaatje<\/p>\n

The way it works is as follows: you switch on the voltage source, to a value of 0.5 x V, when the capacitor C is empty. This results in an increasing current via the coil and ESR to the capacitor, and the current will charge the capacitor (also the current through the ESR R will result in energy loss). This current will not increase at the moment the voltage in the capacitor is equal to 0.5 x V. At that moment, it is the energy in the coil that ensures the current still continues to flow, but it will gradually decrease to zero. Now the charge on the capacitor is almost at a voltage V. We need to switch off the voltage source to prevent the charge in the capacitor (now at V volt) from flowing back to the voltage source. If we put a diode in the circuit, this will ensure that the current will not flow back. The current that has flown has a sinusoidal form (since we work with a capacitor and coil in series). To neglect the influence of the ESR, one has to make the value L of the coil as big as possible, and the value R of the ESR as low as possible, resulting in using thick wires and a lot of turns around a magnetically conducting core material. This scheme is possibly easier to make than a constant current source, when considering the high currents used for the super capacitor.<\/p>\n","protected":false},"excerpt":{"rendered":"

This article explains how charging a capacitor can use a lot of energy, and how your charging method has a big influence on its efficiency. It becomes more important when considering the possible super capacitors that I spoke of in the article Has the supercapacitor been invented?<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[5],"tags":[],"_links":{"self":[{"href":"https:\/\/olino.org\/blog\/us\/wp-json\/wp\/v2\/posts\/60"}],"collection":[{"href":"https:\/\/olino.org\/blog\/us\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/olino.org\/blog\/us\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/olino.org\/blog\/us\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/olino.org\/blog\/us\/wp-json\/wp\/v2\/comments?post=60"}],"version-history":[{"count":0,"href":"https:\/\/olino.org\/blog\/us\/wp-json\/wp\/v2\/posts\/60\/revisions"}],"wp:attachment":[{"href":"https:\/\/olino.org\/blog\/us\/wp-json\/wp\/v2\/media?parent=60"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/olino.org\/blog\/us\/wp-json\/wp\/v2\/categories?post=60"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/olino.org\/blog\/us\/wp-json\/wp\/v2\/tags?post=60"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}