Understanding Varistor Functionality: Choosing Right Parameters for a 12V LED System

Hello and welcome!
I would like to ask for a detailed practical explanation of how a varistor works, because the same description is usually found on the Internet and does not say much.
I found out that its resistance decreases as the voltage increases, which I understand allows it to eliminate small voltage noises, e.g. induction?

I have an LED system powered by a changing voltage of 12V, but a voltage of 4V will also light up the diodes slightly, which I would not like, can I use a varistor for this? What parameters should it have and how to choose it? A smaller current would encounter resistance and not flow to the LED system, when the voltage reaches about 12V the resistance of the varistor would decrease and this would allow the system to operate, am I right? how to choose the right one?

Regards.

Varistors are mainly used as surge protection. I'm not sure, but probably its resistance decreases dramatically when a certain voltage is exceeded. This causes the connector connected between ground and power to short circuit during an overvoltage and dissipate excess energy (voltage) to ground.

There are basically 3 types of varistors:
1. Small electronics - they are mounted in electronic equipment on a board. They cost a few dozen cents to a few zlotys.
2. Protection varistors - they look like a residual current fuse and are installed in fuse boxes. They protect electrical devices against network overvoltages (from lightning), at least up to a certain value and when other requirements are met. They cost from PLN 200 and up.
3. Grid varistors - they can be seen on power lines, they discharge lightning striking the power line. They look like insulators and cost a lot of money.

Example parameters:

varistor voltage = 18V
protected voltage = DC 14V, AC 11V

Does this mean that current above 14V is not subject to resistance? 18v is the maximum voltage? how to understand?

The most I see are several hundred Vs, e.g.:
Type: V20N391K
Maximum operating voltage: 250VAC; 320VDC
Varistor voltage: 390V

What does it mean? maximum operating voltage is 320VDC and voltage is 390V, how to understand this?

Resistance is not found only in superconductors!

The maximum voltage is most likely the one at which the resistance drops rapidly. The varistor voltage is most likely the voltage at which the varistor behaves like a conductor.
But let someone else answer because I'm not a varistor specialist

It's always worth using Google ;)

On the example website with a description of typical elements you can see, among others: characteristics of various types of varistors. The bend of the U/I curve, i.e. the speed of switching on as a function of voltage, varies, but in general there is a clear gap between the voltage at which a significant current begins to flow and the voltage at which the current increases so rapidly that it practically prevents further voltage increase.

The operating voltage is set below the first threshold, then the varistor has practically no effect on the operation of the system, while the protection is effective at the second voltage, because this is how long the system must withstand before the varistor takes over the current for good.

Coming back to your idea of using a varistor connected in series to avoid slight LED illumination at a small voltage lower than the rated one, this is probably not the way to go. Yes, when the voltage is low, the current will not flow and the LEDs will not light up, but later
1) you will have to increase the supply voltage by the value of the varistor voltage to make the current flow
2) the power dissipated by the varistor will not only worsen the efficiency of the system but may also destroy it

If you want to solve the problem of LED lighting at low voltage, you should use a key that will cut off the voltage completely until the voltage is correct.

In general, a varistor is a valve-fuse that protects a given system against overvoltages - the voltage increases, the resistance decreases and, as a result, the current increases ("overvoltage energy"), which flows to the minus of the power supply or to ground, depending on the place of work, and bypasses it by wide margin. protected device or system .

The protected voltage is the maximum voltage at which the varistor can operate without unnecessary increased current consumption or flow, and the varistor voltage is simply the voltage at which a full short circuit of this element occurs, e.g. according to your example, an 18V varistor will work correctly for DC voltages of max. 14V and AC. max 11V, so let's say you want to protect an electronic system operating at 15VDC and if you insert such a varistor there, its resistance may decrease so much that the "valve", e.g., opens halfway and a larger current will flow through it, causing it to constantly heat up (and consequently burn out). ) additionally loading the power supply system, e.g. the battery, and discharging it faster. The given varistor will well protect the device, e.g. 10VDC or AC, because the maximum operating voltage is 14VDC and 11VAC (high resistance and negligible current). Any exceeding of the operating voltage and varistor voltage causes the overvoltage to be extinguished. The higher the voltage, the faster the overvoltage decay time.

This results from the characteristics I=f(U) of the varistors. I hope I explained it in a quite layman's terms and simple way.

Much can be explained by the physical principle of operation of a varistor and its structure. This is an element in which, when a certain threshold voltage is exceeded, a series of miniature electric arcs occur on many "spark gaps".
In simple terms, an electric arc is equivalent to a short circuit, low resistance. No arc, e.g., before exceeding the threshold voltage, and after exceeding the threshold voltage, its decrease, high resistance.
http://www.google.pl/url?url=http://www.zsp1s...KEwjD3Yey7u7IAhXEXBQKHf4fBxc&usg=AFQjCNHqSyG0 tcKDzvV0QpDzHIQOPFRaXw

How to check a varistor with a meter without desoldering it?

Quote:

How to check a varistor with a meter without desoldering it?

It shouldn't be conductive, that's all you can check with a meter.

I have encountered such cases: a) the module controlling the operation of my washing machine activates the door lock through a semiconductor control element protected by an appropriate varistor. Once, after turning on the washing machine, a semiconductor element cracked so much that even one of its legs melted, and the driver on the chip died, and the varistor looks healthy and happy; b) in the shepherd, the S10K320 varistor is to protect the thyristor discharging the capacitor on the HV transformer. I charge the capacitor to 500V and it works, the spark is longer and the varistor does not react. How can this be explained?

Frank-01 wrote:

I have encountered such cases: a) the module controlling the operation of my washing machine activates the door lock through a semiconductor control element protected by an appropriate varistor. Once, after turning on the washing machine, a semiconductor element cracked so much that even one of its legs melted, and the driver on the chip died, and the varistor looks healthy and happy;

With a functional lock and installation, the maximum power in the circuit would be too low for the element to fire, there must have been an overcurrent. The driver in the chip failed due to the lack of driver protection - one less element for the manufacturer and a few cents cheaper, and a greater chance that the washing machine will end up in the trash - only advantages. The varistor only protects against one phenomenon that may cause damage, but there are also other causes of damage.

Frank-01 wrote:

b) in the shepherd, the S10K320 varistor is to protect the thyristor discharging the capacitor on the HV transformer. I charge the capacitor to 500V and it works, the spark is longer and the varistor does not react. How can this be explained?

A varistor designed to operate at 320VAC (450V peak) starts conducting at 510V (1mA), but to flow a specific current, let's say 1A, about 700V is needed, with higher currents it will be even more (data based on the epcos SIOV metal oxide varistors datasheet)

A typical varistor has a power of 2W.
The continuous operating voltage is one that will not damage the varistor under normal power supply.

Parameters of the varistor circuit breaker:
275V AC continuous operation (400V AC will destroy the limiter)
1.5kV limitation threshold.
This means that if there is an impulse, e.g. 2kV will limit to the 1.5kV level
(it must be very short - a longer pulse will destroy the limiter)
Surge protection strip – limit threshold 1.2kV.
(applies to impulse)

I measured the "275L20" varistor.
Diameter 15.3 mm, thickness 2.9 mm.
Measurement of AC current consumption.
275V - 56uA
330V - 3.6mA
350V - 10mA
370V -37mA
400V - approx. 100mA.
The time after which the varistor will short-circuit depends on the heat released. (physics)
In the case of the varistors I measured, 400V AC was about a second.
However, 390V AC is quite long - a few seconds.

Joules – more joules – limits the endurance for longer impulses.
At 400 V AC (instead of 230 V) - more joules = it will take a long time before the varistor short-circuits.
This means that the device will be under 400V AC voltage for longer