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Varistors Introduction
TECHNICAL NOTE
Technical Note
www.vishay.com
Vishay BCcomponents
Revision: 04-Sep-13
9
Document Number: 29079
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Examples, using the following values:
Mains voltage = 220 V
RMS
;
allowable peak voltage = 340 V
Line inductance: L = 20 μH = 20 x 10
-6
H
Line capacitance: C = 300 nF = 0.3 x 10
-6
H
Line resistance: 0.68 
In the event of a short circuit:
Load current:
Energy stored: E = ½ x 20 x 10
-6
x 25 x 10
4
= 2.5 J (Ws)
In the event of a fuse going open circuit:
The energy goes from inductance L towards line
capacitance:
Source of transient

The line impedance becomes high when the fuse goes open
circuit (resistance against high voltage peak in a very short
time).
R
i
= L = 2 f L
Since the rise time of the pulse is 5 μs, the frequency
f = 50 kHz.
R
i
= 6.28  x 50 x 10
3
x 20 x 10
-6
= 6.28 
Z
i
= 6.28  + 0.68  = 6.96 
V
Ri
= 6.96 V x 500 V = 3480 V
V
VDR
= 4082 V - 3480 V = 602 V
VARISTORS APPLICATIONS
Varistors may be used in many applications, including:
• Computers
•Timers
•Amplifiers
• Oscilloscopes
• Medical analysis equipment
• Street lighting
• Tuners
• Televisions
• Controllers
• Industrial power plants
• Telecommunications
• Automotive
• Gas and petrol appliances
• Electronic home appliances
• Relays
• Broadcasting
• Traffic facilities
• Electromagnetic valves
• Railway distribution/vehicles
•Agriculture
• Power supplies
• Line ground (earth protection)
• Microwave ovens
• Toys, etc.
APPLICATION EXAMPLES
For suppression of mains-borne transients in domestic
appliances and industrial equipment, see Suppression via
load, Suppression directly across mains, Switched-mode
power supply protection and Protection of a thyristor bridge
in a washing machine drawings.
Type VDRS05 or VDRS07.
Suppression via load
Suppression directly across mains
I
L
V
R
----
340 V
0.68 
------------------
500 A== =
V
C
2E
C
-------
2 x 2.5
0.3 x 10
-6
-------------------------- 4082 V===
V
peak
5 t (µs)
I
peak
V
peak
R
i
U
LOAD
LOAD
U
U
or motor,
computer,
radio
ELECTRONIC
CIRCUIT