You should know that the voltage of a capacitor cannot change instantaneously, but its current can. When connecting a capacitor with initial voltage value of V0, directly to a voltage source of Vs (Vs#V0) then if no resistance for the wire, a huge (theoretically infinite instantaneous) current will pass through the wires to reach a balanced voltage (Vs, if ideal voltage source), in the system.

In practice, we have no ideal voltage (current) source, neither ideal capacitor nor wires with R=0. However, if you use a wire with very small resistor to connect the capacitor to the voltage source you can see a spark due to the high current. Using a current source in series with capacitor help to limit the charging current of capacitor so reducing the power losses (Resistive Loss ~ I0 squared).

Note that in practice there is always a minimum voltage drop for biasing a current source, and the parasite resistors which consume the energy during the charging (discharging if Vs < V0) of the capacitor.

Dear Huan,

welcome,

I will answer your set of questions one by one:

What is the definition of "current source like load"?

It is a current source acting like a load.

If the voltage between the terminals of the current source is such that it drives a current in the same direction of the current source, it will source  energy. If the voltage between the terminals is such that it drives current in the load in the reverse direction of the current source, the current source will act as sink for the energy like a load.Likewise, this is valid also fore voltage source where it will source energy if its terminal current from its positive terminal to its negative terminate. If wer revered the fow direction of the source it will sink energy acting as load. This will be load like voltage source.

Let us now refer to the current source in your circuit used in the charging circuit of the capacitor, the circuit in the left of the figure, we see that the voltage on the current source drives current in the outer circuit in the reverse direction of the current source, which means by definition that the current source sinks energy acting like a load.

2. What characteristics of current source like loads that help to obtain soft-charging operation?  

Why ONLY "current source like load" obtain the soft-charging operation (almost no loss)? How about other kinds of load?

A known fact when a source acts like a load that it receives electric energy it will not consume this energy it just recycle it. That is ideal sources does not consume energy. Only if they have internal resistances, they will consume a part of the recycled energy. 

The word soft means that the rate of rise of the current in the capacitor is controlled by the current source acting like load.

The charging current of a capacitor i can be expressed by:

i = C dV/dt, if I is constant as the case with constant current load charging process, 

then the rate of charge will be constant dV/dt = I/C, where I is the current of the current source like load. So, the rate of charge is controlled by the current I.

As the rate of rise of the charging gets smaller the charging process will be softer.

At high charging rate through the current source like load the energy loss in the capacitor may increase because of the polarization loss. One would like to decrease the rate of charge to a value such that the polarization losses of the capacitor becomes negligible.

Charging the capacitor through ohmic resistance will cause tremendous energy loss.

In this case, the energy loss in the resistance will be equal to the the stored energy in the capacitor. The rate of charging can be controlled through the value of the charging resistance.

Best wishes

Dear Mohsen Shafiee Khoor and Prof. Abdelhalim Zekry,

Thank you very much for these thoughtful answers. I have some follow-up questions hope you can explain more.

@Prof. Abdelhalim Zekry

"A known fact when a source acts like a load that it receives electric energy it will not consume this energy it just recycle it. That is ideal sources does not consume energy."

When an ideal current source acts like a load, according to passive sign convention the power consumption by the source p = v*i > 0 or it consumes power.

So how is it possible to not consume energy (or power) as you said?

For soft-charging operation, you mentioned about polarization loss. I don't know about this kind of loss, maybe it is related to device physics or material structure.

However, I believe that the soft-charging in these references means no loss (or almost no loss) when charging a capacitor by a voltage source or charging a capacitor from another capacitor with different initial voltages.

Another question:

In addition to current source, is there any load that doesn't consume enery just and recycle it?

Dear Huan,

welcome,

sorry for coming late.

The ideal sources are infinite energy reservoir. So, if they source a limited quantity of energy to the load there energy still infinity and they receive energy from an external source and act as a sink like load then their energy will still infinity. Please look at the ideal cu rent source it can push its current in an infinite resistance for unlimited time, which means it is an infinite source of energy. The same holds for the voltage source which can keep its terminal voltage constant for infinite load for unlimited time.

Practical current and voltage sources behave ideal in a limited  range of loads. They have source resistances to account for their energy dissipation. For ideal voltage source RS=0 and for a current source RS= infinity. So, the power dissipated in the voltage source will be= I RS^2, and in the current source= V^2/RS^2.

Yes, there is a load that source and sink stored energy, it is the inductor. The inductor can be charged by increasing its current  and discharged by decreasing its current. If the voltage on the inductor is kept consonant then  it has a constant rate of change of its current according to the relation dI/dt=  V/L where L is the inductance of the inductor. So if one applies consatnt voltage for a time T, on the inductor it will charge to I(T)= V T/L. This current can be discharged in the capacitor without any losses assuming ideal inductor. So, one can charge a capacitor theoretically without losses from a voltage source by a series indcutor and a shunting switch. So, an inductor and switch can act lossless load for charging a capacitor.

Best wishes

Dear Huan,

I think in practice we can find a voltage source easily (batteries or synchronous generators for example), however, finding a source of energy in form of a ‘constant current source’ is not immediately achievable. Generally, we should make the current sources from voltage sources.

We have three main passive components; R (which consumes energy), L & C. An ideal inductor or capacitor doesn’t consume active energy but, reactive energy in an AC system. The reactive energy of a passive component (L or C) can be interpreted also as the possibility of storing an amount of active energy in one cycle of time and delivering it on the other cycle of time. This cannot be done automatically. We should use switching devices with appropriate controller. You can see the ‘switched mode power supplies’ topic for more information.

Another possibility to made a current source, especially for low power current sources, is using the transistors (Ex: BJT or JFET or MOSFET), without any L or C, but in biasing them in their active region (see the I-V curves of the transistors & related current source circuits and handbooks). In this case, they consume active energy when they are biased, and this energy will show itself in heating up the transistor. However, comparing to the simple resistances which produce power losses with squared of passed though current, the active current source loads consume the active energy in a linear way and proportional to their current, as you have mentioned P = V . I . This is why in the ‘current source like load’ circuits the losses are lower. (Ex: if I= 5A then P_Resistor = 5 * P_Current Source, for a given constant voltage !)

Hope these explanations help,