MLCC discharge curves - Page 1
Forum section: electronics / projects

author: matseng
author_rank: Frequent Contributor
timestamp: August 18, 2015, 06:38:02 am
content: My head hurts just to think abut this, maybe you guys can tell me what happens...As I understand it a MLCC will get a lower capacitance when the voltage over it increases. So charging a cap that has a nominal capacitance of 4.7uF might just be 1uF at 10 volts.If I take a any other capacitor, metal film or whatever, that is 1uF can charge both the MF and MLCC cap to 10 volts they would both hold 50 uJ of energy.When discharging both of them with constant current the voltage curves would be different - the MLCC would fall faster. This caused by the fact that the capacitance increases and the voltage must then drop a bit more to keep the stored energy at the the same level as if the capacitance would be constant as in the MF cap.Is this correct or am I just totally off here?

author: Kleinstein
author_rank: Super Contributor
timestamp: August 18, 2015, 09:25:25 am
content: You are off in the explaination.The key part is to see what is meant with capacitance on a ninlinear / saturating part. Usually it's the differentia capacitance. So inceasing the voltage from 10 V to 10 V plus a bit, will have an efffect like the smaller cap.So during discharging, the MLCC will see an increasing capacitance. So on constant currents discharging voltage will fist drop fast (as for the reduced capacitance) and than slower as to the higher nominal capacitance.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 18, 2015, 02:27:46 pm
content: Yes, if you charge with a constant current, starting from zero, it starts off slow, then accelerates.This is commonly seen in switching waveforms, where the junction capacitance varies with voltage. Example:The "knee" on the bottom left, where the rise really speeds up, is about where the transistor turns off; voltage rises faster and faster, until peak dV/dt is reached in the middle. It slows back down as the voltage approaches the opposite supply rail, where another transistor "catches" the rise, and eventually conducts. (This was measured from a pair of 600V 50A "high speed" IGBTs, with around 5nF = C_oe at low voltages, dropping to ~200pF in the middle.)It's noteworthy that capacitance never rises faster than a certain rate, otherwise the sudden drop in capacitance would increase the voltage, which would violate conservation of energy.How that works is, at a given (nonzero) voltage, the capacitor has some amount of charge on it. Charge is always conserved. If the capacitance drops while no current was applied (i.e., open circuit, no change in charge), then the voltage rises proportionally. Or vice versa. (This can be observed with old-fashioned variable capacitors and a sensitive meter, and is used as the transducer in electrostatic microphones.) Capacitance drops and voltage rises, but energy goes as voltage squared, therefore the energy also rises proportionally. Where does the energy come from? In performing work on the charge: the capacitor was physically moved around, which took some force and distance. (This force can be felt and measured directly with electrostatic apparatus; it is the force which makes cranking a Wimshurst generator difficult just before sparking over, or the force which causes rain drops to deflect in a Kelvin water dropper.)Anyway, as it turns out, MLCCs tend to lose capacitance at a characteristic rate, roughly as sqrt(V), so that E(V) is proportional (V instead of V^2). The energy stored always continues to rise as voltage rises, but the energy density isn't very impressive, and as you get into saturation (typically at or beyond rated voltage), the important part, the bypass capability (how much current flows due to a change in voltage: I = C * dV/dt) disappears, because that's the capacitance alone, not the energy storage.Tim

author: matseng
author_rank: Frequent Contributor
timestamp: August 19, 2015, 03:06:14 am
content: Thanks, now I got at least a little better understanding of this phenomenon.What is the cause of this effect? Is it something similar to what happens in a varicap where the size of the depletion area changes with the bias voltage thus causing the capacitance to change?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 19, 2015, 04:11:21 am
content: Yes, or more precisely, when a ferromagnetic substance goes into saturation.Ferromagnetism and ferroelectricity are duals of the same effect: both are due to the massive, cooperative alignment of dipoles in the material. Saturation occurs when the poles are all pointing the same way, so the material can't be polarized any more except by brute force (sheer electric field pulling on atomic orbits, or whatever). Except each individual dipole has its own resting angle, stiction (hence hysteresis, in part), spring force and so on (not really, but for complicated reasons, it acts like this in bulk). So when you apply a given stimulus, not all respond at once, and saturation is usually fairly gradual.Type 1 dielectrics do not saturate, because they are the ordinary polarizable kind, not the enhanced ferroelectric kind, which are type 2. This is why C0G caps are perfectly linear, and X7R and such are less well behaved.Tim

author: Someone
author_rank: Super Contributor
timestamp: August 19, 2015, 11:32:33 am
content: For fun I swept a low voltage ceramic capacitor I had on hand with a small deviation on top of the DC bias, just the leakage as you go up beyond the operating voltage sure is interesting. Second plot is charging the same capacitor with a constant 1uA source and you can see the change in slope as the effective capacitance drops, when pushing ceramic caps you'd want to integrate the charge with respect to the voltage range utilised. To really extend the ranges out would need some better made fixtures for this specific application, the gigaohm ranging without resorting to guarding was a surprise.

author: matseng
author_rank: Frequent Contributor
timestamp: August 19, 2015, 02:39:26 pm
content: Holy crap.... That charging curve really speaks volumes - it's totally the opposite how a "normal" capacitor acts.It's strange that this is so little mentioned in the regular Electronics 101 texts.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 19, 2015, 08:25:58 pm
content: You generate a family of curves like that when charging a transistor junction (again, for the same reasons, strong nonlinearity). Sometimes these are even specified in the datasheet: newer Infineon MOSFETs usually have a "Coss energy equivalent" or "time equivalent" figure.The energy equivalent is what capacitance you get when you determine the energy stored in the junction capacitance, and assume it was stored in a linear (constant) capacitance. The capacitance at rated voltage might be 100pF, but the energy equivalent might be 150pF, due to the additional charge taken on at lower voltages. However, it's not much difference, because charging at low voltage means the power was low, and therefore even though a lot of time might be taken at those low voltages, the total (integral) energy isn't much higher.The time equivalent is what capacitance you get when you determine the charging time for a particular circuit, and assume it was stored in a linear (constant) capacitance. Typically, the circuit is a 100k resistor supplied from V = Vdss, and the risetime is taken as 0-80% or something like that. An RC conversion factor then gives the effective C. The capacitance equivalent might be 300pF, due to the extra time taken at low voltages, which factors directly into this measurement. The curve slows down at high voltages, but thanks to the arbitrary cutoff level (the 80% or whatever figure), it doesn't spend too much time in the nearly-fully-charged exponential tail.You can define other equivalents quite easily. A constant-current (charge equivalent) figure gives yet another result. You can do an inductive charging equivalent, which would be typical of two transistors in a half-bridge, where the inductance is due to the loop between supply capacitance and the two transistors; the waveform would be representative of hard switching at zero load current. In this case, the extremely high Coss at low voltages looks very much like diode reverse recovery, except that it's taking place at +10-30V rather than -0.5V. The drain voltage "snaps" open in a very similar manner, because during that "recovery" phase, inductor current is rising steadily, therefore charge is rising quadratically, and by the time the capacitance plummets, current is so high that the voltage risetime just explodes out of the way (again, much like my waveform posted earlier).The crappy part about any of these equivalents is: they are extremely sensitive to the particular conditions they were measured at. They vary with supply voltage and threshold range.I have some spreadsheets sitting around where I've both modeled the Coss of some MOSFETs (based on real measurements, datasheets and SPICE models -- none of which really lined up, by the way..), and derived a few of these figures based on them. Lemme see about showing some of that..Tim