0
\$\begingroup\$

What is the meaning of the capacitance parameter in the datasheet of a varistor?

For example: I have noticed that 7mm 430 varistor voltage(07D431K) and 14mm 430 varistor voltage (14D431K) have the same max continuous voltage and the same maximum clamping voltage, but the larger varistor (14D431K) has higher maximum surge current, higher rated power, and higher reference capacitance. The others parameters I understood, but the reference capacitance parameter I didn't understand.

How does its capacitance value influence the circuit?

For example: if I protect a circuit with a varistor between its 220Vac input power source lines, what will be the difference, according to the capacitance aspect, if I use the one with the lower capacitance or the one with the higher capacitance?

\$\endgroup\$
2
\$\begingroup\$

I would expect the capacitance to rise with increase surge current, rated power and size of the varistor. It's just a function of a larger area junction, which has a higher capacitance.

If you are merely protecting 220v mains, then the capacitance figure is purely incidental to your use. Like the physical size of the varistor, it can be measured, it has a value, but it is irrelevant to you.

Other users of varistors may put them on data lines, and there the capacitance determines how fast the data rate can be. In these applications, the user would pick the smallest varistor (and therefore lowest capacitance) that still handled enough power to protect the circuit.

\$\endgroup\$
6
  • \$\begingroup\$ So,if i would like to protect the PCB of the refrigerator,laundry machine,computer etc.by soldering the varistor in parallel to the 220Vac,where it first enters the PCB(against surge voltage).Will the 14D431K varistor(650pF)and the 7D431K varistor(150pF)work the same(capacitance aspect)without disturbing the PCB operation?Isn't any preferability of one of them?(capacitance aspect)If the original soldered varistor 07D431K in the PCB of a loundry machine,for example, failed and i change it to varistor 14D431K,which has more capacitance.won't this capacitance influence the PCB at all? \$\endgroup\$ – xchcui Jul 11 '16 at 5:42
  • \$\begingroup\$ @xchcui The maximum capacitance quoted is 460pF for the 14D431K (no, it's all right, I love tracking things down on google when you as OP could have saved my time and put a link to it in the question). You might like to calculate how much current that impedance will draw at power line frequencies, compare it to how much the appliance draws, and then answer your own question. \$\endgroup\$ – Neil_UK Jul 11 '16 at 7:20
  • \$\begingroup\$ At this address:radiocom.dn.ua/image/data/pdf/VDE_TYEE.pdf the capacitance is 650pF for 14D431K.but it is not the issue here.I just would like to know if changing the small(07D431K)original varistor in the pcb(which contains microcontroller etc.)to a bigger one(14D431K) with the same varistor and clamping voltage,in order to get a better energy absorbent of voltage surge,may disturb,somehow,the pcb circuit appliance,as it has higher capacitance opposed to the smaller varistor,which was original planned and soldered on the pcb by the manufacturer's engineer. \$\endgroup\$ – xchcui Jul 11 '16 at 14:02
  • \$\begingroup\$ I can't say without a circuit diagram of the PCB. If it's connected across the mains terminals, then see my previous comment. If it's connected somewhere else, then we need a circuit diagram. \$\endgroup\$ – Neil_UK Jul 11 '16 at 15:39
  • 1
    \$\begingroup\$ Yes. Given that capacitor impedance =1/jwC, calculate what the capacitive current draw is at your mains frequency and voltage, and what it is for the smaller varistor. \$\endgroup\$ – Neil_UK Jul 11 '16 at 16:54
4
\$\begingroup\$

Metallic oxide varistors (MOV's) behave much like a capacitor except the dielectric insulator between the 2 tin plates has a fixed 'soft' clamp voltage, above which the MOV begins to conduct current in either direction. As the voltage rises so does the current flow.

At twice the MOV's rated voltage they can absorb (briefly) several 10's of thousands of amps. That is why they are so popular in surge suppressors for AC or DC power feeds. Because of their high capacitance they are not used to protect data feeds. Tranzorbs and Sidacs and gas tubes are better for those applications.

An MOV's capacitance is not affected by voltage (the same for Tranzorbs, Sidacs and gas tubes) changes until the voltage exceeds the clamp voltage of the MOV. Often the maximum safe AC/DC voltage allowed is printed on the MOV. Its size and the datasheets offer details about the maximum surge current it can handle one time, and what it can handle with 5,000 or so 'small' surges, so its lifespan can be predicted in real-world conditions.

Because the MOV is basically two metal plates spaced by a dielectric, it acts much like a capacitor in the nF range. The larger the size the more capacitance, but it is not enough to affect AC or DC power feeds, as they have a low drive impedance so MOV's are 'ignored' until a surge event happens. For that reason MOV's must be fused or have a circuit breaker in series with them in case the surge is so intense the MOV fails (shorts out).

For more details and graphs see the following link:https://en.wikipedia.org/wiki/Varistor
This is a few paragraphs from the Wiki link that summarize some important details.

Composition and operation
Varistor current-voltage characteristics for zinc oxide (ZnO) and silicon carbide (SiC) devices:
The most common type of varistor is the metal-oxide varistor (MOV). This type contains a ceramic mass of zinc oxide grains, in a matrix of other metal oxides (such as small amounts of bismuth, cobalt, manganese) sandwiched between two metal plates (the electrodes). The boundary between each grain and its neighbour forms a diode junction, which allows current to flow in only one direction. The mass of randomly oriented grains is electrically equivalent to a network of back-to-back diode pairs, each pair in parallel with many other pairs.

When a small or moderate voltage is applied across the electrodes, only a tiny current flows, caused by reverse leakage through the diode junctions. When a large voltage is applied, the diode junction breaks down due to a combination of thermionic emission and electron tunneling, and a large current flows. The result of this behaviour is a highly nonlinear current-voltage characteristic, in which the MOV has a high resistance at low voltages and a low resistance at high voltages.

Electrical characteristics:
A varistor remains non-conductive as a shunt-mode device during normal operation when the voltage across it remains well below its "clamping voltage", thus varistors are typically used for suppressing line voltage surges. Varistors will almost always eventually fail for either of two reasons.

A catastrophic failure occurs from not successfully limiting a very large surge from an event like a lightning strike, where the energy involved is many orders of magnitude greater than the varistor can handle. Follow-through current resulting from a strike may melt, burn, or even vaporize the varistor. This thermal runaway is due to a lack of conformity in individual grain-boundary junctions, which leads to the failure of dominant current paths under thermal stress when the energy in a transient pulse (normally measured in joules) is too high (i.e. significantly exceeds the manufacture's "Absolute Maximum Ratings"). The probability of catastrophic failure can be reduced by increasing the rating, either by using a single varistor of higher rating or by connecting more devices in parallel.

Cumulative degradation occurs as lesser surges happen. For historical reasons, many MOVs have been incorrectly specified allowing frequent swells to also degrade capacity. In this condition the varistor is not visibly damaged and outwardly appears functional (no catastrophic failure), but it no longer offers protection. Eventually, it proceeds into a shorted circuit condition as the energy discharges create a conductive channel through the oxides.

The main parameter affecting varistor life expectancy is its energy (Joule) rating. Increasing the energy rating raises the number of (defined maximum size) transient pulses that it can accommodate exponentially as well as the cumulative sum of energy from clamping lesser pulses. As these pulses occur, the "clamping voltage" it provides during each event decreases, and a varistor is typically deemed to be functionally degraded when its "clamping voltage" has changed by 10%. Manufacturer's life-expectancy charts relate current, severity and number of transients to make failure predictions based on the total energy dissipated over the life of the part.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.