varistor selection for circuit protection against surge signals

Varistor SelectionFor Circuit Protection Against Surge Signals

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How to select the proper varistor in order to protect the circuit from the surge signals strike using practical, straightforward and simple method

Purpose




Surge Signal

The surge signal is a transient wave of electrical current, voltage, or power propagating along a line or a circuit and characterized by a rapid increase followed by a slower decrease

We will deal with the surge signal as a voltage waveform with the shape as in the figure shown (1.2μs/50μs shape)




Varistor OperationThe varistor operation can be simply briefed to the following:

● As long as the voltage across the varistor is less than its clamping voltage, no current is flowing through it

● If the voltage across the varistor tries to exceed its clamping voltage, large current flows through the varistor keeping the voltage across it approximately equals its clamping voltage




● The varistor is placed in shunt with the circuit to be protected● When surge signal is applied, the varistor draws large current through it trying

to keep the voltage across it to its clamping voltage value absorbing the energy of the surge signal

Varistor Operation

230VRMS Circuit to be protected

Surge

Varistor




Design Examples - Assumptions● The surge signal is modeled by a voltage source with source impedance 2Ω● The surge signal peak value is 4KVolt● The circuit to be protected operates with AC source of 230VRMS ±10%● The surge signal is repeated every 3 minutes

2 Ω

230VRMS

4KVSurge

Circuit to be protected

Varistor

R surge

Design Examples - Varistor Specifications Varistor is selected according to 5 parameters:

1. Operating voltage VM (AC or DC)2. Clamping voltage VC

3. Surge current ITM

4. Energy absorption WTM

5. Rated power PR




Design Examples - Varistor Specifications The desired varistor will be selected from LA series of Littlefuse

A snapshot of the datasheet is shown in figure below

Design Steps - operating voltage VM(AC) The varistor operating AC voltage VM(AC) should be selected to be greater than the maximum allowable AC voltage applied to the circuit

In our design example the varistor operating AC voltage should be greater than 230VRMS + 10% = 253VRMS

Notes

● If the circuit operates with DC source, the varistor operating DC voltage should be greater than the maximum allowable value of the DC source

● If the AC source is not sinusoidal, the varistor operating DC voltage should be greater than peak value of the periodic voltage waveform of the source




Design Steps - operating voltage VM(AC) According to datasheet, we can select any varistor starting from V275LA2P as in figure below. We will select the first varistor in the suggested list: V275LA2P

Design Steps - Varistor Specifications V275LA2P

1. Operating voltage VM (AC or DC) ✓ VM(AC) = 275VRMS > 253VRMS2. Clamping voltage VC

3. Surge current ITM


5. Rated power PR




Design Steps - clamping voltage VC

Once the varistor part number is selected, the clamping voltage is defined as in datasheet. In our example the clamping voltage is 775V

Design Steps - Varistor Specifications V275LA2P

1. Operating voltage VM (AC or DC) ✓ VM(AC) = 275VRMS > 253VRMS2. Clamping voltage VC ✓ VC = 775V3. Surge current ITM


5. Rated power PR




Design Steps - surge current ITM

We should ensure that the maximum current value flowing in the varistor during the surge Isurge is less than the surge current rating ITM of the varistor selected

The current Isurge can be calculated from circuit below

2 Ω

230VRMS

4KVSurge


Varistor

Isurge


The surge current flows completely in the varistor

The voltage across the varistor equals the clamping voltage

Maximum input voltage is the surge voltage peak value + peak value of sine wave of AC input 2 Ω

230VRMS

4KVSurge


Varistor775V

4KV+230*(1+10%) *√2= 4358V

Isurge

Isurge = (4KV+230*(1+10%) *√2 - 775)/ 2 Ω = 1791.5 Amp.

Isurge Should be less than ITM value in datasheet


2 Ω

230VRMS

4KVSurge


Varistor775V

4KV+230*(1+10%) *√2= 4358V

Isurge


As shown in datasheet below, Isurge is greater than ITM value which means that the varistor selected is inappropriate


The varistor surge current ITM should be greater than 1791.5 Amp. Another part number of larger disc size (10mm) can fit the design requirement. Unfortunately, the clamping voltage is changed which means we need to recalculate the surge current again with the new value of clamping voltage

Isurge = (4KV+230*(1+10%) *√2 - 710)/ 2 Ω = 1824 Amp.

Isurge Should be less than ITM value in datasheet


2 Ω

230VRMS

4KVSurge


Varistor710V

4KV+230*(1+10%) *√2= 4358V

Isurge

We can find now that the surge current condition is satisfied

Isurge < ITM


2 Ω

230VRMS

4KVSurge


Varistor710V

4KV+230*(1+10%) *√2= 4358V

Isurge

1824 Amp.

2500 Amp.

Design Steps - Varistor Specifications V275LA2P V275LA10P

1. Operating voltage VM (AC or DC) ✓ VM(AC) = 275VRMS > 253VRMS2. Clamping voltage VC ✓ VC = 775V VC = 710V3. Surge current ITM ✓ ITM = 2500Amp > 1824Amp4. Energy absorption WTM

5. Rated power PR




Design Steps - energy absorption WTM

● Energy absorbed by the varistor is the integration of the instantaneous power consumed by it over the surge period

● The instantaneous power is the multiplication of the instantaneous voltage across the varistor by the instantaneous current flowing through it




● The instantaneous varistor voltage and current can be calculated from the following circuit given that:

○ The surge waveform is known (4KV peak, 1.2μs/50μs shape)○ The source impedance is 2Ω○ The clamping voltage of the varistor is 710V


2 Ω

230VRMS

4KVSurge


Varistor

Isurge

710V


● The relations between voltage and current waveforms are shown in graphs below

Varistor Voltage

Varistor Current





● The overall voltage source in the loop is the maximum value of the AC source (230*[1+10%] *√2 = 358V) plus 4KV surge superimposed on it

● The current starts to flow through the varistor as soon as the overall voltage source exceeds the clamping voltage of the varistor

● The varistor voltage is then fixed at the clamping value (bold black graph)● The varistor current waveform is similar to the voltage waveform after

subtracting the clamping voltage then divided by Rsurge (bold red graph)● The surge current duration Tsurge can be approximated to double the half-

value duration of the 1.2μs/50μs shape. Tsurge ≈100μs




● Voltage and current waveforms can be approximated as figure below● The power waveform can be obtained by the multiplication of the voltage and

current waveforms● Energy absorbed by the varistor is the area under the power waveform


Tsurge

Varistor Voltage

Varistor Current

Varistor Power

Energy=0.5(clamping voltage * peak current * Tsurge)

Clamping Voltage

Peak Current

● The power waveform can be obtained by the multiplication of the voltage and current waveforms

● Energy absorbed by the varistor is the area under the power waveform


Tsurge

Varistor Voltage

Varistor Power

Energy=0.5(clamping voltage * peak current * Tsurge)

100μs1824Amp

710V64.752J

Clamping Voltage

Peak Current

Varistor Current


The varistor rated energy absorption WTM should be greater than 64.752J. Unfortunately the selected varistor can absorb not more than 45J according to datasheet below. Another part number of larger disc size (14mm) can fit the design requirement. Fortunately, the clamping voltage is the same which means we don’t have to repeat the previous calculations. Also ITM is higher

Design Steps - Varistor Specifications V275LA2P V275LA10P V275LA20P

1. Operating voltage VM (AC or DC) ✓ VM(AC) = 275VRMS > 253VRMS2. Clamping voltage VC ✓ VC = 775V VC = 710V3. Surge current ITM ✓ ITM = 2500Amp 4500Amp > 1824Amp4. Energy absorption WTM ✓ WTM = 75J > 64.752J5. Rated power PR




Design Steps - rated power PR

In case of the surge signal is repetitive, the energy absorbed by the varistor per repetition period should not exceed the rated power of the varistor PR

The average power dissipation of the varistor is calculated by dividing the energy absorbed via the varistor in a single surge pulse by the repetition period




Design Steps - rated power PR

In our design example the surge energy absorbed by the varistor equals 64.752J

The surge pulse is repeated every 3 minutes

The average surge signal power = [64.752/(3*60)] = 0.36W

As mentioned in datasheet as shown in figure below the rated power of the varistor is 0.6W which is greater than the average surge power




Design Steps - Varistor Specifications V275LA2P V275LA10P V275LA20P

1. Operating voltage VM (AC or DC) ✓ VM(AC) = 275VRMS > 253VRMS2. Clamping voltage VC ✓ VC = 775V VC = 710V3. Surge current ITM ✓ ITM = 2500Amp 4500Amp > 1824Amp4. Energy absorption WTM ✓ WTM = 75J > 64.752J5. Rated power PR ✓ PR = 0.6W > 0.36W

Selection Completedlinkedin.com/in/mohammedfouly



NotesWe assumed previously that the surge source has an output impedance of 2Ω which is not always guaranteed. In reality the impedance may be more or less this value. This issue will be severe if the impedance is less than the assumed value because that means the peak surge current is increased. It will be better for the designer to evaluate the surge source impedance or even the surge peak current




NotesThe varistor selection method presented above depends on some approximations, so it will be better to select the varistor parameter with suitable margin




NotesDuring the surge duration, the circuit to be protected is being subjected to the clamping voltage (710V), which means the circuit should be able to withstand this voltage level




varistor selection for circuit protection against surge signals

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