Littelfuse Varistor Design Examples
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Littelfuse Varistor Design Examples Application Note January 1998 AN9772 This note is meant to be a guide for the user in selecting a Energy and Current varistor by describing common application examples, and µ Ω [ /Title The 100 H inductor will appear to be about 30 to the illustrating the solution process to determine the appropriate transient. The 30Ω is derived from the inductive reactance at (AN97 varistor. Also described are varistor fusing and the transient generator source frequency of 105π rad. Taking a 72) series/parallel connection rules. first estimate of peak varistor current, 2500V/80Ω = 31A. (This /Sub- first estimate is high, since it assumes varistor clamping ject Applications voltage is zero.) With a tentative selection of a 130V Littelfuse (Har- Power Supply Protection Against Line Transient Varistor, we find that a current of 31A yields a voltage of from ris Damage 325V to 380V, depending on the model size, as shown in Figure 2A and Figure 2B. Varis- PROBLEM Revising the estimate, I ≈ (2500V - 325V)/80Ω = 27.2A. For tor It is desired to prevent failure of the power supply shown in model V130LA20A, 27.2A coincides closely with a 320V Design Figure 1B to be used on residential 117V lines. A AC clamping level. There is no need to further refine the estimate of representative transient generator is to be used for testing, Exam- peak current if model 20A remains the final selection. ples) as shown in Figure 1A. To arrive at an energy figure, assume a sawtooth current /Autho If the transient is applied to the existing circuit, the rectifier will waveform of 27A peak, dropping to zero in two time receive high negative voltages, transmitted through the filter r () constants, or 20µs. /Key- capacitor. The LC network is there to prevent RFI from being µ words transmitted into the power line (as in a TV set), but also serves Energy is then roughly equal to (27A x 320V x 20 s)/2, the to reduce the transient voltage. An analysis shows that the area under the power waveform. The result is 0.086J, well (TVS, transient will be reduced approximately by half, resulting in within the capability of the varistor (70J). Peak current is also Tran- about 2.5kV instead of 5kV at the rectifier. within the 6500A rating. sient This is still too high for any practical rectifier, so some Model Selection Sup- suppression must be added. It is desirable to use the built-in pres- The actual varistor selection is a trade-off between the impedance of the coil to drop the remaining voltage, so the clamping voltage desired and the number of transient sion, suppressor would best be applied as shown. A selection current pulses expected in the life of the equipment. A 70J Pro- process for a Littelfuse Varistor is as follows: rated varistor will clamp at 315V and be capable of handling 6 tec- SOLUTION over 10 such pulses. An 11J unit will clamp to tion, approximately 385V and be capable of handling over 105 Steady-State Voltage ESD, such pulses. Furthermore, the clamping voltage determines The 117V , 110% high line condition is 129V. The closest IEC, AC the cost of the rectifier by determining the voltage rating voltage rating available is 130V. required. A smaller, lower cost varistor may result in a more EMC, expensive higher voltage rectifier diode. Elec- tro- 100µHD 50Ω magne ≈ 5kV tic + VT = 0.1µF 150µF Com- - 5π pati- ± 5kV sin 10 t X bility, e-10-5t Harris Sup- FIGURE 1A. TRANSIENT GENERATOR FIGURE 1B. TYPICAL POWER SUPPLY CIRCUIT FIGURE 1. POWER SUPPLY PROTECTION 10-127 1-800-999-9445 or 1-847-824-1188 | Copyright © Littelfuse, Inc. 1998 Application Note 9772 3000 1000 UL1499 CORD CONNECTED AND DIRECT PLUG-IN 800 2000 CATEGORY 1500 600 V130LA2 500 1000 V130LA5 400 800 V130LA10A V130LA20A 600 300 500 400 MAXIMUM PEAK (V) 200 300 MAXIMUM PEAK (V) IMPULSE GENERATOR LOAD LINES 200 (IMPLIED) UL1449 PERMANENTLY CONNECTED CATEGORY, AND ANSI IEEE C61.41 (IEEE587) CATEGORY B 100 100 101 102 100 27 100 101 102 103 104 31 27 31 PEAK AMPERES 8/20µs WAVESHAPE PEAK AMPERES 8/20µs WAVESHAPE FIGURE 2A. FIGURE 2B. FIGURE 2. V130LA VARISTOR V-I CHARACTERISTICS IV 27 IV 27 APPROXIMATION VARISTOR CURRENT VARISTOR t 10µs 20µs VARISTOR CURRENT VARISTOR t 10µs20µs FIGURE 3A. FIGURE 3B. FIGURE 3. ENERGY APPROXIMATION SCR Motor Control SOLUTION PROBLEM Add a varistor to the transformer secondary to clamp the transformer inductive transient voltage spike. Select the The circuit shown in Figure 4 experiences failures of the lowest voltage Littelfuse Varistor that is equal to or greater rectifiers and SCR when the transformer primary is switched than the maximum high line secondary AC voltage. The off. The manufacturer has tried 600V components with little V130LA types fulfills this requirement. improvement. Determine the peak suppressed transient voltage produced by the transient energy source. This is based on the peak ARMATURE R1 transient current to the suppressor, assuming the worst-case Ω 330k SPEED condition of zero load current. Zero load current is normally CONTROL a valid assumption. Since the dynamic transient impedance 480V R3 of the Littelfuse Varistor is generally quite low, the parallel AC 60Hz FIELD 250kΩ higher impedance load path can be neglected. SCR SUS Since transient current is the result of stored energy in the R2 4:1 core of the transformer, the transformer equivalent circuit 15kΩ CI 0.2µF shown in Figure 5 will be helpful for analysis. The stored inductive energy is: FIGURE 4. SCR MOTOR CONTROL 1 ˆ2 E = --- L I M LM 2 M 10-128 Application Note 9772 After determining voltage and peak current, energy and power ZP ZS N dissipation requirements must be checked. For the given IM example, the single pulse energy is well below the MUTUAL o INDUCTANCE V130LA20B varistor rating of 70J at 85 C maximum ambient VPRIMARY REPRESENTED LM VSECONDARY BY IRON CORE temperature. Average power dissipation requirements over idling power are not needed because of the non-repetitive nature of the expected transient. Should the transient be IDEAL TRANSFORMER repetitive, then the average power is calculated from the product of the repetition rate times the energy of the transient. FIGURE 5. SIMPLIFIED EQUIVALENT CIRCUIT OF A If this value exceeds the V130LA20B capability of 1.0W, TRANSFORMER power varistors of the HA, DA, or DB Series may be required. The designer needs to know the total energy stored and the Should the ambient temperature exceed 85oC or the surface peak current transformed in the secondary circuit due to the temperature exceed 85oC, the single pulse energy ratings mutual inductance, L . At no load, the magnetizing current, M and the average power ratings must be derated by the (I ), is essentially reactive and is equal to I . This assumes NL M appropriate derating factors supplied on the data sheet. that the primary copper resistance, leakage reactance and equivalent core resistive loss components are small compared to LM. This is a valid assumption for all but the 14 smallest control transformers. Since INL is assumed purely reactive, then: 12 10 Vpri XL = ----------- M INL 8 and 6 f = 50...60Hz iM = INL 4 2 INL can be determined from nameplate data. Where nameplate is not available, Figure 6 and Figure 7 can guide the designer. CURRENT PERCENT MAGNETIZING 2 4 6 8 10 12 Assuming a 3.5% value of magnetizing current from Figure 7 TRANSFORMER RATING (kVA) FIGURE 6. MAGNETIZING CURRENT OF TRANSFORMERS for a 20kVA transformer with 480VAC primary, and 120VAC secondary: WITH LOW SILICON STEEL CORE 20kVA L = X /ω i = ()0.035 ------------------- M L M 480V M = 0.872H = 1.46A 6 2 ˆ 0.872() 2.06 iM = 2iM E = ----------------------------------- 5 LM 2 X = 480V/1.46A LM = 1.85J 4 = 329Ω 3 f = 50...60Hz With this information one can select the needed semiconductor voltage ratings and required varistor energy rating. 2 Peak varistor current is equal to transformed secondary 1 magnetizing current, i.e., îM(N), or 8.24A. From Figure 2, the CURRENT PERCENT MAGNETIZING peak suppressed transient voltage is 310V with the 110102 103 104 V130LA10A selection, 295V with the V130LA20B. This TRANSFORMER RATING (kVA) allows the use of 300V rated semiconductors. Safety FIGURE 7. MAGNETIZING CURRENT OF TRANSFORMERS margins exist in the above approach as a result of the WITH HIGH SILICON STEEL CORE OR SQUARE following assumptions: LOOP CORE 1. All of the energy available in the mutual inductance is transferred to the varistor. Because of core hysteresis Contact Arcing Due to Inductive Load and secondary winding capacitance, only a fraction less PROBLEM than two-thirds is available. To extend the life of the relay contacts shown in Figure 8 and 2. The exciting current is not purely reactive. There is a 10% to 20% safety margin in the peak current assumption. reduce radiated noise, it is desired to eliminate the contact arcing. 10-129 Application Note 9772 subsequent welding. To avoid this inrush current, it is RELAY customary to add a series resistor to limit the capacitive L discharge current. However, this additional component + CC reduces the network effectiveness and adds additional cost to R 28V C C = STRAY CAPACITANCE the solution. DC C L = RELAY COIL INDUCTANCE RC = RELAY COIL RESISTANCE BREAKDOWN LEVEL ARCING 100 FIGURE 8. RELAY CIRCUIT When relays or mechanical switches are used to control VOLTAGE CLAMP ABOVE ARC VOLTAGE inductive loads, it is necessary to use the contacts at only about 50 50% of their resistive load current rating to reduce the wear (V) ARC VOLTAGE VOLTAGE CLAMP BELOW caused by arcing of the contacts.